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May 28, 2019 at 16:25 vote accept Deangelo Kingwell
May 28, 2019 at 16:23 comment added Yuval Filmus Right, that's the idea.
May 28, 2019 at 16:18 comment added Deangelo Kingwell We can simply divide $v_i$ to $v_i$ and $v_i'$. Then, the edges from $v_{i-1}$ to $v_i$ and $v_i'$ should be $x_i\;/\;0$, and edges from $v_{i-1}'$ to $v_i$ and $v_i'$ should be $0\;/\;x_i$.
May 28, 2019 at 13:45 history answered Yuval Filmus CC BY-SA 4.0