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

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)

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

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

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.

Subset Sum + clause: 3CNF formula: Make the number table, 1 2 3 4 +x1 –x1 +x2 –x2 +x3 –x3 +x4 –x4 1 3CNF formula: Make the number table, and the ‘target sum’ t dummies + 1 3

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

Subset Sum Note 1: The “1111” in the target forces a proper assignment +x1 –x1 +x2 –x2 +x3 –x3 +x4 –x4 1 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.) 1 3

Subset Sum + is a satisfying assignment +x1 –x1 +x2 –x2 +x3 –x3 +x4 1 3 + is a satisfying assignment 1 3

Subset Sum + is not a satisfying assignment +x1 –x1 +x2 –x2 +x3 –x3 1 ? 2 + is not a satisfying assignment 1 3

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 .

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.

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

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 .

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

Accessing the Clauses stands for 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

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

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.

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

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

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

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

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.

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