MHB Proving Turan's Theorem (Dual Version) and its Implications for $ex(n, K_{p+1})$

  • Thread starter Thread starter joypav
  • Start date Start date
  • Tags Tags
    Dual Theorem
joypav
Messages
149
Reaction score
0
I have already proved that for a graph $G$ with $n$ vertices and $|E(T'(n,q))|$ edges, $\alpha (G) \geq q$. Additionally, if $\alpha (G) = q$ then it must be that $G \cong T'(n,q)$.

Apparently this is the "dual version" of Turan's Theorem. How does this theorem imply Turan's?

That $ex(n, K_{p+1}) = |E(T(n,p))|$

Where:

- $T'(n,q)$ : $q$ disjoint cliques with size as equal as possible
- $\alpha (G)$ : independence number of $G$
- $ex(n, K_{p+1})$ : max number of edges in an n-vertex graph with no $K_{p+1}$ subgraph
 
Physics news on Phys.org
For completeness, here is the proof I wrote.
I'm not sure it is correct! May be some mistakes in the details.

Known Theorem:
Define $T'(n,q)$ to be q disjoint cliques with sizes of vertex sets as equal as possible. Let G be a graph with n vertices and $|E(T'(n,q))|$ edges. Then,
$$\alpha (G) \geq q$$
and if $\alpha (G) = q$ then $G \cong T'(n,q)$.To prove Turan's theorem, it suffices to show that if $|E(G)|>|E(T(n,p))|$ then $\omega (G) \geq p+1$, (where $\omega (G)$ is the size of the largest clique in G).\\
Assume $|E(G)|>|E(T(n,p))|$ then $|E(\overline{G})| \leq |E(T'(n,p))|$. By the in class proof, $\alpha (\overline{G}) \geq p+1$.
Notice that $\omega(G) = \alpha(\overline{G})$. Then we have, $\omega(G) = \alpha(\overline{G}) \geq p+1$, completing the proof.
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top