Exercise Lover

May 10, 2009

Exercise 26.5-5

Filed under: 26.5 — yuhanlyu @ 9:18 am

Because every vertex v, |V|+1 > h[v] > k can not reach sink. Hence, we can set the height to |V|+1.

Exercise 26.5-4

Filed under: 26.5 — yuhanlyu @ 9:05 am

We can prove the times we call the nonsaturating push, and we use the distance funcion d instead of height function h. Let K = \sqrt{|E|}. For a node v, let d’(v) = |{w|d(w) \leq d(v)}| / K. Set the potential function = \sum_{v is overflowing} d'(v). We divide the algorithm into phase, the phase is between consecutive change  of d* = \max d(v). We can classify the phases into two category: cheap and expensive. Cheap phase contains at most K nonsaturating push. We know there are at most |V|^2 cheap phase, and every expensive will decrease the potential function by K. The time complexity will become O(|V|^2\sqrt{E}).

Exercise 26.5-3

Filed under: 26.5 — yuhanlyu @ 9:04 am

It is valid, but the relabel operation may not create admissible edge. Hence the time complexity will increase.

Exercise 26.5-2

Filed under: 26.5 — yuhanlyu @ 8:59 am

Iit is the same as relabel to front algorithm

Exercise 26.5-1

Filed under: 26.5 — yuhanlyu @ 8:57 am

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