Homework 4 Solutions Uploaded 4:00Pm on Dec 6, 2017 Due: Monday Dec 4, 2017

Home , Decision problem, Reduction (complexity)

CS3510 Design & Analysis of Algorithms Section A Homework 4 Solutions Uploaded 4:00pm on Dec 6, 2017 Due: Monday Dec 4, 2017

This homework has a total of 3 problems on 4 pages. Solutions should be submitted to GradeScope before 3:00pm on Monday Dec 4. The problem set is marked out of 20, you can earn up to 21 = 1 + 8 + 7 + 5 points. If you choose not to submit a typed write-up, please write neat and legibly. Collaboration is allowed/encouraged on problems, however each student must independently complete their own write-up, and list all collaborators. No credit will be given to solutions obtained verbatim from the Internet or other sources. Modiﬁcations since version 0

1. 1c: (changed in version 2) rewored to showing NP-hardness.

2. 2b: (changed in version 1) added a hint.

3. 3b: (changed in version 1) added comment about the two loops’ u variables being diﬀerent due to scoping.

0. [1 point, only if all parts are completed]

(a) Submit your homework to Gradescope. (b) Student id is the same as on T-square: it’s the one with alphabet + digits, NOT the 9 digit number from student card. (c) Pages for each question are separated correctly. (d) Words on the scan are clearly readable.

1. (8 points) NP-Completeness Recall that the SAT problem, or the Boolean Satisﬁability problem, is deﬁned as follows:

• Input: A CNF formula F having m clauses in n variables x1, x2, . . . , xn. There is no restriction on the number of variables in each clause. • Output: YES if there is an assignment to the variables which satisﬁes all m clauses, and NO otherwise.

Now consider the Almost-SAT problem which is deﬁned below:

1 • Input: A CNF formula F having m clauses in n variables x1, x2, . . . , xn. There is no restriction on the number of variables in each clause. • Output: YES if there is an assignment to the variables which satisﬁes exactly m − 1 clauses, and NO otherwise.

For example, given the boolean formula F to be

F = (x1 ∨ x2) ∧ (x1 ∨ x¯3 ∨ x¯4 ∨ x5) ∧ (¯x2) ∧ (x2 ∨ x5)

Here is one assignment to the variables which satisﬁes all four clauses: x1 = T, x2 = F, x3 = F, x4 = T, x5 = T . So the expected output of SAT on this example will be YES.

This formula also has an assignment which satisﬁes exactly 3 of the 4 clauses: x1 = T, x2 = T, x3 = F, x4 = T, x5 = T . So the expected output of Almost-SAT on F is also YES.

The goal of this problem is to show that Almost-SAT is NP-complete.

(a) (2 points) We will first prove that Almost-SAT is in NP. A problem is said to be in NP if a solution to the problem can be verified in polynomial time. Give a polynomial time algorithm which takes as input an assignment to each variable, and verifies whether it is a solution to Almost-SAT or not. SOLUTION: Loop through all the clauses and check whether each of them is satisfied. Maintain a global counter, and check whether it equals to m − 1 at the end. This algorithm runs in poly time because there is at most m clauses. GRADING: 1 point for the algo, 1 point for its runtime justification.

(b) (3 points) Let F be a boolean formula having m clauses in variables x1, x2, . . . , xn. Construct a new formula F 0 by taking the conjunction of F with two new clauses (i.e. F 0 should have m + 2 clauses) such that an assignment of x1, . . . , xn which satisﬁes all m clauses in F will satisfy exactly m + 1 clauses in F 0. Explain what your two new clauses should be. There is no restriction on the length of the two clauses, and you can use one or more of the variables from F to construct them. SOLUTION: Add the clauses x1 and ¬x1.

2 As x1 is either true for false, exactly one of them can be satisfied. GRADING: 2 points for clauses. 1 point for justification of why only one of them can be satisfied. (c) (3 points) To complete the proof of NP-completeness, we need to show that Almost-SAT is NP-hard. To do this, we take SAT, a known NP-hard problem, and reduce it to an instance of Almost-SAT. Let F be the boolean formula input to SAT. Let F 0 be the formula obtained from F using the construction in Part (b). We will use F 0 as our input to Almost-SAT. Specifically, show that if F has an assignment which satisfies SAT, F 0 has an assignment which satisfies Almost-SAT; and if F has no assignment which satisfies SAT, F 0 has no assignment which satisfies Almost-SAT. Argue that the construction of F 0 takes polynomial time. SOLUTION: Consider taking F and augmenting it as above in part b).

If an assignment x1 . . . xn satisfies F , then it satisfies exactly one of the two extra constraints, giving exactly m + 1 = m0 − 1 total. Consider an assignment that almost-satisfied F 0. The only unsatisfied clause must be one of x1 or ¬x1, so all the original m clauses are satisfied. Finally, this reduction takes time linear in the input since it simply appends O(1) things to the clauses. GRADING: 1 point for showing the forward direction: F satisfiable implies F 0 almost-satisfiable. 1 point for showing the reverse direction: F 0 almost-satisfiable implies F satisfiable. 1 point for runtime and overall NP-hardness claim. 2. (7 points) Hardness and Approximation of Maximum 2-SAT

Consider a 2-SAT instance: there are n Boolean variables x1 . . . xn and m clauses, each of the form of li1 ∨ li2,

where lij could be xr or ¬xr for some variable xr.

The goal of maximum 2-SAT is to ﬁnd an assignment to the xrs that satisﬁes the maximum number of these clauses.

(a) (1 points) Give a decision version of maximum 2-SAT. SOLUTION: Given a 2-SAT instance, and a parameter k (1 ≤ k ≤ m), output whether at least k of the clauses can be satisﬁed.

3 GRADING: 1 point iﬀ it’s a decision problem that’s inter-reducible to maximum 2-SAT. (b) (3 points) We will show that maximum 2-SAT is NP-complete by showing that

Max-Cut → Max-2-SAT.

Here Max-2-SAT is the decision version of the maximum 2-SAT problem that you defined in part a). The definition of the deterministic version of Max-Cut (which you may assume is NP-complete) is Definition 0.1 (Decision Version of Max-Cut). Given an undirected unweighted graph G and a parameter k, answer whether there is a cut S ⊆ V (G) so that the number of edges leaving S is at least k.

HINT: first figure out the following reduction involving a single edge: for two Boolean variables x and y, give two SAT clauses over them so that they are both satisfied when x 6= y, and only one of them is satisfied when x = y. SOLUTION: For each edge connecting vertices xu and xv, construct the clauses

xu ∨ xv

¬xu∨= 6 xv

Both are satisﬁed when xu 6= xv, and if xu = xv only one of them is satisﬁed. This produces 2m clauses, and we can satisfy m + k of them if and only if there is a cut involving k edges. Therefore this conversion, and adding k to m gives the reduction. GRADING: 1 point for construction per edge. 1 point for the overall SAT instance (one ‘gadget’ per edge). 1 point for conversion of k (if needed). (c) (1 points) Conclude from parts a) and b) that Max-2-SAT is NP-Complete. SOLUTION:

Since MaxCut is NP-hard, part b) implies that Max-2-SAT is also NP-hard. Furthermore, Max-2-SAT is in NP because given an assignment of the variables, we can check whether at least k clauses are satisﬁed. GRADING: 1 point if both the hardness and in NP are there, 0 otherwise.

4 (d) (2 points) Give a 2-approximation algorithm for the optimization version of maximum 2- SAT. Your algorithm should run in O(m) time. SOLUTION: One solution is to emulate maxcut: i. order the variables arbitrarily.

ii. for each variable xi, check among all the unsatisﬁed clauses involving it (with smaller ids) whether more has xi or ¬xi. Set it to the majority.

The choice of xi guarantees that at least m/2 clauses are satisfied. As OPT ≤ m, this is a 2-approximation. Another possible solution is to pick a random assignment to the variables, giving 3/4m clauses satisfied in expectation. This solution is not expected since probability is not a prerequisite. GRADING: 1 point for showing the lower bound on the number of clauses satisfied. 1 point for everything else (mainly inequality between OPT and this bound).

3. (5 points) Approximating Graph Diameter Define the distance between a pair of vertices x and y, Dist(x, y) to be the length of the shortest path between them. The diameter of a graph is then the maximum distance between a pair of vertices D = max Dist (u, v) . u,v In this problem we will give an O(m) time 2-approximation algorithm to the diameter of an undirected, unit weighetd graph with n vertices and m edges. That is, we will try to produce a pair of vertices, x, y, such that 1 Dist(x, y) ≥ D. 2 For the purpose of this problem, you may use the fact that on any weighted (for this problem, unit weights suffice) graph with positive edge weights, for any vertex s, breadth first search (BFS) computes in O(m) time the values of Dist(s, u) for all vertices u. (we did not cover BFS in this class, but you should only use it as a black-box here)

(a) (1 point) Show using the given running time of BFS that we can compute exactly the diameter of a graph in O(nm) time. SOLUTION: Run BFS from every vertex, pick the longest distance computed. GRADING: 1 point if algorithm is correct.

5 (b) (2 points) Use the fact that distances obey the triangle inequality, aka

Dist(u, v) ≤ Dist(u, w) + Dist(v, w)

for any vertices u, v, w to show that for any vertex s,

D ≤ 2 · max Dist(s, u). u (note that ‘iteration-like’ operators such as max or summation work similar to for loops: the u on the RHS is diﬀerent than the u in the deﬁnition of D) SOLUTION: Let u, v be the pair that maximizes the distance between them. We have Dist (u, v) ≤ Dist (s, u) + Dist (s, v) ≤ 2 · max Dist (s, w) . w GRADING: 1 point for invoking triangle inequality.

1 point for the overall bound (Dist(s, u) to maxw Dist(s, w)). (c) (2 points) Use part b) to give an algorithm that on a connected graph with m edges, produces an 2-approximation of the diameter in O(m) time. Speciﬁcally, it should output in O(m) time a pair of vertices, x, y such that 1 Dist(x, y) ≥ D. 2 SOLUTION: Consider teh following algorithm: i. Pick some arbitrary vertex s. ii. Compute Dist(s, u) from s using the routine provided. iii. Find v that maximizes Dist(s, v). iv. Return Dist(s, v). Part b) guarantees that this is a 2-approximation. GRADING: 1 point for algorithm. 1 point for analysis.