A walk of length n in a graph G is an alternating sequence v0; e1; v1...
A walk of length n in a graph G is an alternating sequence v0; e1; v1 : : : ; vn of vertices and edges of G such that for all i is an element or 1; : : : ; n, ei is an edge relating vi-1 to vi. Show that for any finite graph G and walk v0; e1; v1 : : : ; vn in G, there exists a walk from v0 to vn with no repeated edges. (Hint: Use complete (strong) induction on the number of edges in the walk.) PLEASE USE STRONG INDUCTION.
Homework Answers
Let us induct on the number of edges n in the walk.
Base case: P(n) if n = 1 means, the graph has 1 edge and 2 vertices. Therefore, it is trivially true.
Inductive Hypothesis: We will assume that P(1) ... P(n) is true and we will show the strong induction to prove that P(n+1) also holds true.
Inductive Steps: Let us assume that graph G has n edges and n+1 vertices. Let's remove the last edge from Graph G. So, the resultant graph G' has n -1 edge and n vertices. By inductive hypothesis P(n) is true (as per the strong induction). Now, add the removed vertices to v_n in Graph G'. So, G' has a walk from v_0 to v_n and a path from v_n to v_n+1. Which basically implies that there is a walk from v_0 to v_n+1. Therefore P(n+1) also holds true.
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A walk of length n in a graph G is an alternating sequence v0;
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