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| bio | website | dcc.ufrj.br/~viniciussantos |
|---|---|---|
| location | Brazil | |
| age | 28 | |
| visits | member for | 1 year |
| seen | May 21 at 1:41 | |
| stats | profile views | 7 |
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May 15 |
awarded | Caucus |
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May 11 |
answered | tree that approximates the distances and total weight in graphs |
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May 4 |
awarded | Yearling |
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Apr 17 |
comment |
Eccentricity of vertices in a graph Since k >= |V(G)|/2, if you take non-adjacent vertices u and v, you have |N(u)|+|N(v)|>= |V(G)|. Note that the total number of vertices is as big as the vertex set, hence, since they are not adjacent, there must be a repetition between the vertices of N(u) and N(v). |
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Apr 10 |
answered | Eccentricity of vertices in a graph |
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Apr 9 |
comment |
Eccentricity of vertices in a graph There are some trivial restrictions like the degree equals to |V(G)|-1 or |V(G)|-2. Is that the kind of restriction you are looking for? |
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Apr 5 |
revised |
Diameter of $k$-regular graph Fixed a problem pointed by the author of the question. |
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Apr 5 |
comment |
Diameter of $k$-regular graph Sure! Stupid me. I will fix it. |
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Apr 5 |
answered | Diameter of $k$-regular graph |
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Apr 5 |
comment |
Diameter of $k$-regular graph Is this true? Is it bounded by n/k or O(n/k)? |
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Oct 20 |
awarded | Teacher |
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Oct 15 |
answered | Measure the connection between two nodes in a graph |
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Oct 15 |
awarded | Editor |
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Oct 15 |
revised |
Conditional merge of a sequence of DAGs / partial orders Fixed a typo. "if" instead of "it". |
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Oct 15 |
comment |
Conditional merge of a sequence of DAGs / partial orders As I said, I'm not sure the problem can be solved efficiently. It wouldn't surprise me if it's NP-hard. |
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Oct 15 |
answered | Conditional merge of a sequence of DAGs / partial orders |
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Sep 15 |
awarded | Supporter |
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May 4 |
awarded | Autobiographer |
