Prove that set cover is NP-complete with a reduction from vertex cover.
• Set Cover
  Input: a set X of n elements, a family F of subsets of X, and an integer k.
  Question: Does there exist k subsets from F whose union is X?
  
• Vertex Cover
  Input: A graph G = (V, E) and an integer k <= |V|.
  Question: Is there a subset S of at most k vertices such that every edge in E has at least one vertex in S?
  
  
• First show the problem is in NP -
  Given a solution of the set cover problem we can create an array where each element in X has a corresponding element in the array. Search through the subsets in F marking each element found in the array. When finished searching through F, check if all elements were found. This is on the order of O(# of subsets*|X|), and thus polynomial in the input.

• Second find a reduction from vertex cover to set cover.
  VertexCover(G=(V,E), k)
  X = E
  Create sets S_i by adding in each edge adjacent to vertex v_i, for all vertices in G.
  Let F be the collection of all the sets S_i Let k'=k
  SetCover (X, F, k')

• Last show that the reduction preserves the answers.
  • Assume that the graph G has a vertex cover with at most k vertices. Call it set C. For each vertex in C, add the corresponding set S_i to the collection C'. Since every edge in G is adjacent to a vertex in C, then C' must contain a subset that contains that edge. Thus C' is a set cover for X and F with size k. Thus, there is a set cover with k subsets.
  • Assume that a set cover for X and F exists with k subsets. Then, the vertices that correspond to the subsets in the set cover must also be a vertex cover. This follows since all the edges of the graph are covered by the set cover.

• Thus, the problem set cover is NP-complete.

Prove that the baseball card collector problem is NP-hard.
• Let X=the set of players and F be the collection of P₁, P₂, ..., P_m. Then let k be the same k. This is the set cover problem since a collection of packets that contains all of the players is a set cover. Since set cover is NP-complete, the baseball problem is also. Thus, it is NP-hard.

1. Prove that the low degree spanning tree is NP-hard with a reduction from Hamiltonian Path.
   • Low Degree Spanning Tree
     Input: A graph G and an integer k.
     Question: Does G contain a spanning tree such that all vertices in the tree have degree at most k?
     
   • Hamiltonian Path
     Input: A graph G = (V, E).
     Question: Does there exist a path in G that contains all the vertices?
     
   • Reduction:
     Ham Path (G)
     Let G' = G
     let k'=2
     MinSpanTree(G', k')
   • Show the answers are equivalent:
     Assume that there is a hamiltonian path in G. Take the hamiltonian path to be the minimum degree spanning tree (since its maximum degree is 2). Assume that there is a low degree spanning tree of maximum degree 2, then it must be a hamiltonian path since its spanning. Thus, the answers are equivalent.
2. Give an efficient algorithm to solve the high degree spanning tree problem and an analysis of its time complexity.
   • Algorithm HighSpanTree(G, k):
     Calculate d_i = degree of vertex i
     let k'=max {d_i, i = 1, 2, ..., # of vertices}
     if k <= k' return true, else return false
     
   • Analysis: Since it takes linear time (in # of edges + # of vertices) to calculate the degrees and linear time (in # of vertices) to find the maximum of the k_i's, it must take a linear time (in # of vertices and edges) to run the algorithm.

There is no one unique solution.