COMPLETE-SUBGRAPH problem is defined as follows: Given a graph G = (V, E) and an integer k, output yes if and only if there is a subset of vertices S ⊆ V such that |S| = k, and every pair of vertices in S are adjacent (there is an edge between any pair of vertices). How do I show that COMPLETE-SUBGRAPH problem is in NP? How do I show that COMPLETE-SUBGRAPH problem is NP-Complete?

Let G be a graph with n vertices representing a set of gamers. There is an edge between two nodes if the corresponding gamers are friends. You want to partition the gamers into two disjoint groups such that no two gamers in the same group are friends. Is the problem in P or in NP? Give a formal proof for your answer. A 'Yes' or 'No' answer is not sufficient to get a non-zero mark on this question.

In the decision version of the Clique problem you are given a non-empty graph G = (V,E) and a positive integer k and asked whether G has a complete subgraph (or clique) of size k (i.e., k vertices that induce a complete subgraph).  Give the pseudocode of a recursive backtracking algorithm for the problem, and analyze your algorithm by providing a worst-case recurrence. Can you guarantee a worst-case running time in O(1.466^n)? Prove your answer

Please show written work with answer!! An independent set in a graph is a set of vertices no two of which are adjacent to each other. A clique is a complete subgraph of a given graph. This means that there is an edge between any two nodes in the subgraph. The maximal clique is the complete subgraph of a given graph which contains the maximum number of nodes. We know that independent set is a NP complete problem. Transform the independent set to max clique (in polynomial time) to show that Max Clique is also NP

True or False (If your answer to the question is "False", explain why, and provide correction when possible). (a) Let h(n) be the heuristics for the node n, h(m) be the heuristics for the node m, d(m,n) be the actual minimal cost from node m to n in a graph. A* satisfies the monotone restriction iff d(m,n) <= |h(n)-h(m)|. (b) If an A* heuristics is admissible then it satisfies the monotone restriction. (c) Best-first search guarantees optimality in its returned solution. (d) Least-cost-first search guarantees optimality in its returned solution. (e) If all edges are with unit cost, then Breadth-first search guarantees optimality in its returned solution.

The third-clique problem is about deciding whether a given graph G = (V, E) has a clique of cardinality at least |V |/3.Show that this problem is NP-complete.

5. (This question goes slightly beyond what was covered in the lectures, but you can solve it by combining algorithms that we have described.) A directed graph is said to be strongly connected if every vertex is reachable from every other vertex; i.e., for every pair of vertices u, v, there is a directed path from u to v and a directed path from v to u. A strong component of a graph is then a maximal subgraph that is strongly connected. That is all vertices in a strong component can reach each other, and any other vertex in the directed graph either cannot reach the strong component or cannot be reached from the component. (Note that we are considering directed graphs, so for a pair of vertices u and v there could be a path from u to v, but no path path from v back to u; in that case, u and v are not in the same strong component, even though they are connected by a path in one direction.) Given a vertex v in a directed graph D, design an algorithm for com- puting the strong connected…

Can you help me solve this exercise? Please note that the greedy approach described in the advice paragraph does not work.

If A is NP-complete, and A has a polynomial time algorithm, then there is a polynomial time algorithm to find the largest clique in a directed. A clique in a graph G(V, E) is a subset of vertices V’ such that there is an edge (u,v) in E between every two vertices u and v in V’. For example, the four vertices u2, u3,u5, and u6 form the largest clique in the graph below.

A graph is biconnected if every pair of vertices is connectedby two disjoint paths. An articulation point in a connected graph is a vertex that woulddisconnect the graph if it (and its adjacent edges) were removed. Prove that any graphwith no articulation points is biconnected. Hint : Given a pair of vertices s and t and apath connecting them, use the fact that none of the vertices on the path are articulationpoints to construct two disjoint paths connecting s and t.