# 23.1-4 A triangle whose edge weights are all equal is a graph in

## Transcription

23.1-4 A triangle whose edge weights are all equal is a graph in
```23.1-4
A triangle whose edge weights are all equal is a graph in which every edge is a light edge
crossing some cut. But the triangle is cyclic, so it is not a minimum spanning tree.
24-3
(a)
You can first convert the inequality π[π1 , π2 ] β π[π2 , π3 ] βββ π[ππβ1 , ππ ] β π[ππ , π1 ] > 1 into
(β log π[π1 , π2 ]) + (β log π[π2 , π3 ]) + β― + (β log π[ππβ1 , ππ ]) + (β log π[ππ , π1 ]) < 0 , and then
apply the BELLMAN-FORD algorithm in textbook to find the existence of negative cycle (i.e.,
the Arbitrage). Since there are π2 edges for table π , the time complexity can calculate as
π(|π||πΈ|) = π(π3 ).
(b)
After doing BELLMAN-FORD to solve part (a), we go through the edges once again. Once we
find an edge (u, v) for which d(v) > d(u) + Ο(u, v) . We can trace back the parent nodes
starting from vertex u until we get back to π’ , and all vertices in between will constitute a
negative-weight cycle. The overall time complexity still stays at O(n3 ).
```