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Apr
16
comment Proving corollary to Euler's formula by induction
Ah, ok; perfect! Thanks for the clear explanation!
Apr
16
comment Proving corollary to Euler's formula by induction
Ah, ok. So if the walk is not closed then I need to count each edge twice, right? But yes, once this information is clear I should be able to convince myself of that.
Apr
16
accepted Proving corollary to Euler's formula by induction
Apr
16
revised Proving corollary to Euler's formula by induction
added 165 characters in body
Apr
16
revised Proving corollary to Euler's formula by induction
Added LaTeX
Apr
16
asked Proving corollary to Euler's formula by induction
Apr
16
comment Prove that if graph $G$ is a 3-connected planar graph then its dual must be simple.
Since a cycle must return to its starting vertex, a cycle beginning in $G_1$ must end in $G_1$ (it may or may not cross into $G_2$). In other words, the cycle must cross the bridges between $G_1$ and $G_2$ never or an even number of times. So that means such a cycle can only occupy two of the three available edges or none at all. Supposing the first case, if we removed the 2-occupied edges as you suggest, aren't we going to get a self-loop in the dual going across the remaining edge, which is not a simple graph?
Apr
16
asked Prove that if graph $G$ is a 3-connected planar graph then its dual must be simple.
Apr
14
revised Prove that a certain graph and its dual are 4-colorable
edited tags
Apr
14
asked Prove that a certain graph and its dual are 4-colorable
Apr
7
accepted Number of storms in a rainy season
Apr
7
comment Number of storms in a rainy season
Yea, I didn't think the expectation looked quite right either
Apr
7
asked Number of storms in a rainy season
Apr
7
accepted Expectation of a parallel system
Apr
7
accepted Conditioning on a random variable
Apr
6
comment Planar complete tripartite graphs
Let me rephrase that last part. We need not concern ourselves with a $K_5$ because a $K_5$ cannot exist in an $n$-partite graph for $n\leq 4$ by definition. It can exist in a 5-partite graph or above.
Apr
6
comment Planar complete tripartite graphs
Because in the case of $r\geq 3$ and $s+t \geq 3$ there exists a $K_{3,3}$. We need not be concerned with a $K_5$ because by the definition of $n$-partedness a $K_5$ shouldn't exist, correct?
Apr
6
asked Planar complete tripartite graphs
Apr
1
revised number of edges in the complement of a complete bipartite graph as a function of $n$, the toal number of verticies
added 12 characters in body
Apr
1
comment number of edges in the complement of a complete bipartite graph as a function of $n$, the toal number of verticies
Ah yes, I wrote the answer assuming connectedness