Use the restriction method to prove the NP-completeness of the following problem
ID: 3576131 • Letter: U
Question
Use the restriction method to prove the NP-completeness of the following problems: CLUSTER Instance: Graph G and positive integers d and k. Question: Does G contain greaterthanorequalto k vertices V' such that for any two vertices u, v belongs to V' their distance d_G(u, v) in G is at most d? MULTICOMPONENTS Instance: Graph G, integers c and k. Question: Does G contain lessthanorequalto k vertices V such that G - V' contains at least c components? PACKING Instance: A box B of dimension L times W times H, and a set S of n rectangular solid objects of dimension l_i times W_i times h_i, 1 lessthanorequalto i lessthanorequalto n. Question: Can we choose some S' S to put inside B without leaving any space?Explanation / Answer
a)G contains >=k verices for any wo vertices for any two verices u,v belongs to v their distance in G at most D
such that A clique in a graph G is a subset of vertcies V such that V is not equal to w (v,w) is an edge.There is an edge between every pair of vertices in C.we can construct the complement graph bu has the edge(v,w). and only if(v,w) not equal to g.We define the the parameter for clique n-k where k is integer cover for k.T he construction runs in polynomial serves as a reduction for size of n-k if and only if the vertex cover of size k.
b)For each edge E in choose deterministicly if it is to be included in T
check that T is a tree and each vertex has degree less than K.
That means the problem is in NP.As a spanning tree which can also be said to reduce >=2 vertices .It cannot branch such that spanning tree only has two vertices < 2.so the spanning tree is a graph.As this path visits every node and a path is trivially tree.So the required contains all components.Reducing it to spanning tree problem .moreover this problem reduces to spannig tree problem.
c) It can be said as Knap sack problem for np-complete problem .Here the maximum empty set can be considerd as 0.By using their weights w by greatest common divisor
m[0]=0
m[w]=max(v+(w-w1))
The O(nw) complexity does not contract the fact kanpsack problem is not 0/1 knapsack problem.
m[0,w]=0
m[i,w]=m[i-1+w]
It can also be given as dynamic programming problem.It can also be given as subset sum problem which can be further reduced.So without leaving any space it can be reduced 0/1 knap sack problem .No empty space is required
all can be filled without any boundaries.
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