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Next: Corollary 24.2 Up: CSE 2320: Algorithms and Previous: Definitions

Theorem 24.1

Given a connected, undirected graph G = (V, E) with edge weights w, A $\subseteq$ MST(G), cut (S, V-S) that respects A, and light edge (u,v) crossing (S, V-S), then (u,v) is a safe edge.

Proof: Assume T = MST(G) contains edge(x,y) crossing (S, V-S). Note that (x,y) must be on a unique path connecting u to v. Edge (u,v) would form a cycle. Removing (x,y) breaks T in 2 parts, but (u,v) reconnects them.

T' is the new resulting MST.



\psfig{figure=figures/f17-6.ps}

Since (u,v) is a light edge, then T' = T - {(x,y)} $\cup$ {(u,v)} is also MST(G).

Note that this is true because (u,v) and (x,y) cross the same cut and (u,v) is safe, w(u,v) $\leq$ w(x,y), w(T') = w(T) - w(x,y) + w(u,v) $\leq$ w(T).

Since (x,y) \(\not\in\) A ((S, V-S) respects A), then A $\cup${(u,v)} $\subseteq$ T' = MST(G). Thus, (u,v) is a safe edge.



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