Introduction
A matching in an edge-labelled complete graph assigns a label to each vertex it covers: the label of its matching edge. We ask how to choose a matching so that few graph edges have a smaller label than at least one of the matching edges incident to them, that is, smaller than at least one of its endpoints. This extremal problem was introduced by Csókа and Pongrácz [1 ] in their study of algorithms that find cliques and dense subgraphs by querying edges of a random graph. We improve both upper bounds for arbitrary labelings and lower bounds obtained from explicit constructions.
The extremal parameters
Let K N K_N K N N N N λ : E ( K N ) → { 1 , … , ℓ } \lambda:E(K_N)\to\{1,\ldots,\ell\} λ : E ( K N ) → { 1 , … , ℓ } M \mathcal{M} M M M M v v v b M ( v ) b_{\mathcal{M}}(v) b M ( v ) v v v b M ( v ) = 0 b_{\mathcal{M}}(v)=0 b M ( v ) = 0 birth label of v v v u v uv uv critical for M \mathcal{M} M
λ ( u v ) < max { b M ( u ) , b M ( v ) } . \lambda(uv)<\max\{b_{\mathcal{M}}(u),b_{\mathcal{M}}(v)\}. λ ( uv ) < max { b M ( u ) , b M ( v )} . For example, if u u u u u u outward critical .
Write C λ ( M ) C_\lambda(\mathcal{M}) C λ ( M ) [1 ] , define
γ ( ℓ ) = lim sup M → ∞ sup N ≥ 2 M max λ min ∣ M ∣ = M C λ ( M ) ( 2 M 2 ) . (1) \gamma(\ell)=\limsup_{M\to\infty}\sup_{N\ge2M}\max_{\lambda}\min_{|\mathcal{M}|=M}\frac{C_\lambda(\mathcal{M})}{\binom{2M}{2}}. \tag*{(1)} γ ( ℓ ) = M → ∞ lim sup N ≥ 2 M sup λ max ∣ M ∣ = M min ( 2 2 M ) C λ ( M ) . (1) Thus the labeling is chosen first and the matching second. The normalizing denominator counts pairs among the 2 M 2M 2 M
γ ′ ( ℓ ) = lim sup M → ∞ max λ : E ( K 2 M ) → [ ℓ ] min M p e r f e c t C λ ( M ) ( 2 M 2 ) , [ ℓ ] = { 1 , … , ℓ } . \gamma'(\ell)=\limsup_{M\to\infty}\max_{\lambda:E(K_{2M})\to[\ell]}\min_{\mathcal{M}\ \mathrm{perfect}}\frac{C_\lambda(\mathcal{M})}{\binom{2M}{2}},\qquad[\ell]=\{1,\ldots,\ell\}. γ ′ ( ℓ ) = M → ∞ lim sup λ : E ( K 2 M ) → [ ℓ ] max M perfect min ( 2 2 M ) C λ ( M ) , [ ℓ ] = { 1 , … , ℓ } . In particular γ ′ ( ℓ ) ≤ γ ( ℓ ) \gamma'(\ell) \le\gamma(\ell) γ ′ ( ℓ ) ≤ γ ( ℓ ) γ ( ℓ ) \gamma(\ell) γ ( ℓ ) 2 M 2M 2 M
Csóka and Pongrácz [1 ] proved γ ′ ( 2 ) = γ ( 2 ) = 1 / 4 \gamma'(2) = \gamma(2) = 1/4 γ ′ ( 2 ) = γ ( 2 ) = 1/4 γ ′ ( 3 ) = γ ( 3 ) = 3 / 8 \gamma'(3) = \gamma(3) = 3/8 γ ′ ( 3 ) = γ ( 3 ) = 3/8
5 12 ≤ γ ′ ( 4 ) ≤ γ ( 4 ) ≤ 7 16 . \frac{5}{12} \le\gamma'(4) \le\gamma(4) \le\frac{7}{16}. 12 5 ≤ γ ′ ( 4 ) ≤ γ ( 4 ) ≤ 16 7 . For every ℓ ≥ 4 \ell\ge4 ℓ ≥ 4
γ ( ℓ ) ≤ 1 2 − 1 3 ⋅ 2 ℓ − 1 − 4 ℓ + 8 . (2) \gamma(\ell) \le\frac{1}{2} - \frac{1}{3 \cdot2^{\ell-1} - 4\ell+ 8}. \tag*{(2)} γ ( ℓ ) ≤ 2 1 − 3 ⋅ 2 ℓ − 1 − 4 ℓ + 8 1 . (2) They also showed that the corresponding parameter with arbitrarily many labels equals 1 / 2 1/2 1/2
Results
Theorem 1.1. Let α L \alpha_L α L § 17 . Then
0.41977282048624007772 … = α L ≤ γ ′ ( 4 ) ≤ γ ( 4 ) ≤ 866807 2000000 = 0.4334035. 0.41977282048624007772\ldots= \alpha_L \le\gamma'(4) \le\gamma(4) \le\frac{866807}{2000000} = 0.4334035. 0.41977282048624007772 … = α L ≤ γ ′ ( 4 ) ≤ γ ( 4 ) ≤ 2000000 866807 = 0.4334035. The lower bound is realized by a labeling on six vertex parts whose proportions belong to a number field of degree six.
The lower bound extends the search for explicit extremal labelings, whereas the upper bound asserts that every four-label instance has an appropriate matching. The two estimates leave a gap; no equality between γ ′ ( 4 ) \gamma'(4) γ ′ ( 4 ) γ ( 4 ) \gamma(4) γ ( 4 )
Theorem 1.2. For every integer ℓ ≥ 2 \ell\ge2 ℓ ≥ 2
γ ′ ( ℓ ) ≥ 1 2 − 1 4 ( ℓ − 1 ) . \gamma'(\ell) \ge\frac{1}{2} - \frac{1}{4(\ell-1)}. γ ′ ( ℓ ) ≥ 2 1 − 4 ( ℓ − 1 ) 1 . For ℓ ≥ 5 \ell\ge5 ℓ ≥ 5 k = 2 ℓ − 2 k = 2^{\ell-2} k = 2 ℓ − 2 S = 3 k − 2 ℓ + 4 S = 3k - 2\ell+ 4 S = 3 k − 2 ℓ + 4
γ ( ℓ ) ≤ 1 2 − 1 2 S − 1 648 k . \gamma(\ell) \le\frac{1}{2} - \frac{1}{2S} - \frac{1}{648k}. γ ( ℓ ) ≤ 2 1 − 2 S 1 − 648 k 1 . In addition,
γ ( ℓ ) ≤ 1 2 − 1 5 k + 2 2 25 k 3 / 2 + O ( log k k 2 ) ( ℓ → ∞ ) . \gamma(\ell) \le\frac{1}{2} - \frac{1}{5k} + \frac{2\sqrt{2}}{25k^{3/2}} + O\left(\frac{\log k}{k^2}\right) \qquad(\ell\to\infty). γ ( ℓ ) ≤ 2 1 − 5 k 1 + 25 k 3/2 2 2 + O ( k 2 log k ) ( ℓ → ∞ ) . The general lower construction has ℓ \ell ℓ § 2 . The optimality is within the specified labeling family, with one part of mass at least 1 / 2 1/2 1/2 ℓ = 4 \ell= 4 ℓ = 4 5 / 12 5/12 5/12 Theorem 1.1 gives the stronger value above.
The finite general upper bound improves (2) by 1 / ( 162 ⋅ 2 ℓ ) 1/(162 \cdot2^\ell) 1/ ( 162 ⋅ 2 ℓ ) 1 / 2 1/2 1/2 1 / ( 6 k ) 1/(6k) 1/ ( 6 k ) 1 / ( 5 k ) 1/(5k) 1/ ( 5 k ) 6 / 5 6/5 6/5 ℓ \ell ℓ Theorem 16.1 gives a rational two-variable quadratic formula; it is this finite formula, and the bound in Section 15, that produce the following certified estimates.
Table 1.
The lower bounds apply to γ ′ \gamma' γ ′ γ \gamma γ γ \gamma γ γ ′ \gamma' γ ′
Proof strategy and organization
The general lower construction in § 2 provides an elementary example of the problem. Its proof counts critical edges for an arbitrary perfect matching, without a computer calculation.
For the upper bounds, we start with a matching chosen by the order of [1 ] . This order prevents several small exchanges that would reduce its weight. Those restrictions bound the number of critical edges inside and between the sets of vertices covered by matching edges of each label. The original estimate is sharp only if several of these counts are simultaneously large. We show that such a configuration permits other matchings with smaller critical-edge counts.
The four-label argument begins in § 3 . We describe the exchanges explicitly and keep track of the edges that prevent each exchange. This leads to a finite family of quadratic inequalities. § 13 combines these inequalities to prove the numerical bound. Each combination has rational coefficients, and its validity is checked by an exact matrix factorization. The combinatorial derivations are given in the text; the supplied programs check the resulting identities, signs, and coverage of the parameter ranges.
For general ℓ \ell ℓ § 16 derive the corresponding finite and asymptotic upper bounds. The estimates account for outward critical edges, so the number of uncovered vertices does not enter the error terms.
§ 17 proves the six-part lower bound. For a labeling constant on pairs of vertex parts, a matching is described by the amount it matches between each pair of parts. These amounts satisfy linear constraints, and the noncritical-edge count is a quadratic function of them. We bound this function over the entire feasible set using exact algebraic arithmetic. The algebraic number and the part proportions are specified explicitly. Appendix § 19 examine small perturbations of this construction. § 18 gives the verification commands and identifies the data needed for each computational claim.
Two normalizations occur in the proofs. Upper-bound counts are divided by M 2 M^2 M 2 γ \gamma γ 1 / 2 1/2 1/2 ( 2 M 2 ) = 2 M 2 − M \binom{2M}{2}=2M^2-M ( 2 2 M ) = 2 M 2 − M γ ′ \gamma' γ ′
Application to clique queries
Consider an Erdős–Rényi graph G ( n , 1 / 2 ) G(n,1/2) G ( n , 1/2 ) n δ n^\delta n δ ℓ \ell ℓ α ∗ ( δ , ℓ ) \alpha_*(\delta,\ell) α ∗ ( δ , ℓ ) α \alpha α ( α + o ( 1 ) ) log 2 n (\alpha+o(1))\log_2 n ( α + o ( 1 )) log 2 n
Feige, Gamarnik, Neeman, Rácz, and Tetali [2 ] obtained upper bounds in this query model, and Feige and Ferster [3 ] sharpened the analysis for two and three rounds. The clique-query theorem of Csóka and Pongrácz [1 ] gives, for 1 ≤ δ ≤ 2 1\le\delta\le2 1 ≤ δ ≤ 2 ℓ ≥ 3 \ell\ge3 ℓ ≥ 3
α ∗ ( δ , ℓ ) ≤ 1 + 1 − ( 2 − δ ) 2 4 γ ( ℓ ) . \alpha_*(\delta,\ell) \le1+\sqrt{1-\frac{(2-\delta)^2}{4\gamma(\ell)}}. α ∗ ( δ , ℓ ) ≤ 1 + 1 − 4 γ ( ℓ ) ( 2 − δ ) 2 . Consequently any upper bound G ℓ G_\ell G ℓ γ ( ℓ ) \gamma(\ell) γ ( ℓ ) γ ( ℓ ) \gamma(\ell) γ ( ℓ ) G 4 = 0.4334035 G_4=0.4334035 G 4 = 0.4334035
A lower bound for every number of labels
The following construction extends the four-part labeling of Csókа and Pongrácz [1 ] , which gives the lower bound 5 / 12 5/12 5/12
Theorem 2.1. For every integer ℓ ≥ 2 \ell\ge2 ℓ ≥ 2
γ ( ℓ ) ≥ γ ′ ( ℓ ) ≥ 1 2 − 1 4 ( ℓ − 1 ) . \gamma(\ell) \ge\gamma'(\ell) \ge\frac{1}{2} - \frac{1}{4(\ell-1)}. γ ( ℓ ) ≥ γ ′ ( ℓ ) ≥ 2 1 − 4 ( ℓ − 1 ) 1 . In particular these lower bounds tend to 1 / 2 1/2 1/2 ℓ → ∞ \ell\to\infty ℓ → ∞
Put m = ℓ − 1 m=\ell-1 m = ℓ − 1 A 1 , … , A m , R A_1,\ldots,A_m,R A 1 , … , A m , R
a 1 = 1 4 m , a i = 1 2 m ( 2 ≤ i ≤ m ) , r = 2 m + 1 4 m . (3) a_1=\frac{1}{4m},\qquad a_i=\frac{1}{2m}\quad(2\le i\le m),\qquad r=\frac{2m+1}{4m}. \tag*{(3)} a 1 = 4 m 1 , a i = 2 m 1 ( 2 ≤ i ≤ m ) , r = 4 m 2 m + 1 . (3) The part R R R
λ ( A i , A j ) = max { 1 , min ( i , j ) − 1 } , λ ( A i , R ) = i , λ ( R , R ) = m + 1. (4) \lambda(A_i,A_j)=\max\{1,\min(i,j)-1\},\qquad\lambda(A_i,R)=i,\qquad\lambda(R,R)=m+1. \tag*{(4)} λ ( A i , A j ) = max { 1 , min ( i , j ) − 1 } , λ ( A i , R ) = i , λ ( R , R ) = m + 1. (4) The first rule includes i = j i=j i = j ℓ = 4 \ell=4 ℓ = 4 ( 1 , 2 , 2 , 7 ) / 12 (1,2,2,7)/12 ( 1 , 2 , 2 , 7 ) /12 [1 ] .
Proof of Theorem 2.1. We prove an exact finite statement. For any positive integer q q q N = 4 m q N=4mq N = 4 m q
∣ A 1 ∣ = q , ∣ A i ∣ = 2 q ( 2 ≤ i ≤ m ) , ∣ R ∣ = ( 2 m + 1 ) q . |A_1|=q,\qquad|A_i|=2q\ (2\le i\le m),\qquad|R|=(2m+1)q. ∣ A 1 ∣ = q , ∣ A i ∣ = 2 q ( 2 ≤ i ≤ m ) , ∣ R ∣ = ( 2 m + 1 ) q . Write A = ⋃ i = 1 m A i A=\bigcup_{i=1}^{m}A_i A = ⋃ i = 1 m A i s = ∣ A ∣ = ( 2 m − 1 ) q s=|A|=(2m-1)q s = ∣ A ∣ = ( 2 m − 1 ) q r N = ∣ R ∣ r_N=|R| r N = ∣ R ∣
( 4 m 2 − 2 m ) q 2 − ( m − 1 ) q . (5) (4m^2-2m)q^2-(m-1)q. \tag*{(5)} ( 4 m 2 − 2 m ) q 2 − ( m − 1 ) q . (5) Fix a perfect matching. Let x i x_i x i A i A_i A i R R R
X = ∑ i = 2 m x i , p i = ∑ j < i ∣ A j ∣ . X=\sum_{i=2}^{m}x_i,\qquad p_i=\sum_{j<i}|A_j|. X = i = 2 ∑ m x i , p i = j < i ∑ ∣ A j ∣. Thus p 1 = 0 p_1=0 p 1 = 0 p i = ( 2 i − 3 ) q p_i=(2i-3)q p i = ( 2 i − 3 ) q i ≥ 2 i\ge2 i ≥ 2 A i A_i A i i i i A i A_i A i i i i A i A_i A i R R R i i i R R R i i i r N − ∑ j ≤ i x j r_N-\sum_{j\le i}x_j r N − ∑ j ≤ i x j R R R A A A R R R
∑ i ∣ A i ∣ ( r N − ∑ j ≤ i x j ) = s r N − ∑ i x i ( s − p i ) . (6) \sum_i |A_i|\left(r_N-\sum_{j\le i}x_j\right)=sr_N-\sum_i x_i(s-p_i). \tag*{(6)} i ∑ ∣ A i ∣ ( r N − j ≤ i ∑ x j ) = s r N − i ∑ x i ( s − p i ) . (6) Let U U U A i A_i A i i ≥ 2 i\ge2 i ≥ 2 R R R A A A U U U A i ∩ U A_i\cap U A i ∩ U i i i i i i ∣ U ∣ = X |U|=X ∣ U ∣ = X
( s 2 ) − ( s − X 2 ) = s X − X 2 + X 2 . \binom{s}{2}-\binom{s-X}{2}=sX-\frac{X^2+X}{2}. ( 2 s ) − ( 2 s − X ) = s X − 2 X 2 + X . Adding this to (6) yields
C λ ( M ) ≥ s ( r N − x 1 ) + ∑ i = 2 m p i x i − X 2 + X 2 . (7) C_{\lambda}(\mathcal{M}) \ge s(r_N-x_1)+\sum_{i=2}^{m}p_i x_i-\frac{X^2+X}{2}. \tag*{(7)} C λ ( M ) ≥ s ( r N − x 1 ) + i = 2 ∑ m p i x i − 2 X 2 + X . (7) All 0 ≤ x i ≤ 2 q 0\le x_i\le2q 0 ≤ x i ≤ 2 q i ≥ 2 i\ge2 i ≥ 2
2 ∑ i = 2 m p i x i − X 2 = ∑ i = 2 m x i ( 2 q − x i ) + 2 ∑ 2 ≤ i < j ≤ m x j ( 2 q − x i ) ≥ 0 2\sum_{i=2}^{m}p_i x_i-X^2=\sum_{i=2}^{m}x_i(2q-x_i)+2\sum_{2\le i<j\le m}x_j(2q-x_i)\ge0 2 i = 2 ∑ m p i x i − X 2 = i = 2 ∑ m x i ( 2 q − x i ) + 2 2 ≤ i < j ≤ m ∑ x j ( 2 q − x i ) ≥ 0 therefore applies. Moreover x 1 ≤ q x_1\le q x 1 ≤ q X ≤ 2 ( m − 1 ) q X\le2(m-1)q X ≤ 2 ( m − 1 ) q (7) now gives
C λ ( M ) ≥ s ( r N − q ) − ( m − 1 ) q = ( 4 m 2 − 2 m ) q 2 − ( m − 1 ) q . C_{\lambda}(\mathcal{M})\ge s(r_N-q)-(m-1)q=(4m^2-2m)q^2-(m-1)q. C λ ( M ) ≥ s ( r N − q ) − ( m − 1 ) q = ( 4 m 2 − 2 m ) q 2 − ( m − 1 ) q . Empty sums are interpreted as zero, so the argument includes m = 1 m=1 m = 1
Equality is attained by matching every vertex of A A A R R R 2 q 2q 2 q R R R x 1 = q x_1=q x 1 = q x i = 2 q x_i=2q x i = 2 q i ≥ 2 i\ge2 i ≥ 2 A A A U U U R R R (5) .
Dividing by ( 4 m q 2 ) \binom{4mq}{2} ( 2 4 m q ) q → ∞ q\to\infty q → ∞
lim q → ∞ ( 4 m 2 − 2 m ) q 2 − ( m − 1 ) q ( 4 m q 2 ) = 2 m − 1 4 m , \lim_{q\to\infty}\frac{(4m^2-2m)q^2-(m-1)q}{\binom{4mq}{2}}=\frac{2m-1}{4m}, q → ∞ lim ( 2 4 m q ) ( 4 m 2 − 2 m ) q 2 − ( m − 1 ) q = 4 m 2 m − 1 , which proves the theorem by the definition of γ ′ ( ℓ ) \gamma'(\ell) γ ′ ( ℓ )
The theorem gives 1 / 4 , 3 / 8 , 5 / 12 , 7 / 16 , 9 / 20 1/4,3/8,5/12,7/16,9/20 1/4 , 3/8 , 5/12 , 7/16 , 9/20 ℓ = 2 , 3 , 4 , 5 , 6 \ell=2,3,4,5,6 ℓ = 2 , 3 , 4 , 5 , 6 Theorem 17.1 gives a stronger bound. The present family supplies a simple bound for every ℓ \ell ℓ
Proposition 2.2. Keep the labels in (4) , and allow arbitrary nonnegative satellite proportions a 1 , … , a m a_1,\ldots,a_m a 1 , … , a m r = 1 − ∑ i a i ≥ 1 / 2 r=1-\sum_i a_i\ge1/2 r = 1 − ∑ i a i ≥ 1/2 ( 2 m − 1 ) / ( 4 m ) (2m-1)/(4m) ( 2 m − 1 ) / ( 4 m ) (3) .
Proof. Put S = ∑ i a i ≤ 1 / 2 S=\sum_i a_i\le1/2 S = ∑ i a i ≤ 1/2
G ( a ) = 2 S − 2 S 2 − ( 2 a 1 2 + ∑ i = 2 m a i 2 ) . G(a)=2S-2S^2-\left(2a_1^2+\sum_{i=2}^{m}a_i^2\right). G ( a ) = 2 S − 2 S 2 − ( 2 a 1 2 + i = 2 ∑ m a i 2 ) . For clarity, its satellite contribution is S 2 − a 1 2 S^2-a_1^2 S 2 − a 1 2 2 ∑ i a i ( r − ∑ j < i a j ) 2\sum_i a_i(r-\sum_{j<i}a_j) 2 ∑ i a i ( r − ∑ j < i a j ) G ( a ) G(a) G ( a )
Weighted Cauchy–Schwarz gives
2 a 1 2 + ∑ i = 2 m a i 2 ≥ 2 S 2 2 m − 1 . 2a_1^2+\sum_{i=2}^{m}a_i^2\ge\frac{2S^2}{2m-1}. 2 a 1 2 + i = 2 ∑ m a i 2 ≥ 2 m − 1 2 S 2 . It follows that
G ( a ) ≤ 2 S − 4 m 2 m − 1 S 2 = 2 m − 1 4 m − 4 m 2 m − 1 ( S − 2 m − 1 4 m ) 2 . G(a)\le2S-\frac{4m}{2m-1}S^2=\frac{2m-1}{4m}-\frac{4m}{2m-1}\left(S-\frac{2m-1}{4m}\right)^2. G ( a ) ≤ 2 S − 2 m − 1 4 m S 2 = 4 m 2 m − 1 − 2 m − 1 4 m ( S − 4 m 2 m − 1 ) 2 . Equality requires S = ( 2 m − 1 ) / ( 4 m ) S=(2m-1)/(4m) S = ( 2 m − 1 ) / ( 4 m ) a 2 = ⋯ = a m = 2 a 1 a_2=\cdots=a_m=2a_1 a 2 = ⋯ = a m = 2 a 1 (3) . For m = 1 m=1 m = 1 S = a 1 S=a_1 S = a 1
This optimality statement concerns the specified labels and the condition r ≥ 1 / 2 r\ge1/2 r ≥ 1/2 1 / 2 1/2 1/2
The initial matching and its structural properties
In the four-label problem, fix a labeling of K N K_N K N M ≤ N / 2 M \le N/2 M ≤ N /2 M M M M 0 \mathcal{M}_0 M 0
( n 2 + 2 n 3 + 4 n 4 , n 3 + 2 n 4 , n 4 ) , (n_2+2n_3+4n_4,\ n_3+2n_4,\ n_4), ( n 2 + 2 n 3 + 4 n 4 , n 3 + 2 n 4 , n 4 ) , where n i n_i n i i i i [1 ] . Formal weights
w 1 = 0 , w 2 = 1 , w 3 = 2 + ε , w 4 = 4 + 2 ε + ε 2 w_1=0,\qquad w_2=1,\qquad w_3=2+\varepsilon,\qquad w_4=4+2\varepsilon+\varepsilon^2 w 1 = 0 , w 2 = 1 , w 3 = 2 + ε , w 4 = 4 + 2 ε + ε 2 give an equivalent notation for this order. We compare polynomials at their first differing coefficient. Alternatively one may choose a sufficiently small positive ε \varepsilon ε ε \varepsilon ε M M M
The coefficients of 1 , ε , ε 2 1,\varepsilon,\varepsilon^2 1 , ε , ε 2 M 0 \mathcal{M}_0 M 0
Let X i X_i X i i i i outward edge joins a covered vertex to an uncovered vertex. An outward critical edge could replace its incident matching edge and strictly decrease the weight, so none exists. Between two differently labeled matching edges, the four cross edges form two complementary pairs. Both edges of a complementary pair cannot be critical: if the larger original label is j j j j j j 2 w j − 1 < w j 2w_{j-1}<w_j 2 w j − 1 < w j X 1 X_1 X 1 X 4 X_4 X 4
Decomposing the label-1 graph inside X 2 X_2 X 2
Color the original X 2 X_2 X 2
The following special case of the peeling argument in [1 ] gives the decomposition we need. We include the proof.
Lemma 3.1. Let a graph consist of a red perfect matching and a disjoint set of blue edges. Suppose there is no alternating cycle and no alternating path with four blue edges. Remove every red pair having an endpoint of blue degree at most three. The remaining graph has no alternating path with two blue edges, and each of its red pairs has an endpoint of residual blue degree at most one.
Proof. For a longest red-ended alternating path with p p p p p p p p p
Let G ′ G' G ′ P P P G ′ G' G ′ u , v u,v u , v p p p p < 3 p<3 p < 3 G ′ G' G ′ p = 3 p=3 p = 3 u u u P P P p = 2 p=2 p = 2 u u u v v v u ′ , v ′ u',v' u ′ , v ′ P P P u ′ v ′ u'v' u ′ v ′ P P P p ≤ 1 p \le1 p ≤ 1
Finally, if both ends of a residual red edge had blue degree at least two, choose distinct blue neighbors of its endpoints. The two neighbors cannot be red mates, by the alternating-cycle condition, so adding their red mates gives a red-ended alternating path with two blue edges. This contradiction proves the last claim.
Apply the lemma by removing every red pair having an endpoint of blue degree at most three in the original blue graph. Call the low-degree endpoints B B B A A A D D D C C C O ( M ) O(M) O ( M )
A A , A C , A D , C C . AA,\quad AC,\quad AD,\quad CC. AA , A C , A D , C C . The A B AB A B C D CD C D d d d q q q M M M d + q = x = ∣ X 2 ∣ / ( 2 M ) d+q=x=|X_2|/(2M) d + q = x = ∣ X 2 ∣/ ( 2 M )
d 2 2 + 2 d q + q 2 2 = 3 x 2 4 − ( d − q ) 2 4 . \frac{d^2}{2}+2dq+\frac{q^2}{2}=\frac{3x^2}{4}-\frac{(d-q)^2}{4}. 2 d 2 + 2 d q + 2 q 2 = 4 3 x 2 − 4 ( d − q ) 2 . in units of M 2 M^2 M 2 O ( 1 / M ) O(1/M) O ( 1/ M ) E A A E_{AA} E AA E A C E_{AC} E A C E A D E_{AD} E A D E C C E_{CC} E C C template . A missing blue pair in a region is a vertex pair whose edge has label at least two. At most a constant times M M M
Inside X 3 X_3 X 3 z M zM z M z = ∣ X 3 ∣ / ( 2 M ) z=|X_3|/(2M) z = ∣ X 3 ∣/ ( 2 M ) z M zM z M z 2 M 2 / 2 + O ( M ) z^2M^2/2+O(M) z 2 M 2 /2 + O ( M )
Matching facts used below
We use three classical results with their finite forms. Kőnig’s theorem [4 ] identifies the maximum matching size with the minimum vertex-cover size in a bipartite graph. In a bipartite graph with sides of sizes m ≤ D m \le D m ≤ D a a a m − a / D m-a/D m − a / D t t t t D tD t D
If an n n n k k k [5 ] is
e ( G ) ≤ max { ( 2 k + 1 2 ) , k n − ( k + 1 2 ) } . e(G) \le\max\left\{\binom{2k+1}{2},\; kn-\binom{k+1}{2}\right\}. e ( G ) ≤ max { ( 2 2 k + 1 ) , k n − ( 2 k + 1 ) } . In particular e ( G ) ≤ k n + O ( n ) e(G) \le kn+O(n) e ( G ) ≤ k n + O ( n ) [6 , 7 , 8 ] gives
2 ν ( G ) = min S ⊆ V ( G ) ( ∣ V ( G ) ∣ + ∣ S ∣ − o ( G − S ) ) . 2\nu(G)=\min_{S\subseteq V(G)}\left(|V(G)|+|S|-o(G-S)\right). 2 ν ( G ) = S ⊆ V ( G ) min ( ∣ V ( G ) ∣ + ∣ S ∣ − o ( G − S ) ) . where o ( G − S ) o(G-S) o ( G − S )
We will also use the following dense-graph consequence of the Erdős–Gallai inequality. Suppose a graph has t M tM tM c M 2 + O ( M ) cM^2+O(M) c M 2 + O ( M ) c ≤ 9 t 2 / 50 c \le9t^2/50 c ≤ 9 t 2 /50 R M RM R M
R 2 ≥ t 2 − 2 c − o ( 1 ) , R ≥ 4 t / 5 − o ( 1 ) . (8) R^2 \ge t^2-2c-o(1),\qquad R \ge4t/5-o(1). \tag*{(8)} R 2 ≥ t 2 − 2 c − o ( 1 ) , R ≥ 4 t /5 − o ( 1 ) . (8) Indeed the two asymptotic branches of the edge bound are R 2 / 2 R^2/2 R 2 /2 R t / 2 − R 2 / 8 Rt/2-R^2/8 R t /2 − R 2 /8 R ≤ 4 t / 5 R \le4t/5 R ≤ 4 t /5 8 t 2 / 25 8t^2/25 8 t 2 /25 t 2 / 2 − c ≥ 8 t 2 / 25 t^2/2-c \ge8t^2/25 t 2 /2 − c ≥ 8 t 2 /25 R ≥ 4 t / 5 R \ge4t/5 R ≥ 4 t /5 (8) . Equality at the branch boundary and the O ( M ) O(M) O ( M ) o ( 1 ) o(1) o ( 1 )
We repeatedly augment a matching while preserving all vertices it already covers. This is possible because an augmenting path changes only the matching status of its two previously uncovered endpoints; the internal vertices stay covered [7 ] . A related elementary count will be useful. If U U U ∣ U ∣ + 1 |U|+1 ∣ U ∣ + 1 U U U O ( M ) O(M) O ( M )
A matching replacement on alternating paths
We construct several M M M M M M M 2 M^2 M 2 C 0 = C λ ( M 0 ) / M 2 C_0=C_\lambda(\mathcal{M}_0)/M^2 C 0 = C λ ( M 0 ) / M 2 C ≤ 2 G + o ( 1 ) C\le2G+o(1) C ≤ 2 G + o ( 1 ) γ ( 4 ) ≤ G \gamma(4)\le G γ ( 4 ) ≤ G U V = 3 UV=3 U V = 3 λ ( U V ) = 3 \lambda(UV)=3 λ ( U V ) = 3 U V ≤ 3 UV\le3 U V ≤ 3 λ ( U V ) ≤ 3 \lambda(UV)\le3 λ ( U V ) ≤ 3 o ( 1 ) o(1) o ( 1 ) N ≥ 2 M N\ge2M N ≥ 2 M O ( M ) O(M) O ( M )
Write y , x , z , w y,x,z,w y , x , z , w 1 , 2 , 3 , 4 1,2,3,4 1 , 2 , 3 , 4 M 0 \mathcal{M}_0 M 0 ∣ X 1 ∣ / M = 2 y |X_1|/M=2y ∣ X 1 ∣/ M = 2 y X 4 X_4 X 4 top vertices . The decomposition in § 3 gives label-2 pairs A B AB A B C D CD C D d , q d,q d , q d + q = x d+q=x d + q = x
b = ∣ d − q ∣ / 2 , m = min ( d , q ) = x / 2 − b , B = 1 − y 2 − x 2 / 4 − z 2 / 2 − w 2 , B 0 = B − b 2 . b=|d-q|/2,\qquad m=\min(d,q)=x/2-b,\qquad B=1-y^2-x^2/4-z^2/2-w^2,\qquad B_0=B-b^2. b = ∣ d − q ∣/2 , m = min ( d , q ) = x /2 − b , B = 1 − y 2 − x 2 /4 − z 2 /2 − w 2 , B 0 = B − b 2 . We write h = ∣ d − q ∣ = 2 b h=|d-q|=2b h = ∣ d − q ∣ = 2 b E A A , E A C , E A D , E C C E_{AA},E_{AC},E_{AD},E_{CC} E AA , E A C , E A D , E C C E E E C i j C_{ij} C ij X i X_i X i X j X_j X j i ≠ j i\ne j i = j X i X_i X i i = j i=j i = j i < j i<j i < j e i j e_{ij} e ij X i X_i X i X j X_j X j C 12 M 2 C_{12}M^2 C 12 M 2 e 12 = 2 x y − C 12 ≥ 0 e_{12}=2xy-C_{12}\ge0 e 12 = 2 x y − C 12 ≥ 0
a = E A D , c = E C C , j = E A C . a=E_{AD},\qquad c=E_{CC},\qquad j=E_{AC}. a = E A D , c = E C C , j = E A C . The original count satisfies
C 0 ≤ B − b 2 − E − ∑ i < j e i j + o ( 1 ) . (9) C_0\le B-b^2-E-\sum_{i<j}e_{ij}+o(1). \tag*{(9)} C 0 ≤ B − b 2 − E − i < j ∑ e ij + o ( 1 ) . (9) Here the X 2 X_2 X 2 3 x 2 / 4 − b 2 − E + o ( 1 ) 3x^2/4-b^2-E+o(1) 3 x 2 /4 − b 2 − E + o ( 1 ) X 3 X_3 X 3 z 2 / 2 + o ( 1 ) z^2/2+o(1) z 2 /2 + o ( 1 ) (9) .
The retained alternating paths
Choose a blue matching from some A A A D D D C C C D D D
B 1 A 1 D 1 C 1 C 2 D 2 A 2 B 2 . B_1A_1D_1C_1C_2D_2A_2B_2. B 1 A 1 D 1 C 1 C 2 D 2 A 2 B 2 . The edges B 1 A 1 , D 1 C 1 , C 2 D 2 , A 2 B 2 B_1A_1,D_1C_1,C_2D_2,A_2B_2 B 1 A 1 , D 1 C 1 , C 2 D 2 , A 2 B 2 A 1 D 1 , C 1 C 2 , D 2 A 2 A_1D_1,C_1C_2,D_2A_2 A 1 D 1 , C 1 C 2 , D 2 A 2 B 1 B 2 B_1B_2 B 1 B 2 B 1 B 2 B_1B_2 B 1 B 2 neutral core . The four sets of vertices retained in it each have the same mass r r r 0 ≤ r ≤ min ( d , q ) 0 \le r \le\min(d,q) 0 ≤ r ≤ min ( d , q )
Fix a neutral core with maximum possible r r r A D AD A D C C CC C C A D AD A D C C CC C C B B BB B B M 0 \mathcal{M}_0 M 0 C r C_r C r
Let H 3 M 2 H_3M^2 H 3 M 2 B B B X 3 X_3 X 3 h 3 = r z − H 3 h_3 = rz - H_3 h 3 = r z − H 3 good if it can be oriented P Q PQ P Q P P P B B B exceptional . The switching argument below shows that h 3 ≥ − o ( 1 ) h_3 \ge-o(1) h 3 ≥ − o ( 1 ) h 3 / r + o ( 1 ) h_3/r + o(1) h 3 / r + o ( 1 ) r r r f R = r ( x + z − 3 y ) − 3 r 2 f_R = r(x+z-3y)-3r^2 f R = r ( x + z − 3 y ) − 3 r 2
For z , w > 0 z,w>0 z , w > 0 § 4.2 gives
C r ≤ B − b 2 + r ( x + z − 3 y ) − 3 r 2 − ( 1 − y / z ) h 3 − e 13 − e 34 + ( r / ( 2 w ) − 1 ) e 14 + ( y / z − 1 ) e 23 + ( y / w − 1 ) e 24 + o ( 1 ) . (10) C_r \le B-b^2+r(x+z-3y)-3r^2-(1-y/z)h_3-e_{13}-e_{34}+(r/(2w)-1)e_{14}+(y/z-1)e_{23}+(y/w-1)e_{24}+o(1). \tag*{(10)} C r ≤ B − b 2 + r ( x + z − 3 y ) − 3 r 2 − ( 1 − y / z ) h 3 − e 13 − e 34 + ( r / ( 2 w ) − 1 ) e 14 + ( y / z − 1 ) e 23 + ( y / w − 1 ) e 24 + o ( 1 ) . (10) Derivation of the core comparison
We derive (10) by counting the changes on disjoint edge classes. The retained sets A , B , C , D A,B,C,D A , B , C , D r r r A D AD A D C C CC C C B B BB B B A B ( 2 ) AB(2) A B ( 2 ) C D ( 2 ) CD(2) C D ( 2 ) B B B B B B
For each original label-3 edge P Q PQ P Q B , B ′ B,B' B , B ′ B B B B B ′ ( 4 ) , P Q ( 3 ) → B P ( 3 ) , B ′ Q ( 3 ) BB'(4),PQ(3)\mathbin{\to}BP(3),B'Q(3) B B ′ ( 4 ) , P Q ( 3 ) → B P ( 3 ) , B ′ Q ( 3 ) B B B r M + O ( 1 ) rM+O(1) r M + O ( 1 ) H 3 ≤ r z + o ( 1 ) H_3 \le rz+o(1) H 3 ≤ r z + o ( 1 ) h 3 / r + o ( 1 ) h_3/r+o(1) h 3 / r + o ( 1 )
For each retained D D D A A A B B B A A A B ′ B' B ′
B B ′ ( 4 ) , A D ( 1 ) → B ′ D ( ≤ 3 ) , A B ( 2 ) BB'(4),AD(1)\mathbin{\to}B'D(\le3),AB(2) B B ′ ( 4 ) , A D ( 1 ) → B ′ D ( ≤ 3 ) , A B ( 2 ) would reduce primary weight. Hence there are at most r M rM r M B D BD B D D T = 4 DT=4 D T = 4 T T T
Let P Q ( 3 ) PQ(3) P Q ( 3 ) P P P B B B P Q PQ P Q B B ′ BB' B B ′ Q B = 4 QB=4 QB = 4 B ′ B' B ′ B B B A A A D D D
B B ′ ( 4 ) , A D ( 1 ) , P Q ( 3 ) → P B ′ ( 3 ) , Q D ( ≤ 3 ) , A B ( 2 ) BB'(4),AD(1),PQ(3)\mathbin{\to}PB'(3),QD(\le3),AB(2) B B ′ ( 4 ) , A D ( 1 ) , P Q ( 3 ) → P B ′ ( 3 ) , Q D ( ≤ 3 ) , A B ( 2 ) shows Q D = 4 QD=4 Q D = 4 C C ′ CC' C C ′ D D D C ′ C' C ′ C C ′ CC' C C ′ Q C ( ≤ 2 ) , C ′ D ( 2 ) QC(\le2),C'D(2) QC ( ≤ 2 ) , C ′ D ( 2 ) Q C ≥ 3 QC\ge3 QC ≥ 3
Put h 1 = 2 r y − H 1 h_1=2ry-H_1 h 1 = 2 r y − H 1 H 1 H_1 H 1 B B B X 1 X_1 X 1
Table 4 (PDF p. 11).
For the first row, at least 3 r 2 − E − o ( 1 ) 3r^2-E-o(1) 3 r 2 − E − o ( 1 ) B A BA B A B C BC B C O ( M ) O(M) O ( M ) B D BD B D 2 r 2 + o ( 1 ) 2r^2+o(1) 2 r 2 + o ( 1 ) B B B B B ( 4 ) BB(4) B B ( 4 )
For the third row, the old selected A B AB A B C D CD C D 4 r y − e 12 4ry-e_{12} 4 r y − e 12 X 1 X_1 X 1 H 1 H_1 H 1 B B B X 3 X_3 X 3 B B B X 4 X_4 X 4
C r ≤ B − b 2 + r ( x + z − 2 y ) − 3 r 2 − h 1 − h 3 − ∑ ( i , j ) ≠ ( 1 , 2 ) e i j + o ( 1 ) . C_r \le B-b^2+r(x+z-2y)-3r^2-h_1-h_3-\sum_{(i,j)\ne(1,2)}e_{ij}+o(1). C r ≤ B − b 2 + r ( x + z − 2 y ) − 3 r 2 − h 1 − h 3 − ( i , j ) = ( 1 , 2 ) ∑ e ij + o ( 1 ) . It remains to relate h 1 h_1 h 1 h 3 h_3 h 3 P Q PQ P Q g P g_P g P B B B b Q , d P b_Q,d_P b Q , d P A , D A,D A , D Q B = Q D = 4 QB=QD=4 QB = Q D = 4 Q C ≥ 3 QC\ge3 QC ≥ 3 b Q + d P ≤ e P + r − g P b_Q+d_P\le e_P+r-g_P b Q + d P ≤ e P + r − g P e P e_P e P e 23 e_{23} e 23 V ∈ X 1 V\in X_1 V ∈ X 1 T ∈ X 4 T\in X_4 T ∈ X 4 a V , c T a_V,c_T a V , c T V B ≤ 3 VB\le3 V B ≤ 3 T C ≤ 3 TC\le3 T C ≤ 3 ∑ V a V = h 1 \sum_V a_V=h_1 ∑ V a V = h 1 ∑ T c T ≤ e 24 \sum_T c_T\le e_{24} ∑ T c T ≤ e 24
For original matching edges U V ( 1 ) UV(1) U V ( 1 ) T T ′ ( 4 ) TT'(4) T T ′ ( 4 ) A B ( 2 ) AB(2) A B ( 2 ) C D ( 2 ) CD(2) C D ( 2 ) P Q ( 3 ) PQ(3) P Q ( 3 )
U T ( ≤ 3 ) , T ′ C ( ≤ 3 ) , D P ( ≤ 2 ) , Q A ( ≤ 2 ) , B V ( ≤ 3 ) UT(\le3),\quad T'C(\le3),\quad DP(\le2),\quad QA(\le2),\quad BV(\le3) U T ( ≤ 3 ) , T ′ C ( ≤ 3 ) , D P ( ≤ 2 ) , Q A ( ≤ 2 ) , B V ( ≤ 3 ) has weight at most 8 + 3 ε 8+3\varepsilon 8 + 3 ε 8 + 3 ε + ε 2 8+3\varepsilon+\varepsilon^2 8 + 3 ε + ε 2 U T UT U T a V + c T ′ ≥ g P − e P a_V+c_{T'}\ge g_P-e_P a V + c T ′ ≥ g P − e P O ( 1 ) O(1) O ( 1 ) B B B o ( 1 ) o(1) o ( 1 ) 2 y w − e 14 2yw-e_{14} 2 y w − e 14 X 1 X_1 X 1 X 4 X_4 X 4 2 w 2w 2 w 2 y 2y 2 y
( 2 y w − e 14 ) ( r z − h 3 − e 23 ) ≤ 2 z ( w h 1 + y e 24 ) + o ( 1 ) . (2yw-e_{14})(rz-h_3-e_{23})\le2z(wh_1+ye_{24})+o(1). ( 2 y w − e 14 ) ( r z − h 3 − e 23 ) ≤ 2 z ( w h 1 + y e 24 ) + o ( 1 ) . On the localized domain z , w > 0 z,w>0 z , w > 0
h 1 ≥ r y − y z h 3 − r 2 w e 14 − y z e 23 − y w e 24 − o ( 1 ) . h_1\ge ry-\frac{y}{z}h_3-\frac{r}{2w}e_{14}-\frac{y}{z}e_{23}-\frac{y}{w}e_{24}-o(1). h 1 ≥ r y − z y h 3 − 2 w r e 14 − z y e 23 − w y e 24 − o ( 1 ) . Substituting in the count table proves (10) . The same proof retains − I 3 -I_3 − I 3 X 3 X_3 X 3 z 2 / 2 − I 3 z^2/2-I_3 z 2 /2 − I 3
In particular, on 2 y ≤ z 2y\le z 2 y ≤ z y ≤ w y\le w y ≤ w r ≤ 2 w r\le2w r ≤ 2 w
C r ≤ B − b 2 + r ( x + z − 3 y ) − 3 r 2 − h 3 / 2 − e 13 − e 34 + o ( 1 ) . (11) C_r\le B-b^2+r(x+z-3y)-3r^2-h_3/2-e_{13}-e_{34}+o(1). \tag*{(11)} C r ≤ B − b 2 + r ( x + z − 3 y ) − 3 r 2 − h 3 /2 − e 13 − e 34 + o ( 1 ) . (11) If y ≤ 3 z / 8 y\le3z/8 y ≤ 3 z /8 1 / 2 1/2 1/2 h 3 h_3 h 3 (11) can be replaced by 5 / 8 5/8 5/8 1 − y / z ≥ 5 / 8 1-y/z\ge5/8 1 − y / z ≥ 5/8 (10) .
Bounds on the size of the neutral core
Let a = E A D a=E_{AD} a = E A D c = E C C c=E_{CC} c = E C C m = min ( d , q ) m=\min(d,q) m = min ( d , q ) D = max ( d , q ) D=\max(d,q) D = max ( d , q ) M 2 M^2 M 2 M M M o ( 1 ) o(1) o ( 1 ) 1 / m 1/m 1/ m 1 / D 1/D 1/ D m > 0 m>0 m > 0 § 8 .
First find a maximum blue A D AD A D t M tM tM t ≥ m − a / D t\ge m-a/D t ≥ m − a / D § 3 . Then find a maximum blue C C CC C C A D AD A D E ≤ 9 m 2 / 50 E \le9m^2/50 E ≤ 9 m 2 /50 (8) gives
r 2 ≥ t 2 − 2 c ≥ ( m − a / D ) 2 − 2 c . r^2 \ge t^2 - 2c \ge(m-a/D)^2 - 2c. r 2 ≥ t 2 − 2 c ≥ ( m − a / D ) 2 − 2 c . Indeed c ≤ 9 m 2 / 50 − a ≤ ( 9 / 50 ) ( m − a / D ) 2 c \le9m^2/50-a \le(9/50)(m-a/D)^2 c ≤ 9 m 2 /50 − a ≤ ( 9/50 ) ( m − a / D ) 2
m D a + c ≥ m 2 − r 2 2 + a 2 2 D 2 . (12) \frac{m}{D}a+c \ge\frac{m^2-r^2}{2}+\frac{a^2}{2D^2}. \tag*{(12)} D m a + c ≥ 2 m 2 − r 2 + 2 D 2 a 2 . (12) Missing A A AA AA A C AC A C a a a c c c
In the larger-C D CD C D d = m d=m d = m q = d + h q=d+h q = d + h h ≥ 3 d / 10 h \ge3d/10 h ≥ 3 d /10 E ≤ 9 d 2 / 50 E \le9d^2/50 E ≤ 9 d 2 /50 C C C
q 2 − 2 c ≥ ( 13 d / 10 ) 2 − 9 d 2 / 25 = 133 / 100 d > d . \sqrt{q^2-2c}\ge\sqrt{(13d/10)^2-9d^2/25}=\sqrt{133/100}\,d>d. q 2 − 2 c ≥ ( 13 d /10 ) 2 − 9 d 2 /25 = 133/100 d > d . Here (8) is applicable since c ≤ 9 d 2 / 50 ≤ 9 q 2 / 50 c \le9d^2/50 \le9q^2/50 c ≤ 9 d 2 /50 ≤ 9 q 2 /50 d M + O ( 1 ) dM+O(1) d M + O ( 1 ) D D D A A A ( a / d ) M + O ( 1 ) (a/d)M+O(1) ( a / d ) M + O ( 1 ) C C CC C C D D D A D AD A D C C CC C C
r ≥ d − 2 a d , a ≥ d ( d − r ) 2 . (13) r \ge d-\frac{2a}{d},\qquad a\ge\frac{d(d-r)}{2}. \tag*{(13)} r ≥ d − d 2 a , a ≥ 2 d ( d − r ) . (13) Choose a core of maximum size among all joint blue A D AD A D C C CC C C (12) and (13) . All switching and counting statements used subsequently apply to this actual core.
In particular, if r < d r<d r < d (13) forces a positive number of missing A D AD A D C C CC C C
Averaging the reverse construction
We can also average the choice of the d M dM d M C C CC C C L M LM L M L ≥ q 2 − 2 c − o ( 1 ) L\ge\sqrt{q^2-2c}-o(1) L ≥ q 2 − 2 c − o ( 1 ) d M / 2 dM/2 d M /2 A D AD A D D D D ( d / L ) a M 2 (d/L)aM^2 ( d / L ) a M 2 A D AD A D ( a / L ) M + O ( 1 ) (a/L)M+O(1) ( a / L ) M + O ( 1 ) C C CC C C
2 a ≥ q 2 − 2 c ( d − r ) − o ( 1 ) . (14) 2a\ge\sqrt{q^2-2c}(d-r)-o(1). \tag*{(14)} 2 a ≥ q 2 − 2 c ( d − r ) − o ( 1 ) . (14) For a ratio interval v 0 ≤ h / d ≤ v 1 v_0\le h/d\le v_1 v 0 ≤ h / d ≤ v 1 v 0 ≥ 3 / 10 v_0\ge3/10 v 0 ≥ 3/10
α = 1 1 + v 0 , β = 1 9 ( 1 + v 0 ) , K = d 2 − r 2 2 , J = q ( d − r ) 2 . \alpha=\frac{1}{1+v_0},\qquad\beta=\frac{1}{9(1+v_0)},\qquad K=\frac{d^2-r^2}{2},\qquad J=\frac{q(d-r)}{2}. α = 1 + v 0 1 , β = 9 ( 1 + v 0 ) 1 , K = 2 d 2 − r 2 , J = 2 q ( d − r ) . The elementary inequality 1 − 2 t ≥ 1 − 10 t / 9 \sqrt{1-2t}\ge1-10t/9 1 − 2 t ≥ 1 − 10 t /9 0 ≤ t ≤ 9 / 50 0\le t\le9/50 0 ≤ t ≤ 9/50 ( d − r ) / q ≤ 1 / [ 5 ( 1 + v 0 ) ] (d-r)/q\le1/[5(1+v_0)] ( d − r ) / q ≤ 1/ [ 5 ( 1 + v 0 )]
α a + c ≥ K , a + β c ≥ J . (15) \alpha a+c\ge K,\qquad a+\beta c\ge J. \tag*{(15)} α a + c ≥ K , a + β c ≥ J . (15) These inequalities apply to the fixed maximum core of § 4 . A core constructed later on a restricted set of vertices will have its own size parameter.
Restrictions on labels near the retained vertices
The count bounds will depend on missing edges both inside and outside the retained sets. This section relates those quantities and proves label restrictions that permit later matching replacements without creating critical edges to uncovered vertices.
Edges incident to the Q Q Q
The core replacements imply
Q v ≠ 1 for every good P Q and every old v ∈ X 2 . (16) Qv \ne1 \qquad\text{for every good } PQ \text{ and every old } v \in X_2. \tag*{(16)} Q v = 1 for every good P Q and every old v ∈ X 2 . (16) Indeed the neutral core is a union of disjoint alternating paths, each consisting of four original red label-2 edges and three blue links, with two B B B B B B P P P u v uv uv u v uv uv B , B ′ B,B' B , B ′ P B = 3 PB=3 P B = 3 Q u = 1 Qu=1 Q u = 1 u v ( 2 ) uv(2) uv ( 2 ) P Q ( 3 ) PQ(3) P Q ( 3 ) P B ( 3 ) PB(3) P B ( 3 ) Q u ( 1 ) Qu(1) Q u ( 1 ) v B ′ ( ≤ 4 ) vB'(\le4) v B ′ ( ≤ 4 ) (16) using the original matching, with no assumption that u u u
A good Q Q Q B B ′ ( 4 ) BB'(4) B B ′ ( 4 ) P B = 3 PB=3 P B = 3 B B ′ ( 4 ) BB'(4) B B ′ ( 4 ) P Q ( 3 ) → P B ( 3 ) PQ(3) \to PB(3) P Q ( 3 ) → P B ( 3 ) Q u ( ≤ 3 ) Qu(\le3) Q u ( ≤ 3 ) B ′ B' B ′ u u u Q T = 4 QT=4 QT = 4 Q Q Q
A Tutte–Berge packing stability inequality
Let c ∗ c_* c ∗ C C C K = ( q 2 − r 2 ) / 2 K=(q^2-r^2)/2 K = ( q 2 − r 2 ) /2 d ≥ q d\ge q d ≥ q
( 5 − 4 q r ) c ∗ ≤ c − K + 2 r d a + 2 a 2 d 2 + o ( 1 ) ≤ c − K + 59 25 a + o ( 1 ) . (17) \left(5-\frac{4q}{r}\right)c_* \le c-K+\frac{2r}{d}a+\frac{2a^2}{d^2}+o(1) \le c-K+\frac{59}{25}a+o(1). \tag*{(17)} ( 5 − r 4 q ) c ∗ ≤ c − K + d 2 r a + d 2 2 a 2 + o ( 1 ) ≤ c − K + 25 59 a + o ( 1 ) . (17) Its left coefficient is nonnegative on r ≥ 4 q / 5 r\ge4q/5 r ≥ 4 q /5
Here is the argument. Extend the retained blue C C CC C C R R R R ≥ r R\ge r R ≥ r D D D A A A a M / d + O ( 1 ) aM/d+O(1) a M / d + O ( 1 ) D D D C C CC C C 2 a M / d + O ( 1 ) 2aM/d+O(1) 2 a M / d + O ( 1 ) R ≤ r + 2 a / d + o ( 1 ) R\le r+2a/d+o(1) R ≤ r + 2 a / d + o ( 1 ) c R c_R c R δ = q − R \delta=q-R δ = q − R S S S s s s S S S δ s + o ( 1 ) \delta s+o(1) δ s + o ( 1 ) S S S O ( M ) O(M) O ( M ) S S S s M sM s M
c R ≥ s ( R − 3 s / 2 ) − o ( 1 ) ≥ s R / 4 − o ( 1 ) , s ≤ R / 2. c_R \ge s(R-3s/2)-o(1)\ge sR/4-o(1), \qquad s\le R/2. c R ≥ s ( R − 3 s /2 ) − o ( 1 ) ≥ s R /4 − o ( 1 ) , s ≤ R /2. Since the unmatched set is independent,
c ≥ q 2 − R 2 2 + c R − δ s − o ( 1 ) ≥ q 2 − R 2 2 + ( 5 − 4 q R ) c R − o ( 1 ) . c\ge\frac{q^2-R^2}{2}+c_R-\delta s-o(1)\ge\frac{q^2-R^2}{2}+\left(5-\frac{4q}{R}\right)c_R-o(1). c ≥ 2 q 2 − R 2 + c R − δ s − o ( 1 ) ≥ 2 q 2 − R 2 + ( 5 − R 4 q ) c R − o ( 1 ) . Use c ∗ ≤ c R c_\ast\le c_R c ∗ ≤ c R R R R a ≤ 9 d 2 / 50 a \le9d^2/50 a ≤ 9 d 2 /50 (17) . The only standard input is the Tutte–Berge formula [6 , 7 , 8 ] ; the displayed stability estimate follows from the component count above.
In particular, at a = 0 a = 0 a = 0 c = K c = K c = K r > 4 q / 5 r > 4q/5 r > 4 q /5 c ∗ = o ( 1 ) c_\ast= o(1) c ∗ = o ( 1 ) C C CC C C
A stronger bound when the retained graph is dense
In the proof of (17) , the barrier set S S S s ≤ R / 2 s \le R/2 s ≤ R /2 s s s
Lemma 6.1. Suppose r ≥ 7 q / 8 r \ge7q/8 r ≥ 7 q /8 c ≤ q 2 / 8 c \le q^2/8 c ≤ q 2 /8
7 − 4 q / r 3 c ∗ ≤ c − K + 2 r d a + 2 a 2 d 2 + o ( 1 ) , (18) \frac{7-4q/r}{3}c_\ast\le c-K+\frac{2r}{d}a+\frac{2a^2}{d^2}+o(1), \tag*{(18)} 3 7 − 4 q / r c ∗ ≤ c − K + d 2 r a + d 2 2 a 2 + o ( 1 ) , (18) U ≤ 2 r d a + 4 ( q − r ) 7 r − 4 q ( c − K + 2 r d a + 2 a 2 d 2 ) + o ( 1 ) . (19) U \le\frac{2r}{d}a+\frac{4(q-r)}{7r-4q}\left(c-K+\frac{2r}{d}a+\frac{2a^2}{d^2}\right)+o(1). \tag*{(19)} U ≤ d 2 r a + 7 r − 4 q 4 ( q − r ) ( c − K + d 2 r a + d 2 2 a 2 ) + o ( 1 ) . (19) Here c ∗ c_\ast c ∗ C C C K = ( q 2 − r 2 ) / 2 K=(q^2-r^2)/2 K = ( q 2 − r 2 ) /2 U U U C C C
Proof. Extend the retained blue matching to a maximum one covering mass R R R r ≤ R ≤ r + 2 a / d + o ( 1 ) r \le R \le r+2a/d+o(1) r ≤ R ≤ r + 2 a / d + o ( 1 ) c R c_R c R (17) gives
c ≥ q 2 − R 2 2 + ( 5 − 4 q R ) c R − o ( 1 ) . c \ge\frac{q^2-R^2}{2}+\left(5-\frac{4q}{R}\right)c_R-o(1). c ≥ 2 q 2 − R 2 + ( 5 − R 4 q ) c R − o ( 1 ) . If c R ≥ R 2 / 8 c_R \ge R^2/8 c R ≥ R 2 /8
q 2 8 + ( q − R ) ( 3 q − R ) 8 − o ( 1 ) . \frac{q^2}{8}+\frac{(q-R)(3q-R)}{8}-o(1). 8 q 2 + 8 ( q − R ) ( 3 q − R ) − o ( 1 ) . Consequently, in the limiting region R < q R<q R < q c ≤ q 2 / 8 c \le q^2/8 c ≤ q 2 /8 c R < R 2 / 8 c_R<R^2/8 c R < R 2 /8 s ( R − 3 s / 2 ) s(R-3s/2) s ( R − 3 s /2 ) R 2 / 8 R^2/8 R 2 /8 s ∈ [ R / 6 , R / 2 ] s \in[R/6,R/2] s ∈ [ R /6 , R /2 ] s < R / 6 s<R/6 s < R /6
c R ≥ 3 R 4 s − o ( 1 ) . c_R \ge\frac{3R}{4}s-o(1). c R ≥ 4 3 R s − o ( 1 ) . Its unmatched-to-covered blue density is at most ( q − R ) s + o ( 1 ) ≤ 4 ( q − R ) c R / ( 3 R ) + o ( 1 ) (q-R)s+o(1) \le4(q-R)c_R/(3R)+o(1) ( q − R ) s + o ( 1 ) ≤ 4 ( q − R ) c R / ( 3 R ) + o ( 1 )
c ≥ q 2 − R 2 2 + 7 − 4 q / R 3 c R − o ( 1 ) . c \ge\frac{q^2-R^2}{2}+\frac{7-4q/R}{3}c_R-o(1). c ≥ 2 q 2 − R 2 + 3 7 − 4 q / R c R − o ( 1 ) . Use c ∗ ≤ c R c_\ast\le c_R c ∗ ≤ c R ( R 2 − r 2 ) / 2 ≤ 2 r a / d + 2 a 2 / d 2 + o ( 1 ) (R^2-r^2)/2 \le2ra/d+2a^2/d^2+o(1) ( R 2 − r 2 ) /2 ≤ 2 r a / d + 2 a 2 / d 2 + o ( 1 ) (18) .
To obtain the cross-blue bound, contacts with the newly covered vertices cost at most r ( R − r ) ≤ 2 r a / d + o ( 1 ) r(R-r) \le2ra/d+o(1) r ( R − r ) ≤ 2 r a / d + o ( 1 ) 4 ( q − R ) c R / ( 3 R ) 4(q-R)c_R/(3R) 4 ( q − R ) c R / ( 3 R ) 4 ( q − R ) / ( 7 R − 4 q ) 4(q-R)/(7R-4q) 4 ( q − R ) / ( 7 R − 4 q ) R R R (19) .
For completeness the boundary and finite errors do not require a strict hypothesis c < q 2 / 8 c<q^2/8 c < q 2 /8 R < q R<q R < q R = q R=q R = q o ( 1 ) o(1) o ( 1 ) r ( R − r ) r(R-r) r ( R − r ) U U U o ( 1 ) o(1) o ( 1 ) O ( M ) O(M) O ( M ) O ( M ) O(M) O ( M ) N N N
The complementary region c ≥ q 2 / 8 c \ge q^2/8 c ≥ q 2 /8
C 0 ≤ B 0 − q 2 / 8 + o ( 1 ) , C r ≤ B 0 + f R + o ( 1 ) . C_0 \le B_0-q^2/8+o(1), \qquad C_r \le B_0+f_R+o(1). C 0 ≤ B 0 − q 2 /8 + o ( 1 ) , C r ≤ B 0 + f R + o ( 1 ) . A rational positive-definite certificate on 7 q / 8 ≤ r ≤ q 7q/8 \le r \le q 7 q /8 ≤ r ≤ q γ ( 4 ) ≤ 0.43321 \gamma(4) \le0.43321 γ ( 4 ) ≤ 0.43321 ( x + y + z + w ) ( r − 7 q / 8 ) (x+y+z+w)(r-7q/8) ( x + y + z + w ) ( r − 7 q /8 ) ( r − 7 q / 8 ) ( q − r ) (r-7q/8)(q-r) ( r − 7 q /8 ) ( q − r ) h / q ∈ [ v 0 , v 1 ] h/q \in[v_0,v_1] h / q ∈ [ v 0 , v 1 ] r / q ∈ [ ρ 0 , ρ 1 ] r/q \in[\rho_0,\rho_1] r / q ∈ [ ρ 0 , ρ 1 ]
L = 4 ( 1 − ρ 0 ) 7 ρ 0 − 4 , β = 2 ρ 1 1 + v 0 + 9 / 25 ( 1 + v 0 ) 2 . L=\frac{4(1-\rho_0)}{7\rho_0-4}, \qquad\beta=\frac{2\rho_1}{1+v_0}+\frac{9/25}{(1+v_0)^2}. L = 7 ρ 0 − 4 4 ( 1 − ρ 0 ) , β = 1 + v 0 2 ρ 1 + ( 1 + v 0 ) 2 9/25 . Using maximality of the joint choice of matchings
The core was chosen to have maximum size among all feasible joint blue A D AD A D C C CC C C r r r a a a A D AD A D c c c C C CC C C d ≥ q d \ge q d ≥ q r > 4 q / 5 r > 4q/5 r > 4 q /5
u 0 = a / d , K = ( q 2 − r 2 ) / 2 , E p a c k = c − K + q u 0 − u 0 2 / 2. u_0=a/d, \qquad K=(q^2-r^2)/2, \qquad E_{\mathrm{pack}}=c-K+qu_0-u_0^2/2. u 0 = a / d , K = ( q 2 − r 2 ) /2 , E pack = c − K + q u 0 − u 0 2 /2. Lemma 6.2. With uniform vanishing finite errors,
E p a c k ≥ 0 , c ∗ ≤ E p a c k 5 − 4 q / r + o ( 1 ) , U ≤ r u 0 + 4 ( q − r ) 5 r − 4 q E p a c k + o ( 1 ) . (20) E_{\mathrm{pack}} \ge0,\qquad c_* \le\frac{E_{\mathrm{pack}}}{5-4q/r}+o(1),\qquad U \le ru_0+\frac{4(q-r)}{5r-4q}E_{\mathrm{pack}}+o(1). \tag*{(20)} E pack ≥ 0 , c ∗ ≤ 5 − 4 q / r E pack + o ( 1 ) , U ≤ r u 0 + 5 r − 4 q 4 ( q − r ) E pack + o ( 1 ) . (20) Here c ∗ c_* c ∗ C C C U U U C C C
Proof. Augment the core’s blue A D AD A D A D AD A D D D D t ≥ q − a / d − o ( 1 ) t \ge q-a/d-o(1) t ≥ q − a / d − o ( 1 ) T T T C C C C C CC C C G [ T ] G[T] G [ T ] T T T A D AD A D T T T c T c_T c T T T T s s s T T T
0 ≤ s ≤ r / 2 + o ( 1 ) , c ∗ ≥ s ( r − 3 s / 2 ) − o ( 1 ) ≥ r s / 4 − o ( 1 ) , 0 \le s \le r/2+o(1),\qquad c_* \ge s(r-3s/2)-o(1) \ge rs/4-o(1), 0 ≤ s ≤ r /2 + o ( 1 ) , c ∗ ≥ s ( r − 3 s /2 ) − o ( 1 ) ≥ r s /4 − o ( 1 ) , and the blue density from its unmatched set to its covered set is at most ( t − r ) s + o ( 1 ) (t-r)s+o(1) ( t − r ) s + o ( 1 )
c T ≥ t 2 − r 2 2 + c ∗ − ( t − r ) s − o ( 1 ) ≥ t 2 − r 2 2 + ( 5 − 4 t / r ) c ∗ − o ( 1 ) . c_T \ge\frac{t^2-r^2}{2}+c_*-(t-r)s-o(1) \ge\frac{t^2-r^2}{2}+(5-4t/r)c_*-o(1). c T ≥ 2 t 2 − r 2 + c ∗ − ( t − r ) s − o ( 1 ) ≥ 2 t 2 − r 2 + ( 5 − 4 t / r ) c ∗ − o ( 1 ) . Since c T ≤ c c_T \le c c T ≤ c t ≤ q t \le q t ≤ q t ≥ q − u 0 − o ( 1 ) t \ge q-u_0-o(1) t ≥ q − u 0 − o ( 1 ) C C C T T T u 0 + o ( 1 ) u_0+o(1) u 0 + o ( 1 ) r u 0 + o ( 1 ) ru_0+o(1) r u 0 + o ( 1 ) 4 ( t − r ) c ∗ / r + o ( 1 ) 4(t-r)c_*/r+o(1) 4 ( t − r ) c ∗ / r + o ( 1 )
The maximum-core choice preserves all label restrictions and count inequalities proved for arbitrary neutral cores, as well as the lower bounds on r r r § 5 . All auxiliary graphs have O ( M ) O(M) O ( M )
The same argument also gives inequalities retaining the barrier size. For this purpose retain an actual u = q − t ∈ [ 0 , u 0 ] u=q-t \in[0,u_0] u = q − t ∈ [ 0 , u 0 ] s ∈ [ 0 , r / 2 ] s \in[0,r/2] s ∈ [ 0 , r /2 ]
c ∗ ≥ s ( r − 3 s / 2 ) , c ∗ ≤ c − K + q u − u 2 / 2 + ( q − u − r ) s , c_* \ge s(r-3s/2),\qquad c_* \le c-K+qu-u^2/2+(q-u-r)s, c ∗ ≥ s ( r − 3 s /2 ) , c ∗ ≤ c − K + q u − u 2 /2 + ( q − u − r ) s , U ≤ r u + ( q − u − r ) s . U \le ru + (q-u-r)s. U ≤ r u + ( q − u − r ) s . Thus
c − K + q u − u 2 / 2 ≥ s ( 2 r − q + u − 3 s / 2 ) . c-K+qu-u^2/2 \ge s(2r-q+u-3s/2). c − K + q u − u 2 /2 ≥ s ( 2 r − q + u − 3 s /2 ) . In particular, at equality in the packing constraint, E p a c k = 0 E_{\mathrm{pack}}=0 E pack = 0 U ≤ r a / d U \le ra/d U ≤ r a / d
The internal count on the label-3 class
Write Z Z Z P i Q i P_iQ_i P i Q i β = z − Z \beta=z-Z β = z − Z I 3 = z 2 / 2 − C 33 ≥ 0 I_3=z^2/2-C_{33}\ge0 I 3 = z 2 /2 − C 33 ≥ 0 E P P E_{PP} E P P P P P
E P P ≤ I 3 + o ( 1 ) . (21) E_{PP}\le I_3+o(1). \tag*{(21)} E P P ≤ I 3 + o ( 1 ) . (21) First, a good Q i Q_i Q i R R R R S ( 3 ) RS(3) R S ( 3 )
B B ′ ( 4 ) , P i Q i ( 3 ) , R S ( 3 ) ⟶ P i B ( 3 ) , Q i R ( ≤ 2 ) , B ′ S ( ≤ 4 ) BB'(4),\ P_iQ_i(3),\ RS(3)\longrightarrow P_iB(3),\ Q_iR(\le2),\ B'S(\le4) B B ′ ( 4 ) , P i Q i ( 3 ) , R S ( 3 ) ⟶ P i B ( 3 ) , Q i R ( ≤ 2 ) , B ′ S ( ≤ 4 ) would reduce primary cost from 8 to at most 7. Second, if R P i ≤ 2 RP_i\le2 R P i ≤ 2 S P j ≤ 2 SP_j\le2 S P j ≤ 2 original label-3 matching edges by
R P i ( ≤ 2 ) , S P j ( ≤ 2 ) , Q i Q j ( ≤ 4 ) . RP_i(\le2),\ SP_j(\le2),\ Q_iQ_j(\le4). R P i ( ≤ 2 ) , S P j ( ≤ 2 ) , Q i Q j ( ≤ 4 ) . Primary cost is at most 6; if it equals 6, the secondary cost decreases from 3 to 2. This contradicts the choice of M 0 \mathcal{M}_0 M 0 Z M + O ( 1 ) ZM+O(1) Z M + O ( 1 )
C 33 ≤ Z 2 / 2 − E P P + Z β + β 2 / 2 + o ( 1 ) = z 2 / 2 − E P P + o ( 1 ) , C_{33}\le Z^2/2-E_{PP}+Z\beta+\beta^2/2+o(1)=z^2/2-E_{PP}+o(1), C 33 ≤ Z 2 /2 − E P P + Z β + β 2 /2 + o ( 1 ) = z 2 /2 − E P P + o ( 1 ) , which proves (21) .
Rematching through the label-4 class
We next construct a matching by pairing retained C C C X 4 X_4 X 4 D D D reservoir comparison ; the auxiliary set of D D D
Every retained D D D A A A B B B
B B ′ ( 4 ) , A D ( 1 ) ⟶ D u ( ≤ 3 ) , A B ( 2 ) BB'(4),\ AD(1)\longrightarrow Du(\le3),\ AB(2) B B ′ ( 4 ) , A D ( 1 ) ⟶ D u ( ≤ 3 ) , A B ( 2 ) would reduce primary weight from 4 to at most 3. The matching size is unchanged; B ′ B' B ′ u u u C ′ C' C ′ C C C D D D C ′ C' C ′
B B ′ ( 4 ) , A D ( 1 ) , C C ′ ( 1 ) ⟶ C u ( ≤ 2 ) , C ′ D ( 2 ) , A B ( 2 ) BB'(4),\ AD(1),\ CC'(1)\longrightarrow Cu(\le2),\ C'D(2),\ AB(2) B B ′ ( 4 ) , A D ( 1 ) , C C ′ ( 1 ) ⟶ C u ( ≤ 2 ) , C ′ D ( 2 ) , A B ( 2 ) proves that retained C C C T T ′ ( 4 ) TT'(4) T T ′ ( 4 ) D T = 4 DT=4 D T = 4 C T ≥ 3 CT\ge3 C T ≥ 3 e 24 M 2 e_{24}M^2 e 24 M 2 C C C
Choose a label-3 matching from retained C C C t M tM tM t M tM tM t t t C C C t ≥ [ r − e 24 / ( 2 w ) ] + − o ( 1 ) t\ge[r-e_{24}/(2w)]_+-o(1) t ≥ [ r − e 24 / ( 2 w ) ] + − o ( 1 ) r ≤ 2 w r\le2w r ≤ 2 w D D D s s s u D = t − s u_D=t-s u D = t − s A B AB A B C D CD C D X 4 X_4 X 4 M M M C T C_T C T
A matched edge has at most u D + o ( 1 ) u_D+o(1) u D + o ( 1 ) D D D D D DD D D
u D s / 2 + u D ( q − t ) = u D ( q − t / 2 ) − u D 2 / 2 ≤ ( q − t / 2 ) 2 / 2. u_Ds/2+u_D(q-t)=u_D(q-t/2)-u_D^2/2\le(q-t/2)^2/2. u D s /2 + u D ( q − t ) = u D ( q − t /2 ) − u D 2 /2 ≤ ( q − t /2 ) 2 /2. Every new label-2 D D DD D D C C CC C C C C CC C C D D ( 2 ) DD(2) D D ( 2 ) C D ( 2 ) CD(2) C D ( 2 ) D D DD D D D D DD D D ( q − t / 2 ) 2 / 2 + E C C (q-t/2)^2/2+E_{CC} ( q − t /2 ) 2 /2 + E C C
The remaining direct contact count, with a = E A D a=E_{AD} a = E A D c = E C C c=E_{CC} c = E C C j = E A C j=E_{AC} j = E A C
C T − C 0 ≤ t ( 3 2 q − r − z + 2 y + 3 h 3 r ) − 7 8 t 2 + q 2 2 + 2 c + j + ( 2 + t / r ) a + e 12 + e 23 + e 24 + e 34 + o ( 1 ) . C_T-C_0\le t\left(\frac{3}{2}q-r-z+2y+\frac{3h_3}{r}\right)-\frac{7}{8}t^2+\frac{q^2}{2}
+2c+j+(2+t/r)a+e_{12}+e_{23}+e_{24}+e_{34}+o(1). C T − C 0 ≤ t ( 2 3 q − r − z + 2 y + r 3 h 3 ) − 8 7 t 2 + 2 q 2 + 2 c + j + ( 2 + t / r ) a + e 12 + e 23 + e 24 + e 34 + o ( 1 ) . Indeed C D CD C D t q + t ( q − t ) tq+t(q-t) tq + t ( q − t ) t r − e 24 tr-e_{24} t r − e 24 C C C t ( z − h 3 / r ) − e 34 t(z-h_3/r)-e_{34} t ( z − h 3 / r ) − e 34 P P P 2 t h 3 / r 2th_3/r 2 t h 3 / r X 1 X_1 X 1 B B B A A A D D D ( t / r ) a (t/r)a ( t / r ) a c c c C C C D D ( 2 ) DD(2) D D ( 2 )
The derivative of the displayed t t t − ( 5 r / 4 + z ) -(5r/4+z) − ( 5 r /4 + z ) 0 ≤ t ≤ r ≤ q 0\le t\le r\le q 0 ≤ t ≤ r ≤ q (9) gives
C T ≤ B − b 2 + 3 2 q r − 15 8 r 2 − r z + 2 r y + q 2 2 + 3 h 3 C_T\le B-b^2+\frac{3}{2}qr-\frac{15}{8}r^2-rz+2ry+\frac{q^2}{2}+3h_3 C T ≤ B − b 2 + 2 3 q r − 8 15 r 2 − r z + 2 r y + 2 q 2 + 3 h 3 − E A A + 2 a + c − e 13 − e 14 + 5 r / 4 + z 2 w e 24 + o ( 1 ) . (22) -E_{AA}+2a+c-e_{13}-e_{14}+\frac{5r/4+z}{2w}e_{24}+o(1). \tag*{(22)} − E AA + 2 a + c − e 13 − e 14 + 2 w 5 r /4 + z e 24 + o ( 1 ) . (22) All selected C C C D D D D D D
Missing edges wholly outside the retained set
Here q = m ≤ d q=m\le d q = m ≤ d C C C D D D A A A C C CC C C a / d a/d a / d D D D C C CC C C r C r_C r C r C ≤ r + 2 a / d r_C\le r+2a/d r C ≤ r + 2 a / d
Outside the actual core’s C C C a M / d + o ( M ) aM/d+o(M) a M / d + o ( M ) k k k k ∣ V ∣ + O ( ∣ V ∣ ) k|V|+O(|V|) k ∣ V ∣ + O ( ∣ V ∣ ) C C CC C C
c o u t ≥ ( q − r ) 2 / 2 − a ( q − r ) / d − o ( 1 ) . c_{\mathrm{out}}\ge(q-r)^2/2-a(q-r)/d-o(1). c out ≥ ( q − r ) 2 /2 − a ( q − r ) / d − o ( 1 ) . Both positive c c c C C C c o u t c_{\mathrm{out}} c out 2 c o u t 2c_{\mathrm{out}} 2 c out (22) is valid. Since q − r ≤ q / 5 ≤ d / 5 q-r\le q/5\le d/5 q − r ≤ q /5 ≤ d /5 12 a / 5 + c − E A A 12a/5+c-E_{AA} 12 a /5 + c − E AA
f T = 7 2 q r − 23 8 r 2 − r z + 2 r y − q 2 2 . (23) f_T=\frac{7}{2}qr-\frac{23}{8}r^2-rz+2ry-\frac{q^2}{2}. \tag*{(23)} f T = 2 7 q r − 8 23 r 2 − r z + 2 r y − 2 q 2 . (23) On r + z ≤ 6 w r+z\le6w r + z ≤ 6 w r ≤ 2 w r\le2w r ≤ 2 w e 24 e_{24} e 24 (22) is at most $13/4.
Averaging the choice of top vertices
Select uniformly a subset T 0 T_0 T 0 r r r 2 w 2w 2 w C C C T 0 T_0 T 0 t = r − δ t=r-\delta t = r − δ r e 24 / ( 2 w ) r e_{24}/(2w) r e 24 / ( 2 w ) t M tM tM r δ M 2 r\delta M^2 r δ M 2
E δ ≤ e 24 2 w + o ( 1 ) . \mathbb{E}\delta\le\frac{e_{24}}{2w}+o(1). E δ ≤ 2 w e 24 + o ( 1 ) . Now choose the maximum low-label matching on the corresponding D D D T 0 T_0 T 0 C T CT C T
Every old X 3 X_3 X 3 2 z w − e 34 2zw-e_{34} 2 z w − e 34 δ M \delta M δ M
L T C ≥ r 2 − r E δ − r 2 w e 24 − o ( 1 ) , L T 3 ≥ r z − 2 z E δ − r 2 w e 34 − o ( 1 ) . L_{TC} \ge r^2-r\mathbb{E}\delta-\frac{r}{2w}e_{24}-o(1),\qquad L_{T3} \ge rz-2z\mathbb{E}\delta-\frac{r}{2w}e_{34}-o(1). L T C ≥ r 2 − r E δ − 2 w r e 24 − o ( 1 ) , L T 3 ≥ r z − 2 z E δ − 2 w r e 34 − o ( 1 ) . This counts contacts with all original label-3 pairs, so there is no exceptional-pair allowance for these top-vertex losses. The remaining D D D 2 t h 3 / r ≤ 2 h 3 2th_3/r\le2h_3 2 t h 3 / r ≤ 2 h 3
After these losses are separated, the t t t F ( t ) = t ( 3 q / 2 + 2 y ) − 7 t 2 / 8 F(t)=t(3q/2+2y)-7t^2/8 F ( t ) = t ( 3 q /2 + 2 y ) − 7 t 2 /8 F ′ ( t ) ≥ − r / 4 F'(t)\ge-r/4 F ′ ( t ) ≥ − r /4 t t t r r r 3 h 3 3h_3 3 h 3 2 h 3 2h_3 2 h 3 e 24 , e 34 e_{24},e_{34} e 24 , e 34
9 r / 4 + 2 z 2 w − 1 ≤ 21 4 , r 2 w − 1 ≤ 0. \frac{9r/4+2z}{2w}-1\le\frac{21}{4},\qquad\frac{r}{2w}-1\le0. 2 w 9 r /4 + 2 z − 1 ≤ 4 21 , 2 w r − 1 ≤ 0. The first bound uses r + z ≤ 6 w r+z\le6w r + z ≤ 6 w r ≤ 2 w r\le2w r ≤ 2 w C C CC C C
C T ≤ B − b 2 + f T + 2 h 3 + 12 5 a + c − E A A − e 13 − e 14 + 21 4 e 24 + o ( 1 ) . (24) C_T\le B-b^2+f_T+2h_3+\frac{12}{5}a+c-E_{AA}-e_{13}-e_{14}+\frac{21}{4}e_{24}+o(1). \tag*{(24)} C T ≤ B − b 2 + f T + 2 h 3 + 5 12 a + c − E AA − e 13 − e 14 + 4 21 e 24 + o ( 1 ) . (24) The averaging selects among valid finite matchings. Its error terms are uniform in the number of originally uncovered vertices.
Separating the two labels in the D D DD D D
We can bound together two types of newly critical D D DD D D C C CC C C D D DD D D C C CC C C U o u t U_{\mathrm{out}} U out C C C I I I t ≤ r t\le r t ≤ r c I c_I c I C C CC C C I I I u u u D D D
Every newly critical label-2 D D DD D D C C CC C C D D DD D D C C CC C C c I c_I c I D D DD D D u ( t − u ) / 2 u(t-u)/2 u ( t − u ) /2 C C CC C C u ( r − t ) + U o u t u(r-t)+U_{\mathrm{out}} u ( r − t ) + U out D D DD D D
c I + u ( r − t / 2 ) − u 2 / 2 + U o u t ≤ c I + ( r − t / 2 ) 2 / 2 + U o u t + o ( 1 ) . c_I+u(r-t/2)-u^2/2+U_{\mathrm{out}}\le c_I+(r-t/2)^2/2+U_{\mathrm{out}}+o(1). c I + u ( r − t /2 ) − u 2 /2 + U out ≤ c I + ( r − t /2 ) 2 /2 + U out + o ( 1 ) . The corresponding bound used in (24) was c I + ( q − t / 2 ) 2 / 2 c_I+(q-t/2)^2/2 c I + ( q − t /2 ) 2 /2 t ≤ r t\le r t ≤ r
( r − t / 2 ) 2 − ( q − t / 2 ) 2 2 = − ( q − r ) ( q + r − t ) 2 ≤ − q ( q − r ) 2 . \frac{(r-t/2)^2-(q-t/2)^2}{2}
=-\frac{(q-r)(q+r-t)}{2}\le-\frac{q(q-r)}{2}. 2 ( r − t /2 ) 2 − ( q − t /2 ) 2 = − 2 ( q − r ) ( q + r − t ) ≤ − 2 q ( q − r ) . This inequality holds before averaging, so it incurs no additional e 24 e_{24} e 24 C C C
The blue resource U o u t U_{\mathrm{out}} U out C C CC C C R R R r ≤ R ≤ r + 2 a / d + o ( 1 ) r \le R \le r + 2a/d + o(1) r ≤ R ≤ r + 2 a / d + o ( 1 ) c R c_R c R (17) gives
blue ( unmatched , covered ) ≤ 4 ( q − R ) R c R + o ( 1 ) , ( 5 − 4 q R ) c R ≤ c − q 2 − R 2 2 + o ( 1 ) . \operatorname{blue}(\text{unmatched},\text{ covered}) \le\frac{4(q-R)}{R}c_R+o(1), \qquad\left(5-\frac{4q}{R}\right)c_R \le c-\frac{q^2-R^2}{2}+o(1). blue ( unmatched , covered ) ≤ R 4 ( q − R ) c R + o ( 1 ) , ( 5 − R 4 q ) c R ≤ c − 2 q 2 − R 2 + o ( 1 ) . Contacts with the newly covered vertices cost at most r ( R − r ) r(R-r) r ( R − r ) r > 4 q / 5 r>4q/5 r > 4 q /5
K = q 2 − r 2 2 , L = 4 ( q − r ) 5 r − 4 q , β = 2 r d + 2 a d 2 . K=\frac{q^2-r^2}{2}, \qquad L=\frac{4(q-r)}{5r-4q}, \qquad\beta=\frac{2r}{d}+\frac{2a}{d^2}. K = 2 q 2 − r 2 , L = 5 r − 4 q 4 ( q − r ) , β = d 2 r + d 2 2 a . Then
U o u t ≤ 2 r d a + L ( c − K + β a ) + o ( 1 ) . (25) U_{\mathrm{out}} \le\frac{2r}{d}a+L(c-K+\beta a)+o(1). \tag*{(25)} U out ≤ d 2 r a + L ( c − K + β a ) + o ( 1 ) . (25) Here K ( r ) − K ( R ) ≤ 2 r a / d + 2 a 2 / d 2 ≤ β a K(r)-K(R)\le2ra/d+2a^2/d^2\le\beta a K ( r ) − K ( R ) ≤ 2 r a / d + 2 a 2 / d 2 ≤ β a L ≥ 0 L\ge0 L ≥ 0 o ( 1 ) o(1) o ( 1 ) r = q r=q r = q L = 0 L=0 L = 0
Before rounding its coefficient to 12 / 5 12/5 12/5 2 + 2 ( q − r ) / d 2+2(q-r)/d 2 + 2 ( q − r ) / d (25) therefore gives
C T ≤ B − b 2 + f T − q ( q − r ) 2 − L K + 2 h 3 + ( 2 + 2 q d + L β ) a + ( 1 + L ) c − E A A − e 13 − e 14 + 21 4 e 24 + o ( 1 ) . (26) \begin{aligned}
C_T &\le B-b^2+f_T-\frac{q(q-r)}{2}-LK+2h_3 \\
&\quad+\left(2+\frac{2q}{d}+L\beta\right)a+(1+L)c-E_{AA}-e_{13}-e_{14}+\frac{21}{4}e_{24}+o(1). \tag*{(26)}
\end{aligned} C T ≤ B − b 2 + f T − 2 q ( q − r ) − L K + 2 h 3 + ( 2 + d 2 q + L β ) a + ( 1 + L ) c − E AA − e 13 − e 14 + 4 21 e 24 + o ( 1 ) . (26) On a ratio box h / q ≥ v 0 h/q\ge v_0 h / q ≥ v 0 r / q ≥ ρ 0 > 4 / 5 r/q\ge\rho_0>4/5 r / q ≥ ρ 0 > 4/5 L L L 4 ( 1 − ρ 0 ) / ( 5 ρ 0 − 4 ) 4(1-\rho_0)/(5\rho_0-4) 4 ( 1 − ρ 0 ) / ( 5 ρ 0 − 4 ) β \beta β 2 / ( 1 + v 0 ) + ( 9 / 25 ) / ( 1 + v 0 ) 2 2/(1+v_0)+(9/25)/(1+v_0)^2 2/ ( 1 + v 0 ) + ( 9/25 ) / ( 1 + v 0 ) 2 a ≤ 9 q 2 / 50 a\le9q^2/50 a ≤ 9 q 2 /50 L L L c − K + β a c-K+\beta a c − K + β a
Including the exceptional label-3 pairs
Write b 3 b_3 b 3 b 3 ≤ h 3 / r + o ( 1 ) b_3\le h_3/r+o(1) b 3 ≤ h 3 / r + o ( 1 ) C T CT C T t t t D D D 2 b 3 2b_3 2 b 3 D D D u u u
The low-label matching covers mass t + 2 b 3 − u t+2b_3-u t + 2 b 3 − u D D D u ( t + 2 b 3 − u ) / 2 + o ( 1 ) u(t+2b_3-u)/2+o(1) u ( t + 2 b 3 − u ) /2 + o ( 1 ) D D ( 3 ) DD(3) D D ( 3 ) C C CC C C
c I + u ( r − t / 2 + b 3 ) − u 2 / 2 + U o u t ≤ c I + ( r − t / 2 + b 3 ) 2 / 2 + U o u t + o ( 1 ) . c_I+u(r-t/2+b_3)-u^2/2+U_{\mathrm{out}}\le c_I+(r-t/2+b_3)^2/2+U_{\mathrm{out}}+o(1). c I + u ( r − t /2 + b 3 ) − u 2 /2 + U out ≤ c I + ( r − t /2 + b 3 ) 2 /2 + U out + o ( 1 ) . There is no separate 2 t b 3 2tb_3 2 t b 3
Before top-demotion losses, the target-dependent polynomial is
g ( t ) = t ( 2 q + 2 y ) − t 2 + ( r − t / 2 + b 3 ) 2 / 2 , g ′ ( t ) ≥ − r / 4 − b 3 / 2. g(t)=t(2q+2y)-t^2+(r-t/2+b_3)^2/2,\qquad g'(t)\ge-r/4-b_3/2. g ( t ) = t ( 2 q + 2 y ) − t 2 + ( r − t /2 + b 3 ) 2 /2 , g ′ ( t ) ≥ − r /4 − b 3 /2. The same random sampling therefore adds at most b 3 e 24 / ( 4 w ) b_3e_{24}/(4w) b 3 e 24 / ( 4 w ) t = r t=r t = r r 2 / 8 + r b 3 / 2 + b 3 2 / 2 r^2/8+rb_3/2+b_3^2/2 r 2 /8 + r b 3 /2 + b 3 2 /2 (26) strengthens to
C T ≤ B − b 2 + f T − q ( q − r ) 2 − L K + h 3 2 + h 3 2 2 r 2 + ( 2 + 2 q d + L β ) a + ( 1 + L ) c − E A A − e 13 − e 14 + 13 2 e 24 + o ( 1 ) . (27) \begin{aligned}
C_T &\le B-b^2+f_T-\frac{q(q-r)}{2}-LK+\frac{h_3}{2}+\frac{h_3^2}{2r^2} \\
&\quad+\left(2+\frac{2q}{d}+L\beta\right)a+(1+L)c-E_{AA}-e_{13}-e_{14}+\frac{13}{2}e_{24}+o(1). \tag*{(27)}
\end{aligned} C T ≤ B − b 2 + f T − 2 q ( q − r ) − L K + 2 h 3 + 2 r 2 h 3 2 + ( 2 + d 2 q + L β ) a + ( 1 + L ) c − E AA − e 13 − e 14 + 2 13 e 24 + o ( 1 ) . (27) The last coefficient follows from b 3 ≤ z b_3 \le z b 3 ≤ z r + z ≤ 6 w r+z \le6w r + z ≤ 6 w ( 9 r / 4 + 2 z + b 3 / 2 ) / ( 2 w ) − 1 ≤ 13 / 2 (9r/4+2z+b_3/2)/(2w)-1 \le13/2 ( 9 r /4 + 2 z + b 3 /2 ) / ( 2 w ) − 1 ≤ 13/2 § 13 use its complete defect terms.
On the additional low-c c c (19) improves (27) by replacing L L L 4 ( q − r ) / ( 7 r − 4 q ) 4(q-r)/(7r-4q) 4 ( q − r ) / ( 7 r − 4 q ) h / q ≥ v 0 h/q \ge v_0 h / q ≥ v 0 r / q ∈ [ ρ 0 , ρ 1 ] r/q \in[\rho_0,\rho_1] r / q ∈ [ ρ 0 , ρ 1 ]
L ≤ 4 ( 1 − ρ 0 ) 7 ρ 0 − 4 , β ≤ 2 ρ 1 1 + v 0 + 9 / 25 ( 1 + v 0 ) 2 . L \le\frac{4(1-\rho_0)}{7\rho_0-4}, \qquad\beta\le\frac{2\rho_1}{1+v_0}+\frac{9/25}{(1+v_0)^2}. L ≤ 7 ρ 0 − 4 4 ( 1 − ρ 0 ) , β ≤ 1 + v 0 2 ρ 1 + ( 1 + v 0 ) 2 9/25 . As before, the coefficient L L L c − K + β a c-K+\beta a c − K + β a − L K -LK − L K
For the final bound we use maximality of the joint A D AD A D C C CC C C U o u t U_{\mathrm{out}} U out
C T ≤ B − b 2 + f T − q ( q − r ) 2 − L K + h 3 2 + h 3 2 2 r 2 + ( 2 + 2 q − r d + L q d ) a + ( 1 + L ) c − E A A − e 13 − e 14 + 13 2 e 24 − I 3 + o ( 1 ) , (28) \begin{aligned}
C_T &\le B-b^2+f_T-\frac{q(q-r)}{2}-LK+\frac{h_3}{2}+\frac{h_3^2}{2r^2}\\
&\quad+\left(2+\frac{2q-r}{d}+L\frac{q}{d}\right)a+(1+L)c-E_{AA}-e_{13}-e_{14}+\frac{13}{2}e_{24}-I_3+o(1), \tag*{(28)}
\end{aligned} C T ≤ B − b 2 + f T − 2 q ( q − r ) − L K + 2 h 3 + 2 r 2 h 3 2 + ( 2 + d 2 q − r + L d q ) a + ( 1 + L ) c − E AA − e 13 − e 14 + 2 13 e 24 − I 3 + o ( 1 ) , (28) L = 4 ( q − r ) 5 r − 4 q , K = q 2 − r 2 2 . L=\frac{4(q-r)}{5r-4q}, \qquad K=\frac{q^2-r^2}{2}. L = 5 r − 4 q 4 ( q − r ) , K = 2 q 2 − r 2 . The favorable quadratic packing term − a 2 / ( 2 d 2 ) -a^2/(2d^2) − a 2 / ( 2 d 2 ) r > 4 q / 5 r>4q/5 r > 4 q /5 C C C
A preliminary bound and restriction of the parameter domain
We first combine only the original and neutral-core counts. Their quadratic bounds give both the preliminary estimate 23 / 53 23/53 23/53
f = y 2 + x 2 4 + z 2 2 + w 2 , B 0 = 1 − f − b 2 , m = x / 2 − b . f=y^2+\frac{x^2}{4}+\frac{z^2}{2}+w^2,\qquad B_0=1-f-b^2,\qquad m=x/2-b. f = y 2 + 4 x 2 + 2 z 2 + w 2 , B 0 = 1 − f − b 2 , m = x /2 − b . where x + y + z + w = 1 x+y+z+w=1 x + y + z + w = 1
Entry into the dense region
If E ≥ 9 m 2 / 50 E\ge9m^2/50 E ≥ 9 m 2 /50
x 2 4 + b 2 + 9 50 ( x / 2 − b ) 2 = 17 x 2 59 + 59 50 ( b − 9 x / 118 ) 2 . \frac{x^2}{4}+b^2+\frac{9}{50}(x/2-b)^2=\frac{17x^2}{59}+\frac{59}{50}(b-9x/118)^2. 4 x 2 + b 2 + 50 9 ( x /2 − b ) 2 = 59 17 x 2 + 50 59 ( b − 9 x /118 ) 2 . Weighted Cauchy–Schwarz consequently bounds the original count by
C 0 ≤ 1 − y 2 − 17 x 2 59 − z 2 2 − w 2 + o ( 1 ) ≤ 110 127 + o ( 1 ) . C_0\le1-y^2-\frac{17x^2}{59}-\frac{z^2}{2}-w^2+o(1)\le\frac{110}{127}+o(1). C 0 ≤ 1 − y 2 − 59 17 x 2 − 2 z 2 − w 2 + o ( 1 ) ≤ 127 110 + o ( 1 ) . Indeed the reciprocals of the four coefficients sum to 127 / 17 127/17 127/17
Otherwise the packing construction gives an actual core with
r 2 ≥ m 2 − 2 E − o ( 1 ) ≥ 16 25 m 2 − o ( 1 ) , E ≥ m 2 − r 2 2 − o ( 1 ) . r^2\ge m^2-2E-o(1)\ge\frac{16}{25}m^2-o(1),\qquad E\ge\frac{m^2-r^2}{2}-o(1). r 2 ≥ m 2 − 2 E − o ( 1 ) ≥ 25 16 m 2 − o ( 1 ) , E ≥ 2 m 2 − r 2 − o ( 1 ) . The maximum core fixed in § 4 is at least as large as the core used to establish these inequalities, so both remain valid for its size r r r 2 G 2G 2 G m m m C 0 > 2 G C_0 > 2G C 0 > 2 G
1 − f = 7 8 − Q , Q = ( y − 1 / 8 ) 2 + ( x − 1 / 2 ) 2 / 4 + ( z − 1 / 4 ) 2 / 2 + ( w − 1 / 8 ) 2 , 1-f=\frac{7}{8}-Q,\qquad Q=(y-1/8)^2+(x-1/2)^2/4+(z-1/4)^2/2+(w-1/8)^2, 1 − f = 8 7 − Q , Q = ( y − 1/8 ) 2 + ( x − 1/2 ) 2 /4 + ( z − 1/4 ) 2 /2 + ( w − 1/8 ) 2 , and
( m − 1 / 4 ) 2 2 ≤ Q + b 2 < 7 / 8 − 2 G + o ( 1 ) . \frac{(m-1/4)^2}{2}\leq Q+b^2<7/8-2G+o(1). 2 ( m − 1/4 ) 2 ≤ Q + b 2 < 7/8 − 2 G + o ( 1 ) . Thus
m ≥ 1 4 − 2 ( 7 / 8 − 2 G ) − o ( 1 ) . (29) m\geq\frac{1}{4}-\sqrt{2(7/8-2G)}-o(1). \tag*{(29)} m ≥ 4 1 − 2 ( 7/8 − 2 G ) − o ( 1 ) . (29) For every final threshold used below the right side exceeds 1 / 9 1/9 1/9 r ≥ 4 m / 5 − o ( 1 ) r\geq4m/5-o(1) r ≥ 4 m /5 − o ( 1 ) m / 2 ≤ r ≤ m m/2\leq r\leq m m /2 ≤ r ≤ m M M M 1 / 9 1/9 1/9 m m m
The other initial absorption conditions follow just as directly. Minimizing f f f
y ≥ 4 w ⟹ f ≥ 17 / 127 , x ≥ 16 w ⟹ f ≥ 65 / 484. y\geq4w\ \Longrightarrow\ f\geq17/127,\qquad x\geq16w\ \Longrightarrow\ f\geq65/484. y ≥ 4 w ⟹ f ≥ 17/127 , x ≥ 16 w ⟹ f ≥ 65/484. For the first calculation the active boundary is y = 4 w y=4w y = 4 w x , z x,z x , z ( 1 − 5 w ) 2 / 6 (1-5w)^2/6 ( 1 − 5 w ) 2 /6 17 w 2 + ( 1 − 5 w ) 2 / 6 17w^2+(1-5w)^2/6 17 w 2 + ( 1 − 5 w ) 2 /6 17 / 127 17/127 17/127 x = 16 w x=16w x = 16 w 65 w 2 + ( 1 − 17 w ) 2 / 3 65w^2+(1-17w)^2/3 65 w 2 + ( 1 − 17 w ) 2 /3 65 / 484 65/484 65/484 y < 4 w y<4w y < 4 w x < 16 w x<16w x < 16 w
2 G > max { 110 / 127 , 1 − 17 / 127 , 1 − 65 / 484 } . (30) 2G>\max\{110/127,\ 1-17/127,\ 1-65/484\}. \tag*{(30)} 2 G > max { 110/127 , 1 − 17/127 , 1 − 65/484 } . (30) These inequalities are checked exactly for the final value of G G G G = 23 / 53 G=23/53 G = 23/53
The two comparison quadratics
Let S = e 14 + e 23 + e 24 S=e_{14}+e_{23}+e_{24} S = e 14 + e 23 + e 24 § 4.2 , retaining the terms h 1 , h 3 h_1,h_3 h 1 , h 3
h 1 + h 3 ≥ [ ( 2 y w − e 14 ) ( r z − e 23 ) − 2 z y e 24 ] + max ( 2 z w , 2 y w − e 14 ) − o ( 1 ) . h_1+h_3\geq\frac{\left[(2yw-e_{14})(rz-e_{23})-2zye_{24}\right]_+}{\max(2zw,2yw-e_{14})}-o(1). h 1 + h 3 ≥ max ( 2 z w , 2 y w − e 14 ) [ ( 2 y w − e 14 ) ( r z − e 23 ) − 2 z y e 24 ] + − o ( 1 ) . Put D ∗ = max ( 2 z w , 2 y w ) D_*=\max(2zw,2yw) D ∗ = max ( 2 z w , 2 y w ) e 14 e 23 e_{14}e_{23} e 14 e 23
r min ( y , z ) − r z D ∗ e 14 − 2 y w D ∗ e 23 − 2 z y D ∗ e 24 − o ( 1 ) . r\min(y,z)-\frac{rz}{D_*}e_{14}-\frac{2yw}{D_*}e_{23}-\frac{2zy}{D_*}e_{24}-o(1). r min ( y , z ) − D ∗ r z e 14 − D ∗ 2 y w e 23 − D ∗ 2 z y e 24 − o ( 1 ) . The inequality remains valid if its untruncated right side is negative. The entry conditions give
r z D ∗ ≤ r 2 w ≤ x 4 w < 4 , 2 y w D ∗ ≤ 1 , 2 z y D ∗ ≤ y w < 4. \frac{rz}{D_*}\leq\frac{r}{2w}\leq\frac{x}{4w}<4,\qquad\frac{2yw}{D_*}\leq1,\qquad\frac{2zy}{D_*}\leq\frac{y}{w}<4. D ∗ r z ≤ 2 w r ≤ 4 w x < 4 , D ∗ 2 y w ≤ 1 , D ∗ 2 z y ≤ w y < 4. Zero-denominator cases are either excluded by the original-count threshold or handled by assigning penalty zero. Consequently h 1 + h 3 ≥ r min ( y , z ) − 4 S − o ( 1 ) h_1+h_3\geq r\min(y,z)-4S-o(1) h 1 + h 3 ≥ r min ( y , z ) − 4 S − o ( 1 )
Define
A = B 0 − m 2 − r 2 2 , B ∗ = B 0 + r ( x + z − 2 y ) − 3 r 2 − r min ( y , z ) . A=B_0-\frac{m^2-r^2}{2},\qquad B_*=B_0+r(x+z-2y)-3r^2-r\min(y,z). A = B 0 − 2 m 2 − r 2 , B ∗ = B 0 + r ( x + z − 2 y ) − 3 r 2 − r min ( y , z ) . The original and core counts obey
C 0 ≤ A − S + o ( 1 ) , C r ≤ B ∗ + 3 S + o ( 1 ) . C_0\leq A-S+o(1),\qquad C_r\leq B_*+3S+o(1). C 0 ≤ A − S + o ( 1 ) , C r ≤ B ∗ + 3 S + o ( 1 ) . Here unused nonpositive deficits have simply been discarded. Taking weights 4 / 5 4/5 4/5 1 / 5 1/5 1/5 − S / 5 -S/5 − S /5
For V = ( x , y , z , w , b , r ) T V=(x,y,z,w,b,r)^T V = ( x , y , z , w , b , r ) T ℓ = ( 1 , 1 , 1 , 1 , 0 , 0 ) T \ell=(1,1,1,1,0,0)^T ℓ = ( 1 , 1 , 1 , 1 , 0 , 0 ) T
Then A = 1 − V T H 0 V A=1-V^T H_0V A = 1 − V T H 0 V B ∗ = 1 − V T H y V B_\ast=1-V^T H_yV B ∗ = 1 − V T H y V y ≤ z y\leq z y ≤ z B ∗ = 1 − V T H z V B_\ast=1-V^T H_zV B ∗ = 1 − V T H z V z ≤ y z\leq y z ≤ y c 0 = ( 1 / 2 , 0 , 0 , 0 , − 1 , − 1 ) T c_0=(1/2,0,0,0,-1,-1)^T c 0 = ( 1/2 , 0 , 0 , 0 , − 1 , − 1 ) T s 0 = ( − 1 / 2 , 0 , 0 , 0 , 1 , 2 ) T s_0=(-1/2,0,0,0,1,2)^T s 0 = ( − 1/2 , 0 , 0 , 0 , 1 , 2 ) T N 0 = ( c 0 s 0 T + s 0 c 0 T ) / 2 N_0=(c_0s_0^T+s_0c_0^T)/2 N 0 = ( c 0 s 0 T + s 0 c 0 T ) /2
Its quadratic form is ( m − r ) ( 2 r − m ) ≥ 0 (m-r)(2r-m)\geq0 ( m − r ) ( 2 r − m ) ≥ 0
A = 4 5 H 0 + 1 5 H y − 7 53 ℓ ℓ T − 1 5 N 0 ⪰ 0 , (31) \mathcal{A}=\frac{4}{5}H_0+\frac{1}{5}H_y-\frac{7}{53}\ell\ell^T-\frac{1}{5}N_0\succeq0, \tag*{(31)} A = 5 4 H 0 + 5 1 H y − 53 7 ℓ ℓ T − 5 1 N 0 ⪰ 0 , (31) Z = 4 5 H 0 + 1 5 H z − 7 53 ℓ ℓ T ≻ 0. (32) \mathcal{Z}=\frac{4}{5}H_0+\frac{1}{5}H_z-\frac{7}{53}\ell\ell^T\succ0. \tag*{(32)} Z = 5 4 H 0 + 5 1 H z − 53 7 ℓ ℓ T ≻ 0. (32) Their exact rational L D L T LDL^T L D L T
( 71 265 , 57 71 , 29 114 , 15 29 , 17 20 , 0 ) , ( 231 1060 , 26 33 , 3 13 , 5 12 , 29 35 , 74 725 ) . \left(\frac{71}{265},\frac{57}{71},\frac{29}{114},\frac{15}{29},\frac{17}{20},0\right),\qquad
\left(\frac{231}{1060},\frac{26}{33},\frac{3}{13},\frac{5}{12},\frac{29}{35},\frac{74}{725}\right). ( 265 71 , 71 57 , 114 29 , 29 15 , 20 17 , 0 ) , ( 1060 231 , 33 26 , 13 3 , 12 5 , 35 29 , 725 74 ) . The zero pivot has zero remaining Schur-complement column. Thus no numerical eigenvalue test is involved. These identities prove
min ( C 0 , C r ) ≤ 46 / 53 + o ( 1 ) , γ ( 4 ) ≤ 23 / 53. \min(C_0,C_r)\leq46/53+o(1),\qquad\gamma(4)\leq23/53. min ( C 0 , C r ) ≤ 46/53 + o ( 1 ) , γ ( 4 ) ≤ 23/53. All matrix entries and decompositions are reconstructed by previous_certificate.py.
Coercivity and exact halfspace losses
The matrix A \mathcal{A} A
V ∗ = ( 26 , 4 , 16 , 7 , 3 , 10 ) T / 53 , ℓ T V ∗ = 1. V_\ast=(26,4,16,7,3,10)^T/53,\qquad\ell^T V_\ast=1. V ∗ = ( 26 , 4 , 16 , 7 , 3 , 10 ) T /53 , ℓ T V ∗ = 1. The kernel claim follows either by multiplication or from the displayed pivots and A V ∗ = 0 \mathcal{A}V_\ast=0 A V ∗ = 0
min ( C 0 , C r ) ≤ 46 53 − ( V − V ∗ ) T A ( V − V ∗ ) − 1 5 ( m − r ) ( 2 r − m ) + o ( 1 ) . (33) \min(C_0,C_r)\leq\frac{46}{53}-(V-V_\ast)^T\mathcal{A}(V-V_\ast)-\frac{1}{5}(m-r)(2r-m)+o(1). \tag*{(33)} min ( C 0 , C r ) ≤ 53 46 − ( V − V ∗ ) T A ( V − V ∗ ) − 5 1 ( m − r ) ( 2 r − m ) + o ( 1 ) . (33) The quadratic form is strictly positive away from V ∗ V_\ast V ∗ W W W e x − e w , e y − e w , e z − e w , e b , e r e_x-e_w,e_y-e_w,e_z-e_w,e_b,e_r e x − e w , e y − e w , e z − e w , e b , e r
K 0 = W T A W = ( 7 / 5 1 1 − 3 / 10 − 1 / 4 1 2 1 0 3 / 10 1 1 3 / 2 0 − 1 / 10 − 3 / 10 0 0 8 / 5 3 / 10 − 1 / 4 3 / 10 − 1 / 10 3 / 10 3 / 5 ) ≻ 0. K_0=W^T\mathcal{A}W=
\begin{pmatrix}
7/5 & 1 & 1 & -3/10 & -1/4\\
1 & 2 & 1 & 0 & 3/10\\
1 & 1 & 3/2 & 0 & -1/10\\
-3/10 & 0 & 0 & 8/5 & 3/10\\
-1/4 & 3/10 & -1/10 & 3/10 & 3/5
\end{pmatrix}\succ0. K 0 = W T A W = 7/5 1 1 − 3/10 − 1/4 1 2 1 0 3/10 1 1 3/2 0 − 1/10 − 3/10 0 0 8/5 3/10 − 1/4 3/10 − 1/10 3/10 3/5 ≻ 0. For a retained linear condition f ( V ) = f ( V ∗ ) + a f T η ≥ 0 f(V)=f(V_\ast)+a_f^T\eta\geq0 f ( V ) = f ( V ∗ ) + a f T η ≥ 0 V − V ∗ = W η V-V_\ast=W\eta V − V ∗ = W η f ( V ∗ ) > 0 f(V_\ast)>0 f ( V ∗ ) > 0 K 0 K_0 K 0
f ( V ) ≤ 0 ⟹ η T K 0 η ≥ f ( V ∗ ) 2 a f T K 0 − 1 a f . f(V)\leq0\quad\Longrightarrow\quad\eta^T K_0\eta\geq\frac{f(V_\ast)^2}{a_f^T K_0^{-1}a_f}. f ( V ) ≤ 0 ⟹ η T K 0 η ≥ a f T K 0 − 1 a f f ( V ∗ ) 2 . This is the exact count loss used for each halfspace exclusion. The checker localize.py evaluates it with rational arithmetic.
The other branch has a convenient uniform loss: even Z − ( 1 / 1000 ) ℓ ℓ T \mathcal{Z} - (1/1000)\ell\ell^{T} Z − ( 1/1000 ) ℓ ℓ T
( 11497 53000 , 180569 229940 , 81827 361138 , 32456 81827 , 3201 4057 , 62923 640200 ) \left(\frac{11497}{53000},\frac{180569}{229940},\frac{81827}{361138},\frac{32456}{81827},\frac{3201}{4057},\frac{62923}{640200}\right) ( 53000 11497 , 229940 180569 , 361138 81827 , 81827 32456 , 4057 3201 , 640200 62923 ) Thus z ≤ y z \le y z ≤ y 46 / 53 − 1 / 1000 + o ( 1 ) 46/53 - 1/1000 + o(1) 46/53 − 1/1000 + o ( 1 )
Finally, when using (33) below at a target G < 23 / 53 G < 23/53 G < 23/53 (29) –(30) with that G G G r ≥ m / 2 r \ge m/2 r ≥ m /2 2 ( 23 / 53 − G ) 2(23/53 - G) 2 ( 23/53 − G )
Rematching when the C D CD C D
Suppose q ≥ d q \ge d q ≥ d h = q − d h = q-d h = q − d a = E A D a = E_{AD} a = E A D c = E C C c = E_{CC} c = E C C j = E A C j = E_{AC} j = E A C h h h C D CD C D § 7 . Throughout this section we assume the following normalized parameter conditions:
3 d / 10 ≤ h ≤ 5 d / 4 , 4 d / 5 ≤ r ≤ d , E ≤ 9 d 2 / 50 , 3d/10 \le h \le5d/4,\qquad4d/5 \le r \le d,\qquad E \le9d^2/50, 3 d /10 ≤ h ≤ 5 d /4 , 4 d /5 ≤ r ≤ d , E ≤ 9 d 2 /50 , y ≤ 13 d / 20 , y ≤ 3 z / 8 , z ≤ 3 d , y ≤ w , d ≤ 2 w . y \le13d/20,\qquad y \le3z/8,\qquad z \le3d,\qquad y \le w,\qquad d \le2w. y ≤ 13 d /20 , y ≤ 3 z /8 , z ≤ 3 d , y ≤ w , d ≤ 2 w . Section 13 shows that all instances in this orientation outside these conditions already satisfy the asserted upper bound. No additional condition h ≤ r h \le r h ≤ r h ≤ d h \le d h ≤ d
Which omitted pairs can replace a retained pair
Index the initial retained A D AD A D A i D i A_iD_i A i D i C i C_i C i D i D_i D i σ \sigma σ C i C σ ( i ) C_iC_{\sigma(i)} C i C σ ( i ) C 0 D 0 C_0D_0 C 0 D 0 i i i A i D 0 A_iD_0 A i D 0 C 0 C σ ( i ) C_0C_{\sigma(i)} C 0 C σ ( i ) A i D i A_iD_i A i D i C i C σ ( i ) C_iC_{\sigma(i)} C i C σ ( i ) C i D i C_iD_i C i D i C 0 D 0 C_0D_0 C 0 D 0 U U U u u u a U a_U a U A D AD A D A A A D U D_U D U c U c_U c U C C CC C C C U C_U C U C C C
r u ≤ a U + c U , u ≤ a + c r ≤ 9 d 40 < h . ru \le a_U+c_U,\qquad u \le\frac{a+c}{r} \le\frac{9d}{40} < h. r u ≤ a U + c U , u ≤ r a + c ≤ 40 9 d < h . The other C D CD C D V V V q − u q-u q − u a V a_V a V A D AD A D D D D c V c_V c V C C CC C C
a U + a V ≤ a , c U + c V ≤ c . a_U+a_V \le a,\qquad c_U+c_V \le c. a U + a V ≤ a , c U + c V ≤ c . Inserting any eligible pair and applying the outward switches proves that its D D D C C C B B B P Q PQ P Q Q D = 4 QD=4 Q D = 4 Q C ≥ 3 QC \ge3 QC ≥ 3
Transporting the initial core deficit
Let I , J I,J I , J B B B r , s r,s r , s W W W H ( W ) H(W) H ( W ) X 3 X_3 X 3 h ( W ) = ∣ W ∣ z − H ( W ) h(W)=|W|z-H(W) h ( W ) = ∣ W ∣ z − H ( W ) h ( W ) ≥ − o ( 1 ) h(W)\ge-o(1) h ( W ) ≥ − o ( 1 ) B B B B B B B ′ B' B ′ P Q ( 3 ) PQ(3) P Q ( 3 ) B B ′ ( 4 ) BB'(4) B B ′ ( 4 ) P Q ( 3 ) → B P ( 3 ) , B ′ Q ( 3 ) PQ(3) \to BP(3), B'Q(3) P Q ( 3 ) → B P ( 3 ) , B ′ Q ( 3 ) P Q PQ P Q O ( M ) O(M) O ( M )
h ( J ) = h ( I ) − h ( I ∖ J ) + h ( J ∖ I ) ≥ h 3 − z ( d − s ) − o ( 1 ) . h(J) = h(I) - h(I \setminus J) + h(J \setminus I) \ge h_3 - z(d-s) - o(1). h ( J ) = h ( I ) − h ( I ∖ J ) + h ( J ∖ I ) ≥ h 3 − z ( d − s ) − o ( 1 ) . If R ‾ ( s ) \overline{R}(s) R ( s ) (10) without its h 3 h_3 h 3 κ = 1 − y / z \kappa= 1-y/z κ = 1 − y / z
C s ≤ R ‾ ( s ) − κ h 3 + ( z − y ) ( d − s ) + o ( 1 ) . (34) C_s \le\overline{R}(s) - \kappa h_3 + (z-y)(d-s) + o(1). \tag*{(34)} C s ≤ R ( s ) − κ h 3 + ( z − y ) ( d − s ) + o ( 1 ) . (34) The two cores may have different matchings and different good-pair orientations; their union need not be a core.
Completing the target without a repair loss
Put T = h − u > 0 T=h-u>0 T = h − u > 0 d + T d+T d + T H H H
i j ∈ E ( H ) ⟺ C i C j is blue and D i D j has label 3 . ij \in E(H) \quad\Longleftrightarrow\quad C_iC_j \text{ is blue and } D_iD_j \text{ has label 3}. ij ∈ E ( H ) ⟺ C i C j is blue and D i D j has label 3 . A blue C C CC C C D D DD D D C D ( 2 ) CD(2) C D ( 2 ) H H H k k k k ≥ T k \ge T k ≥ T T T T C C C D D D
If k < T k<T k < T k k k d + ( T − k ) d+(T-k) d + ( T − k ) C C CC C C T − k T-k T − k d d d d 2 / 2 − o ( 1 ) d^2/2-o(1) d 2 /2 − o ( 1 ) c ≤ 9 d 2 / 50 c \le9d^2/50 c ≤ 9 d 2 /50 T − k T-k T − k D D DD D D H H H H H H C C CC C C D D D
In the second case only the last T − k T-k T − k D D D D D DD D D H H H T − k + o ( 1 ) T-k+o(1) T − k + o ( 1 ) H H H H H H
k ( T − k ) 2 ≤ T 2 8 . \frac{k(T-k)}{2} \le\frac{T^2}{8}. 2 k ( T − k ) ≤ 8 T 2 . Any other eligible label-3 D D DD D D C C CC C C D D DD D D C C CC C C c V c_V c V c V c_V c V D D D u T uT u T
Counting the new matching and allocating missing edges
Rebuild a separate neutral core on the eligible complement of the target, which has mass exactly d d d s s s C C CC C C D D DD D D M M M C D C_D C D
C D ≤ R ‾ ( s ) − κ h 3 + ( z − y ) ( d − s ) + T ( q − 2 s ) + u T − 3 8 T 2 + j + ( 1 + T / r ) a v + 2 c V + e 12 + e 23 + ( 2 T / r ) h 3 + o ( 1 ) . (35) \begin{aligned}
C_D \le{} & \overline{R}(s) - \kappa h_3 + (z-y)(d-s) + T(q-2s) + uT - \frac{3}{8}T^2 \\
& {} + j + (1+T/r)av + 2c_V + e_{12} + e_{23} + (2T/r)h_3 + o(1). \tag*{(35)}
\end{aligned} C D ≤ R ( s ) − κ h 3 + ( z − y ) ( d − s ) + T ( q − 2 s ) + u T − 8 3 T 2 + j + ( 1 + T / r ) a v + 2 c V + e 12 + e 23 + ( 2 T / r ) h 3 + o ( 1 ) . (35) Here target D D D C C C T q Tq T q C C C 2 T s + T 2 / 2 2Ts+T^2/2 2 T s + T 2 /2 T 2 / 8 T^2/8 T 2 /8 D D DD D D X 1 X_1 X 1 P Q P_Q P Q e 12 , e 23 e_{12},e_{23} e 12 , e 23 2 T h 3 / r 2Th_3/r 2 T h 3 / r B B B ( T / r ) a V (T/r)a_V ( T / r ) a V A D AD A D D D D
The missing-edge terms can be allocated more precisely. Write a J a_J a J A D AD A D D D D c J c_J c J C C CC C C + a V +a_V + a V (35) is actually supported on target D D D a J a_J a J c c c c J c_J c J c c c 2 2 2
Let v = h / d v=h/d v = h / d
P 0 = 10 9 ( 53 / 10 − v ) , P ( T ) = P 0 + 20 T 9 d . P_0=\frac{10}{9}(53/10-v), \qquad P(T)=P_0+\frac{20T}{9d}. P 0 = 9 10 ( 53/10 − v ) , P ( T ) = P 0 + 9 d 20 T . The packing derivative, including (34) , is bounded by
( 6 d − x − z + 3 y ) + + z − y + 2 T = max ( 4 d − h + 2 y , z − y ) + 2 T ≤ 9 10 d P ( T ) . (6d-x-z+3y)_+ + z-y+2T=\max(4d-h+2y,z-y)+2T\leq\frac{9}{10}dP(T). ( 6 d − x − z + 3 y ) + + z − y + 2 T = max ( 4 d − h + 2 y , z − y ) + 2 T ≤ 10 9 d P ( T ) . The last inequality uses y ≤ 13 d / 20 y\leq13d/20 y ≤ 13 d /20 z ≤ 3 d z\leq3d z ≤ 3 d v ≤ 5 / 4 v\leq5/4 v ≤ 5/4 (12) gives s 2 ≥ d 2 − 2 ( a J + c J ) − o ( 1 ) s^2\geq d^2-2(a_J+c_J)-o(1) s 2 ≥ d 2 − 2 ( a J + c J ) − o ( 1 ) s ≥ 4 d / 5 − o ( 1 ) s\geq4d/5-o(1) s ≥ 4 d /5 − o ( 1 ) d 2 − s 2 d^2-s^2 d 2 − s 2 d + s d+s d + s d − s ≤ 10 ( a J + c J ) / ( 9 d ) + o ( 1 ) d-s\leq10(a_J+c_J)/(9d)+o(1) d − s ≤ 10 ( a J + c J ) / ( 9 d ) + o ( 1 ) P ( T ) ≥ 9 / 2 > 2 P(T)\geq9/2>2 P ( T ) ≥ 9/2 > 2
P ( T ) a J + a t a r g e t ≤ P ( T ) a V , P ( T ) c J + 2 c i n c i d e n t ≤ P ( T ) c V . P(T)a_J+a_{\mathrm{target}}\leq P(T)a_V,\qquad P(T)c_J+2c_{\mathrm{incident}}\leq P(T)c_V. P ( T ) a J + a target ≤ P ( T ) a V , P ( T ) c J + 2 c incident ≤ P ( T ) c V . Thus after replacing s s s d d d P ( T ) + T / r P(T)+T/r P ( T ) + T / r P ( T ) P(T) P ( T )
Accounting for the noneligible pairs
Normalize lengths by d d d d 2 d^2 d 2 v = h / d v=h/d v = h / d ρ = r / d \rho=r/d ρ = r / d σ = u / d \sigma=u/d σ = u / d l = T / d = v − σ l=T/d=v-\sigma l = T / d = v − σ
f ( l ) = l ( 2 v − 1 ) − 11 8 l 2 , d f d σ = 1 − 2 v + 11 4 l ≤ 31 / 16. f(l)=l(2v-1)-\frac{11}{8}l^2,\qquad\frac{df}{d\sigma}=1-2v+\frac{11}{4}l\leq31/16. f ( l ) = l ( 2 v − 1 ) − 8 11 l 2 , d σ df = 1 − 2 v + 4 11 l ≤ 31/16. The eligible coefficients P 0 + 20 l / 9 + l / ρ P_0+20l/9+l/\rho P 0 + 20 l /9 + l / ρ P 0 + 20 l / 9 P_0+20l/9 P 0 + 20 l /9 σ \sigma σ P 0 ≥ 9 / 2 P_0\geq9/2 P 0 ≥ 9/2 σ \sigma σ ρ d σ \rho\,d\sigma ρ d σ 18 / 5 18/5 18/5 h 3 h_3 h 3
31 / 16 − 18 / 5 = − 133 / 80 < 0. 31/16-18/5=-133/80<0. 31/16 − 18/5 = − 133/80 < 0. This holds at resource-allocation kinks by comparing optimal allocations at two nearby values of σ \sigma σ u = 0 u=0 u = 0 T = h T=h T = h T = h − u T=h-u T = h − u
Put P h = 10 9 ( 53 / 10 + h / d ) P_h=\frac{10}{9}(53/10+h/d) P h = 9 10 ( 53/10 + h / d )
C D ≤ B − b 2 + d ( x + z − 3 y ) − 3 d 2 − d h + 5 8 h 2 C_D\leq B-b^2+d(x+z-3y)-3d^2-dh+\frac{5}{8}h^2 C D ≤ B − b 2 + d ( x + z − 3 y ) − 3 d 2 − d h + 8 5 h 2 + ( P h + h / r ) a + P h c + j + e 12 + 3 8 e 23 + ( 2 h / r − 5 8 ) h 3 + o ( 1 ) . (36) \qquad+(P_h+h/r)a+P_hc+j+e_{12}+\frac{3}{8}e_{23}+(2h/r-\frac{5}{8})h_3+o(1). \tag*{(36)} + ( P h + h / r ) a + P h c + j + e 12 + 8 3 e 23 + ( 2 h / r − 8 5 ) h 3 + o ( 1 ) . (36) The certificate in § 13 uses these coefficients. The allocation above is essential: the target and complementary packing use disjoint parts of the same missing-edge count.
Using C C C
To bound contacts with omitted B B B C D CD C D C C C C C C g g g A 0 B 0 A_0B_0 A 0 B 0 A 0 D A_0D A 0 D B 0 D ∗ = 4 B_0D_* = 4 B 0 D ∗ = 4 D ∗ D_* D ∗ C ∗ D ∗ C_*D_* C ∗ D ∗ A 0 A_0 A 0 A 0 B 0 A_0B_0 A 0 B 0 B B B B 0 B_0 B 0 B i B_i B i A i A_i A i B 0 B i ( 4 ) , A i D ∗ ( 1 ) → B 0 D ∗ ( ≤ 3 ) , A i B i ( 2 ) B_0B_i(4), A_iD_*(1) \to B_0D_*(\le3), A_iB_i(2) B 0 B i ( 4 ) , A i D ∗ ( 1 ) → B 0 D ∗ ( ≤ 3 ) , A i B i ( 2 ) A D AD A D B B B ( T / g ) a o u t + o ( 1 ) (T/g)a_{\mathrm{out}} + o(1) ( T / g ) a out + o ( 1 )
If the nonrich mass is b C b_C b C b 0 b_0 b 0 c ≥ r b C − b 0 2 / 2 − o ( 1 ) c \ge rb_C - b_0^2/2 - o(1) c ≥ r b C − b 0 2 /2 − o ( 1 ) c ≤ 9 d 2 / 50 c \le9d^2/50 c ≤ 9 d 2 /50 b C < r + o ( 1 ) b_C < r + o(1) b C < r + o ( 1 )
b C ≤ r − r 2 − 2 c + o ( 1 ) ≤ 5 c 4 r + o ( 1 ) , g ≥ q − 9 d 32 − o ( 1 ) . b_C \le r-\sqrt{r^2-2c}+o(1)\le\frac{5c}{4r}+o(1),\qquad g\ge q-\frac{9d}{32}-o(1). b C ≤ r − r 2 − 2 c + o ( 1 ) ≤ 4 r 5 c + o ( 1 ) , g ≥ q − 32 9 d − o ( 1 ) . The last estimate uses r ≥ 4 d / 5 r \ge4d/5 r ≥ 4 d /5 a o u t a_{\mathrm{out}} a out A A A a U a_U a U T / g T/g T / g r ≥ 4 d / 5 r \ge4d/5 r ≥ 4 d /5 g ≥ q − 9 d 2 / ( 40 r ) − o ( 1 ) g \ge q-9d^2/(40r)-o(1) g ≥ q − 9 d 2 / ( 40 r ) − o ( 1 ) v = h / d ∈ [ v 0 , v 1 ] v=h/d\in[v_0,v_1] v = h / d ∈ [ v 0 , v 1 ] r / d ≥ ρ 0 r/d\ge\rho_0 r / d ≥ ρ 0 a a a (36) can consequently be replaced by
A D = 10 9 ( 53 10 + v 1 ) + v 1 1 + v 1 − 9 / ( 40 ρ 0 ) , ρ 0 ≤ r / d . A_D=\frac{10}{9}\left(\frac{53}{10}+v_1\right)+\frac{v_1}{1+v_1-9/(40\rho_0)},\qquad\rho_0\le r/d. A D = 9 10 ( 10 53 + v 1 ) + 1 + v 1 − 9/ ( 40 ρ 0 ) v 1 , ρ 0 ≤ r / d . Also retain the exact deficit term − ( 1 − y / z ) h 3 + 2 h h 3 / r -(1-y/z)h_3+2hh_3/r − ( 1 − y / z ) h 3 + 2 h h 3 / r h 3 h_3 h 3 § 13 .
Auxiliary matchings through good label-3 pairs
This section concerns the orientation d ≥ q d\ge q d ≥ q A D AD A D A A A P P P κ = 1 − y / z \kappa=1-y/z κ = 1 − y / z I 3 = z 2 / 2 − C 33 ≥ 0 I_3=z^2/2-C_{33}\ge0 I 3 = z 2 /2 − C 33 ≥ 0 − I 3 -I_3 − I 3 Z = z − h 3 / r + o ( 1 ) Z=z-h_3/r+o(1) Z = z − h 3 / r + o ( 1 ) E P P ≤ I 3 + o ( 1 ) E_{PP}\le I_3+o(1) E P P ≤ I 3 + o ( 1 ) (21) , and the retained-C C C (17) .
Call an original A A A D D D d − a / r − o ( 1 ) d-a/r-o(1) d − a / r − o ( 1 ) A D AD A D C C CC C C B B B A A A D D D P P P A D AD A D A P AP A P A A A A P = 1 AP=1 A P = 1 A Q = 2 AQ=2 A Q = 2 A D AD A D A A A D D D r + t r+t r + t r r r A D AD A D t t t A P AP A P
Keep the original blue C C CC C C B B B B B B Q Q Q A A A B B B P B ≥ 3 PB\ge3 P B ≥ 3 Q B = 4 QB=4 QB = 4
Any two vertices in the union of new-core and selected B B B A A A D D D o ( 1 ) o(1) o ( 1 ) B B B h n e w h_{\mathrm{new}} h new h s e l h_{\mathrm{sel}} h sel
h n e w + h s e l ≥ h 3 − o ( 1 ) . h_{\mathrm{new}}+h_{\mathrm{sel}}\ge h_3-o(1). h new + h sel ≥ h 3 − o ( 1 ) . The selected B B B t z − h s e l tz-h_{\mathrm{sel}} t z − h sel − κ h n e w − h s e l ≤ − κ h 3 + o ( 1 ) -\kappa h_{\mathrm{new}}-h_{\mathrm{sel}}\le-\kappa h_3+o(1) − κ h new − h sel ≤ − κ h 3 + o ( 1 )
Write R ( r ) R(r) R ( r ) (10) without its − κ h 3 -\kappa h_3 − κ h 3 − I 3 -I_3 − I 3
C A P ≤ R ( r ) + t ( 2 z + d + 2 q − 6 r ) − 2 t 2 + E A A + j + a + e 12 + e 13 + e 23 + I 3 + 3 t r a + ( t / r − κ ) h 3 + o ( 1 ) . (37) C_{AP}\le R(r)+t(2z+d+2q-6r)-2t^2+E_{AA}+j+a+e_{12}+e_{13}+e_{23}+I_3+\frac{3t}{r}a+(t/r-\kappa)h_3+o(1). \tag*{(37)} C A P ≤ R ( r ) + t ( 2 z + d + 2 q − 6 r ) − 2 t 2 + E AA + j + a + e 12 + e 13 + e 23 + I 3 + r 3 t a + ( t / r − κ ) h 3 + o ( 1 ) . (37) Here compatibility A P = 1 , A Q = 2 AP=1,AQ=2 A P = 1 , A Q = 2 C D CD C D P D = 1 PD=1 P D = 1 B C = 4 BC=4 B C = 4 P C = 1 PC=1 P C = 1 B D = 4 BD=4 B D = 4 A B , C D , P Q AB,CD,PQ A B , C D , P Q 2 t ( q − r ) + e 23 2t(q-r)+e_{23} 2 t ( q − r ) + e 23 C C CC C C P P PP P P E P P ≤ I 3 E_{PP}\le I_3 E P P ≤ I 3 X 1 X_1 X 1 e 12 + e 13 e_{12}+e_{13} e 12 + e 13 t h 3 / r th_3/r t h 3 / r (37) with all contacts outside the matching controlled by the exact outward caps.
The saturated alternative
Set H = d − r H=d-r H = d − r τ = ( H − a / r ) + \tau=(H-a/r)_+ τ = ( H − a / r ) + τ \tau τ f ( t ) = t ( 2 z + d + 2 q − 6 r ) − 2 t 2 f(t)=t(2z+d+2q-6r)-2t^2 f ( t ) = t ( 2 z + d + 2 q − 6 r ) − 2 t 2
f ( τ ) ≤ f ( H ) + a r ( 3 d − 2 q + 2 r − 2 z ) + . f(\tau)\le f(H)+\frac{a}{r}(3d-2q+2r-2z)_+. f ( τ ) ≤ f ( H ) + r a ( 3 d − 2 q + 2 r − 2 z ) + . This includes τ = 0 \tau=0 τ = 0 (37) yields the global cap inequality
C A P ≤ R ( r ) + f A + E A A + j + e 12 + e 13 + e 23 + I 3 + A A P a + ( H / r − κ ) h 3 + o ( 1 ) , C_{AP}\le R(r)+f_A+E_{AA}+j+e_{12}+e_{13}+e_{23}+I_3+A_{AP}a+(H/r-\kappa)h_3+o(1), C A P ≤ R ( r ) + f A + E AA + j + e 12 + e 13 + e 23 + I 3 + A A P a + ( H / r − κ ) h 3 + o ( 1 ) , f A = ( d − r ) ( 2 z − d + 2 q − 4 r ) , f_A=(d-r)(2z-d+2q-4r), f A = ( d − r ) ( 2 z − d + 2 q − 4 r ) , A A P = 1 + 3 H / r + 1 r ( 3 d − 2 q + 2 r − 2 z ) + . (38) A_{AP}=1+3H/r+\frac{1}{r}(3d-2q+2r-2z)_+. \tag*{(38)} A A P = 1 + 3 H / r + r 1 ( 3 d − 2 q + 2 r − 2 z ) + . (38) There is no loss from rebuilding a smaller core.
The deficient alternative
Otherwise take a minimum vertex cover A 0 , D 0 , P 0 A_0,D_0,P_0 A 0 , D 0 , P 0 r + t < r + τ r+t<r+\tau r + t < r + τ k = r − ∣ D 0 ∣ k=r-|D_0| k = r − ∣ D 0 ∣ η = a / r \eta=a/r η = a / r 0 ≤ k ≤ r 0\le k\le r 0 ≤ k ≤ r k > η k>\eta k > η A D AD A D t < H − η t<H-\eta t < H − η
a ≥ ( d − ∣ A 0 ∣ ) k ≥ ( d − t − k ) k > ( r + η − k ) k = r η + ( k − η ) ( r − k ) ≥ a , a\ge(d-|A_0|)k\ge(d-t-k)k>(r+\eta-k)k=r\eta+(k-\eta)(r-k)\ge a, a ≥ ( d − ∣ A 0 ∣ ) k ≥ ( d − t − k ) k > ( r + η − k ) k = r η + ( k − η ) ( r − k ) ≥ a , a contradiction. Therefore k ≤ η k\le\eta k ≤ η A 0 ∪ P 0 A_0\cup P_0 A 0 ∪ P 0 A P AP A P t + k < H t+k<H t + k < H W W W A A A P P P
W + e 23 + Z r a ≥ r Z − H ( d − Z ) + . (39) W+e_{23}+\frac{Z}{r}a\ge rZ-H(d-Z)_+. \tag*{(39)} W + e 23 + r Z a ≥ r Z − H ( d − Z ) + . (39) The last allowance accounts for the nonstrong rows. The two-neighbor lemma gives exactly the Z a / r Za/r Z a / r
Let v v v P P P C C C P P P C C CC C C U P C , U P P U_{PC},U_{PP} U P C , U P P c ∗ c_* c ∗ C C C v < r v<r v < r Z ≥ r + v Z\ge r+v Z ≥ r + v
( r − v ) { U P C + 2 ( 1 − r / Z ) U P P − v Z } ≤ 4 Z c ∗ + o ( 1 ) . (40) (r-v)\left\{U_{PC}+2(1-r/Z)U_{PP}-vZ\right\}\le4Zc_*+o(1). \tag*{(40)} ( r − v ) { U P C + 2 ( 1 − r / Z ) U P P − v Z } ≤ 4 Z c ∗ + o ( 1 ) . (40) To prove this, take a maximum P P PP P P p ≤ v p \le v p ≤ v P C PC P C k k k P P P C C C e ( G ) ≤ ν ( G ) ∣ V ( G ) ∣ e(G) \le\nu(G)|V(G)| e ( G ) ≤ ν ( G ) ∣ V ( G ) ∣ p + k ≤ v + 4 c ∗ / ( r − v ) + o ( 1 ) p+k \le v + 4c_\ast/(r-v) + o(1) p + k ≤ v + 4 c ∗ / ( r − v ) + o ( 1 ) U P C ≤ r p + k ( Z − p ) U_{PC} \le rp + k(Z-p) U P C ≤ r p + k ( Z − p ) U P P ≤ p Z / 2 + o ( 1 ) U_{PP} \le pZ/2 + o(1) U P P ≤ pZ /2 + o ( 1 ) (40) . All matchings used in this argument are disjoint on their prescribed vertices.
For a second actual comparison, assign r r r P Q PQ P Q A D AD A D A D ( 1 ) AD(1) A D ( 1 ) P Q ( 3 ) → A Q ( 2 ) PQ(3) \to AQ(2) P Q ( 3 ) → A Q ( 2 ) D P ( 2 ) DP(2) D P ( 2 ) e 23 / Z e_{23}/Z e 23 / Z P D = 1 PD=1 P D = 1 A Q ≤ 2 AQ \le2 A Q ≤ 2 P , Q P,Q P , Q 2 x + 2 y + r 2x+2y+r 2 x + 2 y + r
C A D − C r ≤ r ( 4 y − d ) − r Z W + r Z U P C + ( r Z ) 2 ( U P P + I 3 ) + a + 2 x + 2 y + 3 r Z e 23 + o ( 1 ) . (41) C_{AD}-C_r \le r(4y-d)-\frac{r}{Z}W+\frac{r}{Z}U_{PC}+\left(\frac{r}{Z}\right)^2(U_{PP}+I_3)+a+\frac{2x+2y+3r}{Z}e_{23}+o(1). \tag*{(41)} C A D − C r ≤ r ( 4 y − d ) − Z r W + Z r U P C + ( Z r ) 2 ( U P P + I 3 ) + a + Z 2 x + 2 y + 3 r e 23 + o ( 1 ) . (41) In the formal count, core blue gains are 5 r 2 / 2 5r^2/2 5 r 2 /2 P C , P D PC,PD P C , P D 2 r 2 2r^2 2 r 2 P P PP P P r 2 / 2 r^2/2 r 2 /2 Q A QA Q A A A A r d rd r d X 1 X_1 X 1 4 r y 4ry 4 r y (41) .
For the exact certificate, further retained domains are Z ≥ 3 r / 2 Z \ge3r/2 Z ≥ 3 r /2 d ≤ 23 Z / 20 d \le23Z/20 d ≤ 23 Z /20 r ≥ 7 q / 8 r \ge7q/8 r ≥ 7 q /8 § 13 . Choose 0 < α ≤ 1 / 2 0 < \alpha\le1/2 0 < α ≤ 1/2 v ≥ α r v \ge\alpha r v ≥ α r α r \alpha r α r C P CP C P (42) , whose actual matching and edge count are given in § 10.3 . If v ≤ α r v \le\alpha r v ≤ α r (39) , (40) , (41) , and Lemma 6.2 . Since r / Z ≤ 2 / 3 r/Z \le2/3 r / Z ≤ 2/3 α r 2 + 4 c ∗ / ( 1 − α ) \alpha r^2+4c_\ast/(1-\alpha) α r 2 + 4 c ∗ / ( 1 − α )
r Z H ( d − Z ) + ≤ 3 20 r ( d − r ) , \frac{r}{Z}H(d-Z)_+ \le\frac{3}{20}r(d-r), Z r H ( d − Z ) + ≤ 20 3 r ( d − r ) , since d / Z ≤ 23 / 20 d/Z \le23/20 d / Z ≤ 23/20 r / q ∈ [ ρ 0 , ρ 1 ] r/q \in[\rho_0,\rho_1] r / q ∈ [ ρ 0 , ρ 1 ] h / q ∈ [ v 0 , v 1 ] h/q \in[v_0,v_1] h / q ∈ [ v 0 , v 1 ]
K C = 4 ( 1 − α ) ( 5 − 4 / ρ 0 ) , β 0 = 1 1 + v 0 . K_C=\frac{4}{(1-\alpha)(5-4/\rho_0)},\qquad\beta_0=\frac{1}{1+v_0}. K C = ( 1 − α ) ( 5 − 4/ ρ 0 ) 4 , β 0 = 1 + v 0 1 . The small-v v v
f R + r ( 4 y − d − r ) + 3 20 r ( d − r ) + α r 2 − K C K , K = ( q 2 − r 2 ) / 2 , f_R+r(4y-d-r)+\frac{3}{20}r(d-r)+\alpha r^2-K_CK,\qquad K=(q^2-r^2)/2, f R + r ( 4 y − d − r ) + 20 3 r ( d − r ) + α r 2 − K C K , K = ( q 2 − r 2 ) /2 , and coefficients at most 2 + β 0 K C 2+\beta_0K_C 2 + β 0 K C a a a K C K_C K C c c c 57 / 8 57/8 57/8 e 23 e_{23} e 23 − 5 / 8 -5/8 − 5/8 h 3 h_3 h 3 (39) contributes r / Z r/Z r / Z (41) contributes ( 2 x + 2 y + 3 r ) / Z (2x+2y+3r)/Z ( 2 x + 2 y + 3 r ) / Z y / z − 1 y/z-1 y / z − 1 57 / 8 57/8 57/8 § 13 .
Edge counts for the two auxiliary comparisons
We now construct and count the two matchings used in § 10 . The first extends a blue matching from retained C C C P P P A D AD A D P Q PQ P Q (16) .
For the C P CP C P A D AD A D B B BB B B C C CC C C C C C P P P C C C P P P v v v r + v r+v r + v C C CC C C v M vM v M P Q PQ P Q Q Q Q v M vM v M r M rM r M r M + v M rM+vM r M + v M P Q PQ P Q ( r + v ) M / 2 + v M / 2 = r M / 2 + v M (r+v)M/2+vM/2=rM/2+vM ( r + v ) M /2 + v M /2 = r M /2 + v M Q Q QQ QQ 2 v M 2vM 2 v M
In the C P CP C P P P P Q Q Q A , C , D A,C,D A , C , D − 2 r v -2rv − 2 r v e 23 e_{23} e 23 P P P 3 r v 3rv 3 r v Q Q Q C C C r v rv r v A B AB A B Q B = 4 QB=4 QB = 4 P B ≥ 3 PB\ge3 P B ≥ 3 a / r a/r a / r 2 v a / r 2va/r 2 v a / r C D CD C D v v v ( a + c ) / r (a+c)/r ( a + c ) / r v ( q − r ) + v ( a + c ) / r v(q-r)+v(a+c)/r v ( q − r ) + v ( a + c ) / r
Inside X 3 X_3 X 3 Q Q Q v Z vZ v Z P P P 2 v ( z − Z ) 2v(z-Z) 2 v ( z − Z ) P P P v 2 / 2 − I 3 v^2/2-I_3 v 2 /2 − I 3 E P P ≤ I 3 E_{PP}\le I_3 E P P ≤ I 3 v z − v 2 / 2 + I 3 + v ( z − Z ) vz-v^2/2+I_3+v(z-Z) v z − v 2 /2 + I 3 + v ( z − Z ) X 1 X_1 X 1 e 13 e_{13} e 13 Q Q Q e 23 e_{23} e 23
C C P − C r ≤ v ( z + q − 3 r ) − v 2 / 2 + e 13 + e 23 + I 3 + ( v / r ) h 3 + ( 3 v / r ) a + ( v / r ) c + o ( 1 ) . (42) C_{CP}-C_r\le v(z+q-3r)-v^2/2+e_{13}+e_{23}+I_3+(v/r)h_3+(3v/r)a+(v/r)c+o(1). \tag*{(42)} C C P − C r ≤ v ( z + q − 3 r ) − v 2 /2 + e 13 + e 23 + I 3 + ( v / r ) h 3 + ( 3 v / r ) a + ( v / r ) c + o ( 1 ) . (42) For A D / P Q AD/PQ A D / P Q π = r / Z \pi=r/Z π = r / Z A , D A,D A , D P , Q P,Q P , Q 4 r y 4ry 4 r y X 1 X_1 X 1 5 r 2 / 2 5r^2/2 5 r 2 /2 A A , A C , A D AA,AC,AD AA , A C , A D P A ( 2 ) PA(2) P A ( 2 ) π W \pi W π W P C ( 2 ) , P D ( 2 ) PC(2),PD(2) P C ( 2 ) , P D ( 2 ) 2 r 2 − π U P C 2r^2-\pi U_{PC} 2 r 2 − π U P C P D PD P D P D PD P D A D AD A D A A A A Q > 2 AQ>2 A Q > 2 A D ( 1 ) , P Q ( 3 ) AD(1),PQ(3) A D ( 1 ) , P Q ( 3 ) A B AB A B P Q PQ P Q
The Q A ( 2 ) QA(2) Q A ( 2 ) r d − a rd-a r d − a Q A QA Q A Q B = 4 QB=4 QB = 4 P B ≥ 3 PB\ge3 P B ≥ 3 a / r a/r a / r A B AB A B C D CD C D 2 π e 23 2\pi e_{23} 2 π e 23 P P ( 2 ) PP(2) P P ( 2 ) r 2 / 2 − π 2 ( U P P + I 3 ) + o ( 1 ) r^2/2-\pi^2(U_{PP}+I_3)+o(1) r 2 /2 − π 2 ( U P P + I 3 ) + o ( 1 )
r ( 4 y − d ) − π W + π U P C + π 2 ( U P P + I 3 ) + a + 2 π e 23 + o ( 1 ) . r(4y-d)-\pi W+\pi U_{PC}+\pi^2(U_{PP}+I_3)+a+2\pi e_{23}+o(1). r ( 4 y − d ) − π W + π U P C + π 2 ( U P P + I 3 ) + a + 2 π e 23 + o ( 1 ) . The expected failed assignment mass is at most e 23 / Z e_{23}/Z e 23 / Z A , D A,D A , D P , Q P,Q P , Q 2 x + 2 y 2x+2y 2 x + 2 y X 1 ∪ X 2 X_1\cup X_2 X 1 ∪ X 2 Q Q Q r r r P P P ( 2 x + 2 y + r ) e 23 / Z (2x+2y+r)e_{23}/Z ( 2 x + 2 y + r ) e 23 / Z ( 2 x + 2 y + 3 r ) / Z (2x+2y+3r)/Z ( 2 x + 2 y + 3 r ) / Z A , D A,D A , D P , Q P,Q P , Q
Enlarging the retained sets by re-pairing
We work in the case d ≥ q d\ge q d ≥ q C C C D D D
Index the original C D CD C D C i D i C_iD_i C i D i retained if its pair belongs to the neutral core, and omitted otherwise. For a retained index i i i A i A_i A i D i D_i D i A D AD A D r r r δ = q − r \delta=q-r δ = q − r H = h 3 H=h_3 H = h 3
B 0 = B − b 2 , F ( R ) = 4 q R − 23 8 R 2 − R z + 2 R y − q 2 . B_0=B-b^2,\qquad F(R)=4qR-\frac{23}{8}R^2-Rz+2Ry-q^2. B 0 = B − b 2 , F ( R ) = 4 q R − 8 23 R 2 − R z + 2 R y − q 2 . Here b = ∣ d − q ∣ / 2 b=|d-q|/2 b = ∣ d − q ∣/2 b 3 b_3 b 3
r ≥ 7 q / 8 , q ≤ 2 w , z ≤ 3 q , r + z ≤ 6 w , Z = z − H / r ≥ 3 r / 2. (43) r \ge7q/8,\quad q \le2w,\quad z \le3q,\quad r+z \le6w,\quad Z=z-H/r \ge3r/2. \tag*{(43)} r ≥ 7 q /8 , q ≤ 2 w , z ≤ 3 q , r + z ≤ 6 w , Z = z − H / r ≥ 3 r /2. (43) Section 13 proves that the complementary cases already satisfy the asserted upper bound, using comparisons that do not assume these conditions. All statements here concern finite matchings. Harmless parity adjustments and the bounded-degree discarded blue edges cost O ( M ) O(M) O ( M ) N ≥ 2 M N \ge2M N ≥ 2 M
Let C r e t C_{\mathrm{ret}} C ret C o u t C_{\mathrm{out}} C out C C C
U = blue ( C r e t , C o u t ) , U=\operatorname{blue}(C_{\mathrm{ret}},C_{\mathrm{out}}), U = blue ( C ret , C out ) , where blue ( U 1 , U 2 ) \operatorname{blue}(U_1,U_2) blue ( U 1 , U 2 ) M 2 M^2 M 2
U ≤ r d a + L ( c − K + β a ) + o ( 1 ) , L = 4 ( q − r ) 5 r − 4 q , K = q 2 − r 2 2 , β = q d . (44) U \le\frac{r}{d}a+L(c-K+\beta a)+o(1),\qquad L=\frac{4(q-r)}{5r-4q},\qquad K=\frac{q^2-r^2}{2},\qquad\beta=\frac{q}{d}. \tag*{(44)} U ≤ d r a + L ( c − K + β a ) + o ( 1 ) , L = 5 r − 4 q 4 ( q − r ) , K = 2 q 2 − r 2 , β = d q . (44) This follows from Lemma 6.2 by dropping its favorable − a 2 / ( 2 d 2 ) -a^2/(2d^2) − a 2 / ( 2 d 2 ) U U U
A comparison with specified unmatched mass
Choose a maximum label-3 matching from the initial retained C C C X 4 X_4 X 4 t t t
t ≥ r − e 24 / ( 2 w ) − o ( 1 ) . t\ge r-e_{24}/(2w)-o(1). t ≥ r − e 24 / ( 2 w ) − o ( 1 ) . On the corresponding target D D D s s s D D D t − s t-s t − s D D D σ = s / r ∈ [ 0 , 1 ] \sigma=s/r\in[0,1] σ = s / r ∈ [ 0 , 1 ]
The reservoir counts bound possible gains by counting eligible pairs of vertices. Some of those pairs cannot actually produce a gain. For a covered target index i i i o o o
C i C o C_iC_o C i C o
C i C o C_iC_o C i C o D i D o D_iD_o D i D o
Let ξ \xi ξ M 2 M^2 M 2
ζ = 1 M 2 # { ( i , o ) : i ∈ S , C i D o ≥ 3 } + 1 M 2 # { ( i , o ) : i ∈ S , D i C o ≥ 3 } , V = ξ + ζ , \zeta=\frac{1}{M^2}\#\{(i,o):i\in S,\ C_iD_o\ge3\}+\frac{1}{M^2}\#\{(i,o):i\in S,\ D_iC_o\ge3\},\qquad V=\xi+\zeta, ζ = M 2 1 # {( i , o ) : i ∈ S , C i D o ≥ 3 } + M 2 1 # {( i , o ) : i ∈ S , D i C o ≥ 3 } , V = ξ + ζ , where S S S D D D C i D o ≥ 3 C_iD_o\ge3 C i D o ≥ 3 λ ( C i D o ) ≥ 3 \lambda(C_iD_o)\ge3 λ ( C i D o ) ≥ 3 C D CD C D C i C_i C i D i D_i D i ξ \xi ξ ζ \zeta ζ C C CC C C D D DD D D C D CD C D D C DC D C C D CD C D O ( M ) O(M) O ( M )
Lemma 11.1. Some actual matching satisfies
C I ≤ B 0 + F ( r ) − 1 2 ( s − r / 2 ) 2 + ( 2 − σ ) H − V C_I\le B_0+F(r)-\frac{1}{2}(s-r/2)^2+(2-\sigma)H-V C I ≤ B 0 + F ( r ) − 2 1 ( s − r /2 ) 2 + ( 2 − σ ) H − V + ( 2 + 2 δ d ) a + c − E A A + U + 4 e 24 − e 13 − e 14 − I 3 + o ( 1 ) . (45) +\left(2+\frac{2\delta}{d}\right)a+c-E_{AA}+U+4e_{24}-e_{13}-e_{14}-I_3+o(1). \tag*{(45)} + ( 2 + d 2 δ ) a + c − E AA + U + 4 e 24 − e 13 − e 14 − I 3 + o ( 1 ) . (45) Proof. Let b 3 ≤ H / r + o ( 1 ) b_3 \le H/r + o(1) b 3 ≤ H / r + o ( 1 ) s + 2 b 3 s + 2b_3 s + 2 b 3 D D ( 3 ) DD(3) D D ( 3 ) D D D ( t − s ) ( s / 2 + b 3 ) + o ( 1 ) (t-s)(s/2+b_3)+o(1) ( t − s ) ( s /2 + b 3 ) + o ( 1 ) C C CC C C ( t − s ) ( r − t ) + U (t-s)(r-t)+U ( t − s ) ( r − t ) + U D D DD D D
The direct C D CD C D 2 t q − t 2 2tq-t^2 2 tq − t 2 C C C P P P t r − e 24 tr-e_{24} t r − e 24 t ( z − b 3 ) − e 34 t(z-b_3)-e_{34} t ( z − b 3 ) − e 34 X 1 X_1 X 1 2 t y + e 12 2ty+e_{12} 2 t y + e 12 P P P e 23 e_{23} e 23
G ( t ) = t ( 2 q − r − z + b 3 + 2 y ) − t 2 + ( t − s ) ( r − t + s / 2 + b 3 ) . G(t)=t(2q-r-z+b_3+2y)-t^2+(t-s)(r-t+s/2+b_3). G ( t ) = t ( 2 q − r − z + b 3 + 2 y ) − t 2 + ( t − s ) ( r − t + s /2 + b 3 ) . Keeping s s s b 3 b_3 b 3
G ′ ( t ) = 2 q − z + 2 b 3 + 2 y − 4 t + 3 s / 2 ≥ − 2 r − z . G'(t)=2q-z+2b_3+2y-4t+3s/2\ge-2r-z. G ′ ( t ) = 2 q − z + 2 b 3 + 2 y − 4 t + 3 s /2 ≥ − 2 r − z . Replacing t t t r r r ( 2 r + z ) e 24 / ( 2 w ) ≤ 4 e 24 (2r+z)e_{24}/(2w)\le4e_{24} ( 2 r + z ) e 24 / ( 2 w ) ≤ 4 e 24 t = r t=r t = r P P P
r 2 8 − 1 2 ( s − r / 2 ) 2 + ( 2 r − s ) b 3 . \frac{r^2}{8}-\frac{1}{2}(s-r/2)^2+(2r-s)b_3. 8 r 2 − 2 1 ( s − r /2 ) 2 + ( 2 r − s ) b 3 . Use ( 2 r − s ) b 3 ≤ ( 2 − σ ) H + o ( 1 ) (2r-s)b_3\le(2-\sigma)H+o(1) ( 2 r − s ) b 3 ≤ ( 2 − σ ) H + o ( 1 ) A , B A,B A , B 2 a + j + ( t / r ) a 2a+j+(t/r)a 2 a + j + ( t / r ) a C C CC C C C C C δ 2 / 2 − a δ / d − o ( 1 ) \delta^2/2-a\delta/d-o(1) δ 2 /2 − a δ / d − o ( 1 ) V V V e 12 , e 23 , e 34 e_{12},e_{23},e_{34} e 12 , e 23 , e 34 a a a 2 + 2 δ / d 2+2\delta/d 2 + 2 δ / d F ( r ) F(r) F ( r ) □ \square □
Two ways to re-pair a retained and an omitted pair
On the covered target indices and omitted indices permit either of the links
Table 46.
A pair lacking a straight link is charged to ξ \xi ξ U U U A C AC A C ζ \zeta ζ A D AD A D C D CD C D ( V + U + a + j ) / 2 + o ( 1 ) (V+U+a+j)/2+o(1) ( V + U + a + j ) /2 + o ( 1 ) k M + O ( 1 ) kM+O(1) k M + O ( 1 )
k ≥ min ( s , δ ) − V + U + a + j 2 max ( s , δ ) − o ( 1 ) . (46) k\ge\min(s,\delta)-\frac{V+U+a+j}{2\max(s,\delta)}-o(1). \tag*{(46)} k ≥ min ( s , δ ) − 2 max ( s , δ ) V + U + a + j − o ( 1 ) . (46) If both sides have zero mass, take k = 0 k=0 k = 0 s ≥ δ > 0 s\ge\delta>0 s ≥ δ > 0 k ≥ δ − ( V + U + a + j ) / ( 2 s ) − o ( 1 ) k\ge\delta-(V+U+a+j)/(2s)-o(1) k ≥ δ − ( V + U + a + j ) / ( 2 s ) − o ( 1 ) s < δ s<\delta s < δ D o D_o D o C C C C o C_o C o D D D C o C_o C o C C C D o D_o D o D D D C ∗ C^* C ∗ D ∗ D^* D ∗ R = r + k R=r+k R = r + k
Lemma 11.2. Every vertex of D ∗ D^* D ∗ Q Q Q C ∗ C^* C ∗ Q Q Q Proof. The assertions for initially retained vertices were already proved. For a straight selected pair, the actual core replacement
A i D i ( 1 ) , C o D o ( 2 ) ⟶ A i C o ( 1 ) , D i D o ( 2 ) A_iD_i(1), C_oD_o(2) \longrightarrow A_iC_o(1), D_iD_o(2) A i D i ( 1 ) , C o D o ( 2 ) ⟶ A i C o ( 1 ) , D i D o ( 2 ) is primary-neutral and preserves all initial C C CC C C B B BB B B C o C_o C o A D AD A D B B ( 4 ) BB(4) B B ( 4 ) Q Q Q
For D o D_o D o C o C i = 2 C_oC_i = 2 C o C i = 2 D o u ≤ 2 D_ou \le2 D o u ≤ 2 u u u C o D o ( 2 ) C_oD_o(2) C o D o ( 2 ) D o u ( 2 ) D_ou(2) D o u ( 2 ) C o C_o C o C i C_i C i C i C j ( 1 ) C_iC_j(1) C i C j ( 1 ) C C CC C C A j D j ( 1 ) A_jD_j(1) A j D j ( 1 ) A D AD A D B j B_j B j A j A_j A j B ′ B' B ′ D o T ≤ 2 D_oT \le2 D o T ≤ 2
B j B ′ ( 4 ) , C i C j ( 1 ) , A j D j ( 1 ) , C o D o ( 2 ) , T T ′ ( 4 ) ⟶ D o T ( ≤ 2 ) , C o C i ( 2 ) , C j D j ( 2 ) , A j B j ( 2 ) , B ′ T ′ ( ≤ 4 ) , B_jB'(4), C_iC_j(1), A_jD_j(1), C_oD_o(2), TT'(4)
\longrightarrow D_oT(\le2), C_oC_i(2), C_jD_j(2), A_jB_j(2), B'T'(\le4), B j B ′ ( 4 ) , C i C j ( 1 ) , A j D j ( 1 ) , C o D o ( 2 ) , T T ′ ( 4 ) ⟶ D o T ( ≤ 2 ) , C o C i ( 2 ) , C j D j ( 2 ) , A j B j ( 2 ) , B ′ T ′ ( ≤ 4 ) , reducing primary cost from 9 to at most 8. For a good pair P Q PQ P Q T T ′ ( 4 ) TT'(4) T T ′ ( 4 ) P Q ( 3 ) PQ(3) P Q ( 3 ) B ′ B' B ′ P B ′ = 3 PB' = 3 P B ′ = 3 D o Q ≤ 2 D_oQ \le2 D o Q ≤ 2 □ \square □
Explicit caps for a crossed link. Assume C i D o = D i C o = 2 C_iD_o = D_iC_o = 2 C i D o = D i C o = 2 A i D o = 1 A_iD_o = 1 A i D o = 1 A i D i = 1 A_iD_i = 1 A i D i = 1 A i B i = 2 A_iB_i = 2 A i B i = 2
A i D i ( 1 ) , C o D o ( 2 ) ⟶ A i D o ( 1 ) , D i C o ( 2 ) A_iD_i(1), C_oD_o(2) \longrightarrow A_iD_o(1), D_iC_o(2) A i D i ( 1 ) , C o D o ( 2 ) ⟶ A i D o ( 1 ) , D i C o ( 2 ) keeps all old C C CC C C B B BB B B D o D_o D o A D AD A D C o C_o C o
The following switches prove that D o D_o D o u u u T T T Q Q Q B ′ B' B ′ B i B_i B i P P P
B i B ′ ( 4 ) , A i D o ( 1 ) ⟶ A i B i ( 2 ) , D o u ( ≤ 3 ) , B_iB'(4), A_iD_o(1) \longrightarrow A_iB_i(2), D_ou(\le3), B i B ′ ( 4 ) , A i D o ( 1 ) ⟶ A i B i ( 2 ) , D o u ( ≤ 3 ) , B i B ′ ( 4 ) , A i D o ( 1 ) , T T ′ ( 4 ) ⟶ A i B i ( 2 ) , D o T ( ≤ 3 ) , B ′ T ′ ( ≤ 4 ) , B_iB'(4), A_iD_o(1), TT'(4) \longrightarrow A_iB_i(2), D_oT(\le3), B'T'(\le4), B i B ′ ( 4 ) , A i D o ( 1 ) , T T ′ ( 4 ) ⟶ A i B i ( 2 ) , D o T ( ≤ 3 ) , B ′ T ′ ( ≤ 4 ) , B i B ′ ( 4 ) , A i D o ( 1 ) , P Q ( 3 ) ⟶ A i B i ( 2 ) , D o Q ( ≤ 3 ) , P B ′ ( 3 ) . B_iB'(4), A_iD_o(1), PQ(3) \longrightarrow A_iB_i(2), D_oQ(\le3), PB'(3). B i B ′ ( 4 ) , A i D o ( 1 ) , P Q ( 3 ) ⟶ A i B i ( 2 ) , D o Q ( ≤ 3 ) , P B ′ ( 3 ) . Their primary costs decrease from 4, 8, 6 to at most 3, 7, 5. The first switch leaves B ′ B' B ′ u u u
For C o C_o C o A i D i = 1 A_iD_i = 1 A i D i = 1
A i B i ( 2 ) , C i D i ( 2 ) , C o D o ( 2 ) ⟶ A i D i ( 1 ) , C i D o ( 2 ) , C o u ( ≤ 2 ) , A_iB_i(2), C_iD_i(2), C_oD_o(2) \longrightarrow A_iD_i(1), C_iD_o(2), C_ou(\le2), A i B i ( 2 ) , C i D i ( 2 ) , C o D o ( 2 ) ⟶ A i D i ( 1 ) , C i D o ( 2 ) , C o u ( ≤ 2 ) , A i B i ( 2 ) , C i D i ( 2 ) , C o D o ( 2 ) , T T ′ ( 4 ) ⟶ A i D i ( 1 ) , C i D o ( 2 ) , C o T ( ≤ 2 ) , B i T ′ ( ≤ 4 ) . A_iB_i(2), C_iD_i(2), C_oD_o(2), TT'(4) \longrightarrow A_iD_i(1), C_iD_o(2), C_oT(\le2), B_iT'(\le4). A i B i ( 2 ) , C i D i ( 2 ) , C o D o ( 2 ) , T T ′ ( 4 ) ⟶ A i D i ( 1 ) , C i D o ( 2 ) , C o T ( ≤ 2 ) , B i T ′ ( ≤ 4 ) . The costs decrease from 3, 7 to at most 2, 6. For the good-Q Q Q C i C j = 1 C_iC_j = 1 C i C j = 1 A j D j = 1 A_jD_j = 1 A j D j = 1 B ′ B' B ′ P B ′ = 3 PB' = 3 P B ′ = 3 B j B ′ B_jB' B j B ′
B j B ′ ( 4 ) , C i C j ( 1 ) , A j D j ( 1 ) , C o D o ( 2 ) , P Q ( 3 ) ⟶ C o Q ( ≤ 2 ) , D o C i ( 2 ) , C j D j ( 2 ) , A j B j ( 2 ) , B ′ P ( 3 ) B_jB'(4), C_iC_j(1), A_jD_j(1), C_oD_o(2), PQ(3)
\longrightarrow C_oQ(\le2), D_oC_i(2), C_jD_j(2), A_jB_j(2), B'P(3) B j B ′ ( 4 ) , C i C j ( 1 ) , A j D j ( 1 ) , C o D o ( 2 ) , P Q ( 3 ) ⟶ C o Q ( ≤ 2 ) , D o C i ( 2 ) , C j D j ( 2 ) , A j B j ( 2 ) , B ′ P ( 3 ) decreases primary cost from 7 to at most 6. The fixed good-pair threshold supplies disjoint auxiliary choices. These switches prove all the claimed label restrictions for the crossed orientation. Each restriction follows from the existence of the stated edges and the minimality of the initial matching.
An original B B B A 0 A_0 A 0 D D D D ∗ D^* D ∗ O ( M ) O(M) O ( M ) D D D A 0 A_0 A 0 D D D B B ( 4 ) BB(4) B B ( 4 ) A D ( 1 ) AD(1) A D ( 1 ) a / r + o ( 1 ) a/r+o(1) a / r + o ( 1 )
The enlarged reservoir count
Lemma 11.3. For every R = r + k R=r+k R = r + k (46) , there is an actual matching satisfying
C E ≤ B 0 + F ( R ) + c − E A A + ( 1 + R r + 2 δ d ) a + j + 2 U + R 2 r H + H 2 2 r 2 + 35 4 e 24 − e 13 − e 14 − I 3 + o ( 1 ) . (47) C_E \le B_0+F(R)+c-E_{AA}+\left(1+\frac{R}{r}+\frac{2\delta}{d}\right)a+j+2U
+\frac{R}{2r}H+\frac{H^2}{2r^2}+\frac{35}{4}e_{24}-e_{13}-e_{14}-I_3+o(1). \tag*{(47)} C E ≤ B 0 + F ( R ) + c − E AA + ( 1 + r R + d 2 δ ) a + j + 2 U + 2 r R H + 2 r 2 H 2 + 4 35 e 24 − e 13 − e 14 − I 3 + o ( 1 ) . (47) Proof. Uniformly sample R M RM R M C ∗ C^* C ∗ C ∗ C^* C ∗ e 24 e_{24} e 24 t ′ t' t ′ E ( R − t ′ ) ≤ e 24 / ( 2 w ) + o ( 1 ) E(R-t')\le e_{24}/(2w)+o(1) E ( R − t ′ ) ≤ e 24 / ( 2 w ) + o ( 1 ) D ∗ D^* D ∗
Orient all C D CD C D C n e w , D n e w C_{\mathrm{new}},D_{\mathrm{new}} C new , D new A A A a n e w + j n e w = a + j a_{\mathrm{new}}+j_{\mathrm{new}}=a+j a new + j new = a + j A A A B B B 2 a n e w + j n e w + ( t ′ / r ) a 2a_{\mathrm{new}}+j_{\mathrm{new}}+(t'/r)a 2 a new + j new + ( t ′ / r ) a C C C B B B A D AD A D D D D A A A C C C A A A A D AD A D A C AC A C a + j a+j a + j ( 1 + R / r ) a + j (1+R/r)a+j ( 1 + R / r ) a + j
Partition the originally omitted indices into straight-selected O s O_s O s O c O_c O c O 0 O_0 O 0 O s O_s O s C , D C,D C , D e o u t e_{\mathrm{out}} e out C C C O ( M ) O(M) O ( M )
c n e w = c + U O s + e o u t , i n c i d e n t O s . c_{\mathrm{new}}=c+U_{O_s}+e_{\mathrm{out},\mathrm{incident}\ O_s}. c new = c + U O s + e out , incident O s . Both positive missing-C C CC C C C C C ( q − R ) 2 / 2 − e o u t , O 0 O 0 + o ( 1 ) (q-R)^2/2-e_{\mathrm{out},O_0O_0}+o(1) ( q − R ) 2 /2 − e out , O 0 O 0 + o ( 1 ) D D DD D D U O 0 + e o u t , O c O 0 + o ( 1 ) U_{O_0}+e_{\mathrm{out},O_cO_0}+o(1) U O 0 + e out , O c O 0 + o ( 1 ) − c -c − c
c − ( q − R ) 2 + 2 U O s + U O 0 + 2 e o u t , i n c i d e n t O s + 2 e o u t , O 0 O 0 + e o u t , O c O 0 ≤ c − ( q − R ) 2 + 2 U + 2 e o u t ≤ c − ( q − R ) 2 + 2 U + 2 a δ d + o ( 1 ) . c-(q-R)^2+2U_{O_s}+U_{O_0}+2e_{\mathrm{out},\mathrm{incident}\ O_s}+2e_{\mathrm{out},O_0O_0}+e_{\mathrm{out},O_cO_0}
\le c-(q-R)^2+2U+2e_{\mathrm{out}}
\le c-(q-R)^2+2U+\frac{2a\delta}{d}+o(1). c − ( q − R ) 2 + 2 U O s + U O 0 + 2 e out , incident O s + 2 e out , O 0 O 0 + e out , O c O 0 ≤ c − ( q − R ) 2 + 2 U + 2 e out ≤ c − ( q − R ) 2 + 2 U + d 2 a δ + o ( 1 ) . The three outside-edge supports on the first line are disjoint. This proves that mixing the two orientations introduces no additional resource charge. In particular, this estimate is valid when links of both orientations are selected in the same matching.
The direct C D CD C D 2 t ′ q − t ′ 2 2t'q-t'^2 2 t ′ q − t ′2 D D DD D D ( R − t ′ / 2 + b 3 ) 2 / 2 (R-t'/2+b_3)^2/2 ( R − t ′ /2 + b 3 ) 2 /2 R 2 − R E ( R − t ′ ) − R e 24 / ( 2 w ) R^2-R E(R-t')-R e_{24}/(2w) R 2 − R E ( R − t ′ ) − R e 24 / ( 2 w ) C ∗ C^* C ∗ R z − 2 z E ( R − t ′ ) − R e 34 / ( 2 w ) Rz-2zE(R-t')-R e_{34}/(2w) R z − 2 z E ( R − t ′ ) − R e 34 / ( 2 w ) X 3 X_3 X 3 − R / 4 − b 3 / 2 -R/4-b_3/2 − R /4 − b 3 /2 t ′ = R t'=R t ′ = R R b 3 / 2 + b 3 2 / 2 ≤ R H / ( 2 r ) + H 2 / ( 2 r 2 ) + o ( 1 ) Rb_3/2+b_3^2/2\le RH/(2r)+H^2/(2r^2)+o(1) R b 3 /2 + b 3 2 /2 ≤ R H / ( 2 r ) + H 2 / ( 2 r 2 ) + o ( 1 ) e 24 e_{24} e 24
9 R / 4 + 2 z + b 3 / 2 2 w − 1 ≤ 35 4 , \frac{9R/4+2z+b_3/2}{2w}-1\le\frac{35}{4}, 2 w 9 R /4 + 2 z + b 3 /2 − 1 ≤ 4 35 , using R ≤ q ≤ 2 w R\le q\le2w R ≤ q ≤ 2 w z ≤ 3 q z\le3q z ≤ 3 q b 3 ≤ z b_3\le z b 3 ≤ z e 34 e_{34} e 34 R / ( 2 w ) − 1 ≤ 0 R/(2w)-1\le0 R / ( 2 w ) − 1 ≤ 0 X 1 X_1 X 1 2 R y + e 12 2Ry+e_{12} 2 R y + e 12 P P P e 23 e_{23} e 23 F ( R ) F(R) F ( R )
An optimized tangent envelope
For 0 ≤ ε ≤ 1 0 \le\varepsilon\le1 0 ≤ ε ≤ 1 F F F
F ( R ) ≤ F ( q ) + 23 8 ε 2 q 2 + [ ( 7 4 − 23 4 ε ) q + z − 2 y ] ( q − R ) . F(R) \le F(q) + \frac{23}{8}\varepsilon^2q^2 + \left[\left(\frac{7}{4}-\frac{23}{4}\varepsilon\right)q+z-2y\right](q-R). F ( R ) ≤ F ( q ) + 8 23 ε 2 q 2 + [ ( 4 7 − 4 23 ε ) q + z − 2 y ] ( q − R ) . It is the tangent at R = ( 1 − ε ) q R=(1-\varepsilon)q R = ( 1 − ε ) q R ≤ q R \le q R ≤ q σ = s / r ≥ 1 / 7 \sigma=s/r \ge1/7 σ = s / r ≥ 1/7 s ≥ δ s \ge\delta s ≥ δ
J ε = [ ( 7 / 4 − 23 ε / 4 ) q + z − 2 y ] + 2 σ r . J_{\varepsilon}=\frac{\left[\left(7/4-23\varepsilon/4\right)q+z-2y\right]_{+}}{2\sigma r}. J ε = 2 σ r [ ( 7/4 − 23 ε /4 ) q + z − 2 y ] + . The doubled denominator records the availability of both orientations. Substituting (46) in the extended count yields
C E ≤ B 0 + F ( q ) + 23 8 ε 2 q 2 + c − E A A + ( 1 + q r + 2 δ d + J ε ) a + ( 1 + J ε ) j + ( 2 + J ε ) U + J ε V + q 2 r H + H 2 2 r 2 + 35 4 e 24 − e 13 − e 14 − I 3 + o ( 1 ) . (48) \begin{aligned}
C_E \le{}& B_0+F(q)+\frac{23}{8}\varepsilon^2q^2+c-E_{AA}\\
&+\left(1+\frac{q}{r}+\frac{2\delta}{d}+J_{\varepsilon}\right)a+(1+J_{\varepsilon})j+(2+J_{\varepsilon})U+J_{\varepsilon}V\\
&+\frac{q}{2r}H+\frac{H^2}{2r^2}+\frac{35}{4}e_{24}-e_{13}-e_{14}-I_3+o(1). \tag*{(48)}
\end{aligned} C E ≤ B 0 + F ( q ) + 8 23 ε 2 q 2 + c − E AA + ( 1 + r q + d 2 δ + J ε ) a + ( 1 + J ε ) j + ( 2 + J ε ) U + J ε V + 2 r q H + 2 r 2 H 2 + 4 35 e 24 − e 13 − e 14 − I 3 + o ( 1 ) . (48) The comparison (45) has coefficient − 1 -1 − 1 V V V V V V (48) whenever the weight on (45) is at least J ε J_{\varepsilon} J ε (48) . Equation (44) and the split packing inequality then control the other missing-edge terms. On parameter intervals where this enlargement is not used, (45) remains available with its negative square term.
The choice ε = 0 \varepsilon=0 ε = 0 ε \varepsilon ε ε \varepsilon ε
The two reservoir matchings share a cardinality constraint
The initial and enlarged reservoir constructions use related auxiliary graphs. Comparing their maximum matching sizes yields an additional constraint. Write R = r + k R=r+k R = r + k t ≤ r t \le r t ≤ r t ′ ≤ R t' \le R t ′ ≤ R b 3 b_3 b 3 2 b 3 2b_3 2 b 3 s s s D D D s ′ s' s ′ D ∗ D^* D ∗
Put δ 0 = r − t \delta_0=r-t δ 0 = r − t δ 1 = R − t ′ \delta_1=R-t' δ 1 = R − t ′ I , I ′ I,I' I , I ′
∣ I ′ ∖ I ∣ / M ≤ k + δ 0 + o ( 1 ) . \lvert I'\setminus I\rvert/M \le k+\delta_0+o(1). ∣ I ′ ∖ I ∣ / M ≤ k + δ 0 + o ( 1 ) . Deleting the edges incident to these added vertices from the new maximum matching leaves a matching in the old auxiliary graph. Each deleted vertex destroys at most one matching edge. Because both maximum matchings preserve every exceptional endpoint, this yields
s ′ + 2 b 3 2 ≤ s + 2 b 3 2 + k + δ 0 + o ( 1 ) , s ′ ≤ s + 2 k + 2 δ 0 + o ( 1 ) . (49) \frac{s'+2b_3}{2}\le\frac{s+2b_3}{2}+k+\delta_0+o(1),\qquad s'\le s+2k+2\delta_0+o(1). \tag*{(49)} 2 s ′ + 2 b 3 ≤ 2 s + 2 b 3 + k + δ 0 + o ( 1 ) , s ′ ≤ s + 2 k + 2 δ 0 + o ( 1 ) . (49) This argument compares ordinary maximum matching cardinalities; it does not assume that the new matching contains the old matching’s edges. Starting with the exceptional pairs and augmenting guarantees that a maximum matching preserving their covered vertices exists in each graph.
The unmatched new target mass is u ′ = t ′ − s ′ u'=t'-s' u ′ = t ′ − s ′ u ′ u' u ′
u ′ ( R − t ′ / 2 + b 3 ) − ( u ′ ) 2 2 . u'\left(R-t'/2+b_3\right)-\frac{(u')^2}{2}. u ′ ( R − t ′ /2 + b 3 ) − 2 ( u ′ ) 2 . The unconstrained upper bound is ( R − t ′ / 2 + b 3 ) 2 / 2 (R-t'/2+b_3)^2/2 ( R − t ′ /2 + b 3 ) 2 /2 (49) gives an additional loss at least
1 2 ( r 2 − s − 3 k 2 − b 3 − 2 δ 0 − 3 2 δ 1 ) + 2 . (50) \frac{1}{2}\left(\frac{r}{2}-s-\frac{3k}{2}-b_3-2\delta_0-\frac{3}{2}\delta_1\right)_+^2. \tag*{(50)} 2 1 ( 2 r − s − 2 3 k − b 3 − 2 δ 0 − 2 3 δ 1 ) + 2 . (50) Indeed u ′ ≥ t ′ − s − 2 k − 2 δ 0 u' \ge t'-s-2k-2\delta_0 u ′ ≥ t ′ − s − 2 k − 2 δ 0 R − t ′ / 2 + b 3 R-t'/2+b_3 R − t ′ /2 + b 3
Condition on the old target and the selected swap links, so that r , s , k , δ 0 r,s,k,\delta_0 r , s , k , δ 0
δ 0 ≤ e 24 2 w + o ( 1 ) , E δ 1 ≤ e 24 2 w + o ( 1 ) , b 3 ≤ H / r + o ( 1 ) . \delta_0 \le\frac{e_{24}}{2w}+o(1),\qquad\mathbb{E}\delta_1 \le\frac{e_{24}}{2w}+o(1),\qquad b_3\le H/r+o(1). δ 0 ≤ 2 w e 24 + o ( 1 ) , E δ 1 ≤ 2 w e 24 + o ( 1 ) , b 3 ≤ H / r + o ( 1 ) . The function x ↦ ( x + ) 2 / 2 x\mapsto(x_+)^2/2 x ↦ ( x + ) 2 /2 (50) therefore proves that one may subtract
1 2 ( r 2 − s − 3 k 2 − H r − 7 e 24 4 w ) + 2 (51) \frac{1}{2}\left(\frac{r}{2}-s-\frac{3k}{2}-\frac{H}{r}-\frac{7e_{24}}{4w}\right)_+^2 \tag*{(51)} 2 1 ( 2 r − s − 2 3 k − r H − 4 w 7 e 24 ) + 2 (51) from the right side of the enlarged-reservoir bound (47) . The coefficient 7 / ( 4 w ) 7/(4w) 7/ ( 4 w ) 2 / ( 2 w ) 2/(2w) 2/ ( 2 w ) ( 3 / 2 ) / ( 2 w ) (3/2)/(2w) ( 3/2 ) / ( 2 w ) (51) . All errors are uniform in N ≥ 2 M N\ge2M N ≥ 2 M O ( M ) O(M) O ( M )
Quadratic forms for exact certificates. Let g = r / 2 − s − 3 k / 2 = 2 r − s − 3 R / 2 g=r/2-s-3k/2=2r-s-3R/2 g = r /2 − s − 3 k /2 = 2 r − s − 3 R /2
− 1 2 ( x + ) 2 ≤ 1 2 τ 2 − τ x ( τ ≥ 0 ) -\frac{1}{2}(x_+)^2\le\frac{1}{2}\tau^2-\tau x\qquad(\tau\ge0) − 2 1 ( x + ) 2 ≤ 2 1 τ 2 − τ x ( τ ≥ 0 ) provides useful rational quadratic bounds. Choosing τ = η r \tau=\eta r τ = η r η ≥ 0 \eta\ge0 η ≥ 0 (47)
η r ( s − r / 2 + 3 k / 2 ) + η 2 r 2 2 + η H + 7 η r 4 w e 24 . (52) \eta r(s-r/2+3k/2)+\frac{\eta^2r^2}{2}+\eta H+\frac{7\eta r}{4w}e_{24}. \tag*{(52)} η r ( s − r /2 + 3 k /2 ) + 2 η 2 r 2 + η H + 4 w 7 η r e 24 . (52) The last coefficient is at most 7 η / 2 7\eta/2 7 η /2 r ≤ 2 w r\le2w r ≤ 2 w F ( R ) F(R) F ( R ) F ( R ) + ( 3 η / 2 ) r R F(R)+(3\eta/2)rR F ( R ) + ( 3 η /2 ) r R R = q R=q R = q ε = 0 \varepsilon=0 ε = 0
J η = ( 7 q / 4 + z − 2 y − 3 η r / 2 ) + 2 σ r , σ = s / r , J_\eta=\frac{(7q/4+z-2y-3\eta r/2)_+}{2\sigma r},\qquad\sigma=s/r, J η = 2 σ r ( 7 q /4 + z − 2 y − 3 η r /2 ) + , σ = s / r , and its main polynomial is
F ( q ) + η r ( s − r / 2 + 3 ( q − r ) / 2 ) + η 2 r 2 / 2. F(q)+\eta r(s-r/2+3(q-r)/2)+\eta^2r^2/2. F ( q ) + η r ( s − r /2 + 3 ( q − r ) /2 ) + η 2 r 2 /2. Its H H H η \eta η e 24 e_{24} e 24 7 η / 2 7\eta/2 7 η /2 J η J_\eta J η J J J
Alternatively choose τ = λ q \tau=\lambda q τ = λ q λ ≥ 0 \lambda\ge0 λ ≥ 0 q / r ≤ 8 / 7 q/r\le8/7 q / r ≤ 8/7 q ≤ 2 w q\le2w q ≤ 2 w
8 λ 7 H + 7 λ 2 e 24 . \frac{8\lambda}{7}H+\frac{7\lambda}{2}e_{24}. 7 8 λ H + 2 7 λ e 24 . In the envelope, the main polynomial changes by
( λ 2 2 + 3 λ 2 ) q 2 − λ ( 2 − σ ) q r , \left(\frac{\lambda^2}{2}+\frac{3\lambda}{2}\right)q^2-\lambda(2-\sigma)qr, ( 2 λ 2 + 2 3 λ ) q 2 − λ ( 2 − σ ) q r , and
J λ = ( ( 7 / 4 − 3 λ / 2 ) q + z − 2 y ) + 2 σ r . J_\lambda=\frac{((7/4-3\lambda/2)q+z-2y)_+}{2\sigma r}. J λ = 2 σ r (( 7/4 − 3 λ /2 ) q + z − 2 y ) + . These formulas introduce no new optimization coordinate: a rational η \eta η λ \lambda λ
A coarser consequence, useful when fixed coefficients on the deficits are preferred, is the additional term
− 1 2 g + 2 + H 2 + 7 4 e 24 . -\frac{1}{2}g_+^2+\frac{H}{2}+\frac{7}{4}e_{24}. − 2 1 g + 2 + 2 H + 4 7 e 24 . To verify it, note that g + ≤ r / 2 g_+ \le r/2 g + ≤ r /2 1 2 g + 2 − 1 2 ( g − D ) + 2 ≤ ( r / 2 ) D \frac{1}{2}g_+^2-\frac{1}{2}(g-D)_+^2 \le(r/2)D 2 1 g + 2 − 2 1 ( g − D ) + 2 ≤ ( r /2 ) D D ≥ 0 D \ge0 D ≥ 0 D = b 3 + 2 δ 0 + 3 δ 1 / 2 D=b_3+2\delta_0+3\delta_1/2 D = b 3 + 2 δ 0 + 3 δ 1 /2 r ≤ 2 w r \le2w r ≤ 2 w (52) .
The envelope when the covered side is smaller. The full rectangular bound (46) gives
q − R ≤ ( δ − s ) + + V + U + a + j 2 max ( s , δ ) + o ( 1 ) . q-R \le(\delta-s)_+ + \frac{V+U+a+j}{2\max(s,\delta)}+o(1). q − R ≤ ( δ − s ) + + 2 max ( s , δ ) V + U + a + j + o ( 1 ) . After the rank tangent with parameter η \eta η R = ( 1 − ε ) q R=(1-\varepsilon)q R = ( 1 − ε ) q q − R q-R q − R
[ ( 7 / 4 − 23 ε / 4 ) q + z − 2 y − 3 η r / 2 ] + . \left[(7/4-23\varepsilon/4)q+z-2y-3\eta r/2\right]_+. [ ( 7/4 − 23 ε /4 ) q + z − 2 y − 3 η r /2 ] + . To make this bound uniform on a parameter box, suppose
σ 0 ≤ s / r ≤ σ 1 , ρ 0 ≤ r / q ≤ ρ 1 , ζ 0 ≤ z / q ≤ ζ 1 , υ 0 ≤ y / q ≤ υ 1 . \sigma_0\le s/r\le\sigma_1,\qquad\rho_0\le r/q\le\rho_1,\qquad\zeta_0\le z/q\le\zeta_1,\qquad\upsilon_0\le y/q\le\upsilon_1. σ 0 ≤ s / r ≤ σ 1 , ρ 0 ≤ r / q ≤ ρ 1 , ζ 0 ≤ z / q ≤ ζ 1 , υ 0 ≤ y / q ≤ υ 1 . Put
D b o x = max { σ 0 ρ 0 , 1 − ρ 1 } , A b o x = [ 7 / 4 − 23 ε / 4 + ζ 1 − 2 υ 0 − 3 η ρ 0 / 2 ] + . D_{\mathrm{box}}=\max\{\sigma_0\rho_0,1-\rho_1\},\qquad A_{\mathrm{box}}=[7/4-23\varepsilon/4+\zeta_1-2\upsilon_0-3\eta\rho_0/2]_+. D box = max { σ 0 ρ 0 , 1 − ρ 1 } , A box = [ 7/4 − 23 ε /4 + ζ 1 − 2 υ 0 − 3 η ρ 0 /2 ] + . If D b o x > 0 D_{\mathrm{box}}>0 D box > 0
J = A b o x 2 D b o x , E b o x = A b o x [ 1 − ( 1 + σ 0 ) ρ 0 ] + . (53) J=\frac{A_{\mathrm{box}}}{2D_{\mathrm{box}}},\qquad E_{\mathrm{box}}=A_{\mathrm{box}}[1-(1+\sigma_0)\rho_0]_+. \tag*{(53)} J = 2 D box A box , E box = A box [ 1 − ( 1 + σ 0 ) ρ 0 ] + . (53) The resulting enlarged column has the same resource coefficients as before, with this J J J E b o x q 2 E_{\mathrm{box}}q^2 E box q 2 A b o x q A_{\mathrm{box}}q A box q max ( s , δ ) ≥ D b o x q \max(s,\delta)\ge D_{\mathrm{box}}q max ( s , δ ) ≥ D box q ( δ − s ) + ≤ [ 1 − ( 1 + σ 0 ) ρ 0 ] + q (\delta-s)_+\le[1-(1+\sigma_0)\rho_0]_+q ( δ − s ) + ≤ [ 1 − ( 1 + σ 0 ) ρ 0 ] + q D b o x = 0 D_{\mathrm{box}}=0 D box = 0 σ 0 ρ 0 ≥ 1 − ρ 0 \sigma_0\rho_0\ge1-\rho_0 σ 0 ρ 0 ≥ 1 − ρ 0
The supply of label-4 vertices and the imbalance of the two families
The reservoir construction matches retained C C C 2 w 2w 2 w r r r ∣ d − q ∣ |d-q| ∣ d − q ∣
Write H = h 3 H=h_3 H = h 3 B 0 = B − b 2 B_0=B-b^2 B 0 = B − b 2 r ≤ q r\le q r ≤ q q ≤ 2 w q\le2w q ≤ 2 w P P P
h 0 = min ( r , 2 w ) , D 0 = max ( r , 2 w ) , h_0=\min(r,2w),\qquad D_0=\max(r,2w), h 0 = min ( r , 2 w ) , D 0 = max ( r , 2 w ) , and define the quadratic polynomial, separately on its two capacity regions,
g ( h 0 ) = h 0 ( 3 q / 2 − r − z + 2 y ) − 7 h 0 2 / 8 + q 2 / 2. g(h_0)=h_0(3q/2-r-z+2y)-7h_0^2/8+q^2/2. g ( h 0 ) = h 0 ( 3 q /2 − r − z + 2 y ) − 7 h 0 2 /8 + q 2 /2. A maximum label-3 matching between retained C C C
t ≥ h 0 − e 24 / D 0 − o ( 1 ) . t \ge h_0 - e_{24}/D_0 - o(1). t ≥ h 0 − e 24 / D 0 − o ( 1 ) . This is the rectangular vertex-cover bound and applies regardless of which side is larger. The exact outward caps of retained C , D C,D C , D
The direct reservoir count, with the matched mass t t t
C T − C 0 ≤ t ( 3 q / 2 − r − z + 2 y + 3 H / r ) − 7 t 2 / 8 + q 2 / 2 + 2 c + j + ( 2 + t / r ) a + e 12 + e 23 + e 24 + e 34 + o ( 1 ) . C_T-C_0 \le t(3q/2-r-z+2y+3H/r)-7t^2/8+q^2/2
+2c+j+(2+t/r)a+e_{12}+e_{23}+e_{24}+e_{34}+o(1). C T − C 0 ≤ t ( 3 q /2 − r − z + 2 y + 3 H / r ) − 7 t 2 /8 + q 2 /2 + 2 c + j + ( 2 + t / r ) a + e 12 + e 23 + e 24 + e 34 + o ( 1 ) . For 0 ≤ t ≤ r ≤ q 0\le t\le r\le q 0 ≤ t ≤ r ≤ q H H H − 5 r / 4 − z -5r/4-z − 5 r /4 − z 3 t H / r ≤ 3 H 3tH/r\le3H 3 t H / r ≤ 3 H t / r ≤ 1 t/r\le1 t / r ≤ 1
C T ≤ B 0 + g ( h 0 ) + 3 H + 2 a + c − E A A − e 13 − e 14 + 5 r / 4 + z D 0 e 24 + o ( 1 ) . (54) C_T\le B_0+g(h_0)+3H+2a+c-E_{AA}-e_{13}-e_{14}
+\frac{5r/4+z}{D_0}e_{24}+o(1). \tag*{(54)} C T ≤ B 0 + g ( h 0 ) + 3 H + 2 a + c − E AA − e 13 − e 14 + D 0 5 r /4 + z e 24 + o ( 1 ) . (54) No Z ≥ 3 r / 2 Z\ge3r/2 Z ≥ 3 r /2
On the domain r , w ≥ 1 / 10 r,w\ge1/10 r , w ≥ 1/10 e 24 e_{24} e 24 x ≥ 2 r x\ge2r x ≥ 2 r x + y + z + w = 1 x+y+z+w=1 x + y + z + w = 1 5 r / 4 + z ≤ 1 − 3 r / 4 − w 5r/4+z\le1-3r/4-w 5 r /4 + z ≤ 1 − 3 r /4 − w r ≤ 2 w r\le2w r ≤ 2 w 33 / 8 33/8 33/8 r , w ≥ 1 / 10 ; i f r,w\ge1/10; if r , w ≥ 1/10 ; i f 15 / 4 15/4 15/4 (10) is also valid here. On y ≤ 3 z / 8 y\le3z/8 y ≤ 3 z /8 y ≤ w y\le w y ≤ w − 5 H / 8 -5H/8 − 5 H /8 e 14 e_{14} e 14
( r / ( 2 w ) − 1 ) + ≤ 3 / 2 , (r/(2w)-1)_+\le3/2, ( r / ( 2 w ) − 1 ) + ≤ 3/2 , using r ≤ min ( d , q ) ≤ 1 / 2 r\le\min(d,q)\le1/2 r ≤ min ( d , q ) ≤ 1/2 w ≥ 1 / 10 w\ge1/10 w ≥ 1/10
Let p , b R , b T p,b_R,b_T p , b R , b T (54) , respectively. It suffices to impose
5 8 b R ≥ 3 b T , p ≥ 5 b T , p ≥ 2 b T , p + b T ≥ 3 2 b R . \frac{5}{8}b_R\ge3b_T,\qquad p\ge5b_T,\qquad p\ge2b_T,\qquad p+b_T\ge\frac{3}{2}b_R. 8 5 b R ≥ 3 b T , p ≥ 5 b T , p ≥ 2 b T , p + b T ≥ 2 3 b R . These absorb respectively H , e 24 H,e_{24} H , e 24 a , c a,c a , c e 14 e_{14} e 14 m = min ( d , q ) m=\min(d,q) m = min ( d , q ) K = ( m 2 − r 2 ) / 2 K=(m^2-r^2)/2 K = ( m 2 − r 2 ) /2 E ≥ K − o ( 1 ) E\ge K-o(1) E ≥ K − o ( 1 )
C ≤ B 0 + b R f R + b T g ( h 0 ) − ( p − 2 b T ) K + o ( 1 ) . C\le B_0+b_Rf_R+b_Tg(h_0)-(p-2b_T)K+o(1). C ≤ B 0 + b R f R + b T g ( h 0 ) − ( p − 2 b T ) K + o ( 1 ) . This is a rational quadratic expression on each of the regions r ≤ 2 w r\le2w r ≤ 2 w r ≥ 2 w r\ge2w r ≥ 2 w
Small imbalance. Apply the same comparison to 0 ≤ h ≤ 3 m / 10 0\le h\le3m/10 0 ≤ h ≤ 3 m /10 h = ∣ d − q ∣ h=|d-q| h = ∣ d − q ∣ C D CD C D q q q g ( h 0 ) g(h_0) g ( h 0 ) m = min ( d , q ) m=\min(d,q) m = min ( d , q ) K K K r ≤ 2 w r\le2w r ≤ 2 w r ≥ 2 w r\ge2w r ≥ 2 w h 0 = r h_0=r h 0 = r h 0 = 2 w h_0=2w h 0 = 2 w 3 e 14 / 2 3e_{14}/2 3 e 14 /2
γ ( 4 ) ≤ 433403 / 1000000 = 0.433403 ( 0 ≤ h ≤ 3 m / 10 ) . \gamma(4)\le433403/1000000=0.433403\qquad(0\le h\le3m/10). γ ( 4 ) ≤ 433403/1000000 = 0.433403 ( 0 ≤ h ≤ 3 m /10 ) . The checker explicitly requires all four orientation/capacity cases. It assumes neither m ≤ 2 w m\le2w m ≤ 2 w h ≥ 3 m / 10 h\ge3m/10 h ≥ 3 m /10 q ≥ d ≥ 2 w q\ge d\ge2w q ≥ d ≥ 2 w h ≥ 3 m / 10 h\ge3m/10 h ≥ 3 m /10 h / m h/m h / m
In the case d ≥ q d\ge q d ≥ q ab-boundary-certificate.json prove
γ ( 4 ) ≤ 433 / 1000 when d ≥ q ≥ 2 w \gamma(4)\le433/1000\quad\text{when }d\ge q\ge2w γ ( 4 ) ≤ 433/1000 when d ≥ q ≥ 2 w on the shared localized domain. Its checker verifies the defect budgets, positive definiteness by rational L D L T LDL^T L D L T r r r 2 w 2w 2 w q ≥ 2 w q\ge2w q ≥ 2 w
The larger-C D CD C D The full surplus-target construction uses no top vertices. Its assumption d ≤ 2 w d \le2w d ≤ 2 w ( s / ( 2 w ) − 1 ) e 14 (s/(2w)-1)e_{14} ( s / ( 2 w ) − 1 ) e 14 s , r ≤ d ≤ 1 / 2 s,r \le d \le1/2 s , r ≤ d ≤ 1/2 w ≥ 1 / 10 w \ge1/10 w ≥ 1/10 3 e 14 / 2 3e_{14}/2 3 e 14 /2 s → d s \to d s → d d ≥ 2 w d \ge2w d ≥ 2 w p ≥ 3 ( b R + b D ) / 2 p \ge3(b_R+b_D)/2 p ≥ 3 ( b R + b D ) /2 p , b R , b D p,b_R,b_D p , b R , b D cd-boundary-certificate.json gives γ ( 4 ) ≤ 0.43301 \gamma(4) \le0.43301 γ ( 4 ) ≤ 0.43301 A B AB A B 0.433 0.433 0.433 m ≥ 2 w m \ge2w m ≥ 2 w r ≤ m r \le m r ≤ m r ≥ 2 w r \ge2w r ≥ 2 w
Combining the matching comparisons
We now prove that the preceding inequalities cover every labelled complete graph and imply γ ( 4 ) ≤ 0.4334035 \gamma(4) \le0.4334035 γ ( 4 ) ≤ 0.4334035 h = ∣ d − q ∣ = 2 b h = \lvert d-q\rvert= 2b h = ∣ d − q ∣ = 2 b m = min ( d , q ) m = \min(d,q) m = min ( d , q ) γ \gamma γ 1 / 2 1/2 1/2 M 2 M^2 M 2 ( 2 M 2 ) \binom{2M}{2} ( 2 2 M )
How the quadratic certificates prove an upper bound
For a fixed labelled graph, let C 0 , C 1 , … C_0,C_1,\ldots C 0 , C 1 , … M 2 M^2 M 2 θ j \theta_j θ j
min j C j ≤ ∑ j θ j C j . \min_j C_j \le\sum_j \theta_j C_j. j min C j ≤ j ∑ θ j C j . We choose the weights so that the coefficients of all uncontrolled nonnegative missing-edge counts on the right are nonpositive. Packing inequalities remove the remaining such quantities. The result is an upper bound 1 − Φ ( V ) + o ( 1 ) 1-\Phi(V)+o(1) 1 − Φ ( V ) + o ( 1 ) Φ \Phi Φ V V V
Write Λ = x + y + z + w \Lambda= x+y+z+w Λ = x + y + z + w g i ( V ) ≥ 0 g_i(V) \ge0 g i ( V ) ≥ 0 μ i \mu_i μ i
Φ ( V ) − ( 1 − 2 G ) Λ 2 − ∑ i μ i g i ( V ) = V T A V \Phi(V) - (1-2G)\Lambda^2 - \sum_i \mu_i g_i(V) = V^T A V Φ ( V ) − ( 1 − 2 G ) Λ 2 − i ∑ μ i g i ( V ) = V T A V for a positive semidefinite rational matrix A A A Φ ( V ) ≥ 1 − 2 G \Phi(V) \ge1-2G Φ ( V ) ≥ 1 − 2 G 2 G + o ( 1 ) 2G+o(1) 2 G + o ( 1 ) γ ( 4 ) ≤ G \gamma(4) \le G γ ( 4 ) ≤ G L D L T LDL^T L D L T
The case division proceeds as follows. The initial and core comparisons first restrict the parameter domain. The bounds of § 12 then dispose of small imbalance and insufficient supply of label-4 vertices. We divide the remaining cases into d ≥ q d \ge q d ≥ q q ≥ d q \ge d q ≥ d A P AP A P C P CP C P
Common parameter restrictions
The starting certificate and its entry conditions are proved in § 8 .
The retained conditions are established in two ways. The starting quadratic residual gives the desired bound whenever any of the following inequalities fails:
r , w ≥ 1 / 10 , z ≥ 1 / 5 , y ≥ 187 m / 1000 , y ≤ w , 2 y ≤ z , r,w \ge1/10,\quad z \ge1/5,\quad y \ge187m/1000,\quad y \le w,\quad2y \le z, r , w ≥ 1/10 , z ≥ 1/5 , y ≥ 187 m /1000 , y ≤ w , 2 y ≤ z , r + z ≤ 6 w , h ≤ 5 m / 4 , y ≤ m , z ≤ 3 m . r+z \le6w,\quad h \le5m/4,\quad y \le m,\quad z \le3m. r + z ≤ 6 w , h ≤ 5 m /4 , y ≤ m , z ≤ 3 m . These conditions depend only on m = min ( d , q ) m=\min(d,q) m = min ( d , q ) 2 ( 23 / 53 − 0.4334035 ) 2(23/53-0.4334035) 2 ( 23/53 − 0.4334035 )
The remaining boundary regions are optimized directly, using the initial and core inequalities with weights 1 − β 1-\beta 1 − β β \beta β 0 ≤ β ≤ 1 / 4 0 \le\beta\le1/4 0 ≤ β ≤ 1/4 − 1 -1 − 1 3 3 3
( m − r ) ( 2 r − m ) , Λ ( m − r ) , Λ ( r − 4 m / 5 ) (m-r)(2r-m),\quad\Lambda(m-r),\quad\Lambda(r-4m/5) ( m − r ) ( 2 r − m ) , Λ ( m − r ) , Λ ( r − 4 m /5 ) and the relevant branch and violating-halfspace constraints are subtracted from the quadratic form. Exact rational positive-definite residuals give the following bounds on γ \gamma γ
Table 1 (PDF p. 40).
Their rational weights and multipliers are in exclusion-certificate.json. The checker reconstructs each matrix independently. At this point the retained common conditions include y ≤ 3 z / 8 y \le3z/8 y ≤ 3 z /8 y ≤ 13 m / 20 y \le13m/20 y ≤ 13 m /20 z ≤ 3 m z \le3m z ≤ 3 m h ≤ 5 m / 4 h \le5m/4 h ≤ 5 m /4 h / m h/m h / m
§ 12 supplies the following additional exceptions, in this order:
Table 2.
The first uses the raw reservoir with target min ( r , 2 w ) \min(r,2w) min ( r , 2 w ) h ≥ 3 m / 10 h \ge3m/10 h ≥ 3 m /10 m ≥ 2 w m \ge2w m ≥ 2 w r ≤ m r \le m r ≤ m r ≤ 2 w r \le2w r ≤ 2 w verify_small_imbalance.py, verify_ab_boundary.py, and verify_cd_boundary.py. Their domain forms, weights, and coverage checks are reconstructed from the accompanying rational data. Here h = 2 b h=2b h = 2 b h ≤ r h \le r h ≤ r
Parameter restrictions when d ≥ q d \ge q d ≥ q
In this orientation q = m q=m q = m d = q + h d=q+h d = q + h
Table 3.
This bound is verified by verify_ab_good_domains.py; it uses neither a bound on the missing-edge density within the retained C C C A P AP A P Z = z − h 3 / r Z=z-h_3/r Z = z − h 3 / r
On r ≥ 7 q / 8 r \ge7q/8 r ≥ 7 q /8
Z ≤ 3 r / 2 ⟹ γ ≤ 0.433392. Z \le3r/2 \quad\Longrightarrow\quad\gamma\le0.433392. Z ≤ 3 r /2 ⟹ γ ≤ 0.433392. Five rational seven-variable certificates use the violating form z u − r z + 3 r 2 / 2 ≥ 0 zu-rz+3r^2/2 \ge0 z u − r z + 3 r 2 /2 ≥ 0 u = h 3 / z u=h_3/z u = h 3 / z h / q ∈ [ 3 / 10 , 5 / 4 ] h/q \in[3/10,5/4] h / q ∈ [ 3/10 , 5/4 ] r / q ∈ [ 7 / 8 , 1 ] r/q \in[7/8,1] r / q ∈ [ 7/8 , 1 ] z / q ∈ [ 2 / 5 , 3 ] z/q \in[2/5,3] z / q ∈ [ 2/5 , 3 ] z ≥ 1 / 5 z \ge1/5 z ≥ 1/5 q ≤ 1 / 2 q \le1/2 q ≤ 1/2 P P P verify_z_domain.py verifies the defect budgets, matrices, and full domain coverage from z-domain-certificate.json.
After excluding r < 7 q / 8 r<7q/8 r < 7 q /8 c ≥ q 2 / 8 c \ge q^2/8 c ≥ q 2 /8 B 0 − q 2 / 8 B_0-q^2/8 B 0 − q 2 /8 B 0 + f R B_0+f_R B 0 + f R F = x 2 / 4 + y 2 + z 2 / 2 + w 2 + b 2 F=x^2/4+y^2+z^2/2+w^2+b^2 F = x 2 /4 + y 2 + z 2 /2 + w 2 + b 2 B 0 = 1 − F B_0=1-F B 0 = 1 − F 135283 / 10 6 135283/10^6 135283/1 0 6
F + 864717 10 6 q 2 8 − 135283 10 6 f R − 69682 10 6 Λ ( r − 7 q / 8 ) − 21247 10 6 ( r − 7 q / 8 ) ( q − r ) . F+\frac{864717}{10^6}\frac{q^2}{8}-\frac{135283}{10^6}f_R-\frac{69682}{10^6}\Lambda(r-7q/8)-\frac{21247}{10^6}(r-7q/8)(q-r). F + 1 0 6 864717 8 q 2 − 1 0 6 135283 f R − 1 0 6 69682 Λ ( r − 7 q /8 ) − 1 0 6 21247 ( r − 7 q /8 ) ( q − r ) . Here Λ = x + y + z + w \Lambda=x+y+z+w Λ = x + y + z + w ( 1 − 2 ⋅ 0.43321 ) Λ 2 (1-2\cdot0.43321)\Lambda^2 ( 1 − 2 ⋅ 0.43321 ) Λ 2 verify_high_c.py. Thus this case satisfies γ ≤ 0.43321 \gamma\le0.43321 γ ≤ 0.43321 Lemma 6.1 applies.
After also excluding Z < 3 r / 2 Z<3r/2 Z < 3 r /2
d ≥ 23 Z / 20 ⟹ γ ≤ 0.433378. d \ge23Z/20 \quad\Longrightarrow\quad\gamma\le0.433378. d ≥ 23 Z /20 ⟹ γ ≤ 0.433378. It combines the initial, core, and joint-reservoir bounds and subtracts a nonnegative multiple of
h 3 − r z + 20 r d / 23 = z u − r z + 20 r d / 23 ≥ 0 , u = h 3 / z . h_3-rz+20rd/23=zu-rz+20rd/23 \ge0,\qquad u=h_3/z. h 3 − r z + 20 r d /23 = z u − r z + 20 r d /23 ≥ 0 , u = h 3 / z . The data are in good-domain-new-certificate.json; its checker verifies the resource budgets, both ratio boxes, and rational positive definiteness. It uses the Tutte–Berge inequality (17) , without the additional dense-case estimate.
Both orientations also retain y ≥ 187 m / 1000 y \ge187m/1000 y ≥ 187 m /1000 23138190 / 20583940031 23138190/20583940031 23138190/20583940031 2 ( 23 / 53 − 0.4334035 ) 2(23/53-0.4334035) 2 ( 23/53 − 0.4334035 ) A B AB A B
3 / 10 ≤ h / q ≤ 5 / 4 , 7 / 8 ≤ r / q ≤ 1 , 21 / 16 ≤ z / q ≤ 3 , 187 / 1000 ≤ y / q ≤ 13 / 20 , 3/10 \le h/q \le5/4,\qquad7/8 \le r/q \le1,\qquad21/16 \le z/q \le3,\qquad187/1000 \le y/q \le13/20, 3/10 ≤ h / q ≤ 5/4 , 7/8 ≤ r / q ≤ 1 , 21/16 ≤ z / q ≤ 3 , 187/1000 ≤ y / q ≤ 13/20 , Z ≥ 3 r / 2 , d ≤ 23 Z / 20 , c ≤ q 2 / 8 , Z \ge3r/2,\qquad d \le23Z/20,\qquad c \le q^2/8, Z ≥ 3 r /2 , d ≤ 23 Z /20 , c ≤ q 2 /8 , together with the common conditions already stated.
Three exhaustive alternatives on the larger-A B AB A B
All three alternatives now use seven coordinates V = ( x , y , z , w , b , r , u ) V=(x,y,z,w,b,r,u) V = ( x , y , z , w , b , r , u ) u = h 3 / z u=h_3/z u = h 3 / z ( 1 − y / z ) h 3 (1-y/z)h_3 ( 1 − y / z ) h 3 ( z − y ) u (z-y)u ( z − y ) u
( v , ρ , ζ , σ , υ ) = ( h / q , r / q , z / q , s / r , y / q ) (v,\rho,\zeta,\sigma,\upsilon)=(h/q,r/q,z/q,s/r,y/q) ( v , ρ , ζ , σ , υ ) = ( h / q , r / q , z / q , s / r , y / q ) and their common root box is
[ 3 / 10 , 5 / 4 ] × [ 7 / 8 , 1 ] × [ 21 / 16 , 3 ] × [ 0 , 1 ] × [ 187 / 1000 , 13 / 20 ] . (55) [3/10,5/4]\times[7/8,1]\times[21/16,3]\times[0,1]\times[187/1000,13/20]. \tag*{(55)} [ 3/10 , 5/4 ] × [ 7/8 , 1 ] × [ 21/16 , 3 ] × [ 0 , 1 ] × [ 187/1000 , 13/20 ] . (55) Here s s s D D D § 11 . It is not an independent coordinate of the quadratic form: each box encloses its actual ratio s / r s/r s / r A P AP A P C P CP C P α = 179 / 1000 \alpha=179/1000 α = 179/1000 (55) .
On a box with endpoints v i v_i v i ρ i \rho_i ρ i ζ i \zeta_i ζ i σ i \sigma_i σ i υ i \upsilon_i υ i i = 0 , 1 i = 0,1 i = 0 , 1
L = 4 ( 1 − ρ 0 ) 5 ρ 0 − 4 , β = α 0 = 1 1 + v 0 , K = q 2 − r 2 2 , L = \frac{4(1-\rho_0)}{5\rho_0-4}, \qquad\beta= \alpha_0 = \frac{1}{1+v_0}, \qquad K = \frac{q^2-r^2}{2}, L = 5 ρ 0 − 4 4 ( 1 − ρ 0 ) , β = α 0 = 1 + v 0 1 , K = 2 q 2 − r 2 , A T = 2 + 2 − ρ 0 1 + v 0 + L β , C T = 1 + L , T 0 = f T − q ( q − r ) 2 − L K . A_T = 2+\frac{2-\rho_0}{1+v_0}+L\beta, \qquad C_T = 1+L, \qquad T_0 = f_T-\frac{q(q-r)}{2}-LK. A T = 2 + 1 + v 0 2 − ρ 0 + L β , C T = 1 + L , T 0 = f T − 2 q ( q − r ) − L K . These coefficients use the globally maximal joint-core resource (20), dropping its favorable quadratic term in a a a − L K -LK − L K ϵ ∗ = 5 / 64 \epsilon_* = 5/64 ϵ ∗ = 5/64 η r k = 1 / 10 \eta_{rk} = 1/10 η r k = 1/10 § 11.5 . Put
g = 1 2 dist ( [ σ 0 , σ 1 ] , 1 / 2 ) 2 , A b o x = [ 7 / 4 − 23 ϵ ∗ / 4 + ζ 1 − 2 υ 0 − 3 η r k ρ 0 / 2 ] + . g = \frac{1}{2}\operatorname{dist}([\sigma_0,\sigma_1],1/2)^2, \qquad A_{\mathrm{box}} = [7/4-23\epsilon_*/4+\zeta_1-2\upsilon_0-3\eta_{rk}\rho_0/2]_+. g = 2 1 dist ([ σ 0 , σ 1 ] , 1/2 ) 2 , A box = [ 7/4 − 23 ϵ ∗ /4 + ζ 1 − 2 υ 0 − 3 η r k ρ 0 /2 ] + . For saturated A P AP A P
D b o x = max { σ 0 ρ 0 , 1 − ρ 1 } , J = A b o x 2 D b o x , E b o x = A b o x [ 1 − ( 1 + σ 0 ) ρ 0 ] + D_{\mathrm{box}} = \max\{\sigma_0\rho_0,1-\rho_1\}, \qquad J = \frac{A_{\mathrm{box}}}{2D_{\mathrm{box}}}, \qquad E_{\mathrm{box}} = A_{\mathrm{box}}[1-(1+\sigma_0)\rho_0]_+ D box = max { σ 0 ρ 0 , 1 − ρ 1 } , J = 2 D box A box , E box = A box [ 1 − ( 1 + σ 0 ) ρ 0 ] + when D b o x > 0 D_{\mathrm{box}}>0 D box > 0 D b o x = 0 D_{\mathrm{box}}=0 D box = 0 J = E b o x = 0 J=E_{\mathrm{box}}=0 J = E box = 0 σ 0 ρ 0 ≥ 1 − ρ 0 \sigma_0\rho_0 \ge1-\rho_0 σ 0 ρ 0 ≥ 1 − ρ 0 J = A b o x / ( 2 σ 0 ρ 0 ) J=A_{\mathrm{box}}/(2\sigma_0\rho_0) J = A box / ( 2 σ 0 ρ 0 ) E b o x = 0 E_{\mathrm{box}}=0 E box = 0 J = E b o x = 0 J=E_{\mathrm{box}}=0 J = E box = 0
After removing the common B 0 = 1 − F B_0=1-F B 0 = 1 − F
P R = f R − ( z − y ) u , P T = T 0 + z u / 2 + ( ζ 1 / ρ 0 ) u 2 / 2 , P D = T 0 − g r 2 + ( 2 − σ 0 ) z u − V f a i l , P X = q 2 / 8 − q z + 2 q y + ( 23 ϵ ∗ 2 / 8 + E b o x ) q 2 − ( 2 + J ) L K + z u / ( 2 ρ 0 ) + ( ζ 1 / ρ 0 ) u 2 / 2 + J V f a i l + η r k ( σ 1 − 2 + η r k / 2 ) r 2 + 3 2 η r k q r + η r k z u . \begin{aligned}
P_R &= f_R-(z-y)u,\\
P_T &= T_0+zu/2+(\zeta_1/\rho_0)u^2/2,\\
P_D &= T_0-gr^2+(2-\sigma_0)zu-V_{\mathrm{fail}},\\
P_X &= q^2/8-qz+2qy+(23\epsilon_*^2/8+E_{\mathrm{box}})q^2-(2+J)LK\\
&\quad+zu/(2\rho_0)+(\zeta_1/\rho_0)u^2/2+JV_{\mathrm{fail}}\\
&\quad+\eta_{rk}(\sigma_1-2+\eta_{rk}/2)r^2+\frac{3}{2}\eta_{rk}qr+\eta_{rk}zu.
\end{aligned} P R P T P D P X = f R − ( z − y ) u , = T 0 + z u /2 + ( ζ 1 / ρ 0 ) u 2 /2 , = T 0 − g r 2 + ( 2 − σ 0 ) z u − V fail , = q 2 /8 − q z + 2 q y + ( 23 ϵ ∗ 2 /8 + E box ) q 2 − ( 2 + J ) L K + z u / ( 2 ρ 0 ) + ( ζ 1 / ρ 0 ) u 2 /2 + J V fail + η r k ( σ 1 − 2 + η r k /2 ) r 2 + 2 3 η r k q r + η r k z u . The additional terms in P X P_X P X F ( R ) + ( 3 η r k / 2 ) r R F(R)+(3\eta_{rk}/2)rR F ( R ) + ( 3 η r k /2 ) r R R = ( 1 − ϵ ∗ ) q R=(1-\epsilon_*)q R = ( 1 − ϵ ∗ ) q s = σ r s=\sigma r s = σ r σ 1 \sigma_1 σ 1 V f a i l = ξ + ζ V_{\mathrm{fail}}=\xi+\zeta V fail = ξ + ζ ζ \zeta ζ z / q z/q z / q
A X = 1 + 1 ρ 0 + 2 ( 1 − ρ 0 ) 1 + v 0 + J + ( 2 + J ) ( ρ 1 1 + v 0 + L β ) , C X = 1 + ( 2 + J ) L A_X=1+\frac{1}{\rho_0}+\frac{2(1-\rho_0)}{1+v_0}+J+(2+J)\left(\frac{\rho_1}{1+v_0}+L\beta\right), \qquad C_X=1+(2+J)L A X = 1 + ρ 0 1 + 1 + v 0 2 ( 1 − ρ 0 ) + J + ( 2 + J ) ( 1 + v 0 ρ 1 + L β ) , C X = 1 + ( 2 + J ) L on a a a c c c 1 + J 1+J 1 + J j = E A C j=E_{AC} j = E A C
The remaining auxiliary column depends on the alternative. For saturated A P AP A P
η = 1 + v 1 ρ 0 − 1 , A A P = max { − 2 + 3 ( 1 + v 1 ) ρ 0 , 4 + 6 v 1 − 2 ζ 0 ρ 0 } . \eta=\frac{1+v_1}{\rho_0}-1, \qquad A_{AP}=\max\left\{-2+\frac{3(1+v_1)}{\rho_0},\frac{4+6v_1-2\zeta_0}{\rho_0}\right\}. η = ρ 0 1 + v 1 − 1 , A A P = max { − 2 + ρ 0 3 ( 1 + v 1 ) , ρ 0 4 + 6 v 1 − 2 ζ 0 } . Then
P A = f R + f A − ( z − y ) u + η z u , ( A A , C A , j A , e A ) = ( A A P , 0 , 1 , 3 / 8 ) . P_A=f_R+f_A-(z-y)u+\eta zu, \qquad(A_A,C_A,j_A,e_A)=(A_{AP},0,1,3/8). P A = f R + f A − ( z − y ) u + η z u , ( A A , C A , j A , e A ) = ( A A P , 0 , 1 , 3/8 ) . The last four entries bound its positive coefficients on ( a , c , j , e 23 ) (a,c,j,e_{23}) ( a , c , j , e 23 ) A P AP A P
K C = 4 ( 1 − α ) ( 5 − 4 / ρ 0 ) , α = 179 / 1000. K_C=\frac{4}{(1-\alpha)(5-4/\rho_0)}, \qquad\alpha=179/1000. K C = ( 1 − α ) ( 5 − 4/ ρ 0 ) 4 , α = 179/1000. The large-C P CP C P
P A = f R + α r ( z + q − 3 r ) − α 2 r 2 / 2 + ( α − 5 / 8 ) z u , P_A = f_R + \alpha r(z + q - 3r) - \alpha^{2}r^{2}/2 + (\alpha- 5/8)zu, P A = f R + α r ( z + q − 3 r ) − α 2 r 2 /2 + ( α − 5/8 ) z u , ( A A , C A , j A , e A ) = ( 3 α , α , 0 , 3 / 8 ) . (A_A, C_A, j_A, e_A) = (3\alpha, \alpha, 0, 3/8). ( A A , C A , j A , e A ) = ( 3 α , α , 0 , 3/8 ) . The small-C P CP C P
P A = f R + r ( 4 y − d − r ) + 3 20 r ( d − r ) + α r 2 − K C K − 5 z u / 8 , P_A = f_R + r(4y - d - r) + \frac{3}{20}r(d-r) + \alpha r^{2} - K_C K - 5zu/8, P A = f R + r ( 4 y − d − r ) + 20 3 r ( d − r ) + α r 2 − K C K − 5 z u /8 , ( A A , C A , j A , e A ) = ( 2 + β K C , K C , 0 , 57 / 8 ) . (A_A, C_A, j_A, e_A) = (2 + \beta K_C, K_C, 0, 57/8). ( A A , C A , j A , e A ) = ( 2 + β K C , K C , 0 , 57/8 ) . Here the 3 / 20 3/20 3/20 d ≤ 23 Z / 20 d \le23Z/20 d ≤ 23 Z /20 K C K_C K C
Let p p p b R b_R b R b T b_T b T b A b_A b A b D b_D b D b X b_X b X μ \mu μ α 0 α + c ≥ K \alpha_0\alpha+ c \ge K α 0 α + c ≥ K
p ≥ A T ( b T + b D ) + A A b A + A X b X + α 0 μ , p \ge A_T(b_T + b_D) + A_A b_A + A_X b_X + \alpha_0\mu, p ≥ A T ( b T + b D ) + A A b A + A X b X + α 0 μ , p ≥ C T ( b T + b D ) + C A b A + C X b X + μ , p \ge C_T(b_T + b_D) + C_A b_A + C_X b_X + \mu, p ≥ C T ( b T + b D ) + C A b A + C X b X + μ , p ≥ j A b A + ( 1 + J ) b X , p \ge j_A b_A + (1 + J)b_X, p ≥ j A b A + ( 1 + J ) b X , p ≥ 13 b T / 2 + 4 b D + ( 35 / 4 + 7 η r k / 2 ) b X , p \ge13b_T/2 + 4b_D + (35/4 + 7\eta r_k/2)b_X, p ≥ 13 b T /2 + 4 b D + ( 35/4 + 7 η r k /2 ) b X , p ≥ e A b A , b D ≥ J b X . p \ge e_A b_A,\qquad b_D \ge Jb_X. p ≥ e A b A , b D ≥ J b X . The last inequality absorbs V f a i l V_{\mathrm{fail}} V fail A P AP A P e 23 e_{23} e 23 j j j A P AP A P e 12 e_{12} e 12 E A A E_{AA} E AA
Rational matrices and exhaustive coverage
For clarity we specify the domain multipliers. Let L \mathcal{L} L v v v ρ \rho ρ ζ \zeta ζ υ \upsilon υ
3 z / 8 − y , 13 q / 20 − y , 2 w − q , 6 w − r − z , 3z/8-y,\quad13q/20-y,\quad2w-q,\quad6w-r-z, 3 z /8 − y , 13 q /20 − y , 2 w − q , 6 w − r − z , z − ( ζ 0 / ρ 1 ) u − 3 r / 2 , 23 [ z − ( ζ 0 / ρ 1 ) u ] / 20 − d , z-(\zeta_0/\rho_1)u-3r/2,\quad23[z-(\zeta_0/\rho_1)u]/20-d, z − ( ζ 0 / ρ 1 ) u − 3 r /2 , 23 [ z − ( ζ 0 / ρ 1 ) u ] /20 − d , x , y , z , w , q , h , d − r , w − y . x,\ y,\ z,\ w,\ q,\ h,\ d-r,\ w-y. x , y , z , w , q , h , d − r , w − y . The good-P P P Z = z − ( z / r ) u ≤ z − ( ζ 0 / ρ 1 ) u Z = z - (z/r)u \le z - (\zeta_0/\rho_1)u Z = z − ( z / r ) u ≤ z − ( ζ 0 / ρ 1 ) u z − u z-u z − u Z Z Z Λ = x + y + z + w \Lambda= x + y + z + w Λ = x + y + z + w Λ L \Lambda L Λ L u L uL uL L ∈ L L \in\mathcal{L} L ∈ L u ( r − u ) u(r-u) u ( r − u ) Λ u \Lambda u Λ u q q q + μ K +\mu K + μ K ( 1 − 20.4334035 ) Λ 2 (1 - 20.4334035)\Lambda^{2} ( 1 − 20.4334035 ) Λ 2 L D L T LDL^{T} L D L T
The complete active partitions are
Table 5.
Each checker reads its corresponding JSON certificate, reconstructs the forms from the displayed scalars, checks every resource budget, and verifies root-box containment, disjoint interiors, and exact total five-dimensional volume. The partitions for the two C P CP C P α \alpha α γ ≤ 0.4334035 \gamma\le0.4334035 γ ≤ 0.4334035
The case q ≥ d q \ge d q ≥ d
Use eight coordinates ( x , y , z , w , b , r , u , t ) (x,y,z,w,b,r,u,t) ( x , y , z , w , b , r , u , t ) u = h 3 / z u=h_3/z u = h 3 / z t = h 3 / r t=h_3/r t = h 3 / r z u = r t zu=rt z u = r t
d = x / 2 − b , q = x / 2 + b , h = 2 b , K = ( d 2 − r 2 ) / 2 , J C = q ( d − r ) / 2 , d=x/2-b,\qquad q=x/2+b,\qquad h=2b,\qquad K=(d^2-r^2)/2,\qquad J_C=q(d-r)/2, d = x /2 − b , q = x /2 + b , h = 2 b , K = ( d 2 − r 2 ) /2 , J C = q ( d − r ) /2 , f D = d ( x + z − 3 y ) − 3 d 2 − d h + 5 h 2 / 8. f_D=d(x+z-3y)-3d^2-dh+5h^2/8. f D = d ( x + z − 3 y ) − 3 d 2 − d h + 5 h 2 /8. The 31 boxes in cd-domain-certificate.json partition ( h / d , r / d ) ∈ [ 3 / 10 , 5 / 4 ] × [ 4 / 5 , 1 ] (h/d,r/d)\in[3/10,5/4]\times[4/5,1] ( h / d , r / d ) ∈ [ 3/10 , 5/4 ] × [ 4/5 , 1 ]
P = 10 9 ( 53 / 10 + v 1 ) , A D = P + v 1 1 + v 1 − 9 / ( 40 ρ 0 ) , α 0 = 1 1 + v 0 , β 0 = 1 9 ( 1 + v 0 ) . P=\frac{10}{9}(53/10+v_1),\qquad A_D=P+\frac{v_1}{1+v_1-9/(40\rho_0)},\qquad\alpha_0=\frac{1}{1+v_0},\qquad\beta_0=\frac{1}{9(1+v_0)}. P = 9 10 ( 53/10 + v 1 ) , A D = P + 1 + v 1 − 9/ ( 40 ρ 0 ) v 1 , α 0 = 1 + v 0 1 , β 0 = 9 ( 1 + v 0 ) 1 . The initial, core, and full-target weights p p p b 1 b_1 b 1 c 1 c_1 c 1
p ≥ c 1 , α 0 μ + ν ≤ p − A D c 1 , μ + β 0 ν ≤ p − P c 1 . p\ge c_1,\qquad\alpha_0\mu+\nu\le p-A_Dc_1,\qquad\mu+\beta_0\nu\le p-Pc_1. p ≥ c 1 , α 0 μ + ν ≤ p − A D c 1 , μ + β 0 ν ≤ p − P c 1 . These absorb defects using K ≤ α 0 a + c K\le\alpha_0a+c K ≤ α 0 a + c J C ≤ a + β 0 c J_C\le a+\beta_0c J C ≤ a + β 0 c
F − b 1 f R − c 1 f D + μ K + ν J C + ( b 1 + c 1 ) ( z − y ) u − 2 c 1 h t . F-b_1f_R-c_1f_D+\mu K+\nu J_C+(b_1+c_1)(z-y)u-2c_1ht. F − b 1 f R − c 1 f D + μ K + ν J C + ( b 1 + c 1 ) ( z − y ) u − 2 c 1 h t . The valid linear forms are the four ratio-box faces and
3 z / 8 − y , 13 d / 20 − y , 3 d − z , 2 w − d , 6 w − r − z , 3z/8-y,\quad13d/20-y,\quad3d-z,\quad2w-d,\quad6w-r-z, 3 z /8 − y , 13 d /20 − y , 3 d − z , 2 w − d , 6 w − r − z , w − y , y − 187 d / 1000 , r − u , z − t , u , t , x , y , z , w , d , h . w-y,\quad y-187d/1000,\quad r-u,\quad z-t,\quad u,t,x,y,z,w,d,h. w − y , y − 187 d /1000 , r − u , z − t , u , t , x , y , z , w , d , h . Subtract nonnegative multiples of Λ \Lambda Λ u u u t t t d d d z u − r t = 0 zu-rt=0 z u − r t = 0 Λ ( r − u ) \Lambda(r-u) Λ ( r − u ) Λ ( z − t ) \Lambda(z-t) Λ ( z − t ) h 3 ≤ r z h_3\le rz h 3 ≤ r z u ≤ r u\le r u ≤ r t ≤ z t\le z t ≤ z ( 1 − 20.4334035 ) Λ 2 (1-20.4334035)\Lambda^2 ( 1 − 20.4334035 ) Λ 2 L D L T LDLT L D L T γ ≤ 0.4334035 \gamma\le0.4334035 γ ≤ 0.4334035 A B AB A B
Taking the maximum over both orientations and all exceptional cases proves
γ ( 4 ) ≤ 866807 2000000 = 0.4334035. \gamma(4)\le\frac{866807}{2000000}=0.4334035. γ ( 4 ) ≤ 2000000 866807 = 0.4334035. Upper bounds from alternating paths for general ℓ \ell ℓ
In this section and the next two sections, ℓ ≥ 5 \ell\ge5 ℓ ≥ 5 [1 ] , followed by a second matching obtained by exchanging edges along disjoint alternating paths. The formal edge weights are
w 1 ( ε ) = 0 , w t ( ε ) = ∑ j = 0 t − 2 2 t − 2 − j ε j ( 2 ≤ t ≤ ℓ ) . w_1(\varepsilon)=0,\qquad w_t(\varepsilon)=\sum_{j=0}^{t-2}2^{t-2-j}\varepsilon_j\qquad(2\le t\le\ell). w 1 ( ε ) = 0 , w t ( ε ) = j = 0 ∑ t − 2 2 t − 2 − j ε j ( 2 ≤ t ≤ ℓ ) . Coefficients are compared starting at the constant term. The primary weight is the constant coefficient, so its values are 0 , 1 , 2 , 4 , … , 2 ℓ − 2 0,1,2,4,\ldots,2^{\ell-2} 0 , 1 , 2 , 4 , … , 2 ℓ − 2 M 0 M_0 M 0 M M M X t X_t X t t t t x t = ∣ X t ∣ / ( 2 M ) x_t=|X_t|/(2M) x t = ∣ X t ∣/ ( 2 M )
∑ t x t = 1. \sum_t x_t=1. t ∑ x t = 1. Put
k = 2 ℓ − 2 , S = 3 k − 2 ℓ + 4 , c 1 = c ℓ = 1 , c 2 = k , c t = 2 ℓ − t + 1 − 2 ( 3 ≤ t < ℓ ) . k=2^{\ell-2},\qquad S=3k-2\ell+4,\qquad c_1=c_\ell=1,\qquad c_2=k,\qquad c_t=2^{\ell-t+1}-2\quad(3\le t<\ell). k = 2 ℓ − 2 , S = 3 k − 2 ℓ + 4 , c 1 = c ℓ = 1 , c 2 = k , c t = 2 ℓ − t + 1 − 2 ( 3 ≤ t < ℓ ) . Here and below, an endpoint mass is divided by M M M M 2 M^2 M 2 x t x_t x t t t t M M M C 0 = C λ ( M 0 ) / M 2 C_0=C_\lambda(M_0)/M^2 C 0 = C λ ( M 0 ) / M 2
No critical edge joins a covered vertex to an uncovered vertex: replacing the incident matching edge by such an edge would decrease the primary weight. Between distinct classes X i , X j X_i,X_j X i , X j § 3 gives at most 2 x i x j M 2 2x_ix_jM^2 2 x i x j M 2 M 2 M^2 M 2 e i j ≥ 0 e_{ij} \ge0 e ij ≥ 0 x = x 2 x=x_2 x = x 2
Q = ∑ t = 1 ℓ ( x t − c t / S ) 2 c t . Q=\sum_{t=1}^{\ell}\frac{(x_t-c_t/S)^2}{c_t}. Q = t = 1 ∑ ℓ c t ( x t − c t / S ) 2 . Color the matching edges inside X 2 X_2 X 2 red-ended if its first and last edges are red. There is no alternating cycle, and no red-ended alternating path with k k k [1 ] : a graph on v v v b b b ( 1 − 1 / b ) v 2 + O b ( v ) (1-1/b)v^2+O_b(v) ( 1 − 1/ b ) v 2 + O b ( v ) k k k p = k / 2 p=k/2 p = k /2 O ℓ ( M ) O_\ell(M) O ℓ ( M ) i i i L i R i L_iR_i L i R i L i R j L_iR_j L i R j i < j i<j i < j a i a_i a i i i i M M M ∑ i a i = x \sum_i a_i=x ∑ i a i = x blue template . Let E E E M 2 M^2 M 2
R = 1 2 ∑ i = 1 p ( a i − 2 x / k ) 2 . R=\frac{1}{2}\sum_{i=1}^{p}(a_i-2x/k)^2. R = 2 1 i = 1 ∑ p ( a i − 2 x / k ) 2 . The cross-class bounds just described give 2 x i x j 2x_ix_j 2 x i x j i < j i<j i < j t t t ( 1 − 1 / c t ) x t 2 + o ℓ ( 1 ) (1-1/c_t)x_t^2+o_\ell(1) ( 1 − 1/ c t ) x t 2 + o ℓ ( 1 ) [1 ] ; for t = 1 , ℓ t=1,\ell t = 1 , ℓ X 2 X_2 X 2
x 2 − 1 2 ∑ i a i 2 = x 2 − x 2 / k − R , x^2-\frac{1}{2}\sum_i a_i^2=x^2-x^2/k-R, x 2 − 2 1 i ∑ a i 2 = x 2 − x 2 / k − R , so its missing blue density gives the further loss E E E ∑ t c t = S \sum_t c_t=S ∑ t c t = S ∑ t x t = 1 \sum_t x_t=1 ∑ t x t = 1 ∑ t x t 2 / c t = 1 / S + Q \sum_t x_t^2/c_t=1/S+Q ∑ t x t 2 / c t = 1/ S + Q D D D
C 0 ≤ 1 − 1 S − D + o ℓ ( 1 ) , Q + R + E + ∑ i < j e i j ≤ D + o ℓ ( 1 ) . (56) C_0\le1-\frac{1}{S}-D+o_\ell(1),\qquad Q+R+E+\sum_{i<j}e_{ij}\le D+o_\ell(1). \tag*{(56)} C 0 ≤ 1 − S 1 − D + o ℓ ( 1 ) , Q + R + E + i < j ∑ e ij ≤ D + o ℓ ( 1 ) . (56) One can make D D D M M M O ℓ ( 1 / M ) O_\ell(1/M) O ℓ ( 1/ M ) 2 M 2M 2 M N N N
A constraint from a maximal collection of disjoint paths
Choose an integer b b b k / 2 ≤ b < k k/2\le b<k k /2 ≤ b < k L = b + 1 L=b+1 L = b + 1 X 2 X_2 X 2 b b b L L L μ \mu μ μ M / L \mu M/L μ M / L
The graph on the remaining ( x − μ ) M (x-\mu)M ( x − μ ) M b b b ( 1 − 1 / b ) ( x − μ ) 2 + o ℓ ( 1 ) (1-1/b)(x-\mu)^2+o_\ell(1) ( 1 − 1/ b ) ( x − μ ) 2 + o ℓ ( 1 ) a i ′ a_i' a i ′
( x − μ ) 2 − 1 2 ∑ i ( a i ′ ) 2 − E − o ℓ ( 1 ) . (x-\mu)^2-\frac{1}{2}\sum_i(a_i')^2-E-o_\ell(1). ( x − μ ) 2 − 2 1 i ∑ ( a i ′ ) 2 − E − o ℓ ( 1 ) . Since 0 ≤ a i ′ ≤ a i 0 \le a'_i \le a_i 0 ≤ a i ′ ≤ a i 1 2 ∑ i a i 2 = x 2 / k + R \frac{1}{2}\sum_i a_i^2 = x^2/k + R 2 1 ∑ i a i 2 = x 2 / k + R
R + E ≥ ( x − μ ) 2 b − x 2 k − o ℓ ( 1 ) . (57) R + E \ge\frac{(x-\mu)^2}{b} - \frac{x^2}{k} - o_{\ell}(1). \tag*{(57)} R + E ≥ b ( x − μ ) 2 − k x 2 − o ℓ ( 1 ) . (57) The mass constraint in Q Q Q
Q ≥ S k ( S − k ) ( x − k / S ) 2 . Q \ge\frac{S}{k(S-k)}(x-k/S)^2. Q ≥ k ( S − k ) S ( x − k / S ) 2 . Minimizing the sum of this quadratic and the right side of (57) over x ∈ R x \in\mathbb{R} x ∈ R
D ≥ d b ( μ ) − o ℓ ( 1 ) , d b ( μ ) = ( k − b ) / S − 2 μ + μ 2 S − k + b = ( 1 − μ ) 2 S − k + b − 1 S . (58) D \ge d_b(\mu)-o_{\ell}(1), \qquad d_b(\mu)=\frac{(k-b)/S-2\mu+\mu^2}{S-k+b}=\frac{(1-\mu)^2}{S-k+b}-\frac{1}{S}. \tag*{(58)} D ≥ d b ( μ ) − o ℓ ( 1 ) , d b ( μ ) = S − k + b ( k − b ) / S − 2 μ + μ 2 = S − k + b ( 1 − μ ) 2 − S 1 . (58) The unconstrained minimizer is x = [ b + ( S − k ) μ ] / ( S − k + b ) x=[b+(S-k)\mu]/(S-k+b) x = [ b + ( S − k ) μ ] / ( S − k + b )
Constructing the comparison matching
The closing edge of each packed path has label ℓ \ell ℓ L > k / 2 L>k/2 L > k /2 b b b L L L k / 2 k/2 k /2 each endpoint has label ℓ \ell ℓ : close the path to that vertex instead, leaving its other endpoint uncovered. Matching cardinality is unchanged in this switch.
Replace each path by its blue edges and its label-ℓ \ell ℓ ℓ \ell ℓ ℓ \ell ℓ N ≥ 2 M N \ge2M N ≥ 2 M
Every original red pair has one original left endpoint. Among the μ M \mu M μ M 2 μ M / L 2\mu M/L 2 μ M / L ( 1 − 2 / L ) μ M (1-2/L)\mu M ( 1 − 2/ L ) μ M
1 2 ( 1 − 2 / L ) 2 μ 2 M 2 − E M 2 − O ℓ ( M ) . \frac{1}{2}(1-2/L)^2\mu^2M^2-EM^2-O_{\ell}(M). 2 1 ( 1 − 2/ L ) 2 μ 2 M 2 − E M 2 − O ℓ ( M ) . Only the 2 μ M / L 2\mu M/L 2 μ M / L 4 μ M 2 / L 4\mu M^2/L 4 μ M 2 / L
C p a t h − C 0 ≤ E + 4 μ L − 1 2 ( 1 − 2 / L ) 2 μ 2 + o ℓ ( 1 ) . C_{\mathrm{path}}-C_0 \le E+\frac{4\mu}{L}-\frac{1}{2}(1-2/L)^2\mu^2+o_{\ell}(1). C path − C 0 ≤ E + L 4 μ − 2 1 ( 1 − 2/ L ) 2 μ 2 + o ℓ ( 1 ) . The E E E D D D
C p a t h ≤ 1 − 1 S − h b ( μ ) + o ℓ ( 1 ) , h b ( μ ) = 1 2 ( 1 − 2 / L ) 2 μ 2 − 4 μ L . (59) C_{\mathrm{path}} \le1-\frac{1}{S}-h_b(\mu)+o_{\ell}(1), \qquad h_b(\mu)=\frac{1}{2}(1-2/L)^2\mu^2-\frac{4\mu}{L}. \tag*{(59)} C path ≤ 1 − S 1 − h b ( μ ) + o ℓ ( 1 ) , h b ( μ ) = 2 1 ( 1 − 2/ L ) 2 μ 2 − L 4 μ . (59) We may stop the packing at any smaller prescribed mass, with an O ℓ ( 1 / M ) O_{\ell}(1/M) O ℓ ( 1/ M )
Theorem 14.1. For every integer b b b k / 2 ≤ b < k k/2 \le b < k k /2 ≤ b < k 0 < u < 1 0<u<1 0 < u < 1
γ ( ℓ ) ≤ 1 2 − 1 2 S − 1 2 min { d b ( u ) , h b ( u ) } . \gamma(\ell) \le\frac{1}{2}-\frac{1}{2S}-\frac{1}{2}\min\{d_b(u),h_b(u)\}. γ ( ℓ ) ≤ 2 1 − 2 S 1 − 2 1 min { d b ( u ) , h b ( u )} . In particular one may maximize the last minimum over u u u b b b
Proof. If D ≥ d b ( u ) − o ℓ ( 1 ) D \ge d_b(u)-o_{\ell}(1) D ≥ d b ( u ) − o ℓ ( 1 ) (58) and the strict decrease of d b d_b d b [ 0 , 1 ) [0,1) [ 0 , 1 ) u − o ℓ ( 1 ) u-o_{\ell}(1) u − o ℓ ( 1 ) (59) . Dividing the count bound by 2 proves the assertion. Strict inequalities can first be used with an arbitrarily small fixed margin, followed by passage to the limit, so boundary cases are included.
Exact optimization and asymptotics
Set b = k / 2 b=k/2 b = k /2 L = k / 2 + 1 L=k/2+1 L = k /2 + 1 T = 2 S − k T=2S-k T = 2 S − k
A = 1 2 ( 1 − 2 / L ) 2 − 2 T , B = 4 T − 4 L , d 0 = k S T . A=\frac{1}{2}(1-2/L)^2-\frac{2}{T},\qquad B=\frac{4}{T}-\frac{4}{L},\qquad d_0=\frac{k}{ST}. A = 2 1 ( 1 − 2/ L ) 2 − T 2 , B = T 4 − L 4 , d 0 = S T k . Whenever A > 0 A>0 A > 0 h b ( u ) = d b ( u ) h_b(u)=d_b(u) h b ( u ) = d b ( u )
u ∗ = − B + B 2 + 4 A d 0 2 A , δ k = k / S − 4 u ∗ + 2 u ∗ 2 T . (60) u_*=\frac{-B+\sqrt{B^2+4Ad_0}}{2A},\qquad\delta_k=\frac{k/S-4u_*+2u_*^2}{T}. \tag*{(60)} u ∗ = 2 A − B + B 2 + 4 A d 0 , δ k = T k / S − 4 u ∗ + 2 u ∗ 2 . (60) If 0 < u ∗ < 1 0<u_*<1 0 < u ∗ < 1 δ k > 0 \delta_k>0 δ k > 0 d b d_b d b h b h_b h b
γ ( ℓ ) ≤ 1 2 − 1 2 S − δ k 2 . (61) \gamma(\ell)\le\frac{1}{2}-\frac{1}{2S}-\frac{\delta_k}{2}. \tag*{(61)} γ ( ℓ ) ≤ 2 1 − 2 S 1 − 2 δ k . (61) The hypotheses hold for every ℓ ≥ 10 \ell\ge10 ℓ ≥ 10 k ≥ 256 k\ge256 k ≥ 256 L ≥ 129 L\ge129 L ≥ 129 8 L / ( L − 2 ) 2 8L/(L-2)^2 8 L / ( L − 2 ) 2 h b h_b h b 1 / 12 1/12 1/12 d b ( 1 / 12 ) > 0 d_b(1/12)>0 d b ( 1/12 ) > 0 k / S > 1 / 3 k/S>1/3 k / S > 1/3 A > 0 A>0 A > 0 A ≥ 1 2 ( 127 / 129 ) 2 − 1 / 384 > 0 A\ge\frac{1}{2}(127/129)^2-1/384>0 A ≥ 2 1 ( 127/129 ) 2 − 1/384 > 0 d b d_b d b h b h_b h b d b d_b d b
This choice of path length is optimal within the whole family of Theorem 14.1 when ℓ ≥ 10 \ell\ge10 ℓ ≥ 10 μ \mu μ b ∈ [ k / 2 , k ) b\in[k/2,k) b ∈ [ k /2 , k ) T b = S − k + b T_b=S-k+b T b = S − k + b L = b + 1 L=b+1 L = b + 1 a = 1 − 2 / L a=1-2/L a = 1 − 2/ L h b ( u ) = d b ( u ) > 0 h_b(u)=d_b(u)>0 h b ( u ) = d b ( u ) > 0 h u = a 2 u − 4 / L > a 2 u / 2 h_u=a^2u-4/L>a^2u/2 h u = a 2 u − 4/ L > a 2 u /2 d u = − 2 ( 1 − u ) / T b < 0 d_u=-2(1-u)/T_b<0 d u = − 2 ( 1 − u ) / T b < 0 b b b
( 1 − u ) h u > ( 4 a u 2 + 8 u ) T b L 2 . (1-u)h_u>\frac{(4au^2+8u)T_b}{L^2}. ( 1 − u ) h u > L 2 ( 4 a u 2 + 8 u ) T b . This follows by implicit differentiation, using ∂ b d b = − ( 1 − u ) 2 / T b 2 \partial_b d_b=-(1-u)^2/T_b^2 ∂ b d b = − ( 1 − u ) 2 / T b 2 ∂ b h b = ( 2 a u 2 + 4 u ) / L 2 \partial_b h_b=(2au^2+4u)/L^2 ∂ b h b = ( 2 a u 2 + 4 u ) / L 2 ℓ ≥ 10 \ell\ge10 ℓ ≥ 10 S ≥ 14 k / 5 S\ge14k/5 S ≥ 14 k /5 d b ( u ) d_b(u) d b ( u ) ( 1 − u ) 2 > 23 / 28 > 81 / 100 (1-u)^2>23/28>81/100 ( 1 − u ) 2 > 23/28 > 81/100 u < 1 / 10 u<1/10 u < 1/10 a ≥ 127 / 129 a\ge127/129 a ≥ 127/129 T b / L 2 ≤ 12 / k T_b/L^2\le12/k T b / L 2 ≤ 12/ k
9 20 ( 127 / 129 ) 2 u , \frac{9}{20}(127/129)^2u, 20 9 ( 127/129 ) 2 u , while the right side is at most ( 42 / 5 ) ( 12 / 256 ) u = ( 63 / 160 ) u (42/5)(12/256)u=(63/160)u ( 42/5 ) ( 12/256 ) u = ( 63/160 ) u b = k / 2 b=k/2 b = k /2 general-certificate/verify_general.py.
As ℓ → ∞ \ell\to\infty ℓ → ∞ S = 3 k + O ( log k ) S=3k+O(\log k) S = 3 k + O ( log k ) A = 1 / 2 + O ( k − 1 ) A=1/2+O(k^{-1}) A = 1/2 + O ( k − 1 ) B = O ( k − 1 ) B=O(k^{-1}) B = O ( k − 1 ) d 0 = 1 / ( 15 k ) + O ( ( log k ) / k 2 ) d_0=1/(15k)+O((\log k)/k^2) d 0 = 1/ ( 15 k ) + O (( log k ) / k 2 )
u ∗ = 2 15 k + O ( k − 1 ) , δ k = k S ( 2 S − k ) + O ( k − 3 / 2 ) . u_*=\sqrt{\frac{2}{15k}}+O(k^{-1}),\qquad\delta_k=\frac{k}{S(2S-k)}+O(k^{-3/2}). u ∗ = 15 k 2 + O ( k − 1 ) , δ k = S ( 2 S − k ) k + O ( k − 3/2 ) . Combining the two principal terms in (61) gives
γ ( ℓ ) ≤ 1 2 − 1 2 S − k + O ( k − 3 / 2 ) = 1 2 − 1 5 k + O ( k − 3 / 2 ) . (62) \gamma(\ell)\le\frac{1}{2}-\frac{1}{2S-k}+O(k^{-3/2})=\frac{1}{2}-\frac{1}{5k}+O(k^{-3/2}). \tag*{(62)} γ ( ℓ ) ≤ 2 1 − 2 S − k 1 + O ( k − 3/2 ) = 2 1 − 5 k 1 + O ( k − 3/2 ) . (62) The leading gap in the upper bound of [1 ] is 1 / ( 6 k ) 1/(6k) 1/ ( 6 k ) (62) is 1 / ( 5 k ) 1/(5k) 1/ ( 5 k ) 6 / 5 6/5 6/5
An explicit upper bound for every ℓ ≥ 5 \ell\ge5 ℓ ≥ 5
We continue with the notation and deficit decomposition of Section 14, and put y = x 1 y=x_1 y = x 1 p = k / 2 p=k/2 p = k /2 ℓ ≥ 5 \ell\ge5 ℓ ≥ 5
Selecting layers with few missing edges
Fix an integer j j j k / 4 < j ≤ p k/4<j\le p k /4 < j ≤ p n = p − j + 1 n=p-j+1 n = p − j + 1 h M hM h M h = θ / S h=\theta/S h = θ / S θ > 0 \theta>0 θ > 0 2 E M / h + o ( M ) 2EM/h+o(M) 2 E M / h + o ( M ) j j j
2 x k − 2 ( j − 1 ) p n R − 2 E n h − o ℓ ( 1 ) . (63) \frac{2x}{k}-\sqrt{\frac{2(j-1)}{pn}R}-\frac{2E}{nh}-o_{\ell}(1). \tag*{(63)} k 2 x − p n 2 ( j − 1 ) R − nh 2 E − o ℓ ( 1 ) . (63) Indeed for any n n n
n x p − 2 R n ( p − n ) p . \frac{nx}{p}-\sqrt{2R\frac{n(p-n)}{p}}. p n x − 2 R p n ( p − n ) . This is Cauchy–Schwarz applied to the centered indicator of those layers and the vector ( a i − x / p ) i (a_i-x/p)_i ( a i − x / p ) i n n n x x x Q Q Q
∣ 2 x / k − 2 / S ∣ ≤ 2 S − k k S Q . \left|2x/k-2/S\right|\le2\sqrt{\frac{S-k}{kS}Q}. ∣ 2 x / k − 2/ S ∣ ≤ 2 k S S − k Q . Consequently, with
χ = 4 ( S − k ) k S + 2 ( j − 1 ) p n = 2 n − 4 S , B c = 2 S n θ , \chi=\frac{4(S-k)}{kS}+\frac{2(j-1)}{pn}=\frac{2}{n}-\frac{4}{S},\qquad B_c=\frac{2S}{n\theta}, χ = k S 4 ( S − k ) + p n 2 ( j − 1 ) = n 2 − S 4 , B c = n θ 2 S , the j j j
2 S − χ ( Q + R ) − B c E − o ℓ ( 1 ) . \frac{2}{S}-\sqrt{\chi(Q+R)}-B_cE-o_{\ell}(1). S 2 − χ ( Q + R ) − B c E − o ℓ ( 1 ) . For D ≤ δ D\le\delta D ≤ δ Q + R + E ≤ D + o ( 1 ) Q+R+E\le D+o(1) Q + R + E ≤ D + o ( 1 )
r 0 = 2 S − χ 4 B c − B c δ . (64) r_0=\frac{2}{S}-\frac{\chi}{4B_c}-B_c\delta. \tag*{(64)} r 0 = S 2 − 4 B c χ − B c δ . (64) Explicitly, χ ( δ − E ) + B c E ≤ B c δ + χ / ( 4 B c ) \sqrt{\chi(\delta-E)}+B_cE\le B_c\delta+\chi/(4B_c) χ ( δ − E ) + B c E ≤ B c δ + χ / ( 4 B c ) r M rM r M j j j r = r 0 − o ( 1 ) r=r_0-o(1) r = r 0 − o ( 1 )
r 0 > 2 θ / S , (65) r_0>2\theta/S, \tag*{(65)} r 0 > 2 θ / S , (65) every selected endpoint has fewer than r M / 2 − O ( 1 ) rM/2-O(1) r M /2 − O ( 1 )
Rematching the selected vertices
Number the selected layers 1 , … , j 1,\ldots,j 1 , … , j
1 , 2 , … , j , j , j − 1 , … , t 1,2,\ldots,j,j,j-1,\ldots,t 1 , 2 , … , j , j , j − 1 , … , t and 2 j − t + 1 2j-t+1 2 j − t + 1 R 1 R_1 R 1 R t R_t R t j j j j j j t t t 2 h M + O ℓ ( 1 ) < r M 2hM+O_{\ell}(1)<rM 2 h M + O ℓ ( 1 ) < r M R 1 R_1 R 1 R t R_t R t ℓ \ell ℓ 2 j − t + 1 > k / 2 2j-t+1>k/2 2 j − t + 1 > k /2 R 1 R_1 R 1 ℓ \ell ℓ ℓ \ell ℓ 2 j 2j 2 j 2 j > k / 2 2j>k/2 2 j > k /2 k / 2 k/2 k /2
Make blue perfect matchings between L i L_i L i R i + 1 R_{i+1} R i + 1 L j L_j L j R 1 R_1 R 1 L j L_j L j R 1 R_1 R 1 ℓ \ell ℓ
We also have a conditional cross half bound. Let u v uv uv t < ℓ t<\ell t < ℓ
2 j + 2 t − 2 > k ( t ≥ 2 ) , 2j+2^{t-2}>k \qquad(t\ge2), 2 j + 2 t − 2 > k ( t ≥ 2 ) , two complementary critical contacts from a new R 1 R 1 R_1R_1 R 1 R 1 u v uv uv 2 j 2j 2 j R 1 R_1 R 1 u v uv uv k k k k k k X t X_t X t r x t r x_t r x t 2 r x t 2r x_t 2 r x t
Set
η = { 1 , j = p , 2 , j < p , τ = 2 j − p . \eta=
\begin{cases}
1, & j=p,\\
2, & j<p,
\end{cases}
\qquad\tau=2j-p. η = { 1 , 2 , j = p , j < p , τ = 2 j − p . and for 3 ≤ t < ℓ 3\le t<\ell 3 ≤ t < ℓ α t = 1 \alpha_t=1 α t = 1 2 t − 2 > k − 2 j 2^{t-2}>k-2j 2 t − 2 > k − 2 j α t = 2 \alpha_t=2 α t = 2 X 1 X_1 X 1 2 j r y − ϵ 12 2j r y-\epsilon_{12} 2 j r y − ϵ 12 X 1 X_1 X 1 2 r y 2r y 2 r y
The selected template originally has ( j 2 − j / 2 ) r 2 + O ℓ ( 1 / M ) (j^2-j/2)r^2+O_{\ell}(1/M) ( j 2 − j /2 ) r 2 + O ℓ ( 1/ M ) E E E ( 2 j − τ ) r 2 (2j-\tau)r^2 ( 2 j − τ ) r 2 R 1 R_1 R 1 R 1 , … , R τ R_1,\ldots,R_\tau R 1 , … , R τ ℓ \ell ℓ j r 2 jr^2 j r 2 ( j − τ ) r 2 (j-\tau)r^2 ( j − τ ) r 2 X 2 X_2 X 2 η r ( x − j r ) \eta r(x-jr) η r ( x − j r ) X ℓ X_\ell X ℓ
C j − C 0 ≤ E + ϵ 12 + r Λ − A r 2 + o ℓ ( 1 ) , (66) C_j-C_0\le E+\epsilon_{12}+r\Lambda-Ar^2+o_{\ell}(1), \tag*{(66)} C j − C 0 ≤ E + ϵ 12 + r Λ − A r 2 + o ℓ ( 1 ) , (66) where
A = j 2 + ( η − 5 / 2 ) j + τ , Λ = η x − 2 ( j − 1 ) y + ∑ t = 3 ℓ − 1 α t x t . A=j^2+(\eta-5/2)j+\tau,
\qquad
\Lambda=\eta x-2(j-1)y+\sum_{t=3}^{\ell-1}\alpha_t x_t. A = j 2 + ( η − 5/2 ) j + τ , Λ = η x − 2 ( j − 1 ) y + t = 3 ∑ ℓ − 1 α t x t . Write λ t \lambda_t λ t x t x_t x t λ ℓ = 0 \lambda_\ell= 0 λ ℓ = 0
Λ 0 = 1 S ∑ t c t λ t , V = ∑ t c t λ t 2 − ( ∑ t c t λ t ) 2 S . \Lambda_0 = \frac{1}{S}\sum_t c_t\lambda_t,\qquad V = \sum_t c_t\lambda_t^2-\frac{\left(\sum_t c_t\lambda_t\right)^2}{S}. Λ 0 = S 1 t ∑ c t λ t , V = t ∑ c t λ t 2 − S ( ∑ t c t λ t ) 2 . Weighted Cauchy–Schwarz on the mass hyperplane gives Λ ≤ Λ 0 + V Q ≤ Λ 0 + V δ \Lambda\le\Lambda_0+\sqrt{VQ}\le\Lambda_0+\sqrt{V\delta} Λ ≤ Λ 0 + V Q ≤ Λ 0 + V δ E + e 12 ≤ D + o ( 1 ) E+e_{12}\le D+o(1) E + e 12 ≤ D + o ( 1 )
Theorem 15.1. Let ℓ ≥ 5 \ell\ge5 ℓ ≥ 5 j , θ , δ j,\theta,\delta j , θ , δ r 0 r_0 r 0 (64) . If
r 0 > 2 θ S , r 0 ( A r 0 − Λ 0 − V δ ) > δ , r_0 > \frac{2\theta}{S},\qquad r_0\left(Ar_0-\Lambda_0-\sqrt{V\delta}\right)>\delta, r 0 > S 2 θ , r 0 ( A r 0 − Λ 0 − V δ ) > δ , then
γ ( ℓ ) ≤ 1 2 − 1 2 S − δ 2 . \gamma(\ell)\le\frac{1}{2}-\frac{1}{2S}-\frac{\delta}{2}. γ ( ℓ ) ≤ 2 1 − 2 S 1 − 2 δ . Nonstrict limiting choices are valid by approximation whenever the inequalities hold for a sequence of positive smaller δ \delta δ
Proof. If D ≥ δ − o ( 1 ) D\ge\delta-o(1) D ≥ δ − o ( 1 ) (56) and (66) give C j ≤ 1 − 1 / S + r Λ − A r 2 + o ( 1 ) C_j\le1-1/S+r\Lambda-Ar^2+o(1) C j ≤ 1 − 1/ S + r Λ − A r 2 + o ( 1 ) r = r 0 − o ( 1 ) r=r_0-o(1) r = r 0 − o ( 1 ) N N N
An explicit bound for every finite number of labels
Corollary 15.2. For every ℓ ≥ 5 \ell\ge5 ℓ ≥ 5
γ ( ℓ ) ≤ 1 2 − 1 3 ⋅ 2 ℓ − 1 − 4 ℓ + 8 − 1 162 ⋅ 2 ℓ . \gamma(\ell)\le\frac{1}{2}-\frac{1}{3\cdot2^{\ell-1}-4\ell+8}-\frac{1}{162\cdot2^\ell}. γ ( ℓ ) ≤ 2 1 − 3 ⋅ 2 ℓ − 1 − 4 ℓ + 8 1 − 162 ⋅ 2 ℓ 1 . Proof. Take j = 3 k / 8 + 1 j=3k/8+1 j = 3 k /8 + 1 δ = 1 / ( 324 k ) \delta=1/(324k) δ = 1/ ( 324 k ) 4 ( S − k ) / ( k S ) 4(S-k)/(kS) 4 ( S − k ) / ( k S ) 4 / k 4/k 4/ k χ = 16 / k \chi=16/k χ = 16/ k
θ = { 1 / 2 , k = 8 , 2 / 3 , k = 16 , 4 / 9 , k ≥ 32. \theta=
\begin{cases}
1/2, & k=8,\\
2/3, & k=16,\\
4/9, & k\ge32.
\end{cases} θ = ⎩ ⎨ ⎧ 1/2 , 2/3 , 4/9 , k = 8 , k = 16 , k ≥ 32. The resulting guaranteed layer masses are
r 0 = 11 144 ( k = 8 ) , r 0 = 37 1080 ( k = 16 ) , r 0 = 17 k 2 − S 2 9 S k 2 ( k ≥ 32 ) . r_0=\frac{11}{144}\quad(k=8),\qquad r_0=\frac{37}{1080}\quad(k=16),\qquad r_0=\frac{17k^2-S^2}{9Sk^2}\quad(k\ge32). r 0 = 144 11 ( k = 8 ) , r 0 = 1080 37 ( k = 16 ) , r 0 = 9 S k 2 17 k 2 − S 2 ( k ≥ 32 ) . They strictly exceed 2 θ / S 2\theta/S 2 θ / S ( 9 k 2 − S 2 ) / ( 9 S k 2 ) > 0 (9k^2-S^2)/(9Sk^2)>0 ( 9 k 2 − S 2 ) / ( 9 S k 2 ) > 0 S < 3 k S<3k S < 3 k
For k = 8 k=8 k = 8 A = 14 A=14 A = 14 Λ 0 = 5 / 9 \Lambda_0=5/9 Λ 0 = 5/9 V = 52 V=52 V = 52 k ≥ 16 k\ge16 k ≥ 16
A = 9 k 2 64 + 13 k 16 + 5 2 , Λ 0 = 2 − 3 k / 4 + 12 S , V = 9 k 2 16 + 4 S − 32. A=\frac{9k^2}{64}+\frac{13k}{16}+\frac{5}{2},\qquad
\Lambda_0=2-\frac{3k/4+12}{S},\qquad
V=\frac{9k^2}{16}+4S-32. A = 64 9 k 2 + 16 13 k + 2 5 , Λ 0 = 2 − S 3 k /4 + 12 , V = 16 9 k 2 + 4 S − 32. Here the variance bound drops the nonnegative subtracted term in the formula for V V V Z = A r 0 − Λ 0 − δ / r 0 Z=Ar_0-\Lambda_0-\delta/r_0 Z = A r 0 − Λ 0 − δ / r 0 Z > 0 Z>0 Z > 0 Z 2 > V δ Z^2>V\delta Z 2 > V δ Theorem 15.1 :
Table 6.
All inequalities in the last two columns are strict. They can be checked by substitution and are also verified in general-certificate/verify_general.py.
For k ≥ 64 k \ge64 k ≥ 64 r 0 ≥ 8 / ( 27 k ) r_0 \ge8/(27k) r 0 ≥ 8/ ( 27 k ) Λ 0 ≤ 2 \Lambda_0 \le2 Λ 0 ≤ 2
V δ ≤ k / 576 + 1 / 27 ≤ k / 144. \sqrt{V\delta} \le\sqrt{k/576+1/27} \le k/144. V δ ≤ k /576 + 1/27 ≤ k /144. The last inequality is equivalent to k 2 − 36 k − 768 ≥ 0 k^2-36k-768 \ge0 k 2 − 36 k − 768 ≥ 0 k ≥ 64 k \ge64 k ≥ 64
A r 0 − Λ 0 − V δ ≥ 5 k 144 − 95 54 ≥ 25 54 . Ar_0-\Lambda_0-\sqrt{V\delta} \ge\frac{5k}{144}-\frac{95}{54} \ge\frac{25}{54}. A r 0 − Λ 0 − V δ ≥ 144 5 k − 54 95 ≥ 54 25 . Since 2 δ / r 0 ≤ 1 / 48 < 25 / 54 2\delta/r_0 \le1/48 < 25/54 2 δ / r 0 ≤ 1/48 < 25/54 2 δ 2\delta 2 δ
For small fixed ℓ \ell ℓ ( j , θ , δ ) (j,\theta,\delta) ( j , θ , δ ) general-certificate/cleanup-certificates.json; their verification requires no numerical solver and no floating-point computation.
A weighted comparison of the path matchings
The path comparison in Section 14 bounds the initial matching and the matching obtained from disjoint paths separately. Both bounds depend on the same deviations Q Q Q R R R μ \mu μ
P ( x , μ ) = ( x − μ ) 2 b − x 2 k , h ( μ ) = c μ 2 − d μ , c = 1 2 ( 1 − 2 b + 1 ) 2 , d = 4 b + 1 . P(x,\mu)=\frac{(x-\mu)^2}{b}-\frac{x^2}{k},\qquad h(\mu)=c\mu^2-d\mu,\qquad c=\frac{1}{2}\left(1-\frac{2}{b+1}\right)^2,\qquad d=\frac{4}{b+1}. P ( x , μ ) = b ( x − μ ) 2 − k x 2 , h ( μ ) = c μ 2 − d μ , c = 2 1 ( 1 − b + 1 2 ) 2 , d = b + 1 4 . The two actual matchings satisfy
C 0 ≤ 1 − 1 / S − Q − R − E + o ( 1 ) , C p a t h ≤ 1 − 1 / S − Q − R − h ( μ ) + o ( 1 ) . C_0 \le1-1/S-Q-R-E+o(1),\qquad C_{\mathrm{path}} \le1-1/S-Q-R-h(\mu)+o(1). C 0 ≤ 1 − 1/ S − Q − R − E + o ( 1 ) , C path ≤ 1 − 1/ S − Q − R − h ( μ ) + o ( 1 ) . Moreover R + E ≥ P ( x , μ ) − o ( 1 ) R+E \ge P(x,\mu)-o(1) R + E ≥ P ( x , μ ) − o ( 1 ) (57) . For any 0 ≤ λ ≤ 1 0 \le\lambda\le1 0 ≤ λ ≤ 1 λ \lambda λ 1 − λ 1-\lambda 1 − λ R ≥ 0 R \ge0 R ≥ 0
Q + R + λ E + ( 1 − λ ) h ≥ Q + λ P + ( 1 − λ ) h − o ( 1 ) . Q+R+\lambda E+(1-\lambda)h \ge Q+\lambda P+(1-\lambda)h-o(1). Q + R + λ E + ( 1 − λ ) h ≥ Q + λ P + ( 1 − λ ) h − o ( 1 ) . In particular, the terms Q Q Q R R R
Put
q 0 = S k ( S − k ) , x 0 = k S , K 0 = k S ( S − k ) , q_0=\frac{S}{k(S-k)},\qquad x_0=\frac{k}{S},\qquad K_0=\frac{k}{S(S-k)}, q 0 = k ( S − k ) S , x 0 = S k , K 0 = S ( S − k ) k , and define the symmetric matrix and vector
M λ = ( q 0 + λ ( 1 / b − 1 / k ) − λ / b − λ / b λ / b + ( 1 − λ ) c ) , v λ = ( 1 / ( S − k ) ( 1 − λ ) d / 2 ) . (67) M_\lambda=
\begin{pmatrix}
q_0+\lambda(1/b-1/k) & -\lambda/b\\
-\lambda/b & \lambda/b+(1-\lambda)c
\end{pmatrix},
\qquad
v_\lambda=
\begin{pmatrix}
1/(S-k)\\
(1-\lambda)d/2
\end{pmatrix}. \tag*{(67)} M λ = ( q 0 + λ ( 1/ b − 1/ k ) − λ / b − λ / b λ / b + ( 1 − λ ) c ) , v λ = ( 1/ ( S − k ) ( 1 − λ ) d /2 ) . (67) Then the loss is at least
q 0 ( x − x 0 ) 2 + λ P ( x , μ ) + ( 1 − λ ) h ( μ ) = ( x μ ) M λ ( x μ ) − 2 v λ T ( x μ ) + K 0 . q_0(x-x_0)^2+\lambda P(x,\mu)+(1-\lambda)h(\mu)
=
\begin{pmatrix}x & \mu\end{pmatrix}
M_\lambda
\begin{pmatrix}x\\\mu\end{pmatrix}
-2v_\lambda^{T}
\begin{pmatrix}x\\\mu\end{pmatrix}
+K_0. q 0 ( x − x 0 ) 2 + λ P ( x , μ ) + ( 1 − λ ) h ( μ ) = ( x μ ) M λ ( x μ ) − 2 v λ T ( x μ ) + K 0 . Both endpoint matrices are positive definite: M 0 = diag ( q 0 , c ) M_0=\operatorname{diag}(q_0,c) M 0 = diag ( q 0 , c ) det M 1 = 1 / [ b ( S − k ) ] > 0 \det M_1=1/[b(S-k)]>0 det M 1 = 1/ [ b ( S − k )] > 0 M λ ≻ 0 M_\lambda\succ0 M λ ≻ 0 [ 0 , 1 ] [0,1] [ 0 , 1 ] x x x μ \mu μ
Theorem 16.1. For every integer k / 2 ≤ b < k k/2 \le b < k k /2 ≤ b < k λ ∈ [ 0 , 1 ] \lambda\in[0,1] λ ∈ [ 0 , 1 ]
Δ b ( λ ) = K 0 − v λ T M λ − 1 v λ . \Delta_b(\lambda)=K_0-v_\lambda^{T}M_\lambda^{-1}v_\lambda. Δ b ( λ ) = K 0 − v λ T M λ − 1 v λ . Then
γ ( ℓ ) ≤ 1 2 − 1 2 S − 1 2 Δ b ( λ ) . \gamma(\ell)\le\frac{1}{2}-\frac{1}{2S}-\frac{1}{2}\Delta_b(\lambda). γ ( ℓ ) ≤ 2 1 − 2 S 1 − 2 1 Δ b ( λ ) . All entries are rational when λ \lambda λ N ≥ 2 M N \ge2M N ≥ 2 M
For fixed ℓ \ell ℓ b b b
D ( λ ) = det M λ , N ( λ ) = ( M λ ) 22 ( v λ ) 1 2 − 2 ( M λ ) 12 ( v λ ) 1 ( v λ ) 2 + ( M λ ) 11 ( v λ ) 2 2 . D(\lambda)=\det M_\lambda,\qquad N(\lambda)=(M_\lambda)_{22}(v_\lambda)_1^2-2(M_\lambda)_{12}(v_\lambda)_1(v_\lambda)_2+(M_\lambda)_{11}(v_\lambda)_2^2. D ( λ ) = det M λ , N ( λ ) = ( M λ ) 22 ( v λ ) 1 2 − 2 ( M λ ) 12 ( v λ ) 1 ( v λ ) 2 + ( M λ ) 11 ( v λ ) 2 2 . Here D D D N N N
Δ b ( λ ) = K 0 − N ( λ ) D ( λ ) . \Delta_b(\lambda)=K_0-\frac{N(\lambda)}{D(\lambda)}. Δ b ( λ ) = K 0 − D ( λ ) N ( λ ) . Its stationary equation is the explicit quartic
N ( λ ) D ′ ( λ ) − N ′ ( λ ) D ( λ ) = 0. (68) N(\lambda)D'(\lambda)-N'(\lambda)D(\lambda)=0. \tag*{(68)} N ( λ ) D ′ ( λ ) − N ′ ( λ ) D ( λ ) = 0. (68) The function Δ b \Delta_b Δ b ( x , μ ) (x,\mu) ( x , μ ) λ \lambda λ b b b
The accompanying checker sets b = k / 2 b=k/2 b = k /2 Theorem 15.1 . The optimization concerns only the weight of these two matching comparisons for the fixed path length b b b
The weighted comparison also determines the next term in an asymptotic upper bound. Substitute b = k / 2 b=k/2 b = k /2 λ = 1 − 8 / k \lambda=1-\sqrt{8/k} λ = 1 − 8/ k
Δ k / 2 ( λ ) = 1 15 k − 4 2 25 k 3 / 2 + O ( log k k 2 ) . \Delta_{k/2}(\lambda)=\frac{1}{15k}-\frac{4\sqrt{2}}{25k^{3/2}}+O\left(\frac{\log k}{k^2}\right). Δ k /2 ( λ ) = 15 k 1 − 25 k 3/2 4 2 + O ( k 2 log k ) . Thus
γ ( ℓ ) ≤ 1 2 − 1 5 k + 2 2 25 k 3 / 2 + O ( log k k 2 ) . \gamma(\ell)\le\frac{1}{2}-\frac{1}{5k}+\frac{2\sqrt{2}}{25k^{3/2}}+O\left(\frac{\log k}{k^2}\right). γ ( ℓ ) ≤ 2 1 − 5 k 1 + 25 k 3/2 2 2 + O ( k 2 log k ) . For example, this expansion can be checked after writing ε = 1 − λ \varepsilon=1-\lambda ε = 1 − λ 1 / ( 15 k ) 1/(15k) 1/ ( 15 k ) ε / ( 25 k ) \varepsilon/(25k) ε / ( 25 k ) 8 / ( 25 ε k 2 ) 8/(25\varepsilon k^2) 8/ ( 25 ε k 2 ) ε = 8 / k \varepsilon=\sqrt{8/k} ε = 8/ k γ ( ℓ ) ≤ 1 / 2 − 1 / ( 5 k ) \gamma(\ell)\le1/2-1/(5k) γ ( ℓ ) ≤ 1/2 − 1/ ( 5 k )
An algebraic six-block lower construction
For a symmetric matrix L ∈ { 1 , 2 , 3 , 4 } s × s L\in\{1,2,3,4\}^{s\times s} L ∈ { 1 , 2 , 3 , 4 } s × s a ∈ R s a\in\mathbb{R}^s a ∈ R s ∑ i a i = 1 \sum_i a_i=1 ∑ i a i = 1 a i a_i a i i , j i,j i , j L i j L_{ij} L ij [1 ] ; see [9 ] for general background on graph limits. In this section all vertices are matched, so our construction gives a lower bound on γ ′ ( 4 ) \gamma'(4) γ ′ ( 4 ) γ ( 4 ) \gamma(4) γ ( 4 )
Reducing the matching problem to a quadratic optimization
For i < j i<j i < j x i j x_{ij} x ij N N N x i i x_{ii} x ii i i i
x i i + ∑ j ≠ i x i j = a i , x i j ≥ 0. (69) x_{ii}+\sum_{j\ne i}x_{ij}=a_i,\qquad x_{ij}\ge0. \tag*{(69)} x ii + j = i ∑ x ij = a i , x ij ≥ 0. (69) The convention on x i i x_{ii} x ii P ( a ) \mathcal{P}(a) P ( a ) e = i j e=ij e = ij V ( e ) = { i , j } V(e)=\{i,j\} V ( e ) = { i , j } i = j i=j i = j L e = L i j L_e=L_{ij} L e = L ij
H e f = ∑ u ∈ V ( e ) ∑ v ∈ V ( f ) 1 { L e ≤ L u v , L f ≤ L u v } . H_{ef}=\sum_{u\in V(e)}\sum_{v\in V(f)}1\{L_e\le L_{uv},\,L_f\le L_{uv}\}. H e f = u ∈ V ( e ) ∑ v ∈ V ( f ) ∑ 1 { L e ≤ L uv , L f ≤ L uv } . The asymptotic proportion of noncritical pairs is
Q ( x ) = x T H x . (70) Q(x)=x^T Hx. \tag*{(70)} Q ( x ) = x T H x . (70) Indeed an endpoint born on an edge of label b b b c c c b b b c c c O ( N ) O(N) O ( N ) 1 − max x ∈ P ( a ) Q ( x ) 1-\max_{x\in\mathcal{P}(a)}Q(x) 1 − max x ∈ P ( a ) Q ( x )
A useful reduction removes every cross type i j ij ij L i j ≥ max ( L i i , L j j ) L_{ij}\ge\max(L_{ii},L_{jj}) L ij ≥ max ( L ii , L j j ) (69) and weakly decreases the birth label at every affected endpoint. It therefore cannot decrease Q Q Q
The maximum of a quadratic need not occur at a polytope vertex. We will use a sufficient condition that controls the whole polytope. If v 1 , … , v m v_1,\ldots,v_m v 1 , … , v m
v i T H v j ≤ T ( 1 ≤ i , j ≤ m ) , (71) v_i^T H v_j\le T\qquad(1\le i,j\le m), \tag*{(71)} v i T H v j ≤ T ( 1 ≤ i , j ≤ m ) , (71) then every convex combination x = ∑ i θ i v i x=\sum_i\theta_i v_i x = ∑ i θ i v i Q ( x ) = ∑ i , j θ i θ j v i T H v j ≤ T Q(x)=\sum_{i,j}\theta_i\theta_jv_i^T H v_j\le T Q ( x ) = ∑ i , j θ i θ j v i T H v j ≤ T (71) as the bilinear certificate.
The value and masses
Let τ \tau τ ( 0.359275945611134729 , 0.359275945611134730 ) (0.359275945611134729,0.359275945611134730) ( 0.359275945611134729 , 0.359275945611134730 )
P ( t ) = 34006 t 6 − 148176 t 5 + 211627 t 4 − 226450 t 3 + 422009 t 2 − 423516 t + 105476. P(t)=34006t^6-148176t^5+211627t^4-226450t^3+422009t^2-423516t+105476. P ( t ) = 34006 t 6 − 148176 t 5 + 211627 t 4 − 226450 t 3 + 422009 t 2 − 423516 t + 105476. Set D ( t ) = 694 t 3 − 1512 t 2 + 469 t + 285 D(t)=694t^3-1512t^2+469t+285 D ( t ) = 694 t 3 − 1512 t 2 + 469 t + 285
α L = − 49 τ 4 + 3367 τ 3 − 7130 τ 2 + 2072 τ + 1484 12 D ( τ ) . \alpha_L=\frac{-49\tau^4+3367\tau^3-7130\tau^2+2072\tau+1484}{12D(\tau)}. α L = 12 D ( τ ) − 49 τ 4 + 3367 τ 3 − 7130 τ 2 + 2072 τ + 1484 . Then
α L = 0.419772820486240077721086607635089744919175 … . \alpha_L=0.419772820486240077721086607635089744919175\ldots. α L = 0.419772820486240077721086607635089744919175 … . Use blocks A , B , C , D , E , F A,B,C,D,E,F A , B , C , D , E , F
L = ( 1 2 1 1 1 1 2 4 3 4 4 1 1 3 1 2 2 1 1 4 2 3 3 1 1 4 2 3 4 1 1 1 1 1 1 1 ) L=\begin{pmatrix}
1&2&1&1&1&1\\
2&4&3&4&4&1\\
1&3&1&2&2&1\\
1&4&2&3&3&1\\
1&4&2&3&4&1\\
1&1&1&1&1&1
\end{pmatrix} L = 1 2 1 1 1 1 2 4 3 4 4 1 1 3 1 2 2 1 1 4 2 3 3 1 1 4 2 3 4 1 1 1 1 1 1 1 and masses a ∗ = n ( τ ) / ( 6 D ( τ ) ) a^* = n(\tau)/(6D(\tau)) a ∗ = n ( τ ) / ( 6 D ( τ ))
n ( t ) = ( − 77 t 3 + 312 t 2 − 389 t + 226 1469 t 3 − 2981 t 2 + 675 t + 637 ( t − 1 ) ( 1337 t 2 − 1993 t − 452 ) 222 t 3 − 440 t 2 + 154 t + 32 835 t 3 − 1995 t 2 + 886 t + 250 378 t 3 − 638 t 2 − 53 t + 113 ) . n(t)=\begin{pmatrix}
-77t^3+312t^2-389t+226\\
1469t^3-2981t^2+675t+637\\
(t-1)(1337t^2-1993t-452)\\
222t^3-440t^2+154t+32\\
835t^3-1995t^2+886t+250\\
378t^3-638t^2-53t+113
\end{pmatrix}. n ( t ) = − 77 t 3 + 312 t 2 − 389 t + 226 1469 t 3 − 2981 t 2 + 675 t + 637 ( t − 1 ) ( 1337 t 2 − 1993 t − 452 ) 222 t 3 − 440 t 2 + 154 t + 32 835 t 3 − 1995 t 2 + 886 t + 250 378 t 3 − 638 t 2 − 53 t + 113 . The masses are approximately
( .0705314504 , .3229020494 , .3659074690 , .0234231357 , .2005210862 , .0167148093 ) . (.0705314504,.3229020494,.3659074690,.0234231357,.2005210862,.0167148093). ( .0705314504 , .3229020494 , .3659074690 , .0234231357 , .2005210862 , .0167148093 ) . Theorem 17.1. The minimum asymptotic critical density of this construction is exactly α L \alpha_L α L γ ( 4 ) ≥ γ ′ ( 4 ) ≥ α L \gamma(4) \ge\gamma'(4) \ge\alpha_L γ ( 4 ) ≥ γ ′ ( 4 ) ≥ α L
Proof. We first exhibit three matching flows that attain the proposed value. Write the six masses as ( p , b , z , d , e , f ) (p,b,z,d,e,f) ( p , b , z , d , e , f )
Table 1 (PDF p. 54).
The six nonzero coordinates in each flow are strictly positive at a ∗ a^* a ∗ C 1 , C 2 , C 3 C_1,C_2,C_3 C 1 , C 2 , C 3 (70) . We call these flows active because they attain the maximum noncritical density at the chosen masses. Put K ( t ) = 111 t 3 − 220 t 2 + 77 t + 16 K(t)=111t^3-220t^2+77t+16 K ( t ) = 111 t 3 − 220 t 2 + 77 t + 16
λ 1 = ( τ − 1 ) ( 7 τ 3 − 6 τ 2 − 32 τ − 14 ) 2 K ( τ ) , λ 2 = τ , λ 3 = 1 − λ 1 − λ 2 . \lambda_1=\frac{(\tau-1)(7\tau^3-6\tau^2-32\tau-14)}{2K(\tau)},\qquad\lambda_2=\tau,\qquad\lambda_3=1-\lambda_1-\lambda_2. λ 1 = 2 K ( τ ) ( τ − 1 ) ( 7 τ 3 − 6 τ 2 − 32 τ − 14 ) , λ 2 = τ , λ 3 = 1 − λ 1 − λ 2 . These weights are positive. Exact arithmetic modulo P P P
( a ∗ ) T C k a ∗ = 1 − α L = : T ( k = 1 , 2 , 3 ) , (a^*)^T C_k a^*=1-\alpha_L=:T\qquad(k=1,2,3), ( a ∗ ) T C k a ∗ = 1 − α L =: T ( k = 1 , 2 , 3 ) , ( ∑ k λ k C k ) a ∗ = T 1. \left(\sum_k\lambda_k C_k\right)a^*=T1. ( k ∑ λ k C k ) a ∗ = T 1. Thus the active flows attain noncritical density T T T
For the reverse inequality, the remaining cross types after the reduction above are
A B , A D , A E , B C , B F , C D , C E , D E , D F , E F , AB,\ AD,\ AE,\ BC,\ BF,\ CD,\ CE,\ DE,\ DF,\ EF, A B , A D , A E , B C , B F , C D , C E , D E , D F , E F , together with the six internal types. Let B B B 6 × 16 6\times16 6 × 16 B i , e = 1 B_{i,e}=1 B i , e = 1 e e e i i i B x = a ∗ Bx=a^* B x = a ∗ x ≥ 0 x\ge0 x ≥ 0 B B B J J J B J x J = a ∗ B_Jx_J=a^* B J x J = a ∗ B J B_J B J a ˉ \bar{a} a ˉ
( 0.070531450396389 , 0.322902049431553 , 0.365907468971544 , 0.023423135669300 , 0.200521086237747 , 0.016714809293466 ) . (0.070531450396389,\ 0.322902049431553,\ 0.365907468971544,\ 0.023423135669300,\ 0.200521086237747,\ 0.016714809293466). ( 0.070531450396389 , 0.322902049431553 , 0.365907468971544 , 0.023423135669300 , 0.200521086237747 , 0.016714809293466 ) . The exact algebraic vector a ∗ a^* a ∗ ∣ a i − a ˉ i ∣ ≤ 10 − 13 |a_i-\bar{a}_i| \le10^{-13} ∣ a i − a ˉ i ∣ ≤ 1 0 − 13 B J − 1 a B_J^{-1}a B J − 1 a 240 ⋅ 241 / 2 = 28920 240 \cdot241/2 = 28920 240 ⋅ 241/2 = 28920 (71) , 28915 are strictly below T T T T T T
The computational part of this proof consists of these finite rational sign and polynomial-identity checks. The accompanying lower-certificate/verify_algebraic.py checks normalization, positivity, the three active equalities, and the stationarity identity in Q [ t ] / ( P ) \mathbb{Q}[t]/(P) Q [ t ] / ( P ) verify_flow_bound.py, which reconstructs B B B H H H lower-certificate/stationarity-input.json. No optimization routine is needed to verify the assertions.
Finally, approximate a ∗ a^* a ∗ P ( a ∗ ) \mathcal{P}(a^*) P ( a ∗ ) Q Q Q o ( N ) o(N) o ( N ) o ( N 2 ) o(N^2) o ( N 2 )
The number α L \alpha_L α L Q \mathbb{Q} Q
5754135102948532776960 X 6 − 13507216477071755171328 X 5 + 13208173323262094551248 X 4 − 6886878440990980289832 X 3 + 2019436570431176007853 X 2 − 315750351353916573206 X + 20566231563858477973. 5754135102948532776960X^6 - 13507216477071755171328X^5
+ 13208173323262094551248X^4 - 6886878440990980289832X^3
+ 2019436570431176007853X^2 - 315750351353916573206X + 20566231563858477973. 5754135102948532776960 X 6 − 13507216477071755171328 X 5 + 13208173323262094551248 X 4 − 6886878440990980289832 X 3 + 2019436570431176007853 X 2 − 315750351353916573206 X + 20566231563858477973. Both this polynomial and P P P P P P τ \tau τ
( 0.41977282048624007772 , 0.41977282048624007773 ) . (0.41977282048624007772,\ 0.41977282048624007773). ( 0.41977282048624007772 , 0.41977282048624007773 ) . The polynomial relation and irreducibility checks are included in the symbolic certificate.
Deriving the algebraic parameters
The degree-six equation follows by eliminating the stationarity equations of the three active quadratic forms. Write q = d + e q=d+e q = d + e d d d e e e q q q
d = ( b + q − p − z − f ) / 4. d=(b+q-p-z-f)/4. d = ( b + q − p − z − f ) /4. After substituting this expression, let C ~ 1 , C ~ 2 , C ~ 3 \widetilde{C}_1,\widetilde{C}_2,\widetilde{C}_3 C 1 , C 2 , C 3 ( p , b , z , q , f ) (p,b,z,q,f) ( p , b , z , q , f ) u , v , 1 − u − v u,v,1-u-v u , v , 1 − u − v H ( u , v ) = u C ~ 1 + v C ~ 2 + ( 1 − u − v ) C ~ 3 H(u,v)=u\widetilde{C}_1+v\widetilde{C}_2+(1-u-v)\widetilde{C}_3 H ( u , v ) = u C 1 + v C 2 + ( 1 − u − v ) C 3
T ( u , v ) = 1 1 T H ( u , v ) − 1 1 = N ( u , v ) 4 S ( u , v ) . T(u,v)=\frac{1}{\mathbf{1}^{T}H(u,v)^{-1}\mathbf{1}}=\frac{N(u,v)}{4S(u,v)}. T ( u , v ) = 1 T H ( u , v ) − 1 1 1 = 4 S ( u , v ) N ( u , v ) . where
N = u 2 ( − 20 v 3 + 420 v 2 − 224 v − 128 ) + u ( 7 v 4 + 559 v 3 − 842 v 2 + 164 v + 112 ) + 4 v 4 + 156 v 3 − 248 v 2 + 52 v + 36 , \begin{aligned}
N ={}& u^2(-20v^3 + 420v^2 - 224v - 128) \\
&+ u(7v^4 + 559v^3 - 842v^2 + 164v + 112) \\
&+ 4v^4 + 156v^3 - 248v^2 + 52v + 36,
\end{aligned} N = u 2 ( − 20 v 3 + 420 v 2 − 224 v − 128 ) + u ( 7 v 4 + 559 v 3 − 842 v 2 + 164 v + 112 ) + 4 v 4 + 156 v 3 − 248 v 2 + 52 v + 36 , S = u 2 ( 156 v 2 − 84 v − 56 ) + u ( 234 v 3 − 344 v 2 + 61 v + 49 ) + v 4 + 65 v 3 − 102 v 2 + 20 v + 16. S = u^2(156v^2 - 84v - 56) + u(234v^3 - 344v^2 + 61v + 49)
+ v^4 + 65v^3 - 102v^2 + 20v + 16. S = u 2 ( 156 v 2 − 84 v − 56 ) + u ( 234 v 3 − 344 v 2 + 61 v + 49 ) + v 4 + 65 v 3 − 102 v 2 + 20 v + 16. At the stationary solution T = 1 − α L T = 1 - \alpha_L T = 1 − α L T u = T v = 0 T_u = T_v = 0 T u = T v = 0 F = N − 4 ( 1 − α L ) S F = N - 4(1 - \alpha_L)S F = N − 4 ( 1 − α L ) S u u u
disc u F = ( v − 1 ) ( 104 α L v 2 − 56 α L v − 28 α L + v 3 − 38 v 2 + 20 v + 12 ) × ( 8328 α L v 3 − 18144 α L v 2 + 5628 α L v + 3420 α L + 49 v 4 − 3367 v 3 + 7130 v 2 − 2072 v − 1484 ) . \begin{aligned}
\operatorname{disc}_u F ={}& (v - 1)(104\alpha_L v^2 - 56\alpha_L v - 28\alpha_L + v^3 - 38v^2 + 20v + 12) \\
&\times(8328\alpha_L v^3 - 18144\alpha_L v^2 + 5628\alpha_L v + 3420\alpha_L \\
&\qquad+ 49v^4 - 3367v^3 + 7130v^2 - 2072v - 1484).
\end{aligned} disc u F = ( v − 1 ) ( 104 α L v 2 − 56 α L v − 28 α L + v 3 − 38 v 2 + 20 v + 12 ) × ( 8328 α L v 3 − 18144 α L v 2 + 5628 α L v + 3420 α L + 49 v 4 − 3367 v 3 + 7130 v 2 − 2072 v − 1484 ) . The first two factors do not vanish in the isolating box. The last factor and its v v v α L \alpha_L α L v = τ v = \tau v = τ
d d t ( − 49 t 4 + 3367 t 3 − 7130 t 2 + 2072 t + 1484 12 D ( t ) ) = − P ( t ) 12 D ( t ) 2 . \frac{d}{dt}\left(\frac{-49t^4 + 3367t^3 - 7130t^2 + 2072t + 1484}{12D(t)}\right) = -\frac{P(t)}{12D(t)^2}. d t d ( 12 D ( t ) − 49 t 4 + 3367 t 3 − 7130 t 2 + 2072 t + 1484 ) = − 12 D ( t ) 2 P ( t ) . The double-root formula for u u u λ 1 \lambda_1 λ 1 t t t α L \alpha_L α L lower-certificate/verify_elimination_independent.py reconstructs N / ( 4 S ) N/(4S) N / ( 4 S )
Exact verification and accompanying files
The computer-assisted claims are finite algebraic inequalities. Their inputs are the matching comparisons and constructions proved in the preceding sections. The verification programs reconstruct the relevant quadratic forms and check the signs, identities, and parameter ranges with integer and rational arithmetic. Numerical optimization is not needed to run them. This use of exact certificates after numerical exploration is related to the rational certification methods discussed by Peyrl and Parrilo [10 ] .
Contents and commands
The source archive contains main.tex, the included section files, and an anc/ directory containing all verification programs and data. The bibliography is in references.tex; there is no external bibliography database or downloaded data set. The root file main.tex compiles with PDFLaTeX.
For a local compilation, run it until the log no longer requests another pass:
pdflatex -interaction=nonstopmode -halt-on-error main.tex
pdflatex -interaction=nonstopmode -halt-on-error main.tex
pdflatex -interaction=nonstopmode -halt-on-error main.texThe source uses the standard packages geometry, amsmath, amssymb, amsthm, mathtools, booktabs, microtype, xurl, and hyperref.
For verification, use Python 3.9 or later with its standard library. No additional Python packages, numerical solvers, network access, or external research directories are required. From the root of the extracted source archive, run
python3 anc/check_package.py
python3 anc/verify_all.pyThe first command checks the file inventory and SHA-256 hashes. The second runs every mathematical certificate. Successful completion has exit status zero and ends with
PASS: all exact certificates.Run Python normally: its -O option disables the assertions used by the checkers and must not be used. The driver rejects such an invocation. To run an individual checker, use the paths listed below relative to anc/. The working directory does not matter because paths to data are resolved relative to each script.
The programs print progress as each group of certificates passes. They may take several minutes, particularly for the matching-flow and stability calculations. Three deterministic summaries are regenerated: general-certificate/verified-general-table.json, general-certificate/path-budget-certificates.json, and lower-certificate/neutral-pairs_exact-results.json. These are included in the archive so that their contents can also be inspected without running Python.
What each computation verifies
verify_global.py proves the algebraic part of the four-label upper bound. It reconstructs the quadratic comparisons from § 4 –§ 13 , checks the signs of their rational weights, verifies the matrix factorizations, and checks that the parameter subregions cover the entire required domain. The JSON files beside it contain the weights and subdivisions. The helper modules in the same directory are included in full.
general-certificate/verify_general.py checks the explicit general-label upper bounds, the rational choices giving the numerical table in the introduction, and the finite quadratic formula in Theorem 16.1 . It also brackets the optimizing weight for the fixed path-length choice. The alternating-path counting arguments and the asymptotic expansion are given in the text.
lower-certificate/verify_algebraic.py checks the sextic polynomial, root isolation, mass normalization and positivity, and the identities for the three matching flows in § 17 . It then invokes lower-certificate/verify_flow_bound.py, which enumerates the feasible vertices of the reduced matching-flow polytope and checks all their bilinear pairings. The latter step bounds every feasible flow, not only the vertices. The separate script lower-certificate/verify_elimination_independent.py checks the elimination that produces the algebraic value.
lower-certificate/verify_atom_stability.py, lower-certificate/check_neutral_pairs_exact.py, and lower-certificate/verify_stationarity.py check the finite profile and matrix calculations used in Appendix § 19 . The appendix states precisely which perturbations are covered.
The general lower bound of Theorem 2.1 and its weight-optimality statement have complete elementary proofs in § 2 ; neither depends on a computation.
The verification programs check the finite algebra arising from the proofs. The mathematical justification that each inequality applies to actual matchings is provided in the preceding sections. Classical matching theorems and the alternating-path bound of [1 ] are stated where used and cited to the literature. In particular, computer verification is not being used to infer a universal combinatorial assertion from a finite sample of graphs.
Local optimality under the addition of one or two blocks
We next ask whether the six-part construction can be improved by adjoining parts of small positive mass. Throughout this section the labels between its six existing parts remain fixed. The result is local: neither a change to those labels nor a configuration far from the specified mass vector is covered by the theorem.
Theorem A.1. There is a neighborhood of ( a ∗ , 0 , 0 ) (a^*,0,0) ( a ∗ , 0 , 0 ) L L L { 1 , 2 , 3 , 4 } \{1,2,3,4\} { 1 , 2 , 3 , 4 } α L \alpha_L α L
We use the notation T = 1 − α L T=1-\alpha_L T = 1 − α L C k , λ k C_k,\lambda_k C k , λ k § 17 . To rule out a larger minimum critical density, it suffices to exhibit a matching with noncritical density at least T T T T T T
Write the three active flow maps from the table in the proof of Theorem 17.1 as x = P k a x=P_k a x = P k a H 0 = ∑ k λ k C k H_0=\sum_k\lambda_k C_k H 0 = ∑ k λ k C k L D L T LDL^T L D L T H 0 ≻ 0 H_0\succ0 H 0 ≻ 0 H 0 a ∗ = T 1 H_0a^*=T\mathbf{1} H 0 a ∗ = T 1
a T H 0 a ≥ T whenever 1 T a = 1. (72) a^T H_0 a \ge T \quad\text{whenever } \mathbf{1}^T a=1. \tag*{(72)} a T H 0 a ≥ T whenever 1 T a = 1. (72) Indeed, writing a = a ∗ + h a=a^*+h a = a ∗ + h 1 T h = 0 \mathbf{1}^T h=0 1 T h = 0 a T H 0 a = T + h T H 0 h a^T H_0a=T+h^T H_0h a T H 0 a = T + h T H 0 h a ∗ a^* a ∗
One added block
The label profile of a new block G G G A , B , C , D , E , F A,B,C,D,E,F A , B , C , D , E , F L G G L_{GG} L GG G G G s s s k k k j ∈ { A , B , C , D , E , F , G } j\in\{A,B,C,D,E,F,G\} j ∈ { A , B , C , D , E , F , G } j ≠ G j\ne G j = G j j j a − s e j a-se_j a − s e j j = G j=G j = G G G G a a a s s s a ∗ a^* a ∗ c k , j c_{k,j} c k , j s s s a ∗ a^* a ∗
∑ k λ k ∇ ( a T C k a ) ∣ a = a ∗ = 2 T 1 , \sum_k \lambda_k \nabla(a^T C_k a)\big|_{a=a^*}=2T\mathbf{1}, k ∑ λ k ∇ ( a T C k a ) a = a ∗ = 2 T 1 , a selected set of partners gives a strictly positive first-order increase in the weighted noncritical density, after accounting for the mass removed from the six existing blocks, whenever
∑ k λ k c k , j k − 2 T > 0. (73) \sum_k \lambda_k c_{k,j_k}-2T>0. \tag*{(73)} k ∑ λ k c k , j k − 2 T > 0. (73) The subtraction of 2 T 2T 2 T
It suffices to consider capped profiles, meaning that L G i ≤ max [ L G G , L i i ] L_{Gi}\le\max[L_{GG},L_{ii}] L G i ≤ max [ L GG , L ii ]
48 + 384 + 1296 + 4096 = 5824 48+384+1296+4096=5824 48 + 384 + 1296 + 4096 = 5824 capped profiles. The checker supplies explicit partners for each. For 5801 profiles, the left side of (73) is strictly greater than 19 / 1000 19/1000 19/1000 neutral . They fall into the following classes. The first column specifies a part with which the new mass will be combined in the quadratic identity below; profile entries are ordered A , B , C , D , E , F A, B, C, D, E, F A , B , C , D , E , F
Table 1 (PDF p. 59).
For each neutral profile there are explicit partners such that its extended quadratic in strategy k k k
Q k ext ( a , s ) = Q k ( a + s e role ) + κ k s 2 , κ k ≥ 0. (74) Q^{\operatorname{ext}}_k(a,s)=Q_k(a+se_{\operatorname{role}})+\kappa_k s^2,\qquad\kappa_k\geq0. \tag*{(74)} Q k ext ( a , s ) = Q k ( a + s e role ) + κ k s 2 , κ k ≥ 0. (74) These are identities of rational quadratic matrices. Thus every neutral one-block extension is excluded exactly by (72) , not merely to first order. For a nonneutral profile, positive definiteness on the old mass coordinates and its positive linear s s s ∑ k λ k Q k ext ( a , s ) ≥ T \sum_k \lambda_k Q^{\operatorname{ext}}_k(a,s)\geq T ∑ k λ k Q k ext ( a , s ) ≥ T s s s
Two added blocks
For two neutral profiles of masses s , t s,t s , t
Q 0 ( a + s e i + t e j ) + κ s 2 + κ ′ t 2 + 2 K s t , Q 0 ( x ) = x T H 0 x , κ , κ ′ ≥ 0. Q_0(a+se_i+te_j)+\kappa s^2+\kappa' t^2+2Kst,\qquad Q_0(x)=x^{T}H_0x,\qquad\kappa,\kappa'\geq0. Q 0 ( a + s e i + t e j ) + κ s 2 + κ ′ t 2 + 2 K s t , Q 0 ( x ) = x T H 0 x , κ , κ ′ ≥ 0. There are 979 unordered neutral-profile/contact-label cases after capping the new mutual contact. Exact checks find 453 cases with K = 0 K=0 K = 0 K > 0 K>0 K > 0 K < 0 K<0 K < 0 (72) . In every negative case, at least one active flow admits a strictly positive first-order increase in noncritical density: match equal portions of the two new blocks to each other instead of their designated old partners. Its weighted first-order gain is greater than 19/500 per unit of this matched mass, verified by rational intervals. At old masses near a ∗ a^* a ∗ min ( s , t ) \min(s,t) min ( s , t ) min ( s , t ) \min(s,t) min ( s , t ) ∣ 2 K s t ∣ ≤ 2 ∣ K ∣ ( s + t ) min ( s , t ) |2Kst|\leq2|K|(s+t)\min(s,t) ∣2 K s t ∣ ≤ 2∣ K ∣ ( s + t ) min ( s , t ) s + t s+t s + t
If at least one new profile is nonneutral, its positive linear mass term similarly absorbs all mixed terms involving it. A neutral profile’s remaining contribution is controlled by (75). Since the profile lists are finite, a single sufficiently small neighborhood works for all cases. This completes the proof of Theorem A.1 .
The exact checks for this section are verify_atom_stability.py and check_neutral_pairs_exact.py, with data in atom-certificate.json, all in the lower-certificate directory. They reconstruct the selected flow extensions directly from L L L P k P_k P k
Fixed directions with more added blocks and uncovered vertices
For any fixed finite number of added blocks, a related statement holds along each direction in which their masses tend to zero at positive relative rates. We also allow a small excess of vertices over those covered by the required matching. The threshold for the added mass may depend on the direction; thus this statement differs from the neighborhood valid for all one- and two-block extensions in Theorem A.1 .
Theorem A.2. Keep the old six-by-six label matrix fixed and add a fixed finite collection of blocks with fixed labels. Measure masses per 2 M 2M 2 M M M M u α s + o ( u ) u\alpha_s+o(u) u α s + o ( u ) α s > 0 \alpha_s>0 α s > 0 ∑ s α s = 1 \sum_s\alpha_s=1 ∑ s α s = 1 a ∗ + u h + o ( u ) a^*+uh+o(u) a ∗ + u h + o ( u ) ∑ i h i ≥ − 1 \sum_i h_i\ge-1 ∑ i h i ≥ − 1 u u u ( 2 M 2 ) \binom{2M}{2} ( 2 2 M ) α L + o M ( 1 ) \alpha_L+o_M(1) α L + o M ( 1 )
Proof. First consider total mass one. Use the active flow maps, weights, and T = 1 − α L T=1-\alpha_L T = 1 − α L § 17 . As in the preceding section, an increase in noncritical density provides a matching that prevents an improvement of the lower construction. The check allowing profiles without the capping restriction enumerates all 4 7 = 16384 4^7=16384 4 7 = 16384 > 19 / 1000 >19/1000 > 19/1000 parent , namely an existing part i i i
If a profile with positive first-order noncritical gain has positive relative mass, that gain dominates all quadratic errors. Suppose therefore all profiles are neutral. If a pair of new blocks with parents i , j i,j i , j k < L i j k<L_{ij} k < L ij 9 / 100 9/100 9/100 u u u
It remains to treat neutral profiles whose mutual labels are all at least the corresponding parent labels. Aggregate their masses with their parents, obtaining s s s ∑ i s i = 1 \sum_i s_i=1 ∑ i s i = 1 β i \beta_i β i i i i u u u β i < 1 / 2 \beta_i<1/2 β i < 1/2 P k s P_{ks} P k s t i j t_{ij} t ij β i t i j \beta_i t_{ij} β i t ij i i i j j j β j t i j \beta_j t_{ij} β j t ij i i i j j j ( 1 − β i − β j ) t i j (1-\beta_i-\beta_j)t_{ij} ( 1 − β i − β j ) t ij t i i t_{ii} t ii β i t i i \beta_i t_{ii} β i t ii ( 1 − 2 β i ) t i i (1-2\beta_i)t_{ii} ( 1 − 2 β i ) t ii Q k ( s ) Q_k(s) Q k ( s )
Q max ≥ ∑ k λ k Q k ( s ) = s T H 0 s ≥ T . Q_{\max}\ge\sum_k\lambda_k Q_k(s)=s^T H_0s\ge T. Q m a x ≥ k ∑ λ k Q k ( s ) = s T H 0 s ≥ T . This is exact, and resolves the remaining directions at every order.
Now allow total mass larger than one. The old B B B
L i B ≥ L i j for every old i , j . L_{iB}\ge L_{ij}\qquad\text{for every old }i,j. L i B ≥ L ij for every old i , j . Leave the entire surplus, of mass u ( 1 + ∑ i h i ) + o ( u ) u(1+\sum_i h_i)+o(u) u ( 1 + ∑ i h i ) + o ( u ) B B B u u u a B ∗ > 0 a_B^*>0 a B ∗ > 0
For a strict first-order case, the old–old matching endpoints have no critical edges to the uncovered B B B O ( u ) O(u) O ( u ) O ( u ) O(u) O ( u ) O ( u ) O(u) O ( u ) O ( u 2 ) O(u^2) O ( u 2 )
In the residual neutral case, the proportional lift has no outward critical edges at all. Every lifted old endpoint has a parent birth at most its label to B B B i i i B B B L i B L_{iB} L i B 2 M 2M 2 M
The certificate lower-certificate/verify_stationarity.py checks the 16384 profiles, parent domination, active-edge agreement, and all 81 strict pair responses. Its stationary root lies in the same rational box as the exact algebraic construction of Theorem 17.1 . This result is directional: it does not assert one radius uniform over varying numbers of added blocks or relative masses tending to zero. It also keeps all labels between the six existing parts fixed and does not concern configurations a fixed positive distance away.
Acknowledgment
The author was supported by the NRDI grant KKP 138270.
