Monday, March 10, 2014
Wednesday, March 05, 2014
Friday, January 10, 2014
Wednesday, September 04, 2013
as a control mechanism, they store the minds of their (former) enemies consciously immersed in simulations of their (respective) reality(ies). i had the rare experience of accidentally melding with such a mind---the Mind-Time channel's rotating modulation code glitched or something. the reality was really real, pain and all that, just minor suggestive control over the avatar in space. the rewards were enough to want to exert any control i could...
Saturday, April 13, 2013
Tuesday, March 26, 2013
Tuesday, March 05, 2013
Monday, December 31, 2012
Sunday, October 14, 2012
Friday, June 29, 2012
Wednesday, June 06, 2012
Sunday, May 20, 2012
Wednesday, March 07, 2012
Sunday, October 23, 2011
Tuesday, August 09, 2011
A slightly simpler proof of Kostochka's lemma. It avoids the need for one extra lemma.
For a collection of maximum cliques Q in a graph G, let X_Q be the
intersection graph of Q.
Lemma. If Q is a collection of maximum cliques in a graph G with
\omega(G) > 2/3 (\Delta(G) + 1) such that X_Q is connected, then \cap
Q \neq \emptyset.
Proof. Suppose not and choose a counterexample Q := {Q_1, ..., Q_r}
minimizing r.
Let A be a noncutvertex in X_Q and B a neighbor of A. Put Z := Q -
{A}. Then X_Z is connected and hence by minimality of r, \cap Z \neq
\emptyset. In particular, |\cup Z| \leq \Delta(G) + 1.
Hence |\cup Q| \leq |\cup Z| + |A - B| \leq 2(\Delta(G) + 1) -
\omega(G) < 2\omega(G). This contradicts Hajnal's lemma.
For a collection of maximum cliques Q in a graph G, let X_Q be the
intersection graph of Q.
Lemma. If Q is a collection of maximum cliques in a graph G with
\omega(G) > 2/3 (\Delta(G) + 1) such that X_Q is connected, then \cap
Q \neq \emptyset.
Proof. Suppose not and choose a counterexample Q := {Q_1, ..., Q_r}
minimizing r.
Let A be a noncutvertex in X_Q and B a neighbor of A. Put Z := Q -
{A}. Then X_Z is connected and hence by minimality of r, \cap Z \neq
\emptyset. In particular, |\cup Z| \leq \Delta(G) + 1.
Hence |\cup Q| \leq |\cup Z| + |A - B| \leq 2(\Delta(G) + 1) -
\omega(G) < 2\omega(G). This contradicts Hajnal's lemma.
Tuesday, July 19, 2011
Conjecture 14 from my recent paper is false by some pretty easy examples.

However, i still think Conjecture 13 (which Conjecture 14 implies) is true.

However, i still think Conjecture 13 (which Conjecture 14 implies) is true.
Subscribe to:
Posts (Atom)







