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Two choices suffice: kk-out sampling leaves O(n/k)O(n/k) intercomponent edges for k≥2k\geq 2

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Abstract

Given a simple undirected graph on nn vertices, random kk-out sampling forms a spanning subgraph by having each vertex independently choose kk incident edges uniformly with replacement. We prove that, for every integer k≥2k\geq2, the expected number of original edges joining distinct sampled components is O(n/k)O(n/k). This resolves the conjecture of Holm, King, Thorup, Zamir, and Zwick in its full range, improving their logarithmic sampling requirement to the optimal threshold of two choices per vertex. The bound fails for k=1k=1, and the dependence O(n/k)O(n/k) is tight up to constant factors. Our result holds with replacement and consequently also without replacement. In particular, this implies that contracting just two independent local choices per vertex suffices to reduce an arbitrary simple graph to a quotient with O(n)O(n) edges in expectation, which is crucial in reducing the instance size for many contraction-based algorithms. Our proof uses switching arguments to relate intercomponent edges to bridges and a degree-weighted component potential, whose expectation we also bound with a switching argument.

Introduction

Random edge sampling is a bsimple tool for sparsifying graphs while preserving useful structural information. We study sampling based on local choices at each vertex and ask how many original edges join distinct components of the sample. This property concerns connectivity of the original graph as contracting each sampled component leaves precisely these intercomponent edges and adding them to a spanning forest of the sample preserves the original connected components. Bounds on the number of intercomponent edges are therefore useful for contraction-based minimum-cut algorithms [2], spanning-forest communication protocols [1], and spanning adjacency oracles [13].

Let G=(VG,EG)G=(V_G, E_G) be a simple undirected graph on nn vertices. Holm, King, Thorup, Zamir, and Zwick [1] studied this question in the without-replacement kk-out model, where each vertex vv independently chooses a uniformly random subset of min⁡{k,deg⁡G(v)}\min\{k,\deg_G(v)\} incident edges and the sampled subgraph is the union of these choices. They proved that the expected number of intercomponent edges is O(n/k)O(n/k) for k≥Ω(log⁡n)k\geq \Omega(\log n) and conjectured that the same bound should hold for all k≥2k\geq 2.

We resolve this conjecture in full, for every integer k≥2k\geq 2, in the stronger with-replacement model. Here each vertex independently samples kk incident edges uniformly with replacement, and HGkH_G^k is the union of these choices. Writing xG(H)x_G(H) for the number of intercomponent edges of a spanning subgraph HH, we prove the following result.

Theorem 1. Let GG be a simple graph on nn vertices, and let HGkH_G^k be its random with-replacement kk-out subgraph. For every integer k≥2k\geq 2, E[xG(HGk)]≤O(n/k).\mathop{\mathrm{\mathbb{E}}}\left[x_G(H_G^k)\right] \leq O(n/k).

The bound also holds without replacement as under a natural coupling, the without-replacement subgraph contains HGkH_G^k and has no more intercomponent edges (see § 3.0.0.1). In this work, we use the with-replacement model unless stated otherwise.

The result is optimal in both its sampling rate kk and its intercomponent edge count. The bound fails for k=1k=1, while examples of Holm, King, Thorup, Zamir, and Zwick [1] have Ω(n/k)\Omega(n/k) expected intercomponent edges for 2≤k≤n/32\leq k\leq n/3, even without replacement.

Technical Contribution and Significance

A key obstacle to extending the analysis of Holm, King, Thorup, Zamir, and Zwick [1] to constant kk is their reduction to the expected kk-out model, where each vertex vv independently includes each incident edge with probability kmax⁡{k,deg⁡(v)}\frac{k}{\max\{k,\deg(v)\}}. For constant kk, this model leaves each vertex of a complete graph isolated with constant probability, yielding Ω(n2)\Omega(n^2) expected intercomponent edges instead of O(n)O(n).

Our analysis bypasses this intermediate model and analyzes the kk-out model directly using the switching technique. The switching method is a double-counting technique for estimating the relative sizes of classes of combinatorial objects. We define transformations called switchings between the classes and compare the numbers of forward and inverse operations, thereby obtaining counting and probability bounds. In graph applications, a switching typically makes local modifications to the graphs, replacing a small set of edges with another. This is not necessary, but it often makes counting and reasoning about the number of forward and inverse switchings simpler. We use one switching argument to relate intercomponent edges to a component potential, and another to bound the expectation of this potential.

The case k≥3k\geq3 is outlined in § 2 and proved in § 6. The two-out case in Theorem 12 follows the same two steps but requires more careful treatment.

Implications and Applications

While the distributed sketching application of Holm, King, Thorup, Zamir, and Zwick [1] is already supported by their bound for k≥Ω(log⁡n)k\geq \Omega(\log n), extending the bound to constant kk enables further algorithmic applications.

In particular, our theorem provides the simplification anticipated by Ghaffari, Nowicki, and Thorup  [2] in their contraction-based algorithms for exact minimum cut in simple unweighted graphs. Their random 22-out contraction reduces the graph to O(n/δ)O(n/\delta) vertices, where δ\delta is the minimum degree, while preserving any fixed nontrivial minimum cut with constant probability. However, they have no bound on the number of edges that remain, and so they apply an additional sparsification step to reduce the edge count to O(n)O(n). Our theorem supplies this edge bound directly for their random 22-out contraction with replacement, removing the need for the additional edge sparsification step. Combined with their vertex bound, it yields, with constant probability, a contracted instance with O(n/δ)O(n/\delta) vertices and O(n)O(n) edges that preserves a fixed non-singleton minimum cut. Keeping kk constant is essential here, since it is what gives the constant cut-survival probability.

The same contraction principle improves exact minimum cut in the cut-query model, where each query returns the number of edges crossing a specified vertex subset. Kenneth-Mordoch and Krauthgamer  [6] obtained a two-round algorithm using O~(n4/3)\widetilde O(n^{4/3}) queries by constructing a contraction in the first round and recovering the contracted graph in the second. The recovery cost is nearly linear in the number of remaining vertices and edges. Our expected edge bound lets these sampling and recovery primitives implement random 22-out contraction and reconstruct the contracted graph using O~(n)\widetilde O(n) queries, with the cut-survival guarantee above.

Early work focused on kk-out sampling in the complete graph. Fenner and Frieze  [17] established kk-vertex and kk-edge connectivity for k≥2k\geq 2, Frieze  [20] proved the existence of perfect matchings for even-order graphs when k≥2k\geq 2, and Bohman and Frieze  [18] proved Hamiltonicity for k≥3k\geq 3, all with probability tending to 11. These results can be found in the book by Frieze and Karoński  [21].

Frieze and Johansson  [19] extended connectivity and Hamiltonicity results to graphs GG whose minimum degree is (1/2+ϵ)n(1/2+\epsilon)n for some ϵ>0\epsilon>0. They showed that HGkH^k_G is kk-connected w.h.p. for k=O(1)k = O(1) and is Hamiltonian for sufficiently large kk. However, they remarked that such results are not possible for graphs with smaller minimum degree.

Minimum cut.

Beyond the cut-query application above, random 22-out contraction has also been used in streaming. Assadi and Dudeja [12] gave a two-pass algorithm for exact global minimum cut using O(nlog⁡n)O(n\log n) bits of space, whose first pass builds a with-replacement 22-out sample.

Spanning structures.

The bound of Holm, King, Thorup, Zamir, and Zwick [1] also gives a one-way communication protocol for finding a spanning forest, using private randomness and O(nlog⁡n)O(\sqrt{n}\log n) bits per vertex. Bodwin and Fleischmann [13] also apply this bound to spanning adjacency oracles, which implicitly represent a sparse spanning subgraph and answer edge-membership queries. One construction uses kk-out sampling.

Beyond connectivity.

Saneian and Behnezhad [16] use one-sided random 33-out sampling in bipartite graphs for streaming edge coloring. Each vertex on one side independently chooses three distinct uniform neighbors on the other. When the opposite side is larger by a sufficiently large constant factor, the sample contains a matching covering the sampling side with high probability.

Implementations.

The ConnectIt framework of Dhulipala, Hong, and Shun [14] and the GPU study of Hong, Dhulipala, and Shun [15] evaluate kk-out sampling and its variants for parallel connectivity.

The switching method.

As described in Wormald’s survey [3], McKay introduced switchings into the study of random graphs to bound subgraph probabilities in random graphs with prescribed degrees  [10]. Since then, the switching argument has found many applications, including cycle counts and girth  [7], connectivity  [8], asymptotic estimates of spanning trees  [9], and asymptotic enumeration of graphs with prescribed degrees  [11]. Switchings are also crucial in the analysis of exact uniform generation algorithms, beginning with the work of McKay and Wormald [4] which was later improved by Arman, Gao, and Wormald [5] also using switching.

Technical Overview

We prove the bound by relating intercomponent edges to a degree-sensitive potential of the sampled components, and then showing that a special augmentation of a 22-out sample has expected potential O(n)O(n).

From intercomponent edges to component potential

We first describe a simple switching argument to demonstrate how it can be useful. Fix r<kr<k, and leave the first rr choices at every vertex unchanged. For each intercomponent edge ee of HkH^k, choose its smaller-degree endpoint vv and replace any one of its remaining k−rk-r choices by ee. Each replacement makes ee a bridge of the new sample outside the unchanged HrH^r.

Each intercomponent edge therefore gives k−rk-r forward switchings. Conversely, an image bridge at vv admits at most deg⁡G(v)\deg_G(v) inverse switchings, one for each possible previous choice. Thus forward switchings count intercomponent edges multiplied by k−rk-r, while inverse switchings are bounded by bridges weighted by the smaller degree of their endpoints.

These bridge weights are controlled by the component potential

ΦG(Hr):=∑C∈CC(Hr)max⁡v∈Cdeg⁡G(v).\Phi_G(H^r) :=\sum_{C\in\mathcal{CC}(H^r)}\max_{v\in C}\deg_G(v).

Indeed, after contracting the components of HrH^r, the bridges outside HrH^r form a forest. Rooting its trees and charging each bridge to its child component results in at most one charge per component. Each charge is at most that component’s maximum original degree. Double-counting the switchings therefore gives

(k−r)E[xG(Hk)]≤E[ΦG(Hr)].(k-r)\mathop{\mathrm{\mathbb{E}}}[x_G(H^k)] \leq \mathop{\mathrm{\mathbb{E}}}[\Phi_G(H^r)].

It remains to bound the expected potential of a suitable subgraph by O(n)O(n).

Bounding the component potential

Fix a total order on the vertices by nondecreasing degree, breaking ties arbitrarily. Call the largest vertex of each component its leader, and order the components by their leaders. The potential ΦG(H)\Phi_G(H) is the sum of the leaders’ degrees. Their internal neighbors contribute at most nn because the components are disjoint, so it remains to control their crossing edges. Consider each crossing edge to be owned by the endpoint in the higher component. We prove a bound on the crossing edges owned by leaders using a recurrence.

A recurrence for edges at successive depths.

Write xGj(H2)x_G^j(H^2) for the number of crossing edges whose owner is at distance jj from its leader. For such an edge uvuv with owner uu, replace either choice at vv in the lower component by uvuv. The inserted edge becomes a bridge, and vv is now at depth j+1j+1 from the higher leader. We establish

2E[xGj(H2)]≤n+E[xGj+1(H2)].2\mathop{\mathrm{\mathbb{E}}}[x_G^j(H^2)] \leq n+\mathop{\mathrm{\mathbb{E}}}[x_G^{j+1}(H^2)].

The factor of two comes from the two choices that can be replaced at vv. To reverse a switch, fix an image and recover the previous choice at vv. A valid previous choice either stays inside the branch separated from the leader by the inserted bridge uvuv, or leads to another lower component. In the first case, the branches below bridges at the same depth are disjoint and therefore sum up to nn. In the second case, the previous edge is a crossing edge in the image owned by vv at depth j+1j+1, counted by xGj+1(H2)x_G^{j+1}(H^2). Solving the recurrence gives

E[xG0(H2)]≤n.\mathop{\mathrm{\mathbb{E}}}[x_G^{0}(H^2)]\leq n.
Using one additional choice.

It remains to handle leaders’ edges to higher components. A leader with many such edges is likely to select one of them given an additional choice, allowing us to merge its component into a higher one and remove its degree from the potential.

This motivates the augmentation H2+H^{2+}. Add each H2H^2-leader’s third choice only when it reaches a higher H2H^2-component. Conditioned on H2H^2, a leader survives as a leader in H2+H^{2+} with probability equal to the fraction of its neighbors lying inside its own component or in lower components. Since each surviving leader contributes its degree to the potential, multiplying the surviving probability by its degree gives the expected contribution to the potential, which is exactly the number of these neighbors. Summing the internal neighbors and crossing edges to lower components gives

E[ΦG(H2+)]≤n+E[xG0(H2)]≤2n.\mathop{\mathrm{\mathbb{E}}}[\Phi_G(H^{2+})] \leq n+\mathop{\mathrm{\mathbb{E}}}[x_G^0(H^2)] \leq 2n.

Refining the Switching Argument for k≥3k\geq 3

Since H2+⊆H3H^{2+}\subseteq H^3, the same potential bound holds for H3H^3. Taking r=3r=3 in the first switching argument gives the desired bound for k≥4k\geq4. To reach k=3k=3, we must protect fewer edges, leaving room for switching. We refine the switch so that the inserted crossing edge is a bridge outside H2+H^{2+} in the image, giving

(k−2)E[xG(Hk)]≤3E[ΦG(H2+)],k≥3.(k-2)\mathop{\mathrm{\mathbb{E}}}[x_G(H^k)] \leq 3\mathop{\mathrm{\mathbb{E}}}[\Phi_G(H^{2+})], \qquad k\geq 3.

The refined switching argument is given in § 6.

Preliminary

Let G=(VG,EG)G=(V_G, E_G) be a simple graph, and let NG(v)N_G(v) denote the neighbors of vv in GG. We may assume that GG has no isolated vertices, as deleting them does not change the intercomponent edges and only decreases nn. Define the kk-out sampling process as each node v∈VGv\in V_G independently and uniformly sampling kk edges with replacement. The random kk-out subgraph HGkH_G^{k} is formed by taking the union of these sampled edges. These subgraphs admit a natural coupling. Consider the process in which each vertex independently generates an infinite sequence of independent, uniformly random incident edges

σv=(ev,1, ev,2, … )\sigma_v = (e_{v,1},~ e_{v,2},~ \dots)

The union of the first kk choices at each vertex gives HGk(σ)H_G^k(\sigma), where σ:={σv}v∈VG\sigma:=\{\sigma_v\}_{v\in V_G}. Thus,

HG1(σ)⊆HG2(σ)⊆⋯⊆HGk(σ).H_G^1(\sigma)\subseteq H_G^2(\sigma) \subseteq \cdots \subseteq H_G^k(\sigma).

For fixed kk, the ordered outcomes consisting of the first kk choices, including repetitions, all have probability ∏vdeg⁡G(v)−k\prod_{v} \deg_G(v)^{-k}. Our counting arguments use these finite prefixes as outcome, and containment is defined by prefixes. Outcome Hk(σ)H^k(\sigma) contains outcome Hj(σ′)H^j(\sigma') if σv′\sigma'_v restricted to the first jj positions is a prefix of σv\sigma_v for all v∈VGv\in V_G.

Let CC(G)\mathcal{CC}(G) be the set of connected components of GG and cc(G)=∣CC(G)∣\mathrm{cc}(G)=|\mathcal{CC}(G)|. For a spanning subgraph H=(VG,EH)⊆GH=(V_G,E_H)\subseteq G, define its intercomponent edges by

XG(H):={uv∈EG∣u,v belong to different connected components of H}.X_G(H) :=\{uv\in E_G\mid u, v\text{ belong to different connected components of } H\}.

We also call these edges crossing and write xG(H)=∣XG(H)∣x_G(H)=|X_G(H)|. We omit the subscript GG when it is clear from context.

Comparison of the sampling models.

Let HwrkH^k_{\mathrm{wr}} and HworkH^k_{\mathrm{wor}} denote the random kk-out subgraphs obtained with and without replacement, respectively. Using the sequences σv\sigma_v above, take the first kk draws for HwrkH^k_{\mathrm{wr}} and the first min⁡{k,dG(v)}\min\{k,d_G(v)\} distinct edges for HworkH^k_{\mathrm{wor}}. All incident edges appear almost surely (with probability 11) in the infinite sequence, and their order of first appearance is a uniformly random permutation, so both marginals have the required distributions. Every edge among the first kk draws belongs to the without-replacement sample. Hence,

Hwrk⊆Hworkalmost surely.H^k_{\mathrm{wr}}\subseteq H^k_{\mathrm{wor}} \qquad\text{almost surely}.

Since having more edges can only merge components and reduce crossing edges, we have

xG(Hwork)≤xG(Hwrk)almost surely.x_G(H^k_{\mathrm{wor}}) \leq x_G(H^k_{\mathrm{wr}}) \qquad\text{almost surely}.

Hence, applying Theorem 1, we have E[xG(Hwork)]≤E[xG(Hwrk)]≤O(n/k)\mathop{\mathrm{\mathbb{E}}}[x_G(H^k_{\mathrm{wor}})] \leq \mathop{\mathrm{\mathbb{E}}}[x_G(H^k_{\mathrm{wr}})] \leq O(n/k), for all k≥2k\geq 2.

Bridge Potential to Component Potential

We define two potentials for spanning subgraphs of GG that are important in our main arguments.

Bridge Potential

An edge is a bridge if deleting it increases the number of connected components. Let Br⁡(G)\operatorname{Br}(G) denote the bridges of GG. For nested spanning subgraphs J⊆H⊆GJ\subseteq H\subseteq G, define the bridge potential by

ΨG(J,H):=∑e=uv∈Br⁡(H)∖E(J)min⁡{deg⁡G(u),deg⁡G(v)}\Psi_G(J,H):= \sum_{e=uv\in \operatorname{Br}(H)\setminus E(J)} \min\{\deg_G(u), \deg_G(v)\}

Each bridge’s weight is at most either endpoint’s degree. We can therefore charge it to either endpoint’s component, using that component’s maximum GG-degree.

Max-degree Component Potential

For any spanning subgraph J=(VG,EJ)J=(V_G, E_J) of GG, define its max-degree component potential Φ\Phi by

ΦG(J):=∑C∈CC(J)max⁡v∈Cdeg⁡G(v)\Phi_G(J) := \sum_{C\in \mathcal{CC}(J)} \max_{v\in C} \deg_G(v)

which charges each component of JJ the largest degree of its vertices, with all degrees measured in the original graph GG. Merging components can only decrease ΦG\Phi_G. Hence, if J⊆HJ\subseteq H, ΦG(H)≤ΦG(J)\Phi_G(H)\leq \Phi_G(J). The following lemma relates the two potentials.

Lemma 2. Let GG be a simple graph and J⊆H⊆GJ \subseteq H\subseteq G be nested spanning subgraphs of GG. Then we have the following, ΨG(J,H)≤ΦG(J)\Psi_G(J,H) \leq \Phi_G(J)

Proof. Each edge ee in Br⁡(H)∖E(J)\operatorname{Br}(H)\setminus E(J) joins different connected components of JJ. Otherwise, deleting ee will not create new components in HH, contradicting the definition of a bridge. Contract all the edges in JJ, and the edges in Br⁡(H)∖E(J)\operatorname{Br}(H)\setminus E(J) form a forest joining CC(J)\mathcal{CC}(J). Root each tree arbitrarily and charge the cost min⁡{deg⁡G(u),deg⁡G(v)}\min\{\deg_G(u), \deg_G(v)\} of each edge to its child component. Since each component in CC(J)\mathcal{CC}(J) has at most one parent, the following is true.

∑e=uv∈Br⁡(H)∖E(J)min⁡{deg⁡G(u),deg⁡G(v)}≤∑e=uv∈Br⁡(H)∖E(J)u is in the child component Cdeg⁡G(u)≤∑C∈CC(J)max⁡u∈Cdeg⁡G(u)=ΦG(J)\begin{align*} \sum_{e=uv\in \operatorname{Br}(H)\setminus E(J)} \min\{\deg_G(u), \deg_G(v)\} &\leq \sum_{\substack{e=uv\in \operatorname{Br}(H)\setminus E(J)\\ u\text{ is in the child component } C}} \deg_G(u)\\ &\leq \sum_{C \in \mathcal{CC}(J) } \max_{u\in C} \deg_G(u)\\ &= \Phi_G(J) \end{align*}

◻

Warm-up: Switching Lemma and the near-regular case

In this section, we demonstrate how to reduce the problem of bounding intercomponent edges to bounding the sum of the maximum degrees of each component via a key technique we call the switching argument. We prove the following lemma in this section.

Lemma 3. For any integer k≥1k\geq 1, and integer 0≤r≤k0\leq r\leq k, the following is true. For any graph GG, (k−r)E[xG(HGk)]≤E[ΦG(HGr)](k-r) \mathop{\mathrm{\mathbb{E}}}[x_G(H^{k}_G)]\leq \mathop{\mathrm{\mathbb{E}}}[\Phi_G(H^{r}_G)]

By applying the following lemma on the number of components of 22-out subgraphs due to Ghaffari, Nowicki, and Thorup [2], we can get a simple corollary bounding the number of intercomponent edges in nearly regular graphs.

Lemma 4 ([2]). For any nn-node graph GG with minimum degree δ\delta, E[cc(H2)]=O(n/δ)\mathop{\mathrm{\mathbb{E}}}[\mathrm{cc}(H^{2})] = O(n/\delta)

Corollary 5. For any nn-node graph GG with minimum degree δ\delta and maximum degree Δ\Delta, and any k≥3k\geq 3, E[x(Hk)]=O(Δδ⋅nk)\mathop{\mathrm{\mathbb{E}}}[x(H^k)] = O\left(\frac{\Delta}{\delta} \cdot \frac{n}{k}\right)

Proof. Apply Lemma 3 with r=2r=2, we have (k−2)E[x(Hk)]≤E[Φ(H2)](k-2) \mathop{\mathrm{\mathbb{E}}}[x(H^{k})]\leq \mathop{\mathrm{\mathbb{E}}}[\Phi(H^{2})]. Since E[Φ(H2)]=E[∑C∈CC(H2)max⁡v∈Cdeg⁡G(v)]≤Δ⋅E[cc(H2)]\mathop{\mathrm{\mathbb{E}}}[\Phi(H^{2})] = \mathop{\mathrm{\mathbb{E}}}\left[\sum_{C\in \mathcal{CC}(H^2)} \max_{v\in C} \deg_G(v)\right] \leq \Delta\cdot\mathop{\mathrm{\mathbb{E}}}[\mathrm{cc}(H^2)], applying Lemma 4 gives us our conclusion. ◻

The above corollary can be extended to k≥2k \geq 2 by extending the switching argument to real numbers rr, by selecting the final choice with probability r−⌊r⌋r-\lfloor r\rfloor, and complementing it with a generalization of Lemma 4 to random subgraphs that select the second edge with non-zero probability. However, for simplicity of warm-up, we will not expand on the details.

Motivated by applications, we now move on to the proof of Lemma 3, where we use our key technique, the switching argument, a form of careful double counting.

Proof of Lemma 3. Consider any edge e=(u,v)∈EGe=(u, v)\in E_G. Write Ee\mathcal{E}_e for the event that ee is crossing in HkH^k. Define vev_e to be the designated vertex of ee with smaller degree in GG, breaking ties over a globally fixed vertex order.

Now we define a switching process at the designated vertex vev_e of a crossing edge ee in HkH^k. For a fixed 0≤r≤k0\leq r\leq k, we consider the first rr samples in HkH^k the protected positions and the other samples in HkH^k that are not in HrH^r the switchable positions. The number of switchable positions is (k−r)(k-r).

For a fixed HrH^r outcome Hr(σ)H^r(\sigma), we define the following forward switch operation that maps an HkH^k outcome containing Hr(σ)H^r(\sigma) to another HkH^k outcome containing Hr(σ)H^r(\sigma). For each switchable position ff of vev_e, swap the position of ee and ff in the sequence of vev_e. This will make the edge ee a bridge selected exactly once in the resulting graph. The expected number of forward switches is (k−r)E[x(Hk)](k-r)\mathop{\mathrm{\mathbb{E}}}[x(H^k)].

[proof picture,x=1cm,y=1cm]
  \node[proof heading] at (3.5,2.15) {(a) Source $\sigma$: $e$ is crossing};
  \node[proof heading] at (11.35,2.15) {(b) Image $\sigma'$: $e$ is a bridge};
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    \draw[proof region] (0,-.9) rectangle (2.65,1.65);
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    \node[proof vertex] (w) at (.65,-.25) {$w$};
    \node[proof dot] (a) at (.85,1) {};
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    \node[proof dot] (b) at (6.25,1) {};
    \node[proof dot] (c) at (6.25,-.3) {};
    \draw[proof edge] (v)--(a)--(w) (u)--(b)--(c)--(u);
    \draw[proof choice] (v)--node[below,proof tag] {$f$}(w);
    \draw[proof crossing] (v)--node[above,proof tag] {$e$}(u);
    \node[proof note] at (3.5,-1.35)
      {switch at the smaller-degree endpoint $v_e$};
  \end{scope}
  \begin{scope}[xshift=7.85cm]
    \draw[proof region] (0,-.9) rectangle (7,1.65);
    \node[proof vertex] (v2) at (2,.2) {$v_e$};
    \node[proof vertex] (w2) at (.65,-.25) {$w$};
    \node[proof dot] (a2) at (.85,1) {};
    \node[proof vertex] (u2) at (4.95,.2) {$u$};
    \node[proof dot] (b2) at (6.25,1) {};
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    \draw[proof edge] (v2)--(a2)--(w2) (u2)--(b2)--(c2)--(u2);
    \draw[proof removed] (v2)--node[below,proof tag] {$f$}(w2);
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    \node[proof note] at (3.5,-1.35)
      {one selecting position; $e\notin H^r$};
  \end{scope}

Figure 1. Solid blue edges are protected positions. The dashed gold edge is crossing, hence not sampled initially. The green arrow is its unique selecting occurrence in the image.

On the inverse side, consider a sampling outcome of Hk(σ)H^k(\sigma) and count the number of incoming switches that it can receive. Two conditions have to be met. First, the outcome needs to contain the fixed Hr(σ)H^r(\sigma) outcome, with the same protected choices in the same order. Second, there must exist a bridge edge e∈Br⁡(HGk(σ))∖E(Hr(σ))e \in \operatorname{Br}\left(H_G^k(\sigma)\right) \setminus E(H^r(\sigma)), selected only by its designated vev_e, and selected exactly once.

For a sequence outcome σ\sigma, define Bkr(σ)⊆Br⁡(HGk(σ))∖E(HGr(σ))B_k^r(\sigma)\subseteq \operatorname{Br}\left(H_G^k(\sigma)\right) \setminus E(H_G^r(\sigma)) as the set of valid bridges satisfying the above conditions. The unique occurrence of ee determines the position to restore. There are at most deg⁡G(ve)\deg_G(v_e) possible previous choices that can make e∈Bkr(σ)e\in B_k^r(\sigma) a crossing edge, which is the required condition in the pre-image. Since the ordered kk-choice sequences before and after switching have the same probability, we have,

(k−r)Eσ[x(Hk)]≤Eσ[∑e∈Bkr(σ)deg⁡G(ve)]≤Eσ[Φ(Hr)](k-r)\mathop{\mathrm{\mathbb{E}}}_{\sigma}\left[x(H^k)\right] \leq \mathop{\mathrm{\mathbb{E}}}_{\sigma }\left[\sum_{e \in B_k^r(\sigma) } \deg_G\big(v_e\big)\right] \leq \mathop{\mathrm{\mathbb{E}}}_{\sigma}[\Phi(H^r)]

The last inequality follows from Lemma 2, since Bkr(σ)⊆Br⁡(HGk(σ))∖E(HGr(σ))B_k^r(\sigma)\subseteq \operatorname{Br}\left(H_G^k(\sigma)\right) \setminus E(H_G^r(\sigma)) by definition, and deg⁡G(ve)=min⁡u∈edeg⁡G(u)\deg_G(v_e) = \min_{u\in e}\deg_G(u) by the choice of the designated switching vertex. ◻

Bounding the Component Potential and the Consequence for k≥4k\geq 4

In § 4, we show how the switching argument relates the number of crossing edges xG(Hk)x_G(H^k) to the max-degree component potential ΦG(Hr)\Phi_G(H^r). In this section, we prove the kk-out conjecture for every k≥4k\ge 4 by providing a linear bound on E[Φ(H3)]\mathop{\mathrm{\mathbb{E}}}[\Phi(H^{3})] improving on the crude bound Δ⋅cc(H)\Delta\cdot\mathrm{cc}(H).

Fix an order on the vertices by increasing degrees in GG, breaking ties arbitrarily. We call the largest vertex of a component, which is also the one with the largest original degree, its leader. Denote αC\alpha_C as the leader of a component CC. Then the component potential ΦG(H)\Phi_G(H) is the sum of the degrees of the leaders of each connected component. That is

ΦG(H):=∑C∈CC(H)max⁡v∈Cdeg⁡G(v)=∑C∈CC(H)deg⁡G(αC)\Phi_G(H):= \sum_{C\in \mathcal{CC}(H)} \max_{v\in C} \deg_G(v) = \sum_{C\in \mathcal{CC}(H)} \deg_G(\alpha_C)

We call a component higher if its leader is higher and lower if its leader is lower.

Augmented 22-out Graph

To show a potential bound, our idea is to utilize an additional choice at some of the vertices to merge components together, effectively removing one previous leader from the sum toward the potential. With this idea in mind, we define the augmented 22-out graph H2+H^{2+} in the following way. First, we sample H2H^2. Then for the leader of each connected component in H2H^2, we sample a third choice. If this edge connects to another component with a higher leader, we add that edge to H2+H^{2+}. The sampling coupling applies. For the same sequence σ\sigma,

HG2(σ)⊆HG2+(σ)⊆HG3(σ).H_G^2(\sigma)\subseteq H_G^{2+}(\sigma)\subseteq H_G^3(\sigma).

Lemma 6. For any simple graph GG, E[ΦG(HG2+)]≤2n\mathop{\mathrm{\mathbb{E}}}\left[\Phi_G\left(H_G^{2+}\right)\right] \leq 2n

With this lemma, we can show our desired bound for k≥4k\geq 4 quite simply.

Corollary 7. Given any simple graph GG, and k≥4k\geq 4 E[x(HGk)]≤O(n/k).\mathop{\mathrm{\mathbb{E}}}[x(H_G^k)]\leq O(n/k).

Proof. Since HG2+(σ)⊆HG3(σ)H_G^{2+}(\sigma)\subseteq H_G^3(\sigma), and adding more edges can only merge components and reduce potentials, ΦG(HG2+(σ))≥ΦG(HG3(σ))\Phi_G\left(H_G^{2+}(\sigma)\right)\geq \Phi_G\left(H_G^{3}(\sigma)\right). Applying Lemma 3 with r=3r=3 and Lemma 6, we get

E[x(Hk)]≤E[Φ(H3)]k−3≤E[Φ(H2+)]k−3≤2nk−3≤O(nk).\mathop{\mathrm{\mathbb{E}}}[x(H^{k})] \leq \frac{\mathop{\mathrm{\mathbb{E}}}[\Phi(H^{3})]}{k-3} \leq \frac{\mathop{\mathrm{\mathbb{E}}}[\Phi(H^{2+})]}{k-3} \leq \frac{2n}{k-3} \leq O\left(\frac{n}{k}\right).

◻

Proof of Lemma 6: Bounded Potential on Augmented 22-out graph

The graph H2+H^{2+} is built upon H2H^2 by adding a third randomly sampled edge from a leader of H2H^2 if that edge connects to a higher component. Hence, if a leader of H2H^2 fails to merge to a higher component, it remains a leader of H2+H^{2+} and contributes to both Φ(H2+)\Phi(H^{2+}) and Φ(H2)\Phi(H^2). On the other hand, if a leader of H2H^2 merges with a higher component due to its third choice, it contributes 00 to Φ(H2+)\Phi(H^{2+}). The leaders of H2+H^{2+} is a subset the leaders of H2H^2, so no other vertices can contribute to Φ(H2+)\Phi(H^{2+}).

Conditioned on the first two choices of the sequences at each vertex, we can calculate the expected contribution of the leaders in H2H^2 to Φ(H2+)\Phi(H^{2+}) by bounding the probability of them staying as a leader.

Suppose αC\alpha_C is the leader for a connected component CC in H2H^2. Let dC≤deg⁡G(αC)d_C\leq \deg_G(\alpha_C) denote the number of bad neighbors of αC\alpha_C whose connection fails to merge CC into a higher component. This includes exactly neighbors inside CC and neighbors in lower components. The probability that αC\alpha_C remains a leader in H2+H^{2+} is

Pr⁡[αC is a leader in H2+∣H2]=dCdeg⁡G(αC)\Pr\left[\alpha_C \text{ is a leader in }H^{2+} \mid H^2\right] = \frac{d_C}{\deg_G(\alpha_C)}

If αC\alpha_C remains a leader, it contributes deg⁡G(αC)\deg_G(\alpha_C) to Φ(H2+)\Phi(H^{2+}). Hence, its expected contribution in this case is deg⁡G(αC)⋅dCdeg⁡G(αC)=dC\deg_G(\alpha_C) \cdot \frac{d_C}{\deg_G(\alpha_C)}= d_C. Hence,

E[Φ(H2+)∣H2]=∑C∈CC(H2)deg⁡G(αC)⋅Pr⁡[αC is a leader in H2+∣H2]≤∑C∈CC(H2)dC\mathop{\mathrm{\mathbb{E}}}[\Phi(H^{2+})\mid H^2] = \sum_{C\in \mathcal{CC}(H^2)} \deg_G(\alpha_C) \cdot \Pr\left[\alpha_C \text{ is a leader in }H^{2+} \mid H^2\right]\leq \sum_{C\in \mathcal{CC}(H^2)} d_C

Therefore, it is sufficient to show that the sum of the bad neighbors of the leaders in H2H^2 is small. For each leader αC\alpha_C, we decompose the bad neighbors into internal neighbors in CC, NG(αC)∩CN_G(\alpha_C)\cap C, and external neighbors connecting to lower components in H2H^2, denoted by dC↓d_C^\downarrow. Since the components are disjoint, ∑C∈CC(H2)∣NG(αC)∩C∣≤∑C∈CC(H2)∣C∣≤n\sum_{C\in \mathcal{CC}(H^2)} |N_G(\alpha_C)\cap C|\leq \sum_{C\in \mathcal{CC}(H^2)} |C| \leq n. Then, we can write

∑C∈CC(H2)dC=∑C∈CC(H2)(∣NG(αC)∩C∣ +dC↓)=∑C∈CC(H2)∣NG(αC)∩C∣+∑C∈CC(H2)dC↓≤n+LG(H2)\begin{align*} \sum_{C\in \mathcal{CC}(H^2)} d_C &= \sum_{C\in \mathcal{CC}(H^2)} \left( |N_G(\alpha_C)\cap C| ~+ d_C^\downarrow\right) \\ &= \sum_{C\in \mathcal{CC}(H^2)} |N_G(\alpha_C)\cap C| + \sum_{C\in \mathcal{CC}(H^2)} d_C^\downarrow \\ &\leq n+ L_G(H^2) \end{align*}

where LG(H):=∑C∈CC(H)dC↓L_G(H) := \sum_{C\in \mathcal{CC}(H)} d_C^\downarrow is the sum of the leaders’ external neighbors in lower components. Bounding the expectation of LG(H)L_G(H) allows us to bound the expected potential of H2+H^{2+}.

E[Φ(H2+)]=E[E[Φ(H2+)∣H2]]≤E[n+L(H2)]=n+E[L(H2)](*)\mathop{\mathrm{\mathbb{E}}}[\Phi(H^{2+})] = \mathop{\mathrm{\mathbb{E}}}\left[\mathop{\mathrm{\mathbb{E}}}[\Phi(H^{2+})\mid H^2]\right] \leq \mathop{\mathrm{\mathbb{E}}}[n+ L(H^2)] = n+ \mathop{\mathrm{\mathbb{E}}}[L(H^2)] \tag{*}

Leaders’ external neighbors in lower components

Now we bound L(H2)L(H^2), the total number of external neighbors in lower components for leaders in H2H^2. We consider all crossing edges to be owned by the endpoint in the higher component. Fix a spanning subgraph HH of GG, for each vertex vv, define the depth of vv in HH as its distance to the leader of its component. For any subgraph HH of GG, define xGj(H)x^j_G(H) as the number of crossing edges whose owner is at depth jj of its component. Note that xG0(H)=LG(H)x^0_G(H) = L_G(H). We will show the following relationship between the expected number of crossing edges at adjacent depths via a version of switching argument.

Lemma 8. For any graph GG, and any integer j≥0j\geq 0, 2E[xGj(H2)]≤n+E[xGj+1(H2)]2\mathop{\mathrm{\mathbb{E}}}[x_G^j(H^2)]\leq n + \mathop{\mathrm{\mathbb{E}}}[x_G^{j+1}(H^2)]

Proof. Fix an outcome H2(σ)H^2(\sigma). Define a forward switching operation on the edges counted towards xGj(H2(σ))x_G^j\left(H^2(\sigma)\right). Suppose uvuv is a crossing edge in H2(σ)H^2(\sigma) with uu in the higher component. Suppose uu has depth jj. We perform a switching operation that replaces each of the two existing choices of vv (not the owner) with uvuv. Hence, for each outcome H2(σ)H^2(\sigma), there are 2xGj(H2(σ))2 x_G^j\left(H^2(\sigma)\right) forward switches.

Now consider the inverse switches. Fix an outcome H2(σ)H^2(\sigma) and consider how many forward switches it can receive. A few conditions have to be met. First, after switching, uvuv is a bridge selected only by vv, and selected only once, as it was a crossing edge before the switch. Secondly, vv is at depth j+1j+1. This is because vv was in a lower component before switching, so the leader of uu’s component will remain the leader in the merged component. Since uvuv is a bridge, every path from vv to the leader passes through uvuv. Hence, the depth of vv is exactly one more than the depth of uu.

To count the number of possible pre-images, we need to analyze which neighbors of vv could be a neighbor ww before the switch. Let CC be the component of H2(σ)H^2(\sigma) containing uvuv. We call vv a bridge child and the connected component of vv after the deletion of uvuv, vv’s internal component, denoted by CvC_v. There are two possible cases from which the forward switches could come, illustrated in Figure 2..

  1. The GG-neighbors of vv in vv’s internal component CvC_v. There are at most ∣NG(v)∩Cv∣|N_G(v)\cap C_v| such neighbors. Since uvuv is a bridge, all the internal components of the bridge children at depth j+1j+1 are disjoint. Hence, the total contribution here is at most

    ∑v∣NG(v)∩Cv∣≤∑v∣Cv∣≤n.\sum_v |N_G(v)\cap C_v|\leq \sum_v |C_v|\leq n.
  2. The GG-neighbors of vv that are in another component of H2(σ)H^2(\sigma) lower than CC. Every such edge vwvw is a crossing edge in H2(σ)H^2(\sigma) owned by vv whose depth is j+1j+1. Each such edge gives at most one inverse, since the bridge and its selecting position are uniquely determined by vv. Hence, there are at most xGj+1(H2(σ))x_G^{j+1}\left(H^2(\sigma)\right) such neighbors.

No other sources are possible. Firstly, ww cannot be from C∖CvC\setminus C_v. Otherwise, restoring vwvw after deleting uvuv would reconnect uu and vv, so uvuv would not be a crossing edge in the pre-image. Secondly, ww cannot belong to a component higher than CC. Otherwise, when uvuv is deleted and vwvw is restored, vv lives in a higher component than uu, and vv will become the owner of uvuv, contradicting the definition of the switching operation. Thus, a fixed outcome H2(σ)H^2(\sigma) receives at most

n+xGj+1(H2(σ))n+x_G^{j+1}(H^2(\sigma))

forward switches. All ordered two-choice sequences are equally likely. Summing over these sequences and dividing by the total number, we have

2E[xGj(H2)]≤n+E[xGj+1(H2)].2\mathop{\mathrm{\mathbb{E}}}[x_G^j(H^2)]\leq n + \mathop{\mathrm{\mathbb{E}}}[x_G^{j+1}(H^2)].

◻

[proof picture,x=1cm,y=1cm]
  \node[proof heading] at (3.45,3.3) {(a) Forward: switch at the lower endpoint};
  \node[proof heading] at (11.25,3.3) {(b) Image: the two inverse cases};
  \begin{scope}
    \draw[proof region] (0,-.35) rectangle (2.25,2.8);
    \draw[proof lower] (4,-.35) rectangle (6.9,2.8);
    \node[proof leader] (a) at (1.1,2.1) {$\alpha_C$};
    \node[proof vertex] (u) at (1.1,.55) {$u$};
    \node[proof vertex] (v) at (4.6,.55) {$v$};
    \node[proof vertex] (w) at (6.2,1.9) {$w$};
    \node[proof dot] (t) at (6.2,.2) {};
    \draw[proof path] (a)--node[right,proof note] {$j$}(u);
    \draw[proof edge] (v)--(t)--(w);
    \draw[proof choice] (v)--node[left,proof tag] {$vw$}(w);
    \draw[proof crossing] (u)--node[above,proof tag] {$uv$}(v);
    \node[proof note] at (1.1,-.75) {higher component};
    \node[proof note] at (5.45,-.75) {lower component};
    \node[proof note,text width=6.9cm] at (3.45,-1.45)
      {Replace either retained choice at the lower endpoint $v$.};
  \end{scope}
  \begin{scope}[xshift=7.8cm]
    \draw[proof region] (0,1) rectangle (6.9,2.8);
    \draw[proof child] (3.65,1.1) rectangle (6.8,2.7);
    \draw[proof lower] (3.65,-1) rectangle (6.8,.45);
    \node[proof leader] (a2) at (.65,2) {$\alpha_C$};
    \node[proof vertex] (u2) at (2.25,2) {$u$};
    \node[proof vertex] (v2) at (4.25,2) {$v$};
    \node[proof vertex] (w2) at (6.05,2) {$w_{\rm in}$};
    \node[proof vertex] (z2) at (6.15,-.3) {$w_{\rm out}$};
    \draw[proof path] (a2)--node[above,proof note] {$j$}(u2);
    \draw[proof bridge] (v2)--node[above,proof tag] {$uv$}(u2);
    \draw[proof path] (v2) to[bend left=30] (w2);
    \draw[proof removed] (v2)--(w2);
    \draw[proof crossing] (v2) -- (4.25,.7)
      -- node[below,proof tag] {$vw$} (6.15,.7) -- (z2);
    \node[proof note] at (5.6,1.35) {child side $C_v$};
    \node[proof note,anchor=west,text width=1.7cm] at (3.75,-.3) {lower\\component};
    \node[proof note] at (4.25,2.5) {$j+1$};
    \node[proof note,text width=6.9cm] at (3.45,-1.65)
      {Restore $vw$ either inside $C_v$, or to a lower component.
       In the second case, $v$ owns the residual edge.};
  \end{scope}

Figure 2. Dotted blue lines are retained paths. The green arrow is the inserted choice; dashed rose and gold edges are alternative previous choices.

Lemma 9. For any graph GG, E[LG(H2)]≤n\mathop{\mathrm{\mathbb{E}}}[L_G(H^2)]\leq n.

Proof. Apply Lemma 8, we have E[xGj(H2)]≤n2+E[xGj+1(H2)]2\mathop{\mathrm{\mathbb{E}}}[x_G^j(H^2)]\leq \frac{n}{2} + \frac{\mathop{\mathrm{\mathbb{E}}}[x_G^{j+1}(H^2)]}{2} for every integer j≥0j\geq 0. Hence, by a geometric sum, we have

E[LG(H2)]=E[xG0(H2)]≤n2+E[xG1(H2)]2≤n2+n/2+E[xG2(H2)]/22≤n2+n4+⋯≤n.\begin{align*} \mathop{\mathrm{\mathbb{E}}}[L_G(H^2)] =\mathop{\mathrm{\mathbb{E}}}[x_G^0(H^2)] &\leq \frac{n}{2} + \frac{\mathop{\mathrm{\mathbb{E}}}[x_G^{1}(H^2)]}{2}\\ &\leq \frac{n}{2} + \frac{n/2 + \mathop{\mathrm{\mathbb{E}}}[x_G^{2}(H^2)]/2}{2}\\ &\leq \frac{n}{2}+ \frac{n}{4} + \cdots \\ &\leq n. \end{align*}

◻

With all ingredients needed, we can finally prove Lemma 6.

Proof of Lemma 6. Applying Lemma 9 and (*), we can conclude

E[Φ(H2+)]≤n+E[L(H2)]≤n+n=2n\mathop{\mathrm{\mathbb{E}}}[\Phi(H^{2+})] \leq n+ \mathop{\mathrm{\mathbb{E}}}[L(H^2)] \leq n + n = 2n

◻

Refined Switching to the Augmented 22-out Graph and the k≥3k\geq 3 case

In this section, we will prove a lemma analogous to Lemma 3. If we can relate the crossing edge of HkH^k to the augmented 22-out graph H2+H^{2+}, then applying Lemma 6, we can arrive at the following version of Theorem 1 for k≥3k\geq 3.

Theorem 10. Let GG be a simple graph on nn vertices. For every integer k≥3k\geq 3, E[xG(HGk)]≤6nk−2=O(n/k).\mathop{\mathrm{\mathbb{E}}}\left[x_G(H_G^k)\right] \leq \frac{6n}{k-2}=O(n/k).

Lemma 11. For any graph GG, and any integer k≥3k\geq 3, (k−2)E[x(Hk)]≤3E[Φ(H2+)].(k-2)\mathop{\mathrm{\mathbb{E}}}[x(H^{k})]\leq 3 \mathop{\mathrm{\mathbb{E}}}[\Phi(H^{2+})].

Proof. In this proof, we follow the convention of Lemma 3 that the designated vertex vev_e of an edge ee is the one with smaller degree.

Similar to the strategy in Lemma 3, we define some switching/rotating operation that maps one outcome of HkH^k to another with the following property that allows us to charge the number of switches to the bridges of HkH^{k} that are not in H2+H^{2+} using Lemma 2. The switches should be defined such that in the image Hk(σ′)H^k(\sigma'), ee is a bridge and ee does not belong to H2+(σ′)H^{2+}(\sigma').

Fix a sequence σ\sigma and an outcome Hk(σ)H^k(\sigma). We define a switching operation for all crossing edges ee to the 33-rd to the kk-th position in σve\sigma_{v_e}. There are three cases for the 33-rd position. Let (f1,f2,f3)(f_1, f_2, f_3) be the first three choices in order at vev_e.

  1. If vev_e is a leader in H2+(σ)H^{2+}(\sigma), we do not perform a third-position switch.

  2. If vev_e is a leader in H2(σ)H^2(\sigma) and not a leader in H2+(σ)H^{2+}(\sigma), then we perform a rotation to σve′=(f3,f2,e,… )\sigma'_{v_e} = (f_3, f_2, e, \dots). We consider rotation a special type of switch.

  3. If vev_e is not a leader in H2(σ)H^2(\sigma), then we perform the normal switch, replace the third choice with ee, and let σve′=(f1,f2,e,… )\sigma'_{v_e} = (f_1, f_2, e, \dots).

We perform the normal switches to the 44-th to the kk-th positions. The rotation is illustrated in Figure 3.

[proof picture,x=1cm,y=1cm]
  \node[proof heading] at (3.5,3) {(a) Before: $v_e$ leads an $H^2$-component};
  \node[proof heading] at (11.35,3) {(b) After: $v_e$ is no longer a leader};
  \begin{scope}
    \draw[proof region] (0,-1.6) rectangle (3.05,2.5);
    \draw[proof child] (4.55,.65) rectangle (7,2.5);
    \draw[proof lower] (4.55,-1.6) rectangle (7,.05);
    \node[proof leader] (v) at (2.3,.55) {$v_e$};
    \node[proof vertex] (w) at (.65,1.65) {$w$};
    \node[proof dot] (t) at (.65,-.65) {};
    \node[proof leader] (a) at (5.8,1.6) {$\alpha_D$};
    \node[proof vertex] (u) at (5.8,-.75) {$u$};
    \draw[proof edge] (w)--(t);
    \draw[proof choice] (v)--node[above left,proof tag] {$f_1$}(w);
    \draw[proof edge] (v)--node[below left,proof tag] {$f_2$}(t);
    \draw[proof bridge] (v)--node[above,sloped,proof tag] {$f_3$}(a);
    \draw[proof crossing] (v)--node[below,sloped,proof tag] {$e$}(u);
    \node[proof note] at (5.8,1) {$\alpha_D>v_e$};
    \node[proof note] at (1.5,-1.25) {$H^2$-component};
    \node[proof note,text width=7cm] at (3.5,-2.15)
      {$f_3$ already joins a higher component in $H^{2+}$;
       $e$ crosses the full $H^k$-components.};
  \end{scope}
  \begin{scope}[xshift=7.85cm]
    \draw[proof region] (0,-1.6) -- (3.05,-1.6) -- (3.05,.65)
      -- (7,.65) -- (7,2.5) -- (0,2.5) -- cycle;
    \draw[proof lower] (4.55,-1.6) rectangle (7,.05);
    \node[proof vertex] (v2) at (2.3,.55) {$v_e$};
    \node[proof vertex] (w2) at (.65,1.65) {$w$};
    \node[proof dot] (t2) at (.65,-.65) {};
    \node[proof leader] (a2) at (5.8,1.6) {$\alpha_D$};
    \node[proof vertex] (u2) at (5.8,-.75) {$u$};
    \draw[proof edge] (w2)--(t2);
    \draw[proof removed] (v2)--node[above left,proof tag] {$f_1$}(w2);
    \draw[proof edge] (v2)--node[below left,proof tag] {$f_2$}(t2);
    \draw[proof bridge] (v2)--node[above,sloped,proof tag] {$f_3$: first}(a2);
    \draw[proof choice] (v2)--node[below,sloped,proof tag] {$e$: third}(u2);
    \node[proof note] at (1.55,-1.25) {new $H^2$-component};
    \node[proof note,text width=7cm] at (3.5,-2.15)
      {$f_3$ connects $v_e$ to a higher leader already in $H^2$;
       its third choice $e$ is not added to $H^{2+}$.};
  \end{scope}

Figure 3. Shaded regions are H2H^2-components. The dashed rose edge is the old first choice; labels mark the new positions of f3f_3 and ee.

We count the number of forward switches. Each crossing edge ee whose designated vertex vev_e is a leader in H2+(σ)H^{2+}(\sigma) contributes to (k−3)(k-3) forward switches, while all other crossing edges contribute (k−2)(k-2) forward switches. Denote the number of crossing edges ee whose designated vertex vev_e is a leader in H2+(σ)H^{2+}(\sigma) by η(σ)\eta(\sigma). These are the edges that are omitted from switching. The number of forward switches from outcome sequence σ\sigma is

(k−3)⋅xG(Hk(σ))+[xG(Hk(σ))−η(σ)]=(k−2)⋅xG(Hk(σ))−η(σ).(k-3) \cdot x_G\left(H^k(\sigma)\right)+\left[x_G\left(H^k(\sigma)\right)-\eta(\sigma)\right] =(k-2)\cdot x_G\left(H^k(\sigma)\right)-\eta(\sigma).

Additionally, we have the following bound for η(σ)\eta(\sigma),

η(σ)≤∑v is a leader in H2+(σ)deg⁡G(v)=ΦG(H2+(σ)).\eta(\sigma) \leq \sum_{v \text{ is a leader in } H^{2+}(\sigma)} \deg_G(v) = \Phi_G\left(H^{2+}(\sigma)\right).

Now we check that in the image Hk(σ′)H^k(\sigma'), ee is a bridge and ee does not belong to H2+(σ′)H^{2+}(\sigma'). This is clear for the normal cases except when vev_e is a leader in H2(σ)H^2(\sigma) and not a leader in H2+(σ)H^{2+}(\sigma). In that case, we perform a rotation. The third choice f3f_3 connects vev_e to a higher component, and therefore belongs to H2+(σ)H^{2+}(\sigma). If we just replace f3f_3 with ee to make σ′′\sigma'', we might still connect vev_e to a higher component, making ee belong to H2+(σ′′)H^{2+}(\sigma''). The rotation makes sure that f3f_3 is selected in the first position, so that vev_e is connected to a higher component before the third choice, making vev_e a non-leader in H2(σ′)H^2(\sigma') and hence ee will not be in H2+(σ′)H^{2+}(\sigma').

Hence, the switches map all ee to Br⁡(Hk(σ′))∖H2+(σ′)\operatorname{Br}\left(H^k(\sigma')\right)\setminus H^{2+}(\sigma'), and we can apply Lemma 2. Switches to ee can come from 2(deg⁡G(ve)−1)2(\deg_G(v_e)-1) possibilities if ee is in the third position of σve′\sigma'_{v_e} and (deg⁡G(ve)−1)(\deg_G(v_e)-1) possibilities if ee is in positions 44 onward. Since each sequence σ\sigma has equal positive probability, summing over all sequences and dividing by the number of configurations gives

∑σ((k−2)⋅xG(Hk(σ))−η(σ))≤ ∑σ(∑e∈Br⁡(Hk(σ))∖E(H2+(σ))2(deg⁡G(ve)−1))(k−2)⋅Eσ[xG(Hk)]−Eσ[η]≤ Eσ[∑e∈Br⁡(Hk(σ))∖E(H2+(σ))2(deg⁡G(ve)−1)](k−2)⋅Eσ[xG(Hk)]≤ Eσ[η]+2Eσ[∑e∈Br⁡(Hk(σ))∖E(H2+(σ))min⁡v∈edeg⁡G(v)]≤ E[ΦG(H2+)]+2E[ΦG(H2+)]≤ 3E[ΦG(H2+)].\begin{align*} \sum_\sigma \Bigg( (k-2) \cdot x_G\left(H^k(\sigma)\right) -\eta(\sigma)\Bigg) &\leq ~ \sum_{\sigma}\left(\sum_{e \in \operatorname{Br}\left(H^k(\sigma)\right)\setminus E\left(H^{2+}(\sigma)\right)} 2(\deg_G(v_e)-1)\right)\\ (k-2) \cdot \mathop{\mathrm{\mathbb{E}}}_\sigma\left[x_G(H^k)\right] -\mathop{\mathrm{\mathbb{E}}}_\sigma[\eta] &\leq ~ \mathop{\mathrm{\mathbb{E}}}_{\sigma}\left[\sum_{e \in \operatorname{Br}\left(H^k(\sigma)\right)\setminus E\left(H^{2+}(\sigma)\right)} 2(\deg_G(v_e)-1)\right]\\ (k-2) \cdot \mathop{\mathrm{\mathbb{E}}}_\sigma\left[x_G(H^k)\right] &\leq ~ \mathop{\mathrm{\mathbb{E}}}_\sigma[\eta]+ 2 \mathop{\mathrm{\mathbb{E}}}_{\sigma}\left[\sum_{e \in \operatorname{Br}\left(H^k(\sigma)\right)\setminus E\left(H^{2+}(\sigma)\right)} \min_{v\in e} \deg_G(v) \right]\\ &\leq ~ \mathop{\mathrm{\mathbb{E}}}\left[\Phi_G(H^{2+})\right] + 2\mathop{\mathrm{\mathbb{E}}}\left[\Phi_G(H^{2+})\right]\\ &\leq ~ 3\mathop{\mathrm{\mathbb{E}}}\left[\Phi_G(H^{2+})\right]. \end{align*}

◻

Now we can prove Theorem 10.

Proof of Theorem 10. Apply Lemma 11, and Lemma 6, we get

\mathop{\mathrm{\mathbb{E}}}[x(H^{k})]\leq ~ \frac{3}{k-2} \mathop{\mathrm{\mathbb{E}}}[\Phi(H^{2+})]\leq ~ \frac{6n}{k-2} = O(n/k).\qedhere

◻

The 22-out case

In this section, we extend the result to 22 choices with replacement. We follow the same two steps as before: relate intercomponent edges to a component potential, and then bound the expected potential. For simplicity of presentation, we only show the argument for the 22-out case. For k≥3k\geq3, one can either use Theorem 10 to get the desired bound with a better constant or extend the following argument to fit general kk. The constants below are chosen to simplify the proof.

Theorem 12. For every simple graph GG on nn vertices, its random two-out subgraph with replacement satisfies E[xG(HG2)]≤500n.\mathop{\mathrm{\mathbb{E}}}\bigl[x_G({H_{G}^{2}})\bigr]\le500n.

Technical overview

The 33-out argument protects two choices and some third choices to make the potential small, leaving the other third choices available for switching. With only two choices, these two uses compete for the same second choice. Our idea is to restrict the switching argument to a constant fraction of the crossing edges in expectation. This loses only a constant factor in the reduction, but leads to a weaker potential in which each leader contributes only its relevant degree which counts only the neighbors used in switching.

Since we only need to switch from a random subset of vertices, we can protect two choices at the other vertices. The graph whose potential we need to bound is therefore an augmented 11-out graph that is already close to a 22-out graph. We can then adapt the earlier potential argument to bound this weaker potential in expectation by O(n)O(n).

Some technical details that are different from the 33-out case are as follows. We use a random proper coloring to define the vertices and edges that can perform switchings. As the result, we are no longer working with uniform distributions on the source and image. Hence, we need to correct the expectation calculation with the ratio of the probabilities.

Preliminary

Recall that for the 22-out graph, each vertex vv independently samples two incident edges σv=(ev,1,ev,2){\sigma}_v=(e_{v,1},e_{v,2}) uniformly at random, with replacement. Write si(v)s_{i}(v) for the other endpoint of the ii-th selected edge ev,ie_{v,i}.

The marked set and the base graph.

Conditional on HG1(σ){H_{G}^{1}}(\sigma), choose a uniformly random proper coloring with colors 0,1,…,90,1,\ldots,9. Let SS be its zero set, call its vertices marked, and put U=VG∖SU=V_G\setminus S. The set SS is an independent set in HG1(σ){H_{G}^{1}}(\sigma). Uniform randomness on coloring is crucial in our analysis, since this gives each vertex probability 1/101/10 of being marked and provides the probability estimates needed after switching. An arbitrary proper coloring need not satisfy these estimates. The choice of ten is arbitrary, and any constant larger than 3 would suffice. We only need the information about the SS, so an outcome is the pair (σ,S)({\sigma},S).

We consider the base graph B(σ,S){B}({\sigma},S) that keeps the first position at every vertex and the second position of every vertex in UU. Thus, all second choices at SS are omitted from B{B}. Among these omitted second choices, we reserve those leaving SS for possible use in the augmentation. Denote the set of vertices with reserved choices by

R:={v∈S:s2(v)∈U}.\mathcal R:=\{v\in S:s_{2}(v)\in U\}.
Leaders.

Define the relevant degrees by

dv:={∣NG(v)∩U∣,v∈R,0,v∉R.d_{v}:=\begin{cases} |N_G(v)\cap U|,&v\in\mathcal R,\\ 0,&v\notin\mathcal R. \end{cases}

We assign degree zero to vertices outside R\mathcal R, since they have no reserved choice available for augmentation. Fix a total order π\pi on the vertices, chosen independently of the sample and the coloring. Let ≺\prec be the order that sorts vertices by increasing dvd_{v}, with ties broken according to π\pi. The leader αC\alpha_{C} of a component CC is its largest vertex under ≺\prec, and we compare components by their leaders similar to above.

Augmentation.

Building from B(σ,S){B}({\sigma},S), add the reserved second choice of each leader αC∈R\alpha_{C}\in\mathcal R if it reaches a higher component of B(σ,S){B}({\sigma},S). Call the resulting graph B+(σ,S){{B}^+}({\sigma},S). Then

HG1(σ)⊆B(σ,S)⊆B+(σ,S)⊆HG2(σ).\begin{equation} {H_{G}^{1}}({\sigma})\subseteq{B}({\sigma},S)\subseteq{{B}^+}({\sigma},S)\subseteq{H_{G}^{2}}({\sigma}). \tag*{(1)} \end{equation}
Component Potential.

We define the component potential using dvd_{v} instead of deg⁡G(v)\deg_G(v). Since dv>0d_{v}>0 exactly on R\mathcal R, and αC=arg⁡max⁡v∈Cdv\alpha_{C}=\arg\max_{v\in C} d_{v}, we have

ΦR(J):=∑C∈CC(J)max⁡v∈Cdv=∑C∈CC(J)αC∈RdαC.\begin{equation} \Phi_{\mathcal R}(J):=\sum_{C\in\mathcal{CC}(J)}\max_{v\in C}d_{v} \quad =\sum_{\substack{C\in\mathcal{CC}(J)\\\alpha_{C}\in\mathcal R}}d_{\alpha_{C}}. \tag*{(2)} \end{equation}

From intercomponent edges to component potential

We first prove the following analogue of Lemma 11 using switching.

Lemma 13. For any simple graph GG, 114E[xG(HG2)]≤24180E[ΦR(B+)].\frac1{14}\mathop{\mathrm{\mathbb{E}}}[x_G({H_{G}^{2}})] \leq \frac{241}{80}\mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})].

Proof. Similar to Lemma 11, we define switches that turn crossing edges into bridges outside the augmented graph. Fix (σ,S)({\sigma},S) and an edge e=uv∈XG(HG2)e=uv\in X_G({H_{G}^{2}}). For this procedure, order the full HG2{H_{G}^{2}}-components by their largest vertices under π\pi. Designate vv to be the endpoint in the lower component, and uu the endpoint in the higher component. This designation depends on σ{\sigma} but not on SS. This will allow us to charge the image bridge to the child component which contains the switching vertex, thereby ensuring each component is charged at most once.

The edge ee is eligible for switching when

v∈S,s2(v)∈U,u∈U.\begin{equation} v\in S,\qquad s_{2}(v)\in U,\qquad u\in U. \tag*{(3)} \end{equation}

Thus v∈Rv\in\mathcal R. For an eligible edge, write σv=(f1,f2){\sigma}_v=(f_1,f_2) and apply the following switching operation.

  1. If vv is a leader in B+(σ,S){{B}^+}({\sigma}, S), omit this edge from the switching count.

  2. If vv is a leader in B(σ,S){B}({\sigma}, S) but not in B+(σ,S){{B}^+}({\sigma}, S), perform the rotation σv′=(f2,e).{\sigma}'_v=(f_2,e). Since s2(v)∈Us_{2}(v)\in U, SS is still an independent set in HG1(σ′){H_{G}^{1}}({\sigma}').

  3. If vv is not a leader in B(σ,S){B}({\sigma}, S), perform the normal switch σv′=(f1,e).{\sigma}'_v=(f_1,e).

Leave all other choices and SS unchanged. As a result, the set R\mathcal R, all dvd_{v}, and the leader order ≺\prec are unchanged, since both the old and the new second choices at vv leave SS.

A constant fraction of crossing edges is eligible switching.

We first count eligible crossing edges. For distinct vertices x,yx,y, the random zero set SS satisfies

Pr⁡[x∈S∣HG1(σ)]=110,Pr⁡[y∈S∣x∈S,HG1(σ)]≤18.\begin{equation} \Pr[x\in S\mid{H_{G}^{1}}({\sigma})]=\frac{1}{10}, \qquad \Pr[y\in S\mid x\in S,{H_{G}^{1}}({\sigma})]\leq \frac{1}{8}. \tag*{(4)} \end{equation}

The first assertion follows from color symmetry, as permuting the colors of a proper coloring is also a proper coloring. If x,yx,y are in different components of HG1(σ){H_{G}^{1}}({\sigma}), their colors are independent, so the second probability is 1/101/10. Otherwise, x,yx,y are connected. Since each component of HG1(σ){H_{G}^{1}}({\sigma}) is either a tree or contains exactly one cycle, there are at most two distinct paths from xx to yy. Consider extending a proper coloring from xx to yy along the union of the paths from xx to yy. There are at least 10−2=810-2=8 colors for yy to choose from, and at most 11 of them is color 00. Hence, Pr⁡[y∈S∣x∈S,HG1(σ)]≤18\Pr[y\in S\mid x\in S,{H_{G}^{1}}({\sigma})]\leq \frac{1}{8}.

Now condition on the entire choice sequence σ{\sigma} and a crossing edge e=vue=vu designated at vv. Both s2(v)s_{2}(v) and uu differ from vv. Applying (4) and a union bound, its eligibility probability is

Pr⁡[e is eligible∣σ]=Pr⁡[v∈S∣σ] Pr⁡[u,w∉S∣v∈S,σ]=110(1−Pr⁡[u∈S or w∈S∣v∈S,σ])≥110(1−Pr⁡[u∈S∣v∈S,σ]−Pr⁡[w∈S∣v∈S,σ])≥110(1−18−18)=340≥114.\begin{align*} \Pr[e\text{ is eligible}\mid{\sigma}] &=\Pr[v\in S\mid{\sigma}]\, \Pr[u,w\notin S\mid v\in S,{\sigma}]\\ &=\frac1{10}\left( 1-\Pr[u\in S\text{ or }w\in S \mid v\in S,{\sigma}]\right)\\ &\geq\frac1{10}\left( 1-\Pr[u\in S\mid v\in S,{\sigma}] -\Pr[w\in S\mid v\in S,{\sigma}]\right)\\ &\geq\frac1{10}\left(1-\frac18-\frac18\right) =\frac3{40}\geq\frac1{14}. \end{align*}

Let YY count eligible crossing edges. Conditional on σ{\sigma}, the set of crossing edges is fixed. By linearity of expectation,

E[Y∣σ]=∑e∈XG(HG2)Pr⁡[e is eligible∣σ]≥114xG(HG2),E[Y]=E ⁣[E[Y∣σ]]≥114E[xG(HG2)]\begin{align} \mathop{\mathrm{\mathbb{E}}}[Y\mid{\sigma}] &=\sum_{e\in X_G({H_{G}^{2}})} \Pr[e\text{ is eligible}\mid{\sigma}] \geq\frac1{14}x_G({H_{G}^{2}}),\notag \\ \mathop{\mathrm{\mathbb{E}}}[Y] &=\mathop{\mathrm{\mathbb{E}}}\!\left[\mathop{\mathrm{\mathbb{E}}}[Y\mid{\sigma}]\right] \geq\frac1{14}\mathop{\mathrm{\mathbb{E}}}[x_G({H_{G}^{2}})] \tag*{(5)} \end{align}

Let η\eta count eligible edges omitted because their designated endpoint is an B+{{B}^+}-leader. For each component CC of B+{{B}^+}, any such edge designated at its leader αC\alpha_{C} goes to a neighbor in UU. There are at most dαCd_{\alpha_{C}} such edges. Summing these leader bounds over components gives

η≤∑C∈CC(B+)dαC=ΦR(B+).\begin{equation} \eta \leq\sum_{C\in\mathcal{CC}({{B}^+})} d_{\alpha_{C}} =\Phi_{\mathcal R}({{B}^+}). \tag*{(6)} \end{equation}

The number of performed forward switches is Y−ηY-\eta.

Number of inverse switches.

Fix an image (σ′,S)({\sigma}',S) and an edge ee that can receive a switch. Since ee must be a bridge in HG2(σ′){H_{G}^{2}}({\sigma}'), we call ee a receiving bridge. Each ee must be selected only once in σ′{\sigma}' by one of its endpoints vev_e, called the switching endpoint. Additionally, the switching endpoint vev_e that selects ee must select ee at the second position. If σve′=(f,e){\sigma}'_{v_e}=(f,e), the possible previous rows have the forms

(f,g)for a normal switch,(g,f)for a rotation.(f,g)\quad\hbox{for a normal switch}, \qquad (g,f)\quad\hbox{for a rotation}.

In either case, the other endpoint of gg must lie in UU, so there are at most dved_{v_e} candidates for each type. Repeated choices are allowed, so f=gf=g is possible. To use these counts in the expectation, we must account for the probabilities of the outcomes.

Correcting contributions to the expectation for inverse switches.

The choice sequences σ{\sigma} are equally likely, but the pairs (σ,S)({\sigma},S) need not be, as the conditional probability of obtaining the same zero set SS may be different for HG1{H_{G}^{1}} under different sequences σ{\sigma}. Suppose when we count forward switches from (σ,S)({\sigma}, S) to (σ′,S)({\sigma}', S). Each switch contributes Pr⁡[(σ,S)]\Pr[({\sigma}, S)] to the expected number of forward switches. Thus, an inverse switching should be counted with a correction factor equal to their ratios of probability.

E[#forward switching]=E[Y−η]=∑(σ,S)Pr⁡[(σ,S)]∑σ′,e:(σ,S,e)↦(σ′,S,e)1=∑(σ′,S)∑σ,e:(σ,S,e)↦(σ′,S,e)Pr⁡[(σ,S)]=∑(σ′,S)Pr⁡[(σ′,S)]∑σ,e:(σ,S,e)↦(σ′,S,e)Pr⁡[(σ,S)]Pr⁡[(σ′,S)]=∑(σ′,S)Pr⁡[(σ′,S)]∑σ,e:(σ,S,e)↦(σ′,S,e)Pr⁡[S∣σ]Pr⁡[S∣σ′].\begin{align*} \mathop{\mathrm{\mathbb{E}}}[\text{\#forward switching}]= \mathop{\mathrm{\mathbb{E}}}[Y-\eta] &=\sum_{({\sigma},S)} \Pr[({\sigma},S)] \sum_{\substack{{\sigma}', \, e:\\ ({\sigma},S,e)\mapsto({\sigma}',S,e)}}1\\ &=\sum_{({\sigma}',S)} \sum_{\substack{{\sigma}, \, e:\\ ({\sigma},S,e)\mapsto({\sigma}',S,e)}} \Pr[({\sigma},S)]\\ &=\sum_{({\sigma}',S)} \Pr[({\sigma}',S)] \sum_{\substack{{\sigma}, \, e:\\ ({\sigma},S, e)\mapsto({\sigma}',S,e)}} \frac{\Pr[({\sigma},S)]} {\Pr[({\sigma}',S)]}\\ &=\sum_{({\sigma}',S)} \Pr[({\sigma}',S)] \sum_{\substack{{\sigma}, \, e:\\ ({\sigma},S, e)\mapsto({\sigma}',S,e)}} \frac{\Pr[S\mid {\sigma}]} {\Pr[S\mid {\sigma}']}. \end{align*}

We bound this correction factor now. Let Zq(J)Z_{q}\left(J\right) be the number of proper qq-colorings of JJ. For an independent set SS in HG1{H_{G}^{1}}, fixing its zero class leaves nine colors on UU, so

Pr⁡[S∣HG1(σ)]=Z9(HG1(σ)[U])Z10(HG1(σ)).\begin{equation} \Pr[S\mid{H_{G}^{1}}({\sigma})] =\frac{Z_{9}\left({H_{G}^{1}}({\sigma})[U]\right)}{Z_{10}\left({H_{G}^{1}}({\sigma})\right)}. \tag*{(7)} \end{equation}

In the case of a normal switch, HG1(σ)=HG1(σ′){H_{G}^{1}}({\sigma})={H_{G}^{1}}({\sigma}') and this probability ratio is one.

In a rotation, the first choice at a vertex of SS is redirected to another vertex in UU. This keeps the coloring proper and the induced graph HG1[U]{H_{G}^{1}}[U] unchanged, i.e., HG1(σ)[U]=HG1(σ′)[U]{H_{G}^{1}}({\sigma})[U]={H_{G}^{1}}({\sigma}')[U], so only the denominator in (7) changes. Since HG1{H_{G}^{1}} can be viewed as a directed graph with exactly one directed cycle in each component, the following standard result counts its number of proper colorings. We defer the proof to § 8.

Lemma 14 (Coloring count for a first-choice graph). Let ℓ(Q)\ell(Q) denote the length of directed cycle QQ, where two opposite directed edges form a directed cycle of length two. For every integer q≥2q\ge 2, Zq(HG1(σ))=(q−1)n∏directed cycles Q in HG1(σ)(1+(−1)ℓ(Q)(q−1)ℓ(Q)−1).Z_{q}\left({H_{G}^{1}}({\sigma})\right) = (q-1)^{n} \prod_{\text{directed cycles }Q \text{ in } {H_{G}^{1}}({\sigma})} \left( 1+\frac{(-1)^{\ell(Q)}}{(q-1)^{\ell(Q)-1}} \right).

In a rotation, changing the first choice either preserves all directed cycles or destroys the cycle it was in. It creates no new cycle. This potentially removes one term in the product in Lemma 14. Therefore, for any rotation,

Pr⁡[S∣σ]Pr⁡[S∣σ′]=Z10(HG1(σ′))Z10(HG1(σ))≤1/(min⁡ℓ(1+(−1)ℓ9ℓ−1))≤8180.\begin{equation} \frac{\Pr[S\mid {\sigma}]}{\Pr[S\mid {\sigma}']} =\frac{Z_{10}\left({H_{G}^{1}}({\sigma}')\right)}{Z_{10}\left({H_{G}^{1}}({\sigma})\right)} \leq 1/\left(\min_{\ell} \left( 1+\frac{(-1)^{\ell}}{9^{\ell-1}} \right)\right) \leq \frac{81}{80}. \tag*{(8)} \end{equation}

Hence, each receiving bridge ee contributes at most dved_{v_e} from normal switches and at most 8180dve\frac{81}{80}d_{v_e} from rotations, for a total of at most 16180dve\frac{161}{80}d_{v_e}.

The receiving edge is a bridge of HG2(σ′){H_{G}^{2}}({\sigma}') outside B+(σ′,S){{B}^+}({\sigma}', S).

To use an argument analogous to Lemma 2, we need to show this property. We already know that each receiving edge ee is a bridge of HG2(σ′){H_{G}^{2}}({\sigma}') as it was a crossing edge in HG2(σ){H_{G}^{2}}({\sigma}).

Recall that B{B} only includes second choices from UU. Since ee is selected only in the second position of v∉Uv\notin U and not by any other vertices in σ′{\sigma}', we have e∉E(B(σ′,S))e\notin E({B}({\sigma}',S)). It remains to show that the augmentation does not add ee.

For a normal switch, only the second choice of vv changes, so E(B(σ′,S))=E(B(σ,S))E({B}({\sigma}',S)) = E({B}({\sigma},S)). Since vv is not a B(σ,S){B}({\sigma}, S)-leader, and the leader order did not change, vv is also not a B(σ′,S){B}({\sigma}', S)-leader. Hence, ee will also not be in B+(σ′,S){{B}^+}({\sigma}', S).

For a rotation, f2f_2 reaches a higher B(σ,S){B}({\sigma}, S)-component. Putting it at the first position connects vv to that higher leader even in B(σ′,S){B}({\sigma}', S), making vv not a B(σ′,S){B}({\sigma}', S)-leader. Hence, the new second choice ee is not added to B+(σ′,S){{B}^+}({\sigma}', S). Hence, e∈Br⁡(HG2(σ′))∖E(B+(σ′,S))e\in\operatorname{Br}\left({H_{G}^{2}}({\sigma}')\right)\setminus E\left({{B}^+}({\sigma}', S)\right).

Charging image bridges to components.

Contract every edge of HG2(σ′){H_{G}^{2}}({\sigma}') except those in Br⁡(HG2(σ′))∖E(B+(σ′,S))\operatorname{Br}\left({H_{G}^{2}}({\sigma}')\right)\setminus E\left({{B}^+}({\sigma}', S)\right). The resulting graph is a forest, where each contracted vertex comes from a union of connected components of B+(σ′,S){{B}^+}({\sigma}', S). Root each tree at the contracted vertex that contains the highest original vertex under π\pi. Every receiving bridge ee has its switching endpoint vev_e on the child-side of this tree, as the removal of ee makes the side with uu the side with a higher leader under π\pi. We can charge the contribution of each edge to the child component. Each component of B+(σ′,S){{B}^+}({\sigma}', S) receives at most one charge, since each contracted vertex receives at most one charge from its parent edge. It follows that

E[Y−η]=∑(σ′,S)Pr⁡[(σ′,S)]∑receiving edge e∑σ:(σ,S,e)↦(σ′,S,e)Pr⁡[S∣σ]Pr⁡[S∣σ′]≤∑(σ′,S)Pr⁡[(σ′,S)]∑receiving edge e16180dve≤16180∑(σ′,S)Pr⁡[(σ′,S)]∑C∈CC(B+(σ′,S))max⁡w∈Cdw=16180∑(σ′,S)Pr⁡[(σ′,S)]ΦR(B+(σ′,S))=16180E[ΦR(B+)].\begin{align*} \mathop{\mathrm{\mathbb{E}}}[Y-\eta] &=\sum_{({\sigma}',S)} \Pr[({\sigma}',S)] \sum_{\text{receiving edge } e} \sum_{\substack{{\sigma}:\\ ({\sigma},S,e)\mapsto({\sigma}',S,e)}} \frac{\Pr[S\mid{\sigma}]} {\Pr[S\mid{\sigma}']}\\ &\leq \sum_{({\sigma}',S)} \Pr[({\sigma}',S)] \sum_{\text{receiving edge } e} \frac{161}{80} d_{v_e}\\ &\leq \frac{161}{80} \sum_{({\sigma}',S)} \Pr[({\sigma}',S)] \sum_{C\in\mathcal{CC}\left({{B}^+}({\sigma}',S)\right)} \max_{w\in C}d_{w}\\ &=\frac{161}{80} \sum_{({\sigma}',S)} \Pr[({\sigma}',S)]\Phi_{\mathcal R}\left({{B}^+}({\sigma}',S)\right)\\ &=\frac{161}{80}\mathop{\mathrm{\mathbb{E}}}\left[\Phi_{\mathcal R}\left({{B}^+}\right)\right]. \end{align*}

Combining this with eq:eligible-count,eq:omit-count, we have

114E[xG(HG2)]≤E[Y]=E[Y−η]+E[η]≤16180E[ΦR(B+)]+E[ΦR(B+)]=24180E[ΦR(B+)].\begin{align*} \frac1{14}\mathop{\mathrm{\mathbb{E}}}[x_G({H_{G}^{2}})] &\leq\mathop{\mathrm{\mathbb{E}}}[Y] =\mathop{\mathrm{\mathbb{E}}}[Y-\eta]+\mathop{\mathrm{\mathbb{E}}}[\eta]\\ &\leq\frac{161}{80}\mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})] +\mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})] =\frac{241}{80}\mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})]. \end{align*}

◻

Bounding the component potential

We now bound the expected potential E[ΦR(B+)]\mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})] by examining which B{B}-leaders survive the augmentation. This is analogous to § 5.

Lemma 15. For any simple graph GG, E[ΦR(B+)]≤11n.\mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})]\leq 11 n .

The expected contribution of a B{B}-leader.

Since all leaders of B+{{B}^+} are from B{B}, we bound the expected contribution of a B{B}-leader. We reveal all the randomness that is just enough to reveal B{B}, the relevant degrees, and the leaders. We call these a partial outcome, denoted by

τ=((ev,1)v∈VG, S, R, (ev,2)v∈U).{\tau}= \Bigl( (e_{v,1})_{v\in V_G},\, S,\, \mathcal R,\, (e_{v,2})_{v\in U} \Bigr).

These are all first choices, the sets SS and R\mathcal R, and the second choices at UU. Note that R\mathcal R is needed to define the relevant degrees and the leaders, and it only reveals the vertices in SS whose second choice leaves SS, without revealing its actual second choice.

Condition on τ{\tau}, and consider a B{B}-component CC whose leader αC\alpha_{C} belongs to R\mathcal R. Let bCb_{C} be the number of neighbors of αC\alpha_{C} in UU that lie inside CC or in lower B{B}-components. Since the augmentation adds the reserved choice of αC\alpha_{C} precisely when it reaches a higher component, αC\alpha_{C} remains a leader in B+{{B}^+} with probability,

Pr⁡[αC is a leader in B+∣τ]=bCdαC.\Pr[\alpha_{C}\text{ is a leader in }{{B}^+}\mid{\tau}] =\frac{b_{C}}{d_{\alpha_{C}}}.

When αC\alpha_{C} remains a leader in B+{{B}^+}, it contributes dαCd_{\alpha_{C}} to the potential. Otherwise, it contributes 00 to the potential. Hence, the expected potential conditioned on the partial outcome τ{\tau} is

E[ΦR(B+)∣τ]=∑C∈CC(B)αC∈RPr⁡[αC is a leader in B+∣τ]⋅dαC=∑C∈CC(B)αC∈RbC.\mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})\mid{\tau}] =\sum_{\substack{C\in\mathcal{CC}({B})\\ \alpha_{C}\in\mathcal R}} \Pr[\alpha_{C}\text{ is a leader in }{{B}^+} \mid{\tau}] \cdot d_{\alpha_{C}} \quad=\sum_{\substack{C\in\mathcal{CC}({B})\\ \alpha_{C}\in\mathcal R}}b_{C}.

We split these neighbors into internal neighbors and external neighbors in lower components. The internal neighbors contribute at most ∑C∣C∣=n\sum_C|C|=n. Let L(B)L({B}) denote the total number of neighbors in UU that lie in lower components and are adjacent to B{B}-leaders in R\mathcal R. We have

E[ΦR(B+)∣τ]≤n+L(B),E[ΦR(B+)]=E ⁣[E[ΦR(B+)∣τ]]≤n+E[L(B)].\begin{equation} \mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})\mid{\tau}] \leq n+L({B}), \qquad \mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})] =\mathop{\mathrm{\mathbb{E}}}\!\left[\mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})\mid{\tau}]\right] \leq n+\mathop{\mathrm{\mathbb{E}}}[L({B})]. \tag*{(9)} \end{equation}

It remains to prove E[L(B)]≤10n\mathop{\mathrm{\mathbb{E}}}[L({B})]\leq 10n.

Bounding leaders’ neighbors in lower components with recurrence and switching

Similar to Lemma 8, we assign each crossing edge to its endpoint in the higher component, called its owner. We only switch edges incident to UU. Define this set of allowed edges by

E∘:=EG∖{uv∈EG:u,v∈S}.{E^\circ}:=E_G\setminus\{uv\in E_G:u,v\in S\}.

All selected choices of B{B} are allowed, since SS is independent in HG1{H_{G}^{1}} and only unmarked vertices add their second choices to B{B}.

For a vertex uu in a B{B}-component CC, let hB(u)h_{{B}}(u) be its distance to αC\alpha_{C} in B{B}. For each integer j≥0j\geq0, let Mj(B)M_{j}({B}) and Nj(B)N_{j}({B}) count allowed crossing edges whose owners are at depth jj and are marked and unmarked, respectively. Write

Tj(B):=Mj(B)+Nj(B).T_{j}({B}) :=M_{j}({B})+N_{j}({B}).

Every neighbor counted by L(B)L({B}) corresponds to an allowed crossing edge owned by a leader in R⊆S\mathcal R\subseteq S, at depth zero. Hence,

L(B)≤M0(B).L({B})\leq M_{0}({B}).

Note that M0M_{0} may also include contribution fromleaders in S∖RS\setminus\mathcal R.

We will relate these counts at adjacent depths by switching at the endpoint in the lower component, showing the following recurrences.

Lemma 16. For every integer j≥0j\geq0, E[Tj(B)]≤87n+87E[Mj+1(B)]+47E[Nj+1(B)],E[Mj(B)]≤47n+47E[Nj+1(B)].\begin{aligned} \mathop{\mathrm{\mathbb{E}}}[T_{j}({B})] &\leq\frac87n +\frac87\mathop{\mathrm{\mathbb{E}}}[M_{j+1}({B})] +\frac47\mathop{\mathrm{\mathbb{E}}}[N_{j+1}({B})], \\ \mathop{\mathrm{\mathbb{E}}}[M_{j}({B})] &\leq\frac47n +\frac47\mathop{\mathrm{\mathbb{E}}}[N_{j+1}({B})]. \end{aligned}

Proof. Fix a partial outcome τ{\tau}. Suppose e=uve=uv is an allowed crossing edge whose owner uu has depth jj. We define the forward switch at the lower endpoint vv as follows.

  1. If v∈Sv\in S, replace its first choice by uvuv, giving this switch weight 11.

  2. If v∈Uv\in U, each of its two choices can be replaced by uvuv, giving two switches of weight 1/21/2 each.

In the second case, since v∈Uv\in U, changing its second choice does not change R\mathcal R. Call the image τ′{\tau}' with base graph B′{B}'.Since uvuv is allowed, SS remains independent in the new first-choice graph. The relevant degrees and leader order are therefore fixed.

We first count the forward switches with these weights. Write pv=1p_v=1 for v∈Sv\in S and pv=1/2p_v=1/2 for v∈Uv\in U. Each initial edge gives total weight one, so the total forward weight from τ{\tau} is Tj(B)T_{j}({B}).

Now consider the inverse switches. Fix an image τ′{\tau}' and count how many forward switches it can receive. First, uvuv must be a bridge of B′{B}', selected in exactly one retained position, since it was crossing before the switch. Second, vv must be at depth j+1j+1. This is because the old higher component is unchanged, its leader remains the leader of the merged component, and every path from vv to that leader passes through uvuv. The membership of vv in SS or UU determines the type of switching it can receive. In either case, it receives switching of weight at most 11. Let CvC_{v} be the component containing vv after deleting uvuv. The old choice vzvz has two possible locations, illustrated in Figure 4.

  1. The GG-neighbors of vv inside CvC_{v}. Each such neighbor determines at most one inverse. The sets CvC_{v} for bridge children like vv at depth j+1j+1 are disjoint. Hence the total number of these inverses is at most

    ∑v∣NG(v)∩Cv∣≤∑v∣Cv∣≤n.\sum_v|N_G(v)\cap C_{v}| \leq\sum_v|C_{v}|\leq n.
  2. The GG-neighbors of vv in a lower component of B′{B}'. Since vzvz is a crossing edge owned by vv at depth j+1j+1. There are at most Mj+1(B′)M_{j+1}({B}') such edges when v∈Sv\in S and Nj+1(B′)N_{j+1}({B}') when v∈Uv\in U

[proof picture,x=1cm,y=1cm]
  \node[proof heading] at (3.35,3.65) {(a) Internal inverse: $z\in\ChildSide{v}$};
  \node[proof heading] at (10.65,3.65) {(b) Residual inverse: $z$ is outside};
  \begin{scope}
    \draw[proof region] (0,.9) rectangle (6.7,3.1);
    \draw[proof child] (3.45,1) rectangle (6.6,3);
    \node[proof leader] (a) at (.6,2.05) {$\alpha$};
    \node[proof vertex] (u) at (2.15,2.05) {$u$};
    \node[proof vertex] (v) at (4,2.05) {$v$};
    \node[proof vertex] (z) at (5.9,2.05) {$z$};
    \draw[proof path] (a)--node[above,proof tag] {$j$}(u);
    \draw[proof bridge] (v)--node[above,proof tag] {$uv$}(u);
    \draw[proof path] (v) to[bend left=35] (z);
    \draw[proof removed] (v)--(z);
    \node[proof note] at (4,2.8) {$j+1$};
    \node[proof note] at (5.1,1.3) {child side $\ChildSide{v}$};
    \node[proof note,text width=6.5cm] at (3.35,.2)
      {Restore the old choice $vz$ inside $\ChildSide{v}$.};
  \end{scope}
  \begin{scope}[xshift=7.3cm]
    \draw[proof region] (0,1.4) rectangle (6.7,3.1);
    \draw[proof child] (3.45,1.5) rectangle (6.6,3);
    \draw[proof lower] (3.45,-.75) rectangle (6.6,.6);
    \node[proof leader] (a2) at (.6,2.05) {$\alpha$};
    \node[proof vertex] (u2) at (2.15,2.05) {$u$};
    \node[proof vertex] (v2) at (4,2.05) {$v$};
    \node[proof vertex] (z2) at (5.8,-.05) {$z$};
    \draw[proof path] (a2)--node[above,proof tag] {$j$}(u2);
    \draw[proof bridge] (v2)--node[above,proof tag] {$uv$}(u2);
    \draw[proof crossing] (v2) -- (4,1.05)
      -- node[above,proof tag] {$vz$} (5.8,1.05) -- (z2);
    \node[proof note] at (4,2.8) {$j+1$};
    \node[proof note] at (5.4,2.65) {$\ChildSide{v}$};
    \node[proof note,align=left] at (4.35,-.05) {lower\\component};
    \node[proof note,text width=6.7cm] at (3.35,-1.3)
      {$v$ owns the residual edge $vz$ at depth $j+1$.};
  \end{scope}

Figure 4. Green arrows are inserted retained choices. Dashed rose and gold edges are alternative previous choices; dotted blue lines are retained paths.

As in § 7.3, these outcomes are not equally likely and we need to account for the probability ratio of the partial outcomes. Let ν(τ)\nu({\tau}) be the probability of a partial outcome. A forward switch (τ,e,i)({\tau},e,i) contributes ν(τ)pv\nu({\tau})p_v. Summing these contributions by their images gives

E[Tj(B)]=∑τ′ν(τ′)∑(τ,e,i):(τ,e,i)↦τ′pvν(τ)ν(τ′).\begin{equation} \mathop{\mathrm{\mathbb{E}}}[T_{j}({B})] =\sum_{{\tau}'}\nu({\tau}') \sum_{\substack{({\tau},e,i):\\ ({\tau},e,i)\mapsto{\tau}'}} p_v\frac{\nu({\tau})}{\nu({\tau}')}. \tag*{(11)} \end{equation}

The inner sum is over valid switches. To apply the inverse counts above, we need to bound the ratio ν(τ)/ν(τ′)\nu({\tau})/\nu({\tau}').

Since sampling is with replacement, the possible hidden second choices depend only on SS and R\mathcal R, not on the revealed choices. Since our switching above preserves both sets, so τ{\tau} and τ′{\tau}' have the same number of completions to two choices. All two choice sequences have the same sampling probability, so we only need to compare the probabilities of obtaining SS from the two first-choice graphs. Writing HG1′{H_{G}^{1}}' for the image first-choice graph, (7) gives

ν(τ)ν(τ′)=Z9(HG1[U])Z9(HG1′[U])Z10(HG1′)Z10(HG1).\begin{equation} \frac{\nu({\tau})}{\nu({\tau}')} =\frac{Z_{9}\left({H_{G}^{1}}[U]\right)} {Z_{9}\left({H_{G}^{1}}'[U]\right)} \frac{Z_{10}\left({H_{G}^{1}}'\right)} {Z_{10}\left({H_{G}^{1}}\right)}. \tag*{(12)} \end{equation}

If the switch changes a second position, HG1′=HG1{H_{G}^{1}}'={H_{G}^{1}} and the ratio is one. Otherwise, the new edge joins different old B{B}-components. It therefore creates no directed cycle and destroys at most one in HG1{H_{G}^{1}}. Applying Lemma 14 as before gives

Z10(HG1′)Z10(HG1)≤8180.\frac{Z_{10}\left({H_{G}^{1}}'\right)}{Z_{10}\left({H_{G}^{1}}\right)} \leq\frac{81}{80}.

For the nine-coloring factor, we need a lower bound on the number of colorings after the switch. If vv is marked, both its old and new first choices are incident to a vertex outside UU. Neither affects the induced graph on UU, so this factor is one. Now suppose v∈Uv\in U. First remove its old first choice, obtaining HG1−{H_{G}^{1}}^-. Every proper coloring of HG1[U]{H_{G}^{1}}[U] is still proper in HG1−[U]{H_{G}^{1}}^-[U], since removing a choice cannot introduce a coloring constraint. We then insert uvuv. If u∈Su\in S, this edge is absent from the induced graph on UU, so all its colorings remain proper. If u∈Uu\in U, the edge joins distinct components of HG1−[U]{H_{G}^{1}}^-[U], since uu and vv were in different B{B}-components before the switch. The two components can be colored independently, so exactly 1/91/9 of their colorings give uu and vv the same color due to color symmetry. The other 8/98/9 remain proper after inserting uvuv. Thus insertion preserves at least 8/98/9 of the colorings in either case, giving

Z9(HG1′[U])≥89Z9(HG1−[U])≥89Z9(HG1[U]).Z_{9}\left({H_{G}^{1}}'[U]\right) \geq\frac89Z_{9}\left({H_{G}^{1}}^-[U]\right) \geq\frac89Z_{9}\left({H_{G}^{1}}[U]\right).

Combining these bounds, we have

ν(τ)ν(τ′)≤98⋅8180=729640<87.\begin{equation} \frac{\nu({\tau})}{\nu({\tau}')} \leq\frac98\cdot\frac{81}{80} =\frac{729}{640}<\frac87. \tag*{(13)} \end{equation}

After multiplying by the switch weight pvp_v, each inverse contributes at most 8/78/7 when vv is marked and at most 4/74/7 when vv is unmarked. The bounds are listed in Table 1.

Changed positionpvp_vBound on ν(τ)/ν(τ′)\nu(\tau)/\nu(\tau')Bound on pvν(τ)/ν(τ′)p_v\nu(\tau)/\nu(\tau')
v∈Sv\in S, first1181/8081/8081/80<8/781/80<8/7
v∈Uv\in U, first1/21/2729/640729/640729/1280<4/7729/1280<4/7
v∈Uv\in U, second1/21/2111/2<4/71/2<4/7

Table 1. Probability ratios and position weights for the two-out potential switching. The last column bounds the weighted contribution of one inverse.

We can now apply the inverse counts to upper bound the inverse contributions. There are at most nn internal inverses, each contributing at most 8/78/7. Each residual crossing edge contributes at most 8/78/7 if its owner is marked and 4/74/7 if its owner is unmarked. Thus the total inverse contribution at τ′{\tau}' is at most

87n+87Mj+1(B′)+47Nj+1(B′).\frac87n+\frac87M_{j+1}({B}') +\frac47N_{j+1}({B}').

Substituting into (11) and taking expectations proves (10).

For the second recurrence, count only initial edges whose owner uu is marked. Since the edges are allowed, their switching endpoints vv must be unmarked. Now every inverse has weight at most 4/74/7, and every residual edge has an unmarked owner. The same argument gives

E[Mj(B)]≤47n+47E[Nj+1(B)],\mathop{\mathrm{\mathbb{E}}}[M_{j}({B})] \leq\frac47n+\frac47\mathop{\mathrm{\mathbb{E}}}[N_{j+1}({B})],

which proves eq:recurrence-marked. ◻

We now apply the recurrences to bound the depth counts. For every j≥0j\geq0, we will show

E[Tj(B)]≤16n,E[Mj(B)]≤10n.\begin{equation} \mathop{\mathrm{\mathbb{E}}}[T_{j}({B})]\leq16n, \qquad \mathop{\mathrm{\mathbb{E}}}[M_{j}({B})]\leq10n. \tag*{(14)} \end{equation}

We use backward induction on jj. All depth counts are zero for j≥nj\geq n, which gives the base case. For the induction step, use Tj+1=Mj+1+Nj+1T_{j+1}=M_{j+1}+N_{j+1} to get

E[Tj]≤87n+47(E[Tj+1]+E[Mj+1])≤87n+47(16n+10n)=16n,E[Mj]≤47n+47E[Tj+1]≤47(n+16n)=687n≤10n.\begin{align*} \mathop{\mathrm{\mathbb{E}}}[T_{j}] &\leq\frac87n+\frac47\left(\mathop{\mathrm{\mathbb{E}}}[T_{j+1}] +\mathop{\mathrm{\mathbb{E}}}[M_{j+1}]\right) \leq\frac87n+\frac47(16n+10n)=16n,\\ \mathop{\mathrm{\mathbb{E}}}[M_{j}] &\leq\frac47n+\frac47\mathop{\mathrm{\mathbb{E}}}[T_{j+1}] \leq\frac47(n+16n)=\frac{68}{7}n\leq10n. \end{align*}

Thus the bounds at depth j+1j+1 imply those at depth jj, proving (14). In particular,

E[L(B)]≤E[M0(B)]≤10n.\mathop{\mathrm{\mathbb{E}}}[L({B})] \leq\mathop{\mathrm{\mathbb{E}}}[M_{0}({B})]\leq10n.

Proof of Lemma 15. Applying (9) and the preceding bound gives

\mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})] \leq n+\mathop{\mathrm{\mathbb{E}}}[L({B})] \leq n+10n=11n. \qedhere

◻

We can now prove Theorem 12.

Proof of Theorem 12. Combining lem:2crossing-potential,lem:2-potential-exp, we obtain

\mathop{\mathrm{\mathbb{E}}}[x_G({H_{G}^{2}})] \leq14\cdot\frac{241}{80}\mathop{\mathrm{\mathbb{E}}}[\Phi_{\mathcal R}({{B}^+})] \leq14\cdot\frac{241}{80}\cdot11n =\frac{18557}{40}n<500n. \qedhere

◻

Proof of Theorem 1. The case k≥3k\geq3 follows from Theorem 10. For k=2k=2, Theorem 12 gives E[xG(HG2)]≤500n=O(n/k)\mathop{\mathrm{\mathbb{E}}}[x_G(H_G^2)]\leq500n=O(n/k). ◻

Declaration of AI assistance

OpenAI Codex was used to assist with developing and checking parts of the proofs, drafting and organizing parts of the manuscript, finding relevant references, generating and revising TikZ figures, and editing the LaTeX source. The authors are responsible for the mathematical claims, references, and final content of the paper.

Number of proper colorings of pseudo-forest

Proof of Lemma 14. Each component contains exactly one directed cycle, with trees attached to its vertices. Consider a component with mm vertices whose directed cycle has length ℓ\ell.

First suppose ℓ≥3\ell\ge 3. The cycle has

(q−1)ℓ+(−1)ℓ(q−1)(q-1)^\ell+(-1)^\ell(q-1)

proper qq-colorings. Once its colors are fixed, color the remaining vertices in increasing distance from the cycle. Each vertex has exactly q−1q-1 choices, since only the color of its parent is forbidden. Thus the component has

(q−1)m−ℓ((q−1)ℓ+(−1)ℓ(q−1))=(q−1)m(1+(−1)ℓ(q−1)ℓ−1)(q-1)^{m-\ell} \bigl((q-1)^\ell+(-1)^\ell(q-1)\bigr) = (q-1)^m \left( 1+\frac{(-1)^\ell}{(q-1)^{\ell-1}} \right)

proper colorings.

If ℓ=2\ell=2, the two opposite arcs give a single undirected edge, so the underlying component is a tree. Its number of proper colorings is

q(q−1)m−1=(q−1)m(1+1q−1),q(q-1)^{m-1} = (q-1)^m\left(1+\frac{1}{q-1}\right),

which is the same formula with ℓ=2\ell=2.

Finally, colorings of distinct components can be chosen independently. Multiplying their counts and using that their sizes sum to nn proves the identity. ◻

References

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