Odds and Ends

1. Odds and Ends • HP ≤p HC (again) • Turing reductions • Strong NP-completeness versus Weak NP-completeness • Vertex Cover to Hamiltonian Cycle

2. Example: HP ≤p HC • Hamiltonian Path • Input: Undirected Graph G = (V,E) • Y/N Question: Does G contain a Hamiltonian Path? • Hamiltonian Cycle • Input: Undirected Graph G = (V,E) • Y/N Question: Does G contain a Hamiltonian Cycle?

3. Specification of R(x) • Consider any undirected graph G = (V,E) as input x • R(x) will be a graph G’ = (V’, E’) where • V’ = V union {v} where v is not in V • E’ = E union {(v,w) | w in V} • Argument that R(x) has polynomial size • We add exactly 1 node and |V| edges.

4. x is yes  R(x) is yes • Suppose graph G has a Hamiltonian Path • Let this path be v1, v2, …, vn • We now argue that v1, v2, …, vn, v is a Hamiltonian Cycle in G’ • First, all nodes in V’ are included exactly once above or else v1, v2, …, vn would not be a HP in G • Since G’ has all the edges that G has, (vi,vi+1) is an edge in E’ for 1 ≤ i ≤ n-1 • Finally, since E’ contains edge (v,w) for all w in V, it must be the case that E’ contains edges (vn, v) and (v,v1).

5. R(x) is yes x is yes • Suppose graph G’ has a Hamiltonian Cycle • Let this cycle be v1, v2, …, vn, v • We now argue that v1, v2, …, vn is a Hamiltonian Path in G • First, all nodes in V are included exactly once above or else v1, v2, …, vn, v would not be a HC in G’ • Since the only extra edges in E’ compared to E are edges involving node v, it must be the case that E contains edge (vi,vi+1) for 1 ≤ i ≤ n-1

6. Turing Reducibility • Consider the following alternate reduction. • Given graph G, output Q(n2) graphs Gv,w = (V,Ev,w) where • Ev,w = E union {(v,w)} where v,w are nodes in V • This is not a polynomial-time reduction because we are outputting Q(n2) graphs. • However, this idea can be used to show that if HC can be solved in polynomial time, then HP can be solved in polynomial time. • Run each graph Gv,w through our procedure that solves HC. • If HC says yes for any one of these graphs, return yes. • Otherwise return no. • This more general reduction is often called a Turing reduction. • We allow ourselves to use the procedure that solves HC (or P2) a polynomial number of times rather than just once.

7. Number Problems • Problems where the inputs are numbers • Prime number problem: • Input: Integer n • Yes/No Question: Is n prime? • Partition problem • Input: Set S of n numbers {s1, …, sn} • Yes/No Question: Is there an S’ subset of S such that the sum of numbers in S’ = the sum of numbers in S – S’. • What is the input size for these problems?

8. Knapsack Problem • 0-1 Knapsack optimization problem • Input • Capacity K • n items with weights wi and values vi • Yes/No Question • Find a set of items S such that • the sum of weights of items in S is at most K • the sum of values of items in S is maximized • We gave a dynamic programming solution for this problem • We showed that Partition ≤p Knapsack on hw 8 • Is this a contradiction?

9. Definining subproblems • Define P(i,w) to be the problem of choosing a set of objects from the first i objects that maximizes value subject to weight constraint of w. • Impose an arbitrary ordering on the items • V(i,w) is the value of this set of items • Original problem corresponds to V(n, K)

10. Recurrence Relation/Running Time • V(i,w) = max (V(i-1,w-wi) + vi, V(i-1, w)) • A maximal solution for P(i,w) either • uses item i (first term in max) • or does NOT use item i (second term in max) • V(0,w) = 0 (no items to choose from) • V(i,0) = 0 (no weight allowed) • What is the running time of this solution? • Number of table entries: • Time to fill each entry:

11. Example Items wA = 2 vA = $40 wB = 3 vB = $50 wC = 1 vC = $100 wD = 5 vD = $95 wE = 3 vE = $30 Weight

12. Weak NP-completeness • An NP-complete problem is called “weakly NP-complete” if it has in its description one or more integer parameters and the corresponding problem where these parameters are represented in unary is in P. • An NP-complete problem is strongly NP-complete if the problem is still NP-complete even if integer parameters are encoded in unary

13. u v v u Vertex Cover to Ham Cycle Edge Component: For every edge in the Minimum Vertex Cover problem, we create a component in the Hamiltonian Cycle Problem: u v

14. Observations…. u v u v u v v v u u v u There are only three possible ways that a cycle can include all of the vertices in this component. Key property: If a path enters v (u), it leaves on v’ (u’)

15. Node Selection u v All components that represent edges connected to node u are strung together into a chain. If there are V vertices, then we will have V of these chains, all interwoven. Choosing a node u corresponds to traversing such a chain u v w u w u u x x u

16. y v u w y v u w v u y w v u z x v u z v u x Vertex cover = (v,u) x z v u

17. y v u w y v u w v u y w v u z x v u z v u x Vertex cover = (v,w,x) x z v u

18. Tying the Chains Together If we want to know if its possible to cover the original graph using only k vertices, this would be the same as seeing if we can include all of the vertices using only k chains. How can we include exactly k chains in the Hamiltonian Cycle problem? We must add k extra vertices and connect each of them to the beginning and end of every chain. Since each vertex can only be included once, this allows k chains in the final cycle.

19. Beginning a Transform

20. The Final Transform for k=1