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For a finite graph of maximum degree $\Delta \ge 1$, a random independent edge coloring and the Lovász local lemma prove existence of a coloring with no repetitively colored four-edge path when $k\ge 8{\Delta }^{3/2}$, and no repetitively colored six-edge path when $k\ge 5{\Delta }^{5/3}$.

A fixed edge lies in at most $\ell {\Delta }^{\ell -1}$ unoriented paths of length $\ell$. Hence a four-edge event has a closed dependency neighbourhood of size at most $16{\Delta }^3$ and probability $k^{-2}$; the symmetric local-lemma product is at most $16e{\Delta }^3/k^2$. The corresponding six-edge quantities are $36{\Delta }^5$, $k^{-3}$ and $36e{\Delta }^5/k^3$. The stated constants make both products less than one.

Both lengths can be excluded simultaneously with $k\ge C{\Delta }^{5/3}$ for an absolute constant $C$. Apply the asymmetric Lovász local lemma with parameters $x_4=2/k^2$, $x_6=2/k^3$. A path of length $\ell \in \{4,6\}$ meets at most $4\ell {\Delta }^3$ four-edge paths and $6\ell {\Delta }^5$ six-edge paths. The sums of neighbour parameters are therefore bounded by constants times $C^{-2}{\Delta }^{-1/3}+C^{-3}$. Choosing $C$ sufficiently large makes both products in the local-lemma inequalities at least one half, establishing the claim.

For bounded vertex degrees, total degree $M\to \infty$, and a forbidden graph $G_0$ of bounded maximum degree, put $\lambda =M^{-1}{\sum }_i{d}_i(d_i-1)$ and $\mu =M^{-1}{\sum }_{ij\in E(G_0)}{d}_i{d}_j$. Count prescribed loop pairs, double-edge pairs, and forbidden-edge pairs in the configuration model. Their means are respectively $\lambda /2+o(1)$, ${\lambda }^2/4+o(1)$, and $\mu +o(1)$. A fixed collection of $t$ disjoint prescribed pairs occurs with probability $[(M-1)(M-3)\cdots (M-2t+1){]}^{-1}$. Bounded degrees make intersecting bad configurations contribute $O(M^{-1})$ to each fixed factorial moment. Thus the total bad-pattern count has factorial moments $(\lambda /2+{\lambda }^2/4+\mu {)}^j+o(1)$. The Bonferroni inequalities and uniformly bounded means give the displayed zero-pattern probability, without requiring the parameters to converge.

A graph minor compresses connected regions of a graph into single vertices while retaining selected connections between them. For finite simple graphs, $H\preccurlyeq G$ means that some sequence of deleting vertices, deleting edges, and performing edge contractions produces $H$ up to graph isomorphism. Deleting a vertex also deletes all its incident edges; deleting an edge leaves its endpoints. The edge contraction convention here removes resulting self-loops and duplicate edges.^[ref]

For an exact small example, let $G$ be the cycle graph $C_4$ with vertices $u,v,a,b$ and edges $uv,va,ab,bu$. Contract $uv$ to a new vertex $w$. The surviving edges are $wa,ab,bw$, giving the complete graph $K_3$. Thus $K_3\preccurlyeq C_4$, although $C_4$ has no triangle in a graph as a subgraph. The branch sets of a graph minor are $\{u,v\},\{a\},\{b\}$, with the original edges $va,ab,bu$ witnessing all three required adjacencies. The first panel of the edge contraction diagram shows this operation.

A subgraph uses existing vertices and edges; a topological minor permits a graph subdivision of the target as a subgraph, so target edges become graph paths with mutually disjoint interiors and no target vertex in an interior. An ordinary graph minor permits whole connected branch sets of a graph minor in place of target vertices. Every topological minor is a graph minor, by contracting the replacement graph paths.^[ref]

The converse fails. Take the tree with edges $uv,ua,ub,vc,vd$. Contracting $uv$ gives the star graph $K_{1,4}$. Yet the original tree has maximum degree three, whereas every graph subdivision of $K_{1,4}$ retains a centre of degree of a vertex four. Hence $K_{1,4}$ is a graph minor but not a topological minor of this tree. The difference is that a connected branch set of a graph minor may gather connections incident to several original vertices.

The counts of vertices and edges never increase under these operations. However, the $C_4$ example shows that being a bipartite graph is not preserved: its graph minor $K_3$ contains an odd cycle. Its chromatic number also increases from two to three. A connected graph stays connected under an edge contraction, because the image of each graph path still connects its endpoints, but deletion can disconnect it. Thus one must check the particular graph property before treating it as preserved by taking graph minors.

Being a planar graph is preserved: in a crossing-free drawing, shrink a contracted edge and bring its incident edges into the merged vertex; deletion merely removes parts of the drawing. This makes a nonplanar graph minor a certificate that the original graph is nonplanar. More strongly, for a finite simple graph, being a planar graph is equivalent to having neither the complete graph $K_5$ nor the complete bipartite graph $K_{3,3}$ as a graph minor. Finding the corresponding branch sets of a graph minor settles nonplanarity even when no literal $K_5$ or $K_{3,3}$ subgraph is present. The converse is a substantial theorem, rather than a consequence of preservation alone.^[ref]

For every finite bipartite graph, the edge chromatic number equals its maximum degree. Repeatedly remove a matching covering maximum-degree vertices in a bipartite graph: each removal reduces the maximum degree by at least one. The removed matchings in a graph are colour classes of a proper edge colouring. The reverse bound follows from the distinct colours required at a vertex of maximum degree.

A finite bipartite graph of positive maximum degree $\Delta$ has a matching in a graph covering every vertex of vertex degree $\Delta$. To cover the vertices of maximum degree in one side $A$, apply Hall's marriage theorem to that subset: for each subset $S$ of it, double counting gives $\Delta |S|\le \Delta |N(S)|$, where $N(S)$ is its graph neighbourhood. Construct the corresponding matching in a graph for the other side, then use the union of two matchings to cover both specified subsets simultaneously.

A vertex with at least $R(t,t)$ neighbours has either a $t$-clique among them, producing $K_{t+1}$ with the vertex, or an independent $t$-set among them, producing an induced star with that centre.

For the second assertion put $D=R(t,t)-1$. If either of those two outcomes is absent, the preceding result forces maximum degree at most $D$. In a connected graph of this maximum degree, a ball of radius $t-1$ has at most $1+D+\cdots +D^{t-1}$ vertices, using the rough bound on walks at each distance. If the graph has more vertices, some vertex lies at distance at least $t$ from a chosen root. A shortest path to it is induced, since a chord would shorten it; its first $t$ edges give an induced path of the required length. Thus a sufficiently large connected graph has one of the three listed configurations.

For events $A_i$ with a dependency graph, assume each event is independent of the sigma-algebra generated by its nonneighbours. If numbers $x_i\in [0,1)$ satisfy $P(A_i)\le x_i{\prod }_{j\in \Gamma (i)}(1-x_j)$, the general Lovász local lemma givesThe symmetric version says that if every event has probability at most $q$, the dependency graph has maximum degree at most $d$, and $eq(d+1)\le 1$, then none occurs with positive probability. For $d\ge 1$, set $x_i=1/(d+1)$. Since $(1-1/(d+1){)}^d\ge e^{-1}$, the general condition follows from the symmetric numerical hypothesis. The logarithmic inequality $\log (1+1/d)\le 1/d$ proves that estimate. When $d=0$, the events are mutually independent and each has probability below one, so the conclusion is immediate. This is the intended implication from the general to the symmetric form; the source repeats the word symmetric in its final clause.

A proper edge colouring assigns colours to the edges of a graph so that distinct edges sharing a vertex have different colours. The edge chromatic number ${\chi }^{\prime }(G)$ is the least number of colours in such an assignment. Equivalently, it is the least number of matchings in a graph whose disjoint union is the entire collection of edges.

Vizing's theorem states that a finite simple graph of maximum degree $\Delta$ satisfiesThe lower bound holds because all edges incident with a vertex of vertex degree $\Delta$ require different colours.

Hall's marriage theorem states that a finite bipartite graph with vertex sets $A,B$ has a matching in a graph saturating $A$ if and only ifwhere $N(S)\subseteq B$ is the graph neighbourhood of $S$. Here matching saturation means that each vertex of $A$ is incident with an edge of the matching in a graph; it does not require every vertex of $B$ to be met.

Use mathematical induction on the maximum degree $\Delta$, including $\Delta =0$ as the base case. A graph of maximum degree zero has no edges, so zero colours suffice.

For $\Delta >0$, part (ii) supplies a matching in a graph $M$ meeting every vertex of vertex degree $\Delta$. Delete its edges, retaining all vertices. Each vertex formerly of maximum degree loses one incident edge; every other vertex already had vertex degree at most $\Delta -1$. HenceThe resulting graph is still a bipartite graph. By mathematical induction, it has a proper edge colouring using at most $\Delta -1$ colours. Give every edge of $M$ one new colour. Since $M$ is a matching in a graph, this produces a proper edge colouring of $G$ with at most $\Delta$ colours. The edges at a vertex of maximum degree require at least $\Delta$ colours, soThis is Kőnig's line colouring theorem.

Let $A_{\Delta }=\{a\in A:d_G(a)=\Delta \}$. Apply Hall's marriage theorem to the induced subgraph on $A_{\Delta }\cup B$, which is a bipartite graph. For any subset $S\subseteq A_{\Delta }$, double counting the edges from $S$ to its graph neighbourhood givesThe equality uses the fact that every neighbour of $S$ lies in $B$, so all edges incident with $S$ have been counted. The final inequality uses the maximum degree bound on $B$. Since $\Delta >0$, division gives the Hall condition $|S|\le |N(S)|$.

Consequently there is a matching in a graph $M_1$ saturating $A_{\Delta }$. If $A_{\Delta }$ is empty, use the empty matching in a graph. This proves the required matching covering maximum-degree vertices in a bipartite graph on the specified side; other vertices need not be met.

Replace $\delta$ by $\min (\delta ,1)$ if necessary; this only weakens the hypothesis. We prove the result for finite graphs using a random proper partial graph coloring, then a deterministic completion. All constants below may depend on this fixed positive $\delta$.

First record the extension of a partial colouring by neighbour-colour savings. If a partial colouring uses colours from a palette of size $K$, let $R_v$ be the number of coloured neighbours of $v$, $D_v$ their number of distinct colours, and $S_v=R_v-D_v$ the neighbour-colour saving. There are $K-D_v$ available colours and $d(v)-R_v$ uncoloured neighbours. Thusat every vertex suffices for greedy completion: each subsequently coloured neighbour removes at most one available colour while reducing the uncoloured-neighbour count by one.

Call $v$ high-degree if $d(v)\ge (1-\delta /8)\Delta$. Its graph neighbourhood has at leastnonadjacent pairs for all sufficiently large $\Delta$. Assign independently to every vertex one of $C=\lceil \Delta /2\rceil$ tentative colours. Erase the colour of every vertex having a neighbour with the same tentative colour. The remaining colouring is proper; erasure is based on the original assignments and is not an iterative cascade.

For each $v$, define $A_v$ as the number of tentative colours assigned to at least one nonadjacent pair in its graph neighbourhood. Let $B_v$ count those same colours for which at least one neighbour of $v$ loses its colour. Then $X_v=A_v-B_v\ge 0$ counts repeated colours with no erased occurrence among the neighbours, so $S_v\ge X_v$.

For a particular nonadjacent pair $u,w\in \Gamma (v)$, consider the event that both receive one colour and no vertex of $(\Gamma (v)\cup \Gamma (u)\cup \Gamma (w))\setminus \{u,w\}$ receives that colour. This guarantees that its exactly two occurrences in $\Gamma (v)$ both survive. Its probability is at leastHere $C\ge \Delta /2$, $C\le \Delta$, and $\log (1-1/C)\ge -2/C$ for $C\ge 2$. Different pairs cannot be counted by the same colour in this event, since each event specifies exactly two occurrences in the graph neighbourhood. Thereforefor every high-degree vertex. This is the expected colour savings from sparse neighbourhoods.

We need simultaneous concentration, not just a positive expectation. A single tentative-colour change can affect $A_v$ or $B_v$ only in the old and new colour categories, so each function changes by at most two. A witness for $A_v\ge s$ fixes two vertices for each of $s$ colours. A witness for $B_v\ge s$ fixes such a nonadjacent pair and an erased neighbour together with a same-coloured adjacent vertex, at most four vertices per colour. Thus both are certifiable functions with Lipschitz constant $2$ and certificate parameter at most $4$. Both are bounded by $C\le \Delta$.

To see exactly what concentration follows, let an integer-valued $Z$ change by at most $L$ per coordinate and have certificates of size at most $rb$. For an input with $Z\ge b$, fix such a certificate $J$. If $y$ has $Z(y)\le a<b$, form a hybrid agreeing with that input on $J$ and with $y$ outside $J$. The hybrid has value at least $b$, so at least $(b-a)/L$ certificate positions differ from $y$. Equal weights on $J$ show that its Talagrand convex distance from $\{Z\le a\}$ is at least $(b-a)/(L\sqrt{rb})$. Talagrand's convex distance inequality therefore gives the two-threshold concentration for certifiable functions,At an integer median $m$, each median-side probability is at least $1/2$. Taking thresholds $m-t,m$ or $m,m+t$ and using $0\le Z\le \Delta$ givesfor $L=2,r=4$. Summing these tails gives $|\mathbb{E}Z-m|=O(\sqrt{\Delta })$. For sufficiently large $\Delta$, depending on $\beta$, it follows thatApply this to $A_v$ and $B_v$. Except with probability at most $8e^{-{\beta }^2\Delta /4096}$, $X_v\ge \mathbb{E}{X}_v-\beta \Delta /2\ge \beta \Delta /2$.

A bad event at $v$ depends only on tentative colours at distance at most two. It is independent of the entire collection of bad events at distance greater than four. Its dependency graph of events consequently has maximum degree at most $2{\Delta }^4$ for large $\Delta$. The Lovász local lemma now makes all high-degree vertices good simultaneously: exponential decay beats this polynomial dependency bound.

The needed local-lemma step can be checked directly. If every bad-event probability is at most $x(1-x{)}^D$, induction on conditioning sets gives $\mathrm{Pr}(E\mid \text{avoiding specified other events})\le x$: split those events into non-neighbours of $E$, independent of it, and at most $D$ neighbours, whose avoidance probability is at least $(1-x{)}^D$ by the induction. Multiplying the resulting conditional avoidance probabilities makes total avoidance positive. Taking $x=1/(D+1)$ works whenever $ep(D+1)\le 1$, since $(1-1/(D+1){)}^D\ge e^{-1}$. Our bound satisfies this criterion once $\Delta \ge {\Delta }_0(\delta )$.

Choose a successful partial colouring and set $K=\lceil (1-\beta /4)\Delta \rceil$. This contains the original palette. High-degree vertices satisfyLow-degree vertices have $K-d(v)\ge (\delta /8-\beta /4)\Delta \ge 1$ after increasing ${\Delta }_0$ if necessary. Greedy completion therefore gives $\chi (H)\le K$. Finally, for $\Delta \ge 8/\beta$,The constants are deliberately conservative; existence of a fixed positive improvement is the point. The same proof uses an external maximum degree bound, so if infinite locally finite graphs are allowed, every finite subgraph has this uniform colouring bound and compactness extends the colouring to the whole graph.

For the stronger triangle-free result, the appropriate brief route is semi-random palette refinement for triangle-free colouring. Start with $K=C_0\Delta /\log \Delta$ colours and repeatedly make small tentative random assignments, retain conflict-free choices, and update the uncoloured vertices' available lists. Triangle-freeness makes each graph neighbourhood independent. Track list size and average residual neighbour congestion per colour, prune unusually congested colours, and use concentration inequalities and the Lovász local lemma to preserve these invariants through the rounds. The ratio of residual congestion to list size decreases until a final completion using the Lovász local lemma or greedy coloring succeeds. Average control and pruning are important: four-cycles are allowed, so assuming tight uniform concentration for every individual colour load would be unjustified. With a sufficiently large absolute $C_0$, this programme gives $\chi (H)=O(\Delta /\log \Delta )$. The one-round fixed-fraction saving proved above alone does not establish that logarithmic improvement.

To show the order is best possible, let $d\to \infty$ through integers and sample $G(n,d/n)$ with $n=d^7$. A Chernoff bound and union bound give maximum degree at most $2d$ with probability tending to one. The expected triangle count is at most $d^3/6=o(n)$, so fewer than $n/2$ vertices need be deleted to remove all triangles with high probability. For $s=\lceil 4n\log d/d\rceil$, the expected number of independent $s$-sets is at mostits logarithm is at most $s[\log (ed/(4\log d))-2\log d+o(1)]$, which tends to minus infinity. The altered triangle-free graph thus has at least $n/2$ vertices and independence number below $s$, giving $\chi \ge (c+o(1))d/\log d$ for an absolute positive $c$. Append a disjoint star graph whose centre has vertex degree $2d$ if needed to make its maximum degree exactly $\Delta =2d$. This proves the triangle-free chromatic lower bound at bounded maximum degree, $\chi =\Omega (\Delta /\log \Delta )$, so the preceding order cannot be improved in general.

Each vertex of a bipartite graph has a strict preference ordering of its neighbours; every neighbour is preferred to being unmatched. A stable matching in a bipartite graph is a matching in a graph with no blocking edge: an unused edge blocks if both endpoints prefer each other to their assigned partners, interpreting an unmatched endpoint as willing to accept any neighbour. Stability does not require a perfect matching in a graph.

For existence, run the Gale-Shapley algorithm. Whenever a vertex in the first part is unmatched and has an untried neighbour, it proposes to its most preferred untried neighbour. A receiver keeps its preferred proposal and rejects the other, including any previously held proposal. At most $e(G)$ proposals occur, so the procedure stops with a matching in a graph. Suppose an edge $bg$ blocked this final matching in a graph. Since $b$ prefers $g$ to its final partner, or is unmatched, $b$ must have proposed to $g$. But a receiver's held proposal only improves, so $g$'s final partner is preferred to $b$ if $b$ was rejected. If $b$ was not rejected, the edge is in the matching in a graph. Both alternatives contradict blocking. Every finite bipartite preference instance therefore has a stable matching.

For the mutual-worst edge in stable matchings, let $M,N$ be stable matchings with $bg\in M$ and $bg\notin N$. We give the argument using alternating graph components including possible unmatched endpoints. If $b$ were unmatched in $N$, then $g$ would have to prefer its $N$-partner to $b$, since otherwise $bg$ would block $N$. Following the $N$-edge from $g$, stability of $M$ forces its other endpoint to have and prefer an $M$-partner. Stability of $N$ then forces the next vertex to have and prefer an $N$-partner. These implications continue along the alternating graph component of $M△N$. They cannot stop, because at a missing next edge the preceding edge would block the appropriate matching in a graph. They also cannot close into a cycle in a graph, since the graph component already has endpoint $b$. This contradiction proves that $b$ is matched in $N$. The same reasoning applies to $g$.

Now $b$ prefers its $N$-partner to $g$, because $g$ is its worst choice. Along the alternating graph component, stability of $M$ forces that partner to prefer its $M$-partner; stability of $N$ forces the next vertex in $b$'s part to prefer its $N$-partner, and so on. A graph path graph component is impossible by the same endpoint argument, so this is a cycle in a graph. Thus throughout this cycle in a graph the part containing $b$ prefers $N$ and the part containing $g$ prefers $M$. In particular $g$ must prefer $b$ to its $N$-partner. That contradicts $b$ being $g$'s worst choice. Hence an edge ranked last by both endpoints, if present in one stable matching, is present in every stable matching.

For the list edge coloring assertion, let $\Delta$ be the maximum degree. We first prove that the chromatic index is $\Delta$. Use induction on the number of edges, with a proper $\Delta$-edge coloring of $G-uv$. Some color $a$ is missing at $u$ and some color $b$ is missing at $v$. If $a=b$, color $uv$ with it. Otherwise the edges of colors $a,b$ form alternating graph paths and even cycles in a graph. The graph component containing $u$ cannot contain $v$: an alternating graph path from $u$, starting with color $b$, would end at $v$ with color $a$, have even length, and place $u,v$ in the same bipartition class. Swap $a,b$ in the graph component containing $u$. Now $b$ is missing at both ends of $uv$, so it can be colored $b$. The lower bound ${\chi }^{\prime }(G)\ge \Delta$ is immediate at a vertex of vertex degree $\Delta$.

Fix such a proper coloring $c:E(G)\to \{1,\dots ,\Delta \}$. Orient the line graph toward the smaller color at vertices in the first part of $G$, and toward the larger color at vertices in the second part. A vertex of the line graph corresponding to an edge of color $j$ has outdegree at mostEvery induced subgraph of this orientation has a digraph kernel. Indeed, restrict to its edge set in the original bipartite graph, give first-part vertices a preference for smaller colors and second-part vertices a preference for larger colors, and choose a stable matching. Its edges form an independent set in the line graph. Every edge outside the matching in a graph has an endpoint whose partner is preferred to it, by stability; thus it has an outgoing edge to the independent set. This proves that the orientation is a kernel-perfect directed graph.

Here is the kernel list-coloring lemma with its proof. If a kernel-perfect directed graph $D$ has lists satisfying $|L(v)|\ge d_D^{+}(v)+1$, choose any color $a$ appearing in a list and a digraph kernel $K$ of the induced subgraph on vertices whose lists contain $a$. Color $K$ with $a$, delete $K$, and delete $a$ from the remaining lists. Each remaining vertex that loses a color had an outgoing neighbour in $K$, so its outdegree decreases by at least one; the list inequality persists. Vertices not losing a color also retain the inequality. Induction on the number of vertices completes a proper list coloring.

Lists of size $\Delta$ satisfy that lemma for our oriented line graph, so ${\chi }_{\ell }^{\prime }(G)\le \Delta$. Identical lists show ${\chi }_{\ell }^{\prime }(G)\ge {\chi }^{\prime }(G)$. Consequently the Galvin theorem for bipartite list edge coloring givesThe empty graph is immediate; if parallel edges are allowed, orient their line-graph adjacency in both directions when the two endpoints prescribe opposite directions. The outdegree bound still holds, and the same stable matching and digraph kernel proof applies with preferences on incident edges.

Begin with the configuration model for $n$ labelled vertices of degree $r$. Let $Y_{\ell }$ count cycles in a graph with specified paired edges, so parallel-edge choices are counted separately before conditioning. Choose a cyclic ordering of $\ell$ distinct vertices in $(n{)}_{\ell }/(2\ell )$ ways, where division by $2\ell$ removes rotation and reversal. At each vertex choose an ordered pair of distinct half-edges for the entering and leaving edges, in $r(r-1)$ ways. These $\ell$ prescribed pairs appear with the pairing probability from part (i), givingIt remains to justify conditioning on simplicity: merely dividing by a positive simplicity probability would give an incorrect constant.

Unconditionally, part (i) with no forbidden graph has $\lambda =r-1$, soCondition on one specified cycle pairing. The unpaired half-edges form a uniform configuration model with degree $r-2$ at its $\ell$ vertices and degree $r$ elsewhere. A completion of the full graph is simple exactly when the residual graph is simple and has no additional edge along one of the already prescribed cycle edges. This forbidden graph has maximum degree two. Because $r,\ell$ are fixed,After permuting labels if needed to sort the degrees, the assumptions of part (i) still hold, including residual degrees at least one. Its bounded-degree pairing avoidance estimate therefore shows thatThis ratio is the same for each prescribed cycle, by symmetry. Sum over their indicators to obtain $\mathbb{E}(Y_{\ell }\mid \text{simple})=(1+o(1))\mathbb{E}{Y}_{\ell }$. Conditioning is uniform over all simple $r$-regular graphs because each has $(r!{)}^n$ pairing representations. On simplicity, $Y_{\ell }$ is the ordinary cycle count. Hence the short-cycle expectation in a uniform regular graph is

Put $M=2m$ and $a_i=\left(\genfrac{}{}{0ex}{}{d_i}{2}\right)$. Use the configuration model: give vertex $i$ exactly $d_i$ labelled half-edges and choose a uniform perfect matching of all $M$ half-edges. There arepairings, with $(2m{)}_m$ denoting a falling factorial. Each allowed simple graph has exactly ${\prod }_i{d}_i!$ representations: at every vertex, assign its incident graph edges bijectively to its labelled half-edges. Therefore its enumeration reduces to the probability that a pairing is simple and avoids $G_0$.

Count three kinds of prescribed bad patterns. A loop pattern consists of a pair of half-edges at one vertex. A double-edge pattern consists of two half-edges at each of two distinct vertices and one of their two cross-pairings. A forbidden-edge pattern consists of one half-edge at each endpoint of an edge of $G_0$. Let their respective counts in the random pairing be $Z_L,Z_D,Z_F$. Simplicity and avoidance are equivalent to $Z=Z_L+Z_D+Z_F=0$, including exclusion of triple edges because those contain double-edge patterns.

A prescribed collection of $t$ disjoint half-edge pairs occurs with probabilityConsequently the bad-pattern expected values areHere $\lambda =({\sum }_i{a}_i)/m$ and $\mu =M^{-1}{\sum }_{ij\in E(G_0)}{d}_i{d}_j$, as in the PDF. Since $1\le d_i\le \Delta$ and $G_0$ has bounded maximum degree, $M=\Theta (n)$ and all three means are uniformly bounded. For the double-edge estimate, ${\sum }_i{a}_i^2=O(n)$, so its contribution after division by $M^2$ is $o(1)$.

We now give the moment-counting step rather than assume independence. For each fixed $h$, expand the factorial moment $\mathbb{E}(Z{)}_h$ as the sum over ordered distinct bad patterns. Collections on disjoint vertex sets have, by the displayed pairing probability, joint probabilities equal to the product probabilities times $1+O(1/n)$. Their total contribution isDeleting collections with a shared vertex changes the independent-product sum by $O(1/n)$, since all ordinary and forbidden vertex degrees are bounded.

For actual joint probabilities of overlapping patterns, incompatible uses of a half-edge give zero. Compatible collections with no shared paired edge lose at least one free vertex choice, without reducing the number of required pairs, so contribute $O(1/n)$. A shared paired edge can occur only between two double-edge patterns or between a double-edge pattern and a forbidden-edge pattern. In the first case their union requires at least three parallel paired edges: there are $O(n^2)$ endpoint choices and probability $O(n^{-3})$. In the second case the endpoint pair must be an edge of $G_0$: there are only $O(n)$ choices and at least two required pairs, giving $O(n^{-1})$. Adding further patterns in a fixed collection does not remove this deficit: a disjoint addition contributes bounded order, while an addition to an existing cluster has only bounded half-edge choices unless it supplies new free vertices, compensated by its new required pairs. Summing the finitely many overlap types for fixed $h$ therefore gives $o(1)$. Thus, with $\Lambda =\lambda /2+{\lambda }^2/4+\mu$,The Bonferroni inequalities bracket $\mathrm{Pr}(Z=0)$ by the odd and even partial sums of ${\sum }_h(-1{)}^h\mathbb{E}(Z{)}_h/h!$. The means $\Lambda$ are bounded, so the corresponding exponential-series remainders vanish uniformly as the truncation increases. Taking $n\to \infty$ and then the truncation to infinity proves the bounded-degree pairing avoidance estimate $\mathrm{Pr}(Z=0)=e^{-\Lambda }+o(1)$. This does not require $\lambda$ or $\mu$ to converge. The exponential is bounded away from zero, so the additive error is also a relative $o(1)$ error.

Multiplying by the pairing count and dividing by the representation factor givesThe factor $(2m{)}_m$ multiplies the exponential; it is not part of its exponent. The TeX aid misgroups this factor, whereas the original PDF is unambiguous. The bounded-degree proof also covers the stated additional assumption $2m-n\to \infty$.

First suppose $G$ is 3-connected and noncomplete, and write $\Delta =\Delta (G)$. There is an induced path $a,v,b$ with $a,b$ nonadjacent: otherwise the first three vertices of any shortest path between nonadjacent vertices would contradict that assertion. Since $G$ is 3-connected, $G-\{a,b\}$ is a connected graph. Choose a spanning tree of that graph rooted at $v$. Color $a,b$ first with the same color, then color the other vertices in reverse tree order, leaving $v$ last. Every vertex except $v$ has its tree parent uncolored and thus at most $\Delta -1$ previously colored neighbors. At $v$, the two neighbors $a,b$ share a color, so again at most $\Delta -1$ colors are forbidden. This constructs a proper $\Delta$-coloring and proves Brooks' bound for 3-connected noncomplete graphs. A complete graph is exactly the excluded case; a 3-connected graph cannot be an odd cycle.

For the partition claim, assume $d_1,d_2$ are nonnegative integers with $d_1+d_2=\Delta -1$. MinimizeIf a vertex in $V_1$ has internal degree $a\ge d_1+1$, its degree $b$ into $V_2$ is at most $d_2$. Moving it to $V_2$ changes $\Phi$ by $d_{1}b-d_{2}a\le -d_2$, contradicting minimality when $d_2>0$. The analogous move handles a violation in $V_2$ when $d_1>0$. For a zero coefficient, break ties among minimizers by minimizing the total number of internal edges. If $d_2=0<d_1$, minimality of $\Phi =d_{1}e(G[V_2])$ ensures $V_2$ is independent; a violating vertex in $V_1$ has all its $\Delta =d_1+1$ neighbors in $V_1$ and no neighbor in $V_2$, so moving it reduces the tie-break quantity without increasing $\Phi$. The reversed argument handles $d_1=0$. If both are zero then $\Delta =1$, and the graph is a union of isolated vertices and edges, admitting two independent parts. ThusFor $\Delta \ge 5$, put $d_1=3,d_2=\Delta -4$. A maximum-degree-three graph containing no $K_4$ is 3-colorable. Here is why the previously proved 3-connected case suffices: connected graphs of degree at most two are paths or cycles and use at most three colors. Split a larger graph at cut vertices and permute colors when gluing. In a 2-connected cubic graph with a two-vertex cut $\{u,v\}$, a cut edge $uv$ gives colorings of the sides glued with their distinct endpoint colors. If $u,v$ are nonadjacent, the cut components have positive numbers of edges to both endpoints and the total endpoint degrees are at most three. There are at most three components. With three components, adding $uv$ to each side preserves maximum degree three and gives the endpoints degree two. No new $K_4$ is created, so induction supplies side colorings with distinct endpoint colors, which can be permuted to agree. With two components, the components incident twice to $u$ and twice to $v$ either agree or differ. If they differ, adding $uv$ to each side preserves maximum degree three; each augmented side is smaller and cannot be $K_4$, because one endpoint then has degree only two, so induction gives distinct-endpoint colorings. If they agree, call the side with only one edge from each endpoint the small side. Any coloring of the other side can be matched on it: for equal endpoint colors, contract $u,v$ in the small side (the merged vertex has degree at most two); for distinct colors add $uv$ (the endpoints again have degree two). Induction applies in both cases. A 2-connected graph with no two-vertex cut is 3-connected and was handled above. This proves the needed subcubic coloring fact without quoting the unproved full theorem.

Use three colors on $V_1$ and at most $d_2+1=\Delta -3$ fresh colors on $V_2$, by greedy coloring. This gives $\chi (G)\le \Delta$. More usefully, repeatedly remove degree-at-most-three parts: the partition supplies a four-degree reduction for a cost of three colors. If $f(\Delta )=\frac{3}{4}\Delta +\frac{3}{2}$, then $3+f(\Delta -4)=f(\Delta )$. Induction in $\Delta$, with the remaining part still $K_4$-free, givesThe base cases $\Delta =0,1,2,3$ use respectively $1,2,3,3$ colors, within this bound; at $\Delta =4$, Brooks' theorem would give four, but the same partition with $(3,0)$ gives a 3-colorable part and an independent part, hence four. Thus the induction requires no unproved general Brooks theorem.

Define the asymmetric Ramsey number $R(s,t)$ for a red $K_s$ or a blue $K_t$. The base cases $R(1,t)=R(s,1)=1$ are finite. If $n=R(s-1,t)+R(s,t-1)$, choose a vertex of $K_n$. Its red neighbors number at least $R(s-1,t)$, or its blue neighbors number at least $R(s,t-1)$. Apply the appropriate smaller Ramsey number inside that neighborhood: either the opposite-color clique already exists, or a same-color clique extends by the chosen vertex. Thus $R(s,t)\le R(s-1,t)+R(s,t-1)$, proving finiteness by induction, in particular for $R(s)=R(s,s)$.

In a graph of maximum degree $d\ge 2$, a ball of radius $k-1$ about a vertex has at most ${\sum }_{j=0}^{k-1}{d}^j=(d^k-1)/(d-1)<d^k$ vertices. Hence a connected graph of order $d^k$ has a vertex outside this ball, at distance at least $k$. The argument also applies whenever the graph has at least $d^k$ vertices.

Now take a connected graph of order $R(s{)}^s$. If some vertex has at least $R(s)$ neighbors, apply the two-color Ramsey theorem to their pairs, coloring an edge red and a nonedge blue. A red $K_s$ is an induced complete graph, while a blue $K_s$ is an independent set of $s$ neighbors, which with the centre induces $K_{1,s}$. Otherwise the maximum degree satisfies $d<R(s)$. For $s\ge 2$, connectedness and the vertex count rule out $d\le 1$. The ball bound then provides vertices at distance at least $s$. A shortest path between them has no chord, since a chord would shorten it; its first $s$ edges induce $P_s$. The case $s=1$ is immediate. Therefore