Talk:Kernelization

vertex cover - step 3
Sorry for my probably incorrect edit, but I still don't understand why the following is needed in step 3: "In this case, the graph may be replaced by the complete graph $$K_{k+1}$$, which also has no $$k$$-vertex cover". --Erel Segal (talk) 06:34, 19 October 2014 (UTC)
 * First, I am not sure I understand why this is correct. For example, if $$k=2$$ then we have $$K_3$$, which is a triangle and has a 2-vertex cover.
 * Second, this step seems much more complicated than necessary. Why not just replace the graph with any non-empty graph and decrease $k$ to 0, for example?

First, perhaps it should be $$K_{k+2}$$, and second, that would be ok too, as long as it can be described more simply. What I objected to in the earlier edit was replacing step 3 by aborting the algorithm. Of course one would do that in practice, and we should probably say so, but it doesn't fit the actual definition of a kernelization. —David Eppstein (talk) 06:36, 19 October 2014 (UTC)

If you use $$K_{k+2}$$ then at step #1 you remove all vertices one by one (since all of them have degree more than k) until you have 2 vertices with k=0, but then the promise of having less than $$k^2$$ edges is not satisfied! Do you have an idea how to correct this? --Erel Segal (talk) 11:36, 19 October 2014 (UTC)

vertex cover - step 1
"If v is a vertex of degree greater than k, remove v from the graph and decrease k by one" - this seems problematic since k might be 0. --Erel Segal (talk) 11:32, 19 October 2014 (UTC)

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