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For every $\eta >0$, some $\rho >0$ and $N_0$ have this property: a graph on $N\ge N_0$ vertices with fewer than $\rho N^3$ unordered triangles in a graph can be made triangle-free by deleting at most $\eta N^2$ edges. A Szemerédi regularity lemma partition deletes exceptional, intraclass, irregular and sparse-pair edges cheaply. A surviving triangle in a graph would force a positive cubic count by the regular triangle counting lemma.

For $n\ge 3$, put $e_i=e(T_i(n))$, $t_3=k_3(T_3(n))$, and $c_n=(e_3-e_2)/t_3$. Every graph on $n$ vertices satisfiesThus an edge count $e_2+\theta (e_3-e_2)$ forces at least $\theta t_3$ triangles in a graph. This is an exact finite bound, with the rounding in the Turán graph retained. To prove it, use edge-triangle symmetrization. The inequality $c_n\ge 2/n$ allows merging the two smallest classes when there are at least four classes. With three classes, fixing the middle class makes the objective affine in the product of the other two sizes; either merge them or balance them. Only $T_2(n)$ and $T_3(n)$ need remain.

The planar graph with vertices $\{a(1,0)+b(1/2,\sqrt{3}/2):a,b\in \mathbb{Z}\}$ and edges between vertices at Euclidean distance one. Every vertex of a graph has six neighbors. Its site percolation model is a particularly tractable setting for conformal invariance of planar percolation.

Triangulation adds fill edges to an undirected graph to make it a chordal graph. Given an elimination ordering, join all remaining neighbors of each removed vertex. The completed graph has that perfect elimination ordering. Fill edges can enlarge the cliques used for exact probabilistic graphical model inference, without changing the original probability factors; they do not assert new biological relationships.

For a finite graph vertex set $W$ in a locally finite graph, let ${\partial }^{+}W$ be its exterior graph neighbours. The number of trifurcation vertices in percolation lying in $W$ is at most $|{\partial }^{+}W|$.

For each cluster meeting $W$, contract its connected components of a graph outside $W$ to terminal graph vertices, keeping only those adjacent to $W$. The resulting incidence graph is finite and connected. Each trifurcation graph vertex in $W$ separates at least three groups of terminals, because every infinite branch must leave the finite set $W$. Take a minimal subtree connecting all terminals. Each such trifurcation is unavoidable in that subtree and has degree of a vertex at least three. All leaves are terminals. The tree identity ${\sum }_v(\mathrm{deg}(v)-2)=-2$ bounds the number of these branching graph vertices by the number of leaves minus two, hence by the number of terminals. Different terminals, even across different clusters, can be assigned different exterior graph neighbours. Summing proves the bound. For square boxes in ${\mathbb{Z}}^2$, the boundary has order $n$ and the volume has order $n^2$.

A tripartite graph has a set partition of its vertices into three classes, with every edge joining different classes. Each pair of classes induces a bipartite graph, and every triangle in a graph uses one vertex from each class.

For a fixed $\ell$-uniform forbidden hypergraph $F$, its Turán density is the limit of $\mathrm{ex}(n,F)/\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)$. Averaging over smaller vertex subsets shows these normalized extremal numbers are nonincreasing, so the limit exists. For ordinary graphs, the Erdős-Stone theorem determines it from the chromatic number.

How many edges can we retain while excluding a prescribed clique? For integers $r\ge 1$ and $n\ge 0$, every simple graph on $n$ vertices containing no complete subgraph $K_{r+1}$ satisfiesEquality holds exactly for the Turán graph $T_r(n)$, up to graph isomorphism. Thus the extremal number is $\mathrm{ex}(n,K_{r+1})=t_r(n)$.^[ref]

For example, $T_3(8)$ has independent sets of sizes $3,3,2$ and $21$ edges, as counted in Turán graph. It contains triangles in a graph, but no four-vertex clique. Consequently any simple graph on eight vertices with $22$ edges contains $K_4$, while $21$ edges alone do not force one. Excluding larger cliques permits denser graphs; Mantel theorem is precisely the first nontrivial case.

A short proof outline uses Zykov symmetrization. If an extremizer is not a complete multipartite graph, it has vertices $x,y,z$ with $xy,xz$ absent and $yz$ present. If either $y$ or $z$ has larger vertex degree than $x$, clone that vertex in place of $x$. Otherwise replace both $y,z$ by nonadjacent clones of $x$, gaining $2d(x)-d(y)-d(z)+1>0$ edges. Both contradict extremality; balancing the resulting at most $r$ parts gives $T_r(n)$.^[ref]

The vertex cloning in a graph step preserves clique exclusion because any clique using a clone can substitute its original vertex; nonadjacent clones cannot occur together in a clique. At most $r$ parts can be nonempty, since selecting one vertex from each gives a clique. If fewer than $r$ parts are nonempty and one has at least two vertices, splitting it creates more edges, so it cannot be extremal. For $r=1$ the bound is zero. For $r\ge n$, the Turán graph is $K_n$, and the bound is the total $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ possible edges.

A finite graph has a perfect matching if and only if the displayed inequalities hold, where $o(H)$ counts the connected components of a graph $H$ with an odd integer number of vertices. The theorem is associated with William Thomas Tutte.^[ref]

Necessity has a direct counting proof: each such component of $G-S$ must send at least one edge of a perfect matching to $S$, and different components must use different vertices of $S$. The sufficiency direction is the substantive content of the theorem. The criterion is useful because vertex degree and parity estimates can prove existence without constructing the perfect matching, as in the perfect matching theorem for a cubic graph with at most two bridges.

A two-component spanning forest of a finite connected graph is an acyclic spanning subgraph with exactly two connected components of a graph. It has $|V|-2$ edges. For distinct terminals $a,z$, a separating two-component forest places them in different components. Deleting an edge of the $a$-to-$z$ graph path of a spanning tree produces such a forest; adjoining any original edge across its cut reverses this operation. This is the counting correspondence behind the mean spanning-tree path current.

An undirected graph has edges that are unordered pairs of vertices. In a simple undirected graph there are no loops or multiple edges. The skeleton of a directed graph forgets the directions of its edges and is an undirected graph.

A hypergraph is $r$-uniform if each hyperedge has exactly $r$ vertices. Ordinary simple graphs are 2-uniform hypergraphs. Its average hypergraph vertex degree is $r|E|/|V|$.

Start with the empty graph and reveal the edges of the complete graph in a uniformly random order. After $m$ steps its edge set is a uniform $m$-element subset. Independent uniform edge labels couple this process with all binomial random graphs: $G(n,p)=G_{n,M(p)}$, where $M(p)$ has a binomial distribution with parameters $\left(\genfrac{}{}{0ex}{}{n}{2}\right),p$. For a monotone graph property, concentration of $M(p)$ transfers a threshold estimate between edge probability and edge count.

On a finite connected unweighted graph, the graph path between $x$ and $y$ in a uniform spanning tree has the law of loop-erased random walk from $x$ to $y$. Root Wilson's algorithm at $y$ and process $x$ first: the first attached graph path is exactly that loop erasure, and later attachments cannot change it. Changing the root to $x$ proves reversibility of loop-erased random walk, even though chronological loop erasure does not commute with reversing each individual random-walk trajectory.

In the spanning subgraph formed by the union of two matchings in a graph $M_1,M_2$, every vertex has vertex degree at most two. A common edge is an isolated two-vertex connected component of a graph. Every other nontrivial connected component of a graph is a graph path or an even graph cycle, with the edges alternating between the two matchings in a graph.

Suppose the original graph is a bipartite graph with vertex sets $A,B$, $M_1$ covers a specified subset $X\subseteq A$, and $M_2$ covers a specified subset $Y\subseteq B$. Their union contains a matching in a graph covering $X\cup Y$. On an even graph cycle, either alternating choice covers every vertex. On an odd-length graph path, the larger alternating choice covers every vertex. On an even-length graph path, both endpoints lie in the same side: choose $M_1$ if they lie in $A$, or $M_2$ if they lie in $B$. Every required endpoint is covered by that choice, and every internal vertex meets both choices. Select each common edge as well; isolated vertices are not in $X\cup Y$. Different connected components of a graph do not interfere.

A vector $k$-coloring, for $k>1$, assigns a unit vector $v_i$ to each vertex of a graph, such that $⟨v_i,v_j⟩\le -1/(k-1)$ on every edge. A graph colouring with $k$ colors gives a vector $k$-coloring by placing the colors at the vertices of a regular simplex. The Gram matrix of these vectors allows semidefinite programming to search for such a representation.

A vertex cover chooses vertices that collectively touch every edge of a graph. Think of placing monitors at junctions: if a monitor watches every connection incident to its junction, a vertex cover watches all connections. The minimum vertex cover asks how few monitors suffice. This is a model of covering connections through their endpoints, rather than of monitoring distances in the drawing.

For a finite simple graph $G=(V,E)$, a vertex cover is a subset $C\subseteq V$ satisfyingBoth endpoints may be selected; the condition requires at least one. An isolated vertex covers no edge, so a minimum vertex cover never needs it.^[ref]

Take the cycle graph $C_5$ with vertices $0,1,2,3,4$ and edges $01,12,23,34,40$. The set $C=\{1,3,4\}$ is a vertex cover: vertex $1$ covers $01$ and $12$, vertex $3$ covers $23$ and $34$, and vertex $4$ covers $40$ as well as $34$. In particular, the closing edge $40$ is covered even though vertex $0$ is unselected.

The unselected vertices $I=V\setminus C=\{0,2\}$ form an independent set. This is the key connection: an edge missed by $C$ would have both endpoints in $I$, precisely what an independent set forbids. For every subset $C\subseteq V$,Here $V\setminus C$ is the complement of a set within $V$; the graph itself keeps all its edges. The complement graph is a different construction.^[ref]

The endpoints of any maximal matching form a vertex cover: an edge avoiding all endpoints would enlarge that matching in a graph. Consequently a graph with no matching in a graph of size $s$ has a vertex cover of size at most $2(s-1)$. This supplies a convenient vertex cover, but it need not be a minimum vertex cover.

The vertex cover decision problem asks whether a finite simple graph $G$ has a vertex cover containing at most $k$ vertices. For an uncovered edge $uv$, any valid vertex cover contains $u$ or $v$. Branch on these two possibilities, delete the chosen vertex and its incident edges, and decrease $k$ by one. An edgeless graph accepts and a nonempty graph with $k=0$ rejects. The search tree has at most $2^k$ leaves, giving running time $O(2^k(|V|+|E|))$ with straightforward copying. Thus this parameterization is fixed-parameter tractable.

A vertex cycle cover is a collection of cycles whose union contains every vertex of a graph; its cycles can overlap. A cycle cover of a directed graph additionally requires directed, vertex-disjoint cycles, and a spanning circuit family can allow overlaps.

Group random edges by their larger endpoint. These independent coordinates each change edges incident to only one vertex. Removing that vertex leaves the same graph under any two outcomes, so their chromatic numbers differ by at most one. The McDiarmid inequality with $n$ coordinate ranges of length one gives the displayed concentration bound.