Turans theorem and extremal graphs

1
Turans theorem and extremal graphs

• Question How many edges a simple graph must have
  to
• guarantee that the graph contains a
  triangle?
• Since Km,m and Km, m1 do not contain triangles,
  we see that if the graph has n vertices, then
  ?n2/4? edges are not enough.
• Mantel (1907) proved that if there are more edges
  then the graph contains a triangle.

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Proof of Mantels

• Let G have vertex set 1,...,n and no triangle.
• Give vertex i a weight zi ? 0 and ? zi 1 and we
  wish to maximize S ? zi zj , where the sum is
  over all edges i, j.
• Suppose k and l are not joined. Let the
  neighbors of k have total weight x, and those of
  l total weight y, where x ? y.
• Since (zke)x(zl-e)y ? zkxzly, so we can shift
  the weight from l to k to make S larger. Thus S
  is maximized if all the weight concentrates on an
  edge.
• Therefore S ? 1/4. Why?
• On the other hand let all zi1/n, then SE/n2.
• So E ?n2/4. ?

3

• In general, we can ask whether some condition on
  the number of edges guarantees a Kp as a
  subgraph.
• Divide n vertices into p-1 subsets, S1,...,Sp-1,
  of almost equal size, where r of them of size t1
  and p-1-r subsets of size t and nt(p-1)r, 1 ? r
  ? p-1.
• Within each Si there are no edges but every
  vertex in Si is joined to every vertex in Sj if i
  ? j.
• The number of edges is
• M(n,p)(p-2)n2/2(p-1)-r(p-1-r)/2(p-1).

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M(n,p)(p-2)n2/2(p-1)-r(p-1-r)/2(p-1).nt(p-1)r

• Thm 4.1. (Turan 1941) If a simple graph on n
  vertices has more than M(n,p) edges, then it
  contains a Kp.
• Proof
• By induction on t It is obvious, if t0.
• Consider a graph G with n vertices with maximum
  edges but without Kp. Clearly G must have Kp-1,
  say H. Each of the remaining vertices is joined
  to at most p-2 vertices of H.
• The remaining n-p1 vertices do not contain Kp.
• Since n-p1(t-1)(p-1)r, we can apply the
  induction hypothesis to these points.
• So the the number of edges of G is at most
• M(n-p1, p)(n-p1)(p-2)C(p-1, 2), which is
  M(n,p).

5

• The girth of a graph G is the size of a smallest
  polygon (cycle) in G.
• Thm. 4.2. If a graph G on n vertices has more
  than
• edges, then G has girth
  at most 4.
• Pf
• By contradiction, suppose G has girth at least 5.
• Let y1,...,yd be xs neighbors and ddeg(x). No
  two them are adjacent. Why? Also no vertex can
  be adjacent to more than one of y1,...,yd. Why?
• Thus (deg(y1)-1)... (deg(yd)-1)(d1) ? n. Why?

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• Thm 4.3. If a simple graph G on n vertices has
  all vertices of degree at least n/2, then it
  contains a Hamiltonian circuit.
• Pf
• By contradiction, suppose it is not true. Let G
  be such a counterexample with the maximum number
  of edges.
• Let y and z be two non-adjacent vertices. Since
  adding
• y, z creates a Hamiltonian circuit, there
  exists a simple path from y to z with vertices
  yx1, x2,...,xnz, say.
• i y is adjacent to xi1 and i z is adjacent
  to xi each have
• at least n/2 neighbors and are in 1,2,...,n-1,
  so they must
• intersect at some element i0.
• Then yx1, x2,...,xi0, zxn, xn-1 ,...,xi01,
  x1y is Ham circuit.
• Contradiction!

yx1
xnz
x2
xi0
xi01