Timeline for Prove that Ford-Fulkerson can decide if there is more than one min cuts
Current License: CC BY-SA 3.0
4 events
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| Apr 13, 2017 at 12:48 | history | edited | CommunityBot |
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| Jun 10, 2015 at 18:30 | comment | added | D.W.♦ | @user3680924, 1. If you don't understand why it fails, I encourage you to pick an example graph with two or more min cuts, run your algorithm on it (by hand), and add this to the question (add the graph & add the results you get). 2. My answer does not refer to the concept "left-most min-cut", but if you want to know the definition, see the links I gave you earlier in the comments (cs.stackexchange.com/q/42960/755 and stackoverflow.com/q/29418244/781723) -- in particular, if $(S_1,T_1),\dots,(S_n,T_n)$ are a list of all min-cuts, the left-most is $(\cap_i S_i, \cup_i T_i)$. | |
| Jun 10, 2015 at 5:59 | comment | added | Danielyag23 | I still don't understand why the my algorithm fails, but let us leave it. Can you explain the formality of your solution? I realized that if I can define cuts by location, then I can compare left most min cut with right most min cut. But I don't understand the definition of "left-most min-cut"... | |
| Jun 10, 2015 at 5:13 | history | answered | D.W.♦ | CC BY-SA 3.0 |