Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Bookmarks inbookmarks:mine
inbookmarks:1234
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options user 40

Algorithms on graphs, excluding heuristics.

4
votes
The proof of Theorem 4.1 states that: Let $S$ be the connected component of $G$, such that $s \in S$. By the arguments above, $S \times [N]$ is a connected component of $G_{\mathrm{reg}}$, and $ …
answered Dec 28 '14 by Yuval Filmus
6
votes
It is NP-hard to decide whether your problem is soluble. That's bad news from a theoretical point of view, but in practice it is possible that a heuristic algorithm will fare quite well. Also, you're …
answered Sep 17 '12 by Yuval Filmus
5
votes
Here is a counterexample, which I'm too lazy to draw. Let $G_1,G_2$ be diamonds, graphs on four vertices having five of the six edges. Connect matching vertices in $G_1$ and $G_2$ so that you get a co …
answered Nov 1 '12 by Yuval Filmus
4
votes
This problem is very similar to dominating set in bipartite graphs. Indeed, whenever $a_i \lor b_j = 0$, we know that $a_i = b_j = 0$. Remove all these inputs and constraints. Now we are left with set …
answered Oct 15 '12 by Yuval Filmus
13
votes
Johnson graphs are actually easy to recognize. In particular, you can recognize whether an input graph is a Johnson graph in polynomial time, and you can construct an isomorphism between two isomorphi …
answered Jan 27 '17 by Yuval Filmus