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Long paths in undirected graphs In this question m is the number of edges in an...

Long paths in undirected graphs In this question m is the number of edges in an undirected graph.

1. Show that if the degree of every vertex is at least k, then the graph has a simple path of length at least k. Hint: consider the longest simple path in the graph say from x to y. Show that the endpoints x and y do not have edges to vertices outside the path. Thus all the neighbors of x, y are in the path. And recall that all degrees are at least k.

2. Show that every graph has a subgraph with minimum degree m/n. Hint: iteratively remove all vertex of degree strictly smaller than m/n

3. Show that any graph has a path of length at least m/n. Use the two claims proven in the previous questions.

please do not code, explain the algorithm with time complexity

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Answer #1

1. The N vertices are numbered from 1 to N. As there is no self loops or multiple edges, the edge must be present between two different vertices. So the number of ways we can choose two different vertices are NC2 which is equal to (N * (N – 1)) / 2. Assume it P.
Now M edges must be used with these pair of vertices, so the number of ways to choose M pairs of vertices between P pairs will be PCM.
If P < M then the answer will be 0 as the extra edges can not be left alone.

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