SUBSET-SUM

1. SUBSET-SUM Instance: A set of numbers denoted S and a target number t. Problem: To decide if there exists a subset YS, s.t yYy=t. 13 16 8 21 1 3 6 11

2. SUBSET-SUM is in NP On input S,t: • Guess YS • Accept iff yYy=t. The length of the certificate: O(n) (n=|S|) Time complexity: O(n)

3. t SUBSET-SUM is NP-Complete Proof: We’ll show 3SATpSUBSET-SUM.

4. Examples SUBSET-SUM … because 2+8=10 … because 11 cannot be made out of {2,4,8}

5. Reducing 3SAT to SubSet Sum • Proof idea: • Choosing the subset numbers from the set S corresponds to choosing the assignments of the variables in the 3CNF formula. • The different digits of the sum correspond to the different clauses of the formula. • If the target t is reached, a valid and satisfying assignment is found.

6. Subset Sum clause: 3CNF formula: Make the number table, and the ‘target sum’ t dummies +

7. Reducing 3SAT to SubSet Sum • Let 3CNF with k clauses and  variables x1,…,x. • Create a Subset-Sum instance <S,t> by: 2+2k elements of S = {y1,z1,…,y,z,g1,h1,…,gk,hk} • yj indicates positive xj literals in clauses • zj indicates negated xj literals in clauses • gj and hj are dummies • and

8. Subset Sum Note 1: The “1111” in the target forces a proper assignment of the xi variables. Note 2: The target “3333” is only possible if each clause is satisfied. (The dummies can add maximally 2 extra.)

9. Subset Sum + is a satisfying assignment

10. Subset Sum + is not a satisfying assignment

11. Proof 3SAT P Subset Sum • For every 3CNF , take target t=1…13…3 and the corresponding set S. • If 3SAT, then the satisfying assignment defines a subset that reaches the target. • Also, the target can only be obtained via a set that gives a satisfying assignment for .

12. Finding the Solution • If the decision problem Subset-Sum is solvable in polynomial time, then we can also find the subset that reaches the target in poly-time. • How? • By asking smart questions about several variants of the problem. • That way, we can determine which variables xi are involved in the solution.

13. but this graph has no Hamiltonian path this graph has a Hamiltonian path Directed Hamiltonian Path • Given a directed graph G, does there exist a Hamiltonian path, which visits all nodes exactly once? • Two Examples:

14. 3SAT to Hamiltonian Path • Proof idea: Given a 3CNF , make a graph G such that • … a Hamiltonian path is possible if and only if there is a zig-zag path through G, • … where the directions (zig or zag) determine the assignments (true or false) of the variables x1,…,xL of .

15. s For every i, “Zig” means xi=True … while “Zag” means xi=False t “Zig-Zag Graphs” There are 2L different Hamiltonian paths from s to the target t.

16. Accessing the Clauses If  has K clauses,  = C1 Ck, then xi has K “hubs”: Idea: Make K extra nodes Cj that can only be visited via the hub-path that defines an assignment of xi which satisfies the clause (using zig/zag = T/F). stands for

17. If then xi=True=“zig” reaches node Cj If then xi=False=“zag” reaches node Cj Connecting the Clauses Let the clause Cj contain xi:

18. Proof 3SAT P HamPath • Given a 3CNF (x1,…,xL) with K clauses • Make graph G with a zig/zag levels for every xi • Connect the clauses Cj via the proper “hubs” • If SAT, then the satisfying assignment defines a Hamiltonian path, and vice versa.

19. Example 4 variables… 4 clauses… Clauses connected via zig-zag “hubs”

20. Example assignment… …satifies all four clauses; hence it defines a Hamiltonian Path

21. Example assignment… …does not satify first clause; hence the path misses the C1 node

22. More on Hamiltonian Path • Useful for proving NP-completeness of “optimal routing problems” • Typical example: NP-completeness of “Traveling Salesman Problem” • Issue of directed versus undirected graphs

23. Forcing Directions Given a directed graph with s,x1,…,xk,t Replace s with sout, the t with tin ,and every xi with the triplet “xi,in—xi,mid —xi,out” Redraw the original directed edges with edges going from out-nodes to in-nodes. If and only if the directed graph has a HamPath from s to t, then so has this graph.

24. becomes becomes xin yin xin yin xmid ymid xmid ymid sout sout xout yout xout yout Example: Directions x x s s y y “Undirected HamPath” is NP-complete