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Computer Science

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  • $\begingroup$ Why is that "$uv$ has a larger weight than $ab$"? I do not get it. $\endgroup$ Commented May 15, 2016 at 13:39
  • $\begingroup$ @hengxin Well, $\pi$ with $uv$ forms a cycle, with all the edges of this cycle from $T$ except $uv$. Then by the assumption the weight of $uv$ is larger than any weight of the other edges in this cycle. Since $ab$ is in this cycle we get that the weight of $uv$ is larger than the weight of $ab$. $\endgroup$ Commented May 15, 2016 at 14:11
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    $\begingroup$ However, the edges with the maximal weight (on the cycle) may not be unique. I think we can only conclude that $w(uv) \ge w(ab)$. $\endgroup$ Commented May 15, 2016 at 14:16