Question

Suppose we have two weighted graphs G1 = (V, E, w) and G2 = (V, E, w0 ) where w 0 (e) = w(e)+1 and w, w0 are the edge weights in the two graphs. The graphs are identical, except that all the edges weights in G2 are one larger than the corresponding edge in G1. Suppose that in G1, we have computed the shortest weighted path from some fixed node s ∈ V to some other node t ∈ V . Call this path p. Is it always the case that p is the shortest s → t path in G2? If it is, then prove it, otherwise give a counterexample

Answer 1

Yes, it is always the case that, p is the shortest s -> t path in G2.
We will prove this by contradiction.

First, we are given that in graph G1 = (V,E,w)

there is a path p from vertex s to vertex t such that s->t is the shortest path.

Also, it's known that for all edges e' in G2, and corresponding e in G1, the following relation holds true:

w(e') = w(e) + 1

Now, let suppose that path p is NOT the shortest path from s->t in G2, or in other words there exists some other path p' which is shorter than p in G2

Now,

sum of weights in path p in G1 can be represented as:

$\sum_{p}w(e) = S$

over the path p.

Also, the exact corresponding path sum of the weights in G2 is ( which is not the minimum )

$\sum_{p}w_0(e) = S'$

Also, we know that

So,

Here, n(P) is the cardinality of set P ( the path p )
This is a contradiction as S is the minimum path sum in G1 whereas S' is not, and they are related by a constant factor of path length.

This contradiction arose due to the assumption that p is not the minimum sum path in G2.
Hence p is the minimum sum path in G2 as well.