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For a nonempty finite simple graph $G$, the chromatic number $\chi (G)$ is the least positive integer $k$ for which $G$ has a proper vertex graph coloring with at most $k$ colours. To establish an exact chromatic number, give a colouring using that many colours and explain why fewer cannot suffice. A colouring alone gives only an upper bound.^[ref]

These small graphs show why the chromatic number measures more than the presence of triangles in a graph:

For a cycle graph of even length, alternating two colours closes consistently; for an odd cycle, it does not, and a third colour on the final vertex suffices. More generally, the two-colourability criterion for bipartite graphs says that $\chi (G)\le 2$ exactly when $G$ is a bipartite graph, and the odd-cycle characterization of bipartite graphs identifies the obstruction as an odd cycle. A nonempty graph with no edges has chromatic number one; a bipartite graph with at least one edge has chromatic number exactly two.

In the pairwise-conflict scheduling model of graph coloring, $\chi (G)$ is the least number of interchangeable slots needed. This turns a drawing into an optimization question: how many groups of compatible activities are unavoidable? Greedy coloring constructs a valid grouping quickly, but its number of colours can exceed the chromatic number.

A family of triangles in a graph is edge-disjoint if no edge belongs to two members. Any deletion of edges producing a triangle-free graph must remove at least one distinct edge for each member. The triangles in a graph may share vertices.

The Grötzsch graph is the Mycielskian of the five-vertex cycle graph. Take original edges $v_i{v}_{i+1}$, with indices interpreted modulo five; add $u_i{v}_{i-1},u_i{v}_{i+1}$ and $w{u}_i$ for every $i$. There are eleven vertices and twenty edges: five original, ten between originals and Mycielski shadow vertices, and five from the Mycielski apex.

Give $v_1,\dots ,v_5$ colours $(1,2,1,2,3)$, give each $u_i$ the colour of $v_i$, and give $w$ colour 4. This is a proper vertex colouring. The Mycielskian chromatic increase proves that three colours are impossible, so the chromatic number is exactly four. The Mycielskian triangle preservation proves that this is a triangle-free graph. In particular its clique number is only two, even though its chromatic number is four.

The Mycielski apex has an independent set as its graph neighbourhood, yet all three old colours are needed there in any three-colour proper vertex colouring of the subgraph obtained by deleting it. Thus a locally easy group of mutually nonadjacent neighbours can carry globally incompatible colour requirements. Absence of triangles in a graph alone cannot bound the chromatic number.

The complete bipartite graph $K_{3,3}$ has vertex sets $A=\{a_1,a_2,a_3\}$ and $B=\{b_1,b_2,b_3\}$, with every edge between $A$ and $B$ and none inside either part. Its six vertices support $3\cdot 3=9$ edges and no triangles in a graph: among any three vertices, two belong to the same independent set and have no joining edge.

Add the single edge $a_1{a}_2$. Each $b_j$ is a common neighbour of its endpoints, producing the triangle in a graph $a_1{a}_2{b}_j$. These are exactly the three new triangles in a graph, because every triangle in a graph must use the only edge inside a part. The new count is ten edges; nine was the last possible count without a triangle in a graph on six vertices.

In general, every simple graph with a finite set of $n\ge 0$ vertices that is triangle-free has at mostedges. This is the $r=2$ case of Turán's theorem, with forbidden clique $K_3$. Equality holds exactly for the balanced complete bipartite graph $K_{\lfloor n/2\rfloor ,\lceil n/2\rceil }$, up to graph isomorphism; empty parts are allowed for $n<2$.^[ref]

The floor function gives the exact integer bound. Splitting $n$ vertices into two independent sets of sizes $a$ and $n-a$, and putting in every cross edge, givesThe count is largest when $a$ is an integer nearest $n/2$. For $n=2k$ it is $k^2$; for $n=2k+1$ it is $k(k+1)$. Both equal $\lfloor n^2/4\rfloor$. This proves attainability for every $n$, including the empty graph.

A short proof of the upper bound uses degrees of vertices. In a triangle-free graph, the endpoints $u,v$ of an edge have disjoint vertex neighbourhoods: a common neighbour would complete a triangle in a graph. Thus $d(u)+d(v)\le n$. Summing over edges counts $d(v)$ once for each of the $d(v)$ edges incident to $v$, givingFor $n>0$, the Cauchy-Schwarz inequality and handshaking lemma yieldIf $m>0$, division gives $m\le n^2/4$; $m=0$ is immediate. Since $m$ is an integer, apply the floor function. The case $n=0$ is immediate too. This proves the bound; the full equality characterization follows from Turán's theorem.

The useful reversal is that more than $\lfloor n^2/4\rfloor$ edges force a triangle in a graph, whatever their placement. A global count certifies a local pattern. The extremizer is bipartite, but a triangle-free graph need not be bipartite: a five-vertex graph cycle has five edges, below the bound of six. Historically, Willem Mantel posed the triangle-exclusion question in 1907.^[ref]

The Mycielski construction replaces a graph $G$ with a graph $\mu (G)$ having one shadow vertex for each original vertex and one further vertex adjacent to every shadow. It satisfiesand preserves triangle-freeness. Iterating it from the five-cycle gives triangle-free graphs of arbitrarily large chromatic number.

Here $G$ is a finite simple graph with at least one vertex; it need not have an edge. The Mycielskian separates two ways of needing many colours: a large clique forces distinct colours immediately, but a triangle-free graph can impose incompatible choices through its whole pattern of graph adjacency. The Mycielski construction increases that global constraint while preserving the absence of triangles in a graph.

Write the vertex set of $G$ as $\{v_1,\dots ,v_n\}$, with $n\ge 1$. Keep these vertices, add distinct Mycielski shadow vertices $u_1,\dots ,u_n$, and add a Mycielski apex $w$. The vertex set and complete list of edges areThere are no other edges: the Mycielski shadow vertices form an independent set, $w$ has no original vertex as a neighbour, and $u_i$ is not adjacent to $v_i$. Each Mycielski shadow vertex copies its original vertex's graph neighbourhood within the original vertex set, not its position in the drawing. This is the construction in the inspected research paper.^[ref]

For $n$ original vertices and $m$ original edges, direct counting gives $2n+1$ vertices and $3m+n$ edges. The original vertices induce a copy of $G$, so no original graph adjacency is lost. A drawing crossing is not an extra vertex.

Start with the complete graph $K_2$, whose only edge is $v_1{v}_2$, abbreviating $\{v_1,v_2\}$. Its Mycielskian has exactly the edges $v_1{v}_2,u_1{v}_2,u_2{v}_1,w{u}_1,w{u}_2$. Reading them as $v_1,v_2,u_1,w,u_2,v_1$ gives the cycle graph $C_5$. A proper vertex colouring gives $v_1,u_1$ colour 1, $v_2,u_2$ colour 2, and $w$ colour 3. Two colours cannot close an odd cycle, so its chromatic number is exactly three.

A finite simple graph $G$ is a triangle-free graph if and only if its Mycielskian is a triangle-free graph. No triangle in a graph can contain the Mycielski apex, whose neighbours are an independent set, or two Mycielski shadow vertices. A triangle in a graph containing one shadow $u_i$ and two originals $v_j,v_{\ell }$ requires all three original edges $v_i{v}_j,v_i{v}_{\ell },v_j{v}_{\ell }$, hence an original triangle in a graph. Conversely every original triangle in a graph remains. In fact each original triangle in a graph gives four in $\mu (G)$: itself and the three choices of replacing one of its vertices by its shadow. Thus triangle-freeness is preserved, but a positive triangle count is multiplied by four.

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.

In the usual nondegenerate convention, a strongly regular graph of order $n$ and parameters $(d,a,b)$ is a finite simple $d$-regular graph with $0<d<n-1$ in which adjacent vertices have $a$ common neighbours and distinct nonadjacent vertices have $b$. Its adjacency matrix of a graph $A$ obeysFor the usual nondegenerate parameters with $b>0$ the graph is connected, so eigenvalue $d$ has multiplicity one. On the perpendicular complement of the all-ones vector, $J=0$ and the possible eigenvalues areIf $r$ has multiplicity $m$, then zero trace gives $d+mr+(n-1-m)s=0$. Solving provesThe denominator is positive in this nondegenerate convention. Otherwise all eigenvalues on the all-ones complement would coincide, forcing $A=rI+(d-r)J/n$. All off-diagonal entries would then be the same, making the graph complete or edgeless. If complete graphs are admitted with arbitrary unused $b$, the printed claim needs qualification: $K_2$ with $(d,a,b)=(1,0,2)$ satisfies the common-neighbour conditions vacuously but makes the fraction $0/0$. For complete graphs with nonzero denominator, the spectral calculation still works with an unused root of multiplicity zero.

For the triangle-free graph claim, suppose first there are at least two vertices. Nonadjacent vertices have a common neighbour, so the graph is connected. If $u,v$ are adjacent, put $U=N(u)\setminus \{v\}$ and $V=N(v)\setminus \{u\}$. Every member of $U$ is nonadjacent to $v$, and exactly two of its three common neighbours with $v$ lie in $V$, the third being $u$. By symmetry every member of $V$ has exactly two neighbours in $U$. Counting these edges gives $2|U|=2|V|$, hence $d(u)=d(v)$. Connectivity makes the graph $d$-regular. Counting length-two paths from a vertex to non-neighbours givesHere $a=0$, $b=3$, and the larger-root multiplicity becomes $[n-1+d^2/\sqrt{4d-3}]/2$. Thus $t=\sqrt{4d-3}$ is rational and therefore an integer. It is odd, $d=(t^2+3)/4$, and integrality also gives $t\mid d^2$. Since $16d^2\equiv 9(\mathrm{mod}t)$, one obtains $t\mid 9$, henceThe complete bipartite graph $K_{3,3}$ realizes $d=3$: vertices in the same part have all three vertices of the other part as common neighbours, and vertices in opposite parts are adjacent. There are no triangles.

As literally worded, the one-vertex graph is a vacuous exception: it has no triangles or nonadjacent distinct pairs, but has $d=0$. The displayed positive-degree classification therefore requires at least two vertices. This qualification completes the triangle-free graphs with three common neighbours argument rather than silently excluding a counterexample.

For each vertex triple $S$, let $I_S$ indicate that its three edges are present. Then $X={\sum }_S{I}_S$ andEach variance is $p^3(1-p^3)$. Two distinct triples have independent indicators unless they share an edge; in that case their covariance is $p^5-p^6$. There are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\left(\genfrac{}{}{0ex}{}{n-2}{2}\right)=6\left(\genfrac{}{}{0ex}{}{n}{4}\right)$ unordered pairs sharing an edge. Therefore the triangle count in a binomial random graph hasIf $np\to 0$, Markov inequality gives $\mathrm{Pr}(X>0)\le \mathbb{E}X\to 0$. If $np\to \infty$, then $\mathbb{E}X\to \infty$ andThe second term tends to zero since $n^{2}p=n(np)\to \infty$. Chebyshev inequality now gives $\mathrm{Pr}(X=0)\le \mathrm{Var}X/(\mathbb{E}X{)}^2\to 0$, proving the stated threshold. Here the random-graph phrase “almost surely” means asymptotically almost surely, that is, with probability tending to one.

At $p=n^{-1/2}$, let $Y$ count edges. Direct subtraction givesSome graph has $Y-X$ at least this expectation. Successively deleting one edge from each remaining triangle destroys all triangles with at most $X$ deletions, so a triangle-free graph remains with at least $Y-X$ edges. This triangle deletion in the alteration method proves the requested existence bound.

It is not within a constant factor of the optimum. A balanced complete bipartite graph has $\lfloor n^2/4\rfloor$ edges and no triangle. Conversely, in a triangle-free graph each edge $uv$ satisfies $d(u)+d(v)\le n$. Summing over edges gives ${\sum }_{v}d(v{)}^2\le nM$, where $M$ is its edge count. Cauchy-Schwarz inequality gives $4M^2/n\le \sum d(v{)}^2$, hence $M\le n^2/4$. The optimum is therefore $\boxed{\lfloor n^2/4\rfloor }$, quadratic rather than the obtained $n^{3/2}$ bound.

The lower shadow consists of the sets obtained by removing one element from a member:For a nonempty uniform set family, write its size in its unique decreasing binomial coefficient expansionThe Kruskal-Katona theorem states thatA colexicographic initial segment attains equality; thus this is an exact minimum, not just an asymptotic estimate.

Here is a direct proof of the requested Lovász shadow bound, avoiding any unproved numerical interpolation of the Kruskal-Katona theorem. The real binomial coefficient means $\left(\genfrac{}{}{0ex}{}{x}{r}\right)=x(x-1)\cdots (x-r+1)/r!$. It is strictly increasing for $x>r-1$. Since the set family is nonempty and its size is an integer, necessarily $x\ge r$.

Apply the usual left coordinate shifts of a set family until the set family is fixed by every shift. A shift preserves its size and does not increase its lower shadow; the witness argument is supplied in Solution 3. Termination follows because every nontrivial shift reduces ${\sum }_{A\in \mathcal{A}}{\sum }_{i\in A}i$. It suffices to prove the inequality for this shifted set family.

Induct on $r$, and, for each fixed $r$, on $|\mathcal{A}|$. The cases $r=1$ and $|\mathcal{A}|=1$ give, respectively, the single empty shadow member and the $r$ faces of an $r$-set. Split the set family at coordinate one, writing ${\mathcal{A}}_0$ for the members avoiding it and ${\mathcal{A}}_1$ for the sets obtained by deleting it from members containing it. Every $(r-1)$-subset of a member of ${\mathcal{A}}_0$ belongs to ${\mathcal{A}}_1$: replace the one omitted coordinate by coordinate one. ThereforeThe two terms count shadow members avoiding and containing coordinate one.

Put $b=|{\mathcal{A}}_1|$. This is positive. If $b<\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, the Pascal's identity gives $|{\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r}\right)$. In this case $x>r$ and ${\mathcal{A}}_0$ is a smaller nonempty set family. The mathematical induction hypothesis on family size gives $|\partial {\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, contradicting its containment in ${\mathcal{A}}_1$. Consequently $b\ge \left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$. Write $b=\left(\genfrac{}{}{0ex}{}{z}{r-1}\right)$, so $z\ge x-1$. The mathematical induction hypothesis on uniformity and the Pascal's identity now giveFor $r=2$ the degree-zero term is the constant one, which causes no difficulty.

For the graph application, identify a triangle-free graph with a subset of the $N$ possible edges. Removing an edge preserves being triangle-free, so $\partial {\mathcal{G}}_{m+1}\subseteq {\mathcal{G}}_m$. If ${\mathcal{G}}_{m+1}$ is nonempty, the Lovász shadow bound givesThus, for $1\le m<N$ with $p_m>0$,If the next set family is empty its probability is zero, so the same inequality holds without introducing $x_{m+1}$. If $p_m=0$, every later set family is empty; the divided expression is undefined and the correct assertion is this zero-probability conclusion.

Iterate the monotonicity of the effective ground-set parameter of a hereditary uniform layer. Whenever $p_{m+k}>0$, all the intermediate parameters are at least their layer sizes, and all factors below are nonnegative. We obtain the stronger estimateThe second inequality follows from $x_m\le N$ and the fact that $(x_m-t)/(N-t)$ decreases with $t$. If the target set family is empty, the requested bound follows directly when its right side is nonnegative. When $k>x_m+1$, the printed product can have signed factors and should instead be stopped at the first empty layer: its unrestricted signed form is not a meaningful probability bound. In the nonzero target range there is no such ambiguity. A uniformly safe extension, for $0\le k\le N-m$, isIt agrees with the printed expression whenever all its factors are nonnegative. In particular $0\le k\le m$ is always safe when $p_m>0$, and this includes the requested doubling case.

Taking $k=m$ in the nonzero range,If the $2m$-edge set family is empty the conclusion is immediate; if $2m>N$, interpreting the event as impossible gives the same conclusion. These qualifications distinguish the probability assertion from products or ratios outside their domains.

Here $|V(G)|=2n$ and $e(G)=n^2+1$. By Mantel theorem, a triangle-free graph of this order has at mostedges, so $G$ contains a triangle in a graph.

If $n=2$, then $G$ has four vertices and five of the six possible edges, so it is $K_4$ with one edge deleted. Exactly the two triangles containing neither both endpoints of the deleted edge remain. Hence $G$ contains two triangles.

The extremal number $\mathrm{ex}(n,H)$ is the maximum number of edges in an $n$-vertex graph containing no subgraph isomorphic to $H$.

Let $G$ be triangle-free with $e$ edges. For every edge $uv$, the neighbourhoods of $u$ and $v$ are disjoint apart from their endpoints, so
$d(u)+d(v)\le n$. Summing over edges and applying Cauchy-Schwarz inequality giveswhileThus $e\le n^2/4$, proving the required Mantel theorem bound.

Let $X$ count the triangles of the binomial random graph $G(n,p)$. ThenTwo distinct triangle indicators are dependent only when the triangles share an edge. Hence their ordered Janson dependency sum is

If $0<p\le n^{-1/2}$, then $\Delta =O(\mu )$. The first Janson inequality givesFor the reverse bound, the events that individual triangles are absent are decreasing, so Harris' inequality giveswhere $p<1/2$ keeps the logarithmic estimate uniform. Thus the probability is $e^{-\Theta (p^3{n}^3)}$.

If $n^{-1/2}\le p<1/2$, the extended Janson bound givesFor a lower bound, fix a balanced bipartition of the vertices and require every edge inside either part to be absent. The resulting graph is bipartite, hence triangle-free, and this event has probabilityCombining the bounds proves the second regime.

Fix the requested $\epsilon >0$ and choose much smaller constants $\eta ,\delta ,\rho >0$ in that order. Apply the Szemerédi regularity lemma with regularity parameter $\rho$ and a sufficiently large lower bound $m_0$ to a largest triangle-free subgraph $F\subseteq G$. Letbe the resulting equitable partition, and form the reduced graph of a regularity partition $R$ on $[m]$ by joining $i$ and $j$ when $(V_i,V_j)$ is $\rho$-uniform in $F$ and has $F$-density at least $\delta$.

Choose $\rho <\delta /2$. If $R$ contained a triangle, the triangle embedding lemma for regular pairs would give a triangle in $F$. Thus $R$ is triangle-free, and Turan theorem gives

Part (c), with error $\eta$, holds simultaneously for every pair of sets of size at least $n/\log n$. Since $m$ is bounded independently of $n$, all regularity classes and their halves satisfy this size condition for large $n$. It follows that every cross-pair contains at most $(p+\eta )|V_i||V_j|$ edges of $G$. It also givestake a bipartition of $V_i$ carrying at least half its internal edges and apply the cross-pair estimate.

Now count the edges of $F$. Pairs represented in $R$ contribute at mostThe exceptional set, the at most $\rho m^2$ irregular pairs, pairs of $F$-density below $\delta$, and edges inside classes together contribute at mostChoose $\eta$, then $\delta ,ho$, and finally $m_0^{-1}$ small enough in terms of $p\epsilon$. The total is at mostwith high probability, as required.

Apply the Shearer independence bound for a triangle-free graph. It states that an $n$-vertex triangle-free graph of maximum degree at most $d$ has an independent set of sizefor an absolute $c>0$. The result is immediate for bounded $d$ after reducing $c$, while the theorem gives the asserted logarithmic gain for $d$ large. Thus

For completeness, the key input in Shearer's proof is to expose a random independent set one degree scale at a time. Triangle-freeness makes every neighbourhood independent, so conditioning on earlier choices creates no edges inside the available neighbours. The expected gain at degree scale $j$ is $\Omega (n_j/d)$; summing over the $\Theta (\log d)$ nonempty scales gives $\Omega (n\log d/d)$. The entropy, or hard-core-model, form of the argument makes this scale calculation rigorous without losing vertices counted at adjacent scales.

Maintain lists of available colours, tentatively colour small random portions, erase conflicts and prune colours used by neighbours. In a triangle-free graph, a vertex's neighbours form an independent set, which enables local estimates. Track list size and average remaining neighbour congestion per colour; prune unusually congested colours before the next round. Concentration and the Lovász local lemma preserve these invariants until lists have enough slack for completion. Tracking averages matters because triangle-free graphs can still contain many four-cycles and need not have uniform per-colour congestion.

Choose $n=d^7$ and a binomial random graph of edge probability $d/n$. With high probability its maximum degree is at most $2d$, it has $o(n)$ triangles, and every independent set has size at most $4n\log d/d$. Delete one vertex per triangle. The resulting triangle-free graph has chromatic number at least $d/(8\log d)$. Appending a disjoint star of degree $2d$, if necessary, makes the maximum degree exactly $2d$ without lowering the chromatic number. Thus the order $\Delta /\log \Delta$ is unavoidable.

Define $M_2=K_2$ and $M_{k+1}=\mu (M_k)$ for $k\ge 2$. By the Mycielskian chromatic increase and Mycielskian triangle preservation, each $M_k$ is a finite triangle-free graph with chromatic number $k$. The first steps are $K_2,C_5$, the Grötzsch graph, and then a 23-vertex graph needing five colours. The vertex recurrence $n_{k+1}=2n_k+1$ gives $n_k=3\cdot 2^{k-2}-1$. These explicit examples show how arbitrarily strong graph coloring constraints can accumulate without any three-vertex clique.