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# Turán's theorem

In graph theory, Turán's theorem is a result on the number of edges in a Kr+1-free graph.

An n-vertex graph that does not contain any (r + 1)-vertex clique may be formed by partitioning the set of vertices into r parts of equal or nearly-equal size, and connecting two vertices by an edge whenever they belong to two different parts. We call the resulting graph the Turán graph T(n,r). Turán's theorem states that the Turán graph has the largest number of edges among all Kr+1-free 'n''-vertex graphs.

Turán graphs were first described and studied by Hungarian mathematician Paul Turán in 1941, though a special case of the theorem was stated earlier by Mantel in 1907.

## Formal statement of the theorem

Formally, Turán's theorem may be stated as follows.

Let G be any subgraph of Kn such that G is Kr+1 -free. Then the number of edges in G is at most

$frac\left\{r-1\right\}\left\{r\right\}cdotfrac\left\{n^2\right\}\left\{2\right\} = left\left(1-frac\left\{1\right\}\left\{r\right\} right\right) cdotfrac\left\{n^2\right\}\left\{2\right\}.$

An equivalent formulation is the following:

Among the n-vertex simple graphs with no (r + 1)-cliques, T(n,r) has the maximum number of edges.

As a special case, for r = 2, one obtains Mantel's theorem:

The maximum number of edges in an n-vertex triangle-free graph is $lfloor n^2/4 rfloor.$

In other words, one must delete nearly half of the edges in Kn to obtain a triangle-free graph.

## Proof

Let $G$ be an n-vertex simple graph with no (r + 1)-clique and with the maximum number of edges.

Claim 1: Graph $G$ does not contain any three vertices $u,v,w$ such that $G$ contains edge $uv$, but contains neither edge $uw$ nor $vw$.
(This claim is equivalent to the relation x~y iff x not connected to y being an equivalence relation. ~ is always reflexive and symmetric, but only in special cases is it transitive. ~ is not transitive precisely when we have u, v and w with u ~ w and w ~ v without u ~ v.)

Assume the claim is false. Construct a new n-vertex simple graph $G\text{'}$ that contains no (r + 1)-clique but has more edges than $G$, as follows:

Case 1: $d\left(w\right) < d\left(u\right)text\left\{ or \right\}d\left(w\right) < d\left(v\right).$

Assume that $d\left(w\right) < d\left(u\right)$. Delete vertex $w$ and create a copy of vertex $u$ (with all of the same neighbors as $u$); call it $u\text{'}$. Any clique in the new graph contains at most one vertex among $\left\{u,u\text{'}\right\}$. So this new graph does not contain any (r + 1)-clique. However, it contains more edges: $|E\left(G\text{'}\right)| = |E\left(G\right)| - d\left(w\right) + d\left(u\right) > |E\left(G\right)|.$

Case 2: $d\left(w\right)geq d\left(u\right)$ and $d\left(w\right)geq d\left(v\right)$

Delete vertices $u$ and $v$ and create two new copies of vertex $w$. Again, the new graph does not contain any (r + 1)-clique. However it contains more edges: $|E\left(G\text{'}\right)| = |E\left(G\right)| -\left(d\left(u\right) + d\left(v\right) - 1\right) + 2d\left(w\right) geq |E\left(G\right)| + 1$.

This proves Claim 1.

The claim proves that one can partition the vertices of $G$ into equivalence classes based on their nonneighbors; i.e. two vertices are in the same equivalence class if they are nonadjacent. This implies that $G$ is a complete multipartite graph (where the parts are the equivalence classes).

Claim 2: The number of edges in a complete k-partite graph is maximized when the size of the parts differs by at most one.

If G is a complete k-partite graph with parts A and B and $|A| > |B|+1$, then we can increase the number of edges in G by moving a vertex from part A to part B. By moving a vertex from part A to part B, the graph loses $|B|$ edges, but gains $|A|-1$ edges. Thus, it gains at least $|A|-1-|B|geq 1$ edge. This proves Claim 2.

This proof shows that the Turan graph has the maximum number of edges. Additionally, the proof shows that the Turan graph is the only graph that has the maximum number of edges.