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A graph Ramsey number $R(s,t)$ is the least $N$ such that every red-blue edge colouring of the complete graph $K_N$ has a red $K_s$ or a blue $K_t$. At a vertex in $K_{R(s-1,t)+R(s,t-1)}$, either at least $R(s-1,t)$ neighbours have red incident edges or at least $R(s,t-1)$ have blue incident edges. In the first case find a red $K_{s-1}$ and adjoin the vertex, or a blue $K_t$; the second case is symmetric. ThusStarting with $R(2,t)=t$ and $R(s,2)=s$, induction gives the binomial upper bound for a Ramsey numberFor the lower construction, partition $2t-2$ vertices into two classes of size $t-1$, colour inside-class edges red and between-class edges blue. Red cliques have at most $t-1$ vertices, and the blue graph is bipartite, so has no odd cycle.

Conversely, a graph without an odd cycle is bipartite: in each connected component colour vertices by the parity of the length of a path from a root; differing path parities would give an odd closed walk, hence an odd cycle. On $2t-1$ vertices one of the two blue bipartition classes has at least $t$ vertices. Every edge within it is red, giving a red $K_t$. Thus $2t-1$ is the exact threshold for this alternative.

For a $\Delta$-regular undirected graph, its Graph Laplacian is $L=\Delta I-A$, and for any complex vector $v$,If $Av=\Delta v$, the left side is zero, hence $v_i=v_j$ on every edge. In a connected graph all coordinates coincide. Thus the eigenspace is exactly the constant vectors, of dimension one. Since the adjacency matrix of a graph is symmetric and diagonalizable,

If every vertex has degree $\Delta$, then each row of the adjacency matrix of a graph sums to $\Delta$. The nonzero constant vector therefore satisfiesso $\Delta$ is an eigenvalue. This uses the usual finite, nonempty graph convention.

Let $A$ be the adjacency matrix of a graph and $Av=\lambda v$ with $v\ne 0$. Choose a vertex $i$ maximizing $|v_i|>0$. ThenDividing givesThis argument works even with a complex eigenvector; for a finite undirected graph the matrix is real symmetric and all eigenvalues are real.

Write $\alpha =1-1/r$. The Erdős-Stone theorem states that, for a fixed graph $F$ with chromatic number $\chi (F)=r+1\ge 2$,Here $\mathrm{ex}$ is the extremal number, and copies are subgraphs, not necessarily induced subgraphs. We write $K_a(b)$ for the balanced complete multipartite blow-up with $a$ classes of $b$ vertices each. All asymptotic errors below concern $n\to \infty$ with the forbidden graphs fixed.

The stability subgraph. We prove the high-minimum-degree multipartite stability subgraph assertion directly from the Erdős-Stone theorem. For $r=1$, the edgeless spanning subgraph suffices. Suppose $r\ge 2$.

First repeatedly remove a vertex whose degree of a vertex is less than $(\alpha -{\epsilon }_n)$ times the current order, where ${\epsilon }_n\to 0$ sufficiently slowly. The Erdős-Stone theorem gives, uniformly for every $K_{r+1}(t)$-free subgraph $J$ of order at most $n$,Indeed, apply the asymptotic bound at large orders; the contribution from bounded or sufficiently small orders is negligible relative to $n^2$. If $m$ vertices survive, the removed edges number at mostComparing with $e(G_n)=\alpha n^2/2+o(n^2)$ shows that ${\epsilon }_n(n^2-m^2)=o({\epsilon }_n{n}^2)$, provided ${\epsilon }_n$ dominates ${\eta }_n$, the initial relative error, and $1/n$. Consequently $m=n-o(n)$. The surviving induced subgraph $G^{\prime }$ satisfies $\delta (G^{\prime })\ge (\alpha -o(1))m$ and $e(G^{\prime })=\alpha m^2/2+o(m^2)$.

By the Erdős-Stone theorem, $G^{\prime }$ contains $K_r(L)$ for some $L=L_n\to \infty$ as slowly as necessary. This follows because $\alpha$ exceeds the extremal number density for each fixed $K_r(L)$ by the positive gap $1/(r(r-1))$. Choose $L$ so slowly that $rL=o(m)$ and ${\left(\genfrac{}{}{0ex}{}{L}{t}\right)}^r=o(m)$, and denote its root classes by $A_1,\dots ,A_r$.

An outside vertex having at least $t$ neighbours of a vertex in every $A_i$ determines one choice of a $t$-subset in each $A_i$. At most $t-1$ outside vertices can determine any fixed choice, since otherwise these vertices and the chosen root sets form $K_{r+1}(t)$. Thus at most $(t-1){\left(\genfrac{}{}{0ex}{}{L}{t}\right)}^r=o(m)$ outside vertices have this property. Exclude those vertices and the root classes; put each remaining vertex in a class $C_i$ for which it has fewer than $t$ neighbours of a vertex in $A_i$.

Write $\delta (G^{\prime })\ge (\alpha -{\gamma }_n)m$, with ${\gamma }_n\to 0$. Counting missing incidences with $A_i$ givesSince ${\sum }_i|C_i|=m-o(m)$, it follows that $|C_i|=m/r+o(m)$ for every $i$. The total missing incidences from all root vertices to assigned vertices are at most $(1+r{\gamma }_n)Lm$. At least $(L-t+1)(m-o(m))$ of them are incidences with a vertex's own assigned root class. Hence only $o(Lm)$ missing incidences go to other root classes.

Choose ${\rho }_n\to 0$ slowly and discard the $o(m)$ assigned vertices with more than ${\rho }_{n}L$ missing incidences to other root classes. Each remaining vertex of $C_i$ is adjacent to all but at most ${\rho }_{n}L$ vertices of every $A_j$, $j\ne i$. If its class contained a complete bipartite graph $K_{t,t}$, the $2t$ vertices of this copy would have at least $L-2t{\rho }_{n}L\ge t$ common neighbours of a vertex in each other root class. These sets extend the copy to $K_{r+1}(t)$, a contradiction. By the bipartite case of the Erdős-Stone theorem, the number of internal edges in each surviving class is $o(m^2)$.

The surviving graph still has $\alpha m^2/2+o(m^2)$ edges, whereas the total possible cross-class edges, with these balanced classes, is $\alpha m^2/2+o(m^2)$. Thus only $o(m^2)$ cross-class edges are missing. Discard the $o(m)$ vertices missing more than ${\sigma }_{n}m$ cross-class edges, with ${\sigma }_n\to 0$ slowly, and delete all internal edges. The resulting $r$-partite subgraph $H$ has classes $H_i$ of size $n/r+o(n)$ andFor the lower bound, every surviving vertex misses only $o(n)$ vertices outside its class; the upper bound follows from the class sizes. This also records the uniform cross-class error that we need next.

Minimum degree in an extremal graph. Fix $F$ with $\chi (F)=r+1$. Choose $t$ large enough that $F$ embeds in $K_{r+1}(t)$. An extremal $F$-free graph $G$ is therefore $K_{r+1}(t)$-free, and its edge count has the required asymptotics by the Erdős-Stone theorem. For $r\ge 2$ take $H$ as above.

For any vertex $v$, remove $v$ and introduce a new vertex $v^{\ast }$ whose vertex neighbourhood consists precisely of the vertices in $H_2\cup \cdots \cup H_r$ other than $v$. The new degree of a vertex is $\alpha n-o(n)$. This new graph remains $F$-free. Indeed, a new copy of $F$ must use $v^{\ast }$, and all its neighbours of a vertex in that copy lie in the other classes of $H$. There are only boundedly many of them. Each misses only $o(n)$ vertices in $H_1$, so an unused vertex $w\in H_1$ is adjacent to all of them. Replacing $v^{\ast }$ by $w$ would give $F$ in $G-v$, which is impossible.

Extremality now forces $d_G(v)\ge \alpha n-o(n)$, uniformly in $v$. The upper bound for the minimum degree of a graph follows from $\delta (G)\le 2e(G)/n$. For $r=1$, $e(G)=o(n^2)$ already gives $0\le \delta (G)=o(n)$. ThusThis is the minimum degree of an extremal forbidden-subgraph graph principle.

The exact extremal graph for disjoint cliques. DefineThe plus sign denotes the join of graphs. The Turan graph contains no $K_{r+1}$, so every such clique in $J_s(n)$ uses a vertex of $K_{s-1}$. Therefore $J_s(n)$ contains no $s$ vertex-disjoint such cliques, and the extremal number is at least $b_s(n)$.

Induct on $s$, with $s=1$ supplied by the Turan theorem, including its equality characterization. Let $s\ge 2$ and let $G$ be extremal for $s{K}_{r+1}$. The preceding results give $\delta (G)\ge \alpha n-o(n)$ and the $r$-partite subgraph $H$. Keep its classes $H_i$ fixed and assign each of the remaining $o(n)$ vertices to a class in which it has fewest neighbours of a vertex in $H$. Write the resulting classes as $C_i$; all have size $n/r+o(n)$. Choose a positive ${\tau }_n\to 0$ so slowly that ${\tau }_{n}n$ dominates the outside count, every cross-class error in $H$, and every error in the minimum degree of a graph bound.

Suppose some vertex $x\in C_i$ has at least ${\tau }_{n}n$ internal neighbours of a vertex. If $x\notin H$, the assignment rule shows that $x$ has at least ${\tau }_{n}n-o(n)$ neighbours of a vertex in every $H_j$. If $x\in H_i$, it has this many in $H_i$ and $n/r-o(n)$ in the other $H_j$. If $G-x$ contained $s-1$ disjoint $K_{r+1}$, we could choose one vertex in each $N(x)\cap H_j$, successively avoiding those copies and the $o(n)$ missing cross-neighbours of the already chosen vertices. This produces a further $K_{r+1}$ through $x$. For $r=1$, this just chooses an unused neighbour of $x$. Hence $G-x$ is $(s-1)K_{r+1}$-free, and induction givesSince the candidate attains $b_s(n)$, equality holds throughout: $x$ is a universal vertex and $G-x$ is the inductively unique extremal graph. Thus $G$ is isomorphic to $J_s(n)$.

It remains to show that this case must occur. Otherwise all internal vertex degrees are less than ${\tau }_{n}n=o(n)$. Together with the lower bound on $\delta (G)$ and $|C_i|=n/r+o(n)$, this implies that every vertex misses only $o(n)$ vertices outside its own class. If some $C_i$ contains a matching in a graph of $s$ edges, extend its edges one by one to $K_{r+1}$, choosing one vertex from each other class. At every step only $o(n)$ cross-neighbours and boundedly many used vertices are excluded. This gives $s$ disjoint forbidden cliques, a contradiction.

A maximal matching in each class therefore has at most $s-1$ edges, and its at most $2(s-1)$ endpoints form a vertex cover of the internal edges. Their internal vertex degrees are all $o(n)$, so the total internal edge count is $o(n)$. The cross-class edge count is at most $e(T_r(n))$, whenceBut the balanced part sizes of the Turan graph givecontradicting extremality for $s\ge 2$. This argument also works for $r=1$. Consequently, for all sufficiently large $n$,The uniqueness is up to isomorphic graphs, as usual for an extremal graph for disjoint cliques.

Symmetrization for an arbitrary real coefficient. Among graphs maximizing $f(G)=e(G)-c{k}_3(G)$, choose one maximizingFor a vertex $x$, put $L(x)=d(x)-ce(G[N(x)])$. This is the contribution of edges and triangles in a graph containing $x$. If $u,v$ are nonadjacent, Zykov symmetrization replacing $u$ by a clone of $v$ changes $f$ by $L(v)-L(u)$; the opposite replacement changes it by $L(u)-L(v)$. Maximality implies $L(u)=L(v)$, so both changes preserve $f$.

Let $S=N(v)\setminus N(u)$ and $T=N(u)\setminus N(v)$. In the first replacement the vertex degrees in $S$ increase by one and those in $T$ decrease by one; in the opposite replacement these changes reverse. The changes to $d(u{)}^2+d(v{)}^2$ cancel when the two replacements are added. Since $(d+1{)}^2+(d-1{)}^2-2d^2=2$,If the two vertex neighbourhoods differ, one replacement increases $Q$, a contradiction. Thus every pair of nonadjacent vertices has the same vertex neighbourhood. Nonadjacency, with equality allowed, is an equivalence relation: if $u$ and $v$ are nonadjacent and $v$ and $w$ are nonadjacent, their identical vertex neighbourhoods forbid $uw$ as well. Its classes are independent sets and every cross-class edge is present. ThereforeThis edge-triangle symmetrization works for either sign of $c$.

The exact supporting line. For $n\ge 3$, writeWe prove the triangle support line between bipartite and tripartite Turan graphs by maximizing $e-c_{\ast }{k}_3$. First $c_{\ast }\ge 2/n$. For $n=3,4,5$, its values are respectively $1,1/2,1/2$. For $n\ge 6$, the exact balancing of the Turan graph gives $e_3\ge n^2/3-1/3$, while $e_2\le n^2/4$ and the arithmetic-geometric mean inequality gives $t_3\le n^3/27$. Hence

By edge-triangle symmetrization a maximizing graph can be taken complete multipartite graph; among such maximizers use one with the fewest nonempty parts. If there are at least four parts, let $a,b$ be the two smallest sizes. Then $a+b\le n/2$. Merging them loses $ab$ edges and $ab(n-a-b)$ triangles in a graph, so the objective changes byThis contradicts the choice of the number of parts. Thus at most three parts remain.

For three part sizes $a\ge b\ge d>0$, the objective isIf $1-c_{\ast }b\le 0$, merging the parts of sizes $a,d$ does not decrease it, again contradicting minimality. If $1-c_{\ast }b>0$ and $a-d\ge 2$, transferring one vertex from the largest to the smallest part raises $ad$ by $a-d-1>0$ and strictly increases the objective. Thus a three-part maximizer has all sizes differing by at most one, and is $T_3(n)$. A maximizer with at most two parts has objective at most $e_2$, since it has no triangles in a graph and the Mantel theorem bounds its edges. Finally,Therefore every graph satisfies $e(G)-c_{\ast }{k}_3(G)\le e_2$. At the prescribed edge count this givesFor $n\le 2$, the conclusion is simply the nonnegative lower bound zero. All rounding in the Turan graphs has been retained.

One edge above the bipartite threshold. Write $n=2m$. We prove the Triangle lower bound one edge above the Mantel threshold by induction on $m\ge 2$, allowing at least $m^2+1$ edges. For $m=2$, five edges on four vertices give two triangles in a graph, and adding edges preserves this bound.

We will also use the following consequence of an inductive bound: on $2a$ vertices, at least $a^2+q$ edges, for an integer $q\ge 1$, force at least $a+q-1$ triangles in a graph. Delete surplus edges first to leave exactly $a^2+q$. Repeatedly delete an edge in a triangle in a graph until $a^2+1$ edges remain. The Mantel theorem guarantees such a triangle in a graph at every step; each deletion destroys at least one distinct triangle in a graph. Apply the inductive bound to the remaining graph.

Now take exactly $m^2+1$ edges. If every edge lies in a triangle in a graph, the edge-triangle incidence bound gives $3k_3(G)\ge m^2+1$. This is at least $3m$ for $m\ge 3$; the base case was handled separately.

Otherwise choose an edge $uv$ in no triangle in a graph. Its endpoints have disjoint vertex neighbourhoods, so $d(u)+d(v)\le 2m$. Deleting $u,v$ removes exactly $d(u)+d(v)-1$ edges and leaves at least $(m-1{)}^2+1$ edges. If $d(u)+d(v)\le 2m-1$, it leaves at least $(m-1{)}^2+2$ edges, and the strengthened inductive bound gives at least $m$ triangles in a graph already.

If $d(u)+d(v)=2m$, induction gives at least $m-1$ triangles in a graph after deletion. There must be an additional triangle in a graph containing $u$ or $v$. Otherwise both $N(u)$ and $N(v)$ are independent sets; since they are disjoint and cover all vertices, $G$ would be bipartite graph, contradicting $e(G)>m^2$ by the Mantel theorem. HenceThe bound is sharp: add one internal edge to one class of the complete bipartite graph $K_{m,m}$. Its triangles in a graph are precisely that edge together with each of the $m$ vertices in the other class.

Statement and definitions. For disjoint nonempty sets $A,B$ of vertices, their edge density of a bipartite graph is $d(A,B)=e(A,B)/(|A||B|)$. They form a regular pair of vertex sets, or an $\epsilon$-uniform pair, ifwhenever $X\subseteq A$, $Y\subseteq B$, $|X|\ge \epsilon |A|$ and $|Y|\ge \epsilon |B|$.

The Szemerédi regularity lemma states that for every $\epsilon >0$ and positive integer $m_0$, there are integers $M,n_0$ such that every graph of order $n\ge n_0$ has a set partitionwith $m_0\le k\le M$, $|V_0|\le \epsilon n$, equal nonzero sizes $|V_1|=\cdots =|V_k|$, and at most $\epsilon k^2$ unordered pairs $(V_i,V_j)$, $1\le i<j\le k$, that fail to be $\epsilon$-uniform. It is enough to prove this for $0<\epsilon \le 1/2$, since decreasing the parameter strengthens the conclusion.

Energy increment. For the proof, regard each exceptional vertex as a singleton part and use the equitable regularity energyThe sum is over ordered pairs, including diagonal pairs; for those pairs define $d$ as the mean adjacency indicator on $A\times B$. Thus $0\le q\le 1$. Under a refinement, the subrectangle densities have the original density as their weighted mean. The Cauchy-Schwarz inequality, equivalently the nonnegativity of their weighted variance, gives $q({\mathcal{P}}^{\prime })\ge q(\mathcal{P})$.

Suppose there are more than $\epsilon k^2$ irregular pairs, with main cell size $m$. For each such pair choose witnesses $X\subseteq V_i$, $Y\subseteq V_j$ of sizes at least $\epsilon m$ and with $|d(X,Y)-d(V_i,V_j)|>\epsilon$. Refine every main cell by all its witness subsets. Each splits into at most $2^{k-1}$ atoms, and we may use the larger bound $2^k$.

For an ordered irregular rectangle, write its refined densities as $d_{ab}$ and relative areas as $w_{ab}$. The energy gain on that rectangle isThe witness rectangle has relative area at least ${\epsilon }^2$. Applying the Cauchy-Schwarz inequality to the subrectangles it contains shows that this expression is greater than ${\epsilon }^4{m}^2/n^2$. There are two orientations of each irregular pair. Since $km=n-|V_0|\ge n/2$, the total gain is greater than

Restoring equal sizes without losing energy. Put $\ell =\lfloor m/4^k\rfloor$, assuming $m\ge 2\cdot 4^k$. Split every refined atom into sets of size $\ell$, putting its leftover fewer than $\ell$ vertices into the exceptional set as singleton parts. This is a further refinement, so the energy cannot decrease. The number of new exceptional vertices is at mostThe number $k^{\prime }$ of new main cells satisfiesFor the first inequality, in each old cell the leftovers have total size less than $2^k\ell <m$, so at least one full chunk survives. All previous exceptional singleton parts are kept. Treating leftovers as singleton parts is what makes the energy increment survive equitabilization exactly.

Take $S=\lceil 2{\epsilon }^{-5}\rceil +1$ and choose $k_0\ge m_0$ so large that $S{2}^{-k_0}\le \epsilon /2$. Start with $k_0$ equal main cells, with fewer than $k_0$ exceptional vertices. Define a finite bound $M$ by iterating $x\mapsto 2x{4}^x$ a total of $S$ times, starting from $k_0$. Choose $n_0$ sufficiently large that $k_0\le \epsilon n/2$ and $n/(2M)\ge 2\cdot 4^M$ whenever $n\ge n_0$.

At each of the first $S$ refinements, the exceptional set has size at mostthe cell count is between $k_0$ and $M$, and $m\ge n/(2M)$, so all the chunk sizes above are valid. But $S$ unsuccessful refinements would raise $q$ by more than $S{\epsilon }^5/2>1$, which is impossible. Thus the procedure terminates at a partition satisfying the Szemerédi regularity lemma.

Three linearly large uniform sets. Apply the Szemerédi regularity lemma withFor $n\ge n_0$, delete all edges touching $V_0$, all internal main-cell edges, all edges of irregular pairs, and all edges of regular pairs whose edge density of a bipartite graph is less than $d_0$. The respective losses are bounded byThus the total loss is at most $0.00102n^2$, and the remaining graph has at leastedges. Form the reduced graph of a regularity partition $R$, joining two main cells precisely when they are $\epsilon$-uniform and have edge density of a bipartite graph at least $d_0$. If their common size is $m$, the remaining edge count is at most $e(R)m^2$. Since $km\le n$, we obtain $e(R)>k^2/4$. The Mantel theorem gives a triangle in a graph in $R$. Its three cells $U_1,U_2,U_3$ have the required pairwise density and are $10^{-3}$-uniform: an $\epsilon$-uniform pair is uniform for any larger parameter.

Each selected cell has size $m\ge (1-\epsilon )n/M$. ChooseFor $n<n_0$, the Mantel theorem gives a triangle in a graph in the original graph, since its edge count exceeds $n^2/4$; take its three singleton vertices. Their pairs have density one and are uniform for every positive parameter, and $1>cn$. Therefore in all casesThe constant is absolute; the energy proof gives a very large bound for $M$, which does not affect its positivity.

Use natural logarithms. A $K_t$ graph minor is represented by $t$ disjoint nonempty branch sets of a graph minor, each inducing a connected graph, with an edge between every pair. The complete graph minor density threshold is defined over nonempty finite graphs byIt uses $e(G)/|G|$, half the average degree of a vertex; the nonempty convention excludes the vacuous density inequality for the empty graph. The PDF has the intended definition with a universal implication over $G$; the TeX transcription corrupts that definition.

Probabilistic lower bound. Put $N=\lfloor t\sqrt{\log t}/4\rfloor$ and take the binomial random graph $G(N,1/2)$. Its edge count has mean $N(N-1)/4$. The Chernoff bound impliesWe show that the probability of a $K_t$ graph minor tends to zero. Ignore connectedness, which only enlarges the set of possible models, and assign each of the $N$ vertices one of $t$ branch labels or an unused label. There are at most $(t+1{)}^N$ assignments.

For any assignment with $t$ nonempty branch sets, at least $t/2$ of them have size at most $2N/t\le \sqrt{\log t}/2$. Between any two such small sets $A,B$, the probability of no edge isThe absence or presence of edges between different pairs of branch sets depends on disjoint sets of random edges, so these events are independent. For this assignment the probability that all small pairs are adjacent is at mostThe union bound over assignments is therefore at mostsince the positive term is $O(t(\log t{)}^{3/2})$ and the negative term has order $t^{7/4}$. There is consequently a graph with no $K_t$ graph minor and with $e(G)/N\ge N/8$. Any constant having the universal forcing property must exceed this ratio. Since $N/8\ge t\sqrt{\log t}/64$ for large $t$,

The auxiliary density lemma contains a typo. Both the original PDF and the TeX print $e(G)\le 11k|G|$ in the hint. This cannot imply its conclusion: for $k\ge 1$, an edgeless graph has only edgeless graph minors, and none satisfies $2\delta (H)\ge |H|+4k-1$. We use the intended dense minor with bounded order and high minimum degree lemma with the hypothesis $e(G)\ge 11k|G|$ for nonempty $G$. The upper-bound argument below depends on that corrected supplied lemma; the literal printed hint is false.

Probabilistic upper bound and the constant seven. Suppose $G$ is nonempty and $e(G)\ge 7t\sqrt{\log t}|G|$, and setThe corrected lemma gives a graph minor $H$ of order $N\le 11k+2$ with $2\delta (H)\ge N+4k-1$. Since $\delta (H)\le N-1$, we also have $N\ge 4k+1$. For large $k$, every vertex has at most $N/3$ nonneighbours, counting itself, becausethe last inequality uses $N\le 11k+2\le 12k-3$. Moreover any two distinct vertices have at leastcommon neighbours.

Choose $2t$ disjoint random $\ell$-sets $A_1,\dots ,A_{2t}$ from $V(H)$, uniformly, which is possible since $2t\ell <4k-1\le N$ for sufficiently large $t$. For a set $A$, let $U(A)$ be the set of vertices having no neighbour in $A$. For each vertex, the probability that all members of a uniformly sampled $\ell$-set are its nonneighbours is at most $3^{-\ell }$, even when sampling without replacement. By linearity of expected value,Call $A$ good if $|U(A)|\le N{3}^{-19\ell /20}$. The Markov inequality shows that a random set is bad with probability at most $3^{-\ell /20}=o(1)$. Thus the expected number of bad sets among our $2t$ samples is $o(t)$.

Conditional on a fixed good $A_i$, the marginal distribution of $A_j$ is uniform among $\ell$-sets outside $A_i$. For there to be no edge between them, all its members must lie in $U(A_i)$. HenceHere $(N/(N-\ell ){)}^{\ell }=\exp (O({\ell }^2/N))=1+o(1)$. Consequently the expected number of nonadjacent pairs of good sets is at mostsince $(19/20)\log 3>1$. Applying the Markov inequality to both counts shows that there exists a choice with fewer than $t$ bad sets and at most $t$ nonadjacent pairs of good sets. Select $t$ of the good sets; they still have at most $t$ missing adjacencies.

Make each selected set connected by fixing one of its vertices as a root and, for each of its other members, adding one unused common neighbour of that member and the root. This uses at most $t(\ell -1)$ additional vertices. Then for each pair with no original edge, choose one vertex in each set and add an unused common neighbour to one of the sets. The added vertex is joined to that set and to the other set, so it preserves connectedness and repairs their adjacency. At most $t$ such repairs are required.

Throughout this process the total number of occupied vertices is at mostSince every pair has at least $4k-1$ common neighbours, an unused choice always exists. The final sets are disjoint connected branch sets of a graph minor with every pair adjacent. They represent $K_t$ in $H$, and hence in $G$, by transitivity of graph minors. This random branch-set construction of a complete graph minor proves

The sharp linear threshold for a four-vertex complete minor. We prove by induction that a graph on $n\ge 4$ vertices with at least $2n-2$ edges has a $K_4$ graph minor. The case $n=4$ is $K_4$ itself. If some edge $uv$ has at most one common neighbour, an edge contraction reduces the order by one and removes exactly $1+|N(u)\cap N(v)|\le 2$ edges. The contracted graph has at least $2(n-1)-2$ edges, so induction applies.

Otherwise every edge is in at least two triangles in a graph. Choose a vertex $v$ incident with an edge. Each vertex $u\in N(v)$ has at least two neighbours of a vertex in $N(v)$, since these are the common neighbours of $u,v$. Thus $G[N(v)]$ has minimum degree of a graph at least two. A longest path in this finite graph has an endpoint adjacent to an earlier nonconsecutive path vertex, and hence contains a cycle in a graph. Together with $v$, this cycle in a graph gives a wheel graph as a subgraph. Divide its rim into three consecutive nonempty connected arcs; these arcs and $\{v\}$ form four pairwise adjacent branch sets of a graph minor. Therefore

With $2n-3$ edges the answer is no. Take the join of graphs of $K_2$ and an independent set of $n-2$ vertices. It has $1+2(n-2)=2n-3$ edges. Among four disjoint putative branch sets of a graph minor, at most two can contain the two vertices of $K_2$. Any connected branch set avoiding them is a singleton in the independent set. At least two such singleton sets would therefore have no edge between them. This excludes a $K_4$ graph minor for every $n\ge 4$ and proves sharpness.