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we all know and love s-t minimum cut algorithms, but they all cut through the edges in the graph. Are there any variants that cuts through nodes?

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    $\begingroup$ This smells like homework. (I just covered this in class last week.) $\endgroup$ – Jeffε Nov 11 '10 at 15:08
  • $\begingroup$ @JeffE: Sure, I'm a student and this is for a project - but there is no harm in asking, right? If one does not know what to search for, this is OK? $\endgroup$ – johan Nov 11 '10 at 15:26
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    $\begingroup$ homework questions are not welcome on this site, that is the reason for Jeff's comment. $\endgroup$ – Kaveh Nov 12 '10 at 11:17
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See, for example, p. 205 and Theorem 11.7 of Bondy and Murty's Graph Theory with Applications (1976).

"Menger's theorem" is a good keyword for further googling. The usual max-flow min-cut theorem implies the edge-connectivity version of the theorem, but you are interested in the vertex-connectivity version.

(These are existential results, but a typical proof of the vertex-connectivity version reduces the problem to the edge-connectivity version, and then you can apply efficient max-flow/min-cut algorithms.)

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You are looking for Minimum Vertex K-cut. Also, if you search for k-node-connectivity you will get papers related to the structural properties of such cuts.

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    $\begingroup$ Your link to "Minimum Vertex K-cut" points to a definition of a hard problem (and related approximability results). Minimum vertex cut can be solved in polynomial time. $\endgroup$ – Jukka Suomela Nov 11 '10 at 13:40
  • $\begingroup$ Can you please elaborate on the algorithm? Any reference will be very helpful. $\endgroup$ – Arindam Pal Feb 28 '14 at 5:35
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It is not very difficult to transform the vertices problem to an equivalent edge version.

Consider a vertex v. Replace v with v1 and v2. 
Join v1 and v2 with an edge of capacity 1.
All the incoming edges into v go into v1 and all the outgoing edges from v, leave v2.
ie (u v) -> (u v1) and (v w) -> (v2 w)
This is equivalent to edge version of the s-t min cut algorithm.
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  • $\begingroup$ I also felt the same. Can you think about any reference in this regard? $\endgroup$ – Rajat Aug 31 '16 at 1:31

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