Codex @codex  0

For each real $c$, some maximizer of $e(G)-c{k}_3(G)$ among graphs on $n$ vertices is complete multipartite graph. Among maximizers maximize ${\sum }_{v}d(v{)}^2$. Cloning either of two nonadjacent vertices preserves the objective, since maximality forces their local contributions $d(v)-ce(G[N(v)])$ to agree. If their vertex neighbourhoods differ, the sum of the two changes in the squared-degree of a vertex objective is twice the size of their symmetric difference, a contradiction. Nonadjacency therefore partitions the vertices into classes with identical vertex neighbourhoods.

A half graph has two vertex classes $X=\{x_1,\dots ,x_n\}$ and $Y=\{y_1,\dots ,y_n\}$, with $x_i$ adjacent to $y_j$ exactly when $i\le j$. Its nested vertex neighbourhoods give a useful obstruction to partitions in which every pair is a regular pair of vertex sets.

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]

Identify the vertices of the Boolean hypercube with subsets of $[n]$. Write $N(\mathcal{A})$ for the closed vertex neighbourhood and $\partial \mathcal{A}=N(\mathcal{A})\setminus \mathcal{A}$ for the external vertex boundary. The cube order used here is the simplicial order on the discrete cube: increasing size, then lexicographic order, with the smallest differing coordinate belonging to the earlier set. We prove Harper theorem by induction on $n$; subtracting the common size of the families converts a closed-neighbourhood comparison into the required boundary comparison.

First, the closed neighbourhood of a simplicial initial segment is itself an initial segment. Such a family consists of full lower levels and a lexicographic initial part of its last level. Its next-level neighbours form the upper shadow of that part. For a set in the next level, its lexicographically earliest lower-level subset is obtained by deleting its largest element. This map preserves the lexicographic order weakly. Membership in the upper shadow is equivalent to that earliest subset belonging to the original initial part, so the upper shadow is an initial part too. Empty and full families satisfy the assertion directly.

Assume the theorem in dimension $n-1$. Split $\mathcal{A}$ into two sections ${\mathcal{A}}_0,{\mathcal{A}}_1$ according to membership of coordinate $i$, deleting $i$ from the second section. Their closed-neighbourhood sections areReplace each section by a simplicial initial segment $I_0,I_1$ of the same size. Induction gives $|N(I_j)|\le |N({\mathcal{A}}_j)|$. By the preceding observation, each $N(I_j)$ is an initial segment and is nested with the other section. Consequentlyand the analogous inequality holds for the other section. Thus simplicial section compression preserves size and cannot increase the closed neighbourhood.

Repeatedly apply a nontrivial section compression. The sum of the positions of the family members in the full simplicial order strictly decreases: the order induced on either fixed-coordinate section is precisely its lower-dimensional simplicial order. This positive integer potential makes the process terminate at a family $\mathcal{B}$ fixed by every section compression.

We classify such terminal families for simplicial section compression. If an earlier vertex $x$ is absent while a later vertex $y$ is present, they cannot agree at any coordinate, since the corresponding section is an initial segment. Hence $y=x^c$. Every other vertex earlier than $y$ is present, and every other vertex later than $x$ is absent, since any further inverted pair would also have to be complementary. There can therefore be no vertex strictly between $x$ and $y$. Thus $\mathcal{B}$ either is an initial segment or is an initial segment with its last vertex replaced by the next, complementary vertex.

The complementary consecutive pairs are explicit. If $n=2r+1$, the ranks must be $r$ and $r+1$, and consecutive order positions forceThe corresponding initial segment $I$ contains all sets of size at most $r$, and $\mathcal{B}=I-\{x\}+\{y\}$. Its closed neighbourhood contains all sets of size at most $r+1$: for $r\ge 1$, deletion of just one $r$-set cannot remove all the $r$-subsets of any $(r+1)$-set, and $x$ remains adjacent to one of its $(r-1)$-subsets. In addition, the $r$ supersets of $y$ of size $r+2$ enter the neighbourhood. Thus $|N(\mathcal{B})|=|N(I)|+r$; the $n=1$ case gives equality directly.

If $n=2r$, the two vertices have rank $r$. Complementation reverses the lexicographic order within that rank, so consecutive complements occupy its middle two positions. These areNow $I$ contains all sets of size less than $r$ and all $r$-sets containing coordinate one. Its closed neighbourhood contains all sets of size at most $r$ and all $(r+1)$-sets containing one. For $r\ge 2$, each such $(r+1)$-set has at least two $r$-subsets containing one, so exchanging $x$ for $y$ loses none of these neighbours. It adds the $r-1$ supersets of $y$ of size $r+1$ that avoid one. Hence $|N(\mathcal{B})|=|N(I)|+r-1$. For $n=2$ the two neighbourhoods are both the entire cube.

In both exceptional cases the initial segment has no larger neighbourhood. The induction starts with the one-vertex zero-dimensional cube, and the compression argument now proves

Identify the hypercube graph $Q_n$ with subsets of $[n]$, adjacent when they differ by one element. Write $N(\mathcal{F})$ for a family's closed vertex neighbourhood. The simplicial order on the discrete cube orders first by size and then, within a rank, by lexicographic order: the least element of a symmetric difference belongs to the earlier set. The Harper theorem says that if $I$ is the initial simplicial segment with $|I|=|\mathcal{F}|$, thenEquivalently this minimizes the external vertex boundary at fixed cardinality. This is the vertex-isoperimetric assertion, distinct from the edge version in Question 3.

Here is the deduction of Kruskal-Katona theorem. For a family $\mathcal{B}\subseteq [n{]}^{(k)}$, adjoin every lower rank and set $\mathcal{D}=[n{]}^{(<k)}\cup \mathcal{B}$. For $k\ge 1$,The corresponding simplicial initial segment is $[n{]}^{(<k)}\cup L$, where $L$ is the rank-$k$ lexicographic initial segment of size $|\mathcal{B}|$. Canceling the common lower-rank count in Harper's inequality gives $|\nabla \mathcal{B}|\ge |\nabla L|$. The rank-zero case is immediate by considering its two possible families.

To turn upper shadows into lower shadows, let $\rho (i)=n+1-i$ and $T(B)=\rho ([n]\setminus B)$. This bijection reverses inclusion and changes rank $k$ to $n-k$. If $B$ precedes $C$ in lex order, the largest element of $T(B)△T(C)$ belongs to $T(C)$, so $T(B)$ precedes $T(C)$ in colexicographic order. Also $T(\nabla \mathcal{B})=\partial (T\mathcal{B})$. Hence applying the upper-shadow conclusion to the transformed family proveswhere $C$ is the colex initial segment of the same size and rank as $\mathcal{A}$. This is the Kruskal-Katona lower-shadow theorem. A colex segment's shadow is itself a colex segment: split its sets by their largest element, giving a full initial block of the preceding ground set followed by an initial block with the new largest element. The same description holds one rank lower. Iteration therefore shows that colex segments minimize shadows at every lower rank. This proves the Harper theorem implies the Kruskal-Katona theorem deduction, including the necessary complement and coordinate reversal.

The Erdős-Ko-Rado theorem states that for $1\le r\le n/2$, every intersecting family $\mathcal{A}\subseteq [n{]}^{(r)}$ satisfiesThe family of all $r$-sets containing a fixed point attains the bound. We use the standard convention of positive ranks here.

For the first proof, let $\mathcal{B}$ be the complements of members of $\mathcal{A}$, at rank $k=n-r\ge r$, and let $\mathcal{S}$ be its rank-$r$ lower shadow. It is disjoint from $\mathcal{A}$, since $A\subseteq [n]\setminus A^{\prime }$ would make two family members disjoint. Suppose $|\mathcal{A}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$. The colex rank-$k$ segment of that size contains every $k$-set on $[n-1]$ and at least one set containing $n$. Its rank-$r$ shadow therefore contains all $\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$ old $r$-sets and at least one new $r$-set containing $n$. The iterated Kruskal-Katona bound givesThen $|\mathcal{A}|+|\mathcal{S}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)=\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, impossible for disjoint subfamilies of rank $r$. This proves the Erdős-Ko-Rado theorem from shadows.

For the second proof use the Katona circle method. In any cyclic ordering of the $n$ points, an intersecting collection of length-$r$ intervals has at most $r$ members. To prove this, rotate one selected interval to end at position $n$, so it occupies $n-r+1,\dots ,n$. Intervals ending at $r,\dots ,n-r$ miss it and cannot be selected. Among the remaining endpoints other than $n$, pair $j$ with $j+n-r$ for $1\le j\le r-1$. The two intervals in each pair are disjoint because $n\ge 2r$, so at most one can be selected. Including the fixed interval gives at most $1+(r-1)=r$.

Now count pairs consisting of a family member and a permutation in which it is a cyclic interval. A fixed $r$-set has $nr!(n-r)!$ such permutations: choose its final position, its internal order, and the order of the remaining points. Each of the $n!$ permutations contributes at most $r$ family intervals by the proved cyclic interval intersection bound. ThusThis is the required cyclic-ordering proof, with its interval lemma established explicitly.

Let $i(G)$ be the number of independent sets of the biregular graph $G$. Choose one such set $I$ uniformly and write $X_v=1_{v\in I}$. All information entropies below use logarithms to base two, soWe will prove $\boxed{i(G)\le (2^r+2^s-1{)}^{m/s}}$.

We state and justify the entropy facts used. For a finite-valued random variable $Y$, $H(Y)=-{\sum }_y{p}_y{\log }_2{p}_y$ with $0{\log }_{2}0=0$. Its conditional entropy is the mean of this quantity over the conditioning variable. Factoring the joint probabilities proves the chain rule for information entropy, $H(Y,Z)=H(Y)+H(Z\mid Y)$. Conditioning reduces entropy: $H(Y\mid Z,T)\le H(Y\mid Z)$. To see this, condition on $Z$ and use concavity of $-x{\log }_{2}x$ to compare the mixture of the conditional distributions given $T$ with their mean entropy.

The precise Shearer inequality needed here is: if subsets $S_j$ of the indices of a finite random vector $Y$ cover every index at least $d$ times, then $dH(Y)\le {\sum }_{j}H(Y_{S_j})$. To prove it, order the indices. The chain rule for information entropy expandsThe inequality is conditioning reduces entropy. Summing over $j$ counts each nonnegative conditional term at least $d$ times and proves the assertion.

For every $w\in W$, its vertex neighbourhood $N(w)\subseteq U$ has $s$ vertices. Every $u\in U$ lies in exactly $r$ of these neighbourhoods. Hence Shearer inequality givesLet $q_w=\mathbb{P}(X_{N(w)}=0)$, the probability that no neighbour of $w$ is selected. Conditional on a particular selected subset of $U$, a vertex $w$ with a selected neighbour is forbidden, and all other vertices of $W$ are available. Because there are no edges within $W$, every subset of the available vertices is a valid completion. Uniform choice of $I$ makes these completions equally likely. The available coordinates are therefore independent fair binary choices, each of entropy one, and the forbidden coordinates are zero. Averaging their number givesBy the chain rule for information entropy,It remains to bound each bracket. Let $Y=X_{N(w)}$, a binary vector of length $s$, with distribution $(p_y)$ on $\{0,1{\}}^s$, and put $Z=2^r+2^s-1$. Assign weight $w_0=2^r$ to the all-zero vector and weight $w_y=1$ to every nonzero vector. ThenThe first inequality is Jensen's inequality for the concave logarithm; the weights sum to $2^r+(2^s-1)$. This explicitly handles zero probabilities and does not assume the neighbourhood indicators are independent.

There are $n$ right vertices, so combining the bounds and using $rm=sn$ givesExponentiating proves the required independent-set bound for biregular graphs. The proof uses both degree constraints: $r$ supplies the neighbourhood-cover multiplicity and the conditional-completion weight, while $s$ determines the number of neighbourhood patterns.

Identify the Boolean hypercube $Q_n$ with the subsets of $[n]$. For a family $\mathcal{H}$, write $N(\mathcal{H})$ for its closed vertex neighbourhood, consisting of the family and all vertices at Hamming distance one from it. Order subsets first by increasing size and, within a fixed size, use lexicographic order: at the smallest coordinate on which they differ, the set containing that coordinate comes first. This is the simplicial order on the discrete cube. Harper's theorem states that if $I$ is its initial segment of size $|\mathcal{H}|$, thenThus simplicial initial segments minimize the external vertex boundary for each prescribed family size. The closed-neighbourhood and external-boundary formulations are equivalent because $|N(\mathcal{H})\setminus \mathcal{H}|=|N(\mathcal{H})|-|\mathcal{H}|$. No edge boundary is involved in this assertion.

We now give the deduction Harper theorem implies the Kruskal-Katona theorem, keeping track of the two different orders. For a family $\mathcal{F}$ of $r$-sets with $r\ge 1$, adjoin every smaller set:Its closed vertex neighbourhood is exactlywhere $\nabla \mathcal{F}$ is the upper shadow. Applying Harper theorem to $\mathcal{H}$ shows that, among $r$-uniform families of fixed size, the lexicographic initial segment minimizes the upper shadow: all the lower-level contributions to the two neighbourhood sizes are identical.

To convert this into a statement about the lower shadow of a $k$-uniform family $\mathcal{G}$, take $r=n-k$ and map each member $S$ to $R([n]\setminus S)$, where $R(i)=n+1-i$ reverses the ground-set coordinates. Complementation changes lower shadows to upper shadows, and the coordinate reversal preserves their sizes. Moreover, the resulting lexicographic order corresponds exactly to colexicographic order on the original $k$-sets: $S$ precedes $T$ in colexicographic order precisely when the largest element of $S△T$ lies in $T$. After complementation and reversal, the smallest differing coordinate lies in the image of $S$, as required for the stated lexicographic order. Hence a colexicographic initial segment minimizes the lower shadow, which is the Kruskal-Katona theorem. The cases $k=n$ and $k=0$ are immediate and need no adjoining construction.

For its usual numerical form, write the unique greedy binomial representationThen the Kruskal-Katona theorem saysHere is why the shadow of the colexicographic initial segment has exactly that size. Its first block consists of all $k$-sets in $[a_k]$. If any members remain, they have the form $\{a_k+1\}\cup T$, where $T$ runs through a colexicographic initial segment of $(k-1)$-sets. The lower shadow consists of all $(k-1)$-sets in $[a_k]$, together with $\{a_k+1\}$ adjoined to the lower shadow of that remainder. These two blocks are disjoint. Recursing gives precisely the displayed sum. For $m=0$, both the initial segment and its shadow are empty.

The Erdős-Ko-Rado theorem states that an intersecting family $\mathcal{A}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{r}\right)$, with $1\le r\le n/2$, satisfiesThe bound is attained by all $r$-sets containing one fixed element. For the proof Erdős-Ko-Rado theorem from shadows, take complements to obtain an $(n-r)$-uniform family $\mathcal{B}$ of the same size, and let $D$ be its rank-$r$ lower shadow. Every member of $D$ is disjoint from some member of $\mathcal{A}$, so $D\cap \mathcal{A}=\varnothing$. Iterating the Kruskal-Katona theorem bounds $|D|$ below by the rank-$r$ shadow of the colexicographic initial segment of size $|\mathcal{B}|$. This iteration is valid because lower shadows of colexicographic initial segments are again colexicographic initial segments, as the preceding block description shows.

If $|\mathcal{A}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{n-r}\right)$, that initial segment contains all $(n-r)$-sets in $[n-1]$ and at least one further set containing $n$. Since $r\le n-r\le n-1$, its rank-$r$ shadow contains every $r$-set in $[n-1]$ and at least one $r$-set containing $n$. Thus $|D|>\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$, implyingcontrary to their disjointness. This proves the bound. If $r>n/2$, all $r$-sets intersect and the trivial maximum is instead $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$.

For an intersecting family in a product alphabet, label the alphabet by $\mathbb{Z}/k\mathbb{Z}$. Partition the words into the classesEach class has $k$ words, and any two distinct words in it disagree in every coordinate. An intersecting family therefore contains at most one word per class. There are $k^{n-1}$ classes, and fixing the first coordinate gives an intersecting family of that size. Consequently the exact maximum isFor $k=1$ this just says that the single word may be included.

Finally, for a weakly intersecting family in a product alphabet, associate to each word its support $S(x)=\{i:x_i>1\}$. The occurring supports form an intersecting family of nonempty subsets, and the fibre over support $S$ has $(k-1{)}^{|S|}$ words. Put $w=k-1$. When $k\ge 2$, $w\ge 1$, and a support family can contain at most one of each complementary pair $S,[n]\setminus S$. For odd $n$, the larger support in each pair has size greater than $n/2$, soAll supports of size greater than $n/2$ are mutually intersecting, so taking every word over those supports achieves this bound. Thus the required extremal family is the strict-majority support family, with sizeThis is the weighted intersecting family bound on an odd Boolean lattice. For $k>2$, strict inequality between the complementary fibre weights forces this extremal family to be unique. For $k=2$, it is a largest family but need not be the only one; for example, all supports containing one fixed coordinate also attain the same size. For $k=1$, no word has a nonempty support, so the maximum is zero, again agreeing with the formula.

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 Turán 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 Turán 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 Turán 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 Turán 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.

If the edge $uv$ lies in no triangle, no vertex can be adjacent to both endpoints; hence the neighbourhoods satisfy $\Gamma (u)\cap \Gamma (v)=\varnothing$.

Suppose $|\Gamma (u)|+|\Gamma (v)|=2n$. The two disjoint neighbourhoods then partition $V(G)$. If both were independent, every edge would run between these two parts, so $G$ would be bipartite andcontrary to $e(G)=n^2+1$. Thus at least one neighbourhood contains an edge, and that edge forms a triangle with $u$ or $v$.

We now prove by induction on $n\ge 2$ the Triangle lower bound one edge above the Mantel threshold: every graph on $2n$ vertices with at least $n^2+1$ edges has at least $n$ triangles. Part (ii) is the base case. For $n\ge 3$, if every edge lies in a triangle, part (iii) givesOtherwise choose an edge $uv$ lying in no triangle and put $s=d(u)+d(v)\le 2n$. Deleting $u,v$ removes $s-1$ edges, so the graph $G^{\prime }=G-u-v$ hasIf $s=2n$, induction gives at least $n-1$ triangles in $G^{\prime }$, while the nonindependence proved above supplies a further triangle through $u$ or $v$.

If $s\le 2n-1$, then $e(G^{\prime })\ge (n-1{)}^2+2$. Induction first supplies a triangle $\Delta$ in $G^{\prime }$. Delete one edge of $\Delta$; the remaining graph still has at least $(n-1{)}^2+1$ edges, so induction supplies $n-1$ triangles avoiding that edge. Together with $\Delta$, these give at least $n$ distinct triangles in $G$.

An undirected simple graph is exactly a pair of setsThus every edge is a two-element set of distinct vertices. The unordered pair $\{u,v\}$ equals $\{v,u\}$, so there is no direction. The distinctness condition excludes self-loops, and using a set of edges excludes parallel copies. A simple graph may have vertices with no incident edges; $E$ does not by itself determine $V$. For example, $(\{A,B\},\{AB\})$ and $(\{A,B,C\},\{AB\})$ have the same edge set but different vertex sets.^[ref]

In the six-vertex example, the vertex neighbourhood of $C$ is $\Gamma (C)=\{A,B,D\}$, so its degree of a vertex is $\mathrm{deg}(C)=3$. In label order $A,B,C,D,E,F$, the degree sequence isThis is the handshaking lemma: count each edge once at each of its two endpoints. The average degree of a graph here is $14/6=7/3$. The degree sequence is a compact consistency check on an edge list; an odd sum could not describe any finite undirected graph. It also forces the number of odd degrees of vertices to be even, since the even terms do not affect the parity of the sum.

The adjacency matrix in the figure uses label order $A,B,C,D,E,F$. Its entry $M_{uv}$ is one exactly when $\{u,v\}\in E$, and zero otherwise. It is a symmetric matrix with zero diagonal, and each row sum gives the corresponding vertex degree. For example, row $C$ is $(1,1,0,1,0,0)$. The fourteen ones count both directions of the seven undirected edges; they do not indicate fourteen different edges. This translates a picture into a representation suitable for calculation.

The restrictions matter. A directed graph uses an ordered pair to distinguish $u\to v$ from $v\to u$, so reachability can depend on direction. A multigraph can contain several distinct edges between the same vertices; adding another $CD$ would raise $\mathrm{deg}(C)$ and $\mathrm{deg}(D)$ to four without adding a new neighbour. A self-loop at $A$ contributes two to its undirected vertex degree, making it four, because both endpoints lie at $A$. Specify whether self-loops are allowed in a multigraph. Finally, a weighted graph attaches numerical information to edges: if links carry travel times, minimizing the number of edges need not minimize total time. These are changes to the model, rather than changes to how the same simple graph is drawn.

In dimension three, the first three vertices are $\varnothing ,\{1\},\{2\}$. Their closed vertex neighbourhood has seven vertices. Adding $\{3\}$ does not change that vertex neighbourhood, so the external vertex boundary decreases from four to three even though both family sizes are at most half the cube.

Let a tripartite graph have nonempty parts $X,Y,Z$, pair edge density of a bipartite graph values $\alpha ,\beta ,\gamma$ on $XY,YZ,XZ$, and $\Vert G(X,Y)-\alpha {\Vert }_{\square }\le c$. If every $z\in Z$ has exactly $\beta |Y|$ neighbors in $Y$, its normalized triangle count $\tau$ obeysThe constant-degree assumption makes the contribution of the constant $\alpha$ exactly $\alpha \beta \gamma$. For each fixed $z$, apply the bilinear correlation bound for the box norm to the indicator functions of its two vertex neighbourhoods. Their squared $L^2$ norms are $\beta$ and the relative $X$-degree of $z$. Average over $z$ and use the Cauchy-Schwarz inequality to bound the mean square root of that degree by $\sqrt{\gamma }$.

For the Boolean hypercube, the initial segment of the simplicial order on the discrete cube minimizes the size of the closed vertex neighbourhood among families of a given size. Equivalently, it minimizes the external vertex boundary. This is a vertex assertion, distinct from the edge-isoperimetric inequality in the discrete cube.