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visits member for 3 years, 8 months
seen Dec 15 at 8:57

Dec
15
asked Why does Average Log Likelihood
Dec
1
comment Verify that R(p,2) = R(2,p) = p, where R is the Ramsey number
So why doesn't R(2,p) = 2? You can always guarantee that with two nodes you can find a complete graph that is 1 colored? Isn't that what we are looking for: the min number of nodes for a complete graph s.t. we can find a complete subgraph that can be one colored?
Dec
1
accepted Verify that R(p,2) = R(2,p) = p, where R is the Ramsey number
Dec
1
comment Verify that R(p,2) = R(2,p) = p, where R is the Ramsey number
(a) I don't see how this proves R(2,p)=p. (b). I don't know what you mean by $\chi(\{x_0,y_0\} = 2$. I understand this notation to mean that you need two colors to color one edge.
Dec
1
asked Verify that R(p,2) = R(2,p) = p, where R is the Ramsey number
Nov
16
comment Why does adding a vertex $x$ that is adjacent every vertex in $G$ with a subdivision in $K_{3,3}$ or $K_5$ result in subdivison of $K_5$ or $K_{3,3}$
@ShilB. Sorry, I edited the question that changed the meaning of the question.
Nov
16
revised Why does adding a vertex $x$ that is adjacent every vertex in $G$ with a subdivision in $K_{3,3}$ or $K_5$ result in subdivison of $K_5$ or $K_{3,3}$
edited body
Nov
16
asked Prove: if G has exactly one $C$-fragment, then there exists a cycle $C$ in a 3-connected graph that is the boundary of a face in $G$.
Nov
16
revised Why does adding a vertex $x$ that is adjacent every vertex in $G$ with a subdivision in $K_{3,3}$ or $K_5$ result in subdivison of $K_5$ or $K_{3,3}$
edited title
Nov
16
asked Why does adding a vertex $x$ that is adjacent every vertex in $G$ with a subdivision in $K_{3,3}$ or $K_5$ result in subdivison of $K_5$ or $K_{3,3}$
Nov
16
comment How many edges must you remove from Peterson graph to make it planar
What? Can you reword?
Nov
16
asked How many edges must you remove from Peterson graph to make it planar
Nov
16
awarded  Yearling
Nov
16
accepted Proving Konig-Egervary Theorem from Ford-Fulkerson
Nov
6
revised Proving Konig-Egervary Theorem from Ford-Fulkerson
added 150 characters in body; edited title
Nov
6
asked Proving Konig-Egervary Theorem from Ford-Fulkerson
Oct
16
accepted Arguing that Graph, $G$, is 3 regular and pairwise edge-disjoint path can't share internal vertices
Oct
16
comment Arguing that Graph, $G$, is 3 regular and pairwise edge-disjoint path can't share internal vertices
Sorry, I'm being dense. But why would each path would contribute two edges to the internal vertex?
Oct
16
comment Arguing that Graph, $G$, is 3 regular and pairwise edge-disjoint path can't share internal vertices
So why is this a problem?
Oct
15
comment Arguing that Graph, $G$, is 3 regular and pairwise edge-disjoint path can't share internal vertices
At the shared vertex, there is an edge from some path P and another one from P1. How does this help?