Introduction to Computational Complexity Classes

Introduction to Computational Complexity

Introduction

In the lecture on Turing reducibility, we witnessed how an instance of problem A can be solved by successively solving instances of another problem B. This represents a “positive” use of reducibility, in that an algorithm for one problem leads to an algorithm for another. In this lecture we look at a “negative” use of reducibility, in the sense that a reduction from a hard problem A (i.e. a problem for which no known eﬃcient algorithm exists) to a problem B lends evidence to the belief that B is also hard. Despite obtaining a “negative” result, this approach seems quite useful, since many problems encountered in practice are inherently (but often not provably) diﬃcult, and reducibility helps lend supporting evidence.

The logic of negative reducibility proceeds as follows. Suppose it is believed, if not proved, that there is no efficient (e.g. polynomial-time) algorithm for solving problem A. Now suppose A can be efficiently reduced to problem B. Then if there is an efficient algorithm for solving B, then it implies an efficient algorithm for solving A. Therefore, it must be believed, if not proved, that no such algorithm for solving B exists.

Working with Decision Problems

Based on the kind of reducibility that will be employed in this lecture, it will seem more convenient to assume that each problem is a decision problem; i.e. a problem whose solution involves returning a true or false value. For example, SAT, CNF-SAT, k-SAT, 3SAT, 2SAT, and Perfect Matching are all examples of decision problems. Moreover, even a problem whose solution involves determining a structure or numerical value can be defined as a decision problem. For example, the problem of finding a minimum spanning tree for a weighted connected graph G can be defined as the problem of deciding if G has a spanning tree of cost not exceeding k, for some integer k. Thus, by introducing an additional parameter k, the minimum-spanning tree optimization problem is transformed into a decision problem.

1 Example 1. Recall that the Minimum Spanning Tree (MST) problem takes as input a connected weighted graph G = (V, E, w) and returns a spanning tree for G whose edge weights sum to a minimum value. Re-deﬁne this as a decision problem.

2 It should be emphasized that there is little if anything lost by treating a problem as a decision problem. For example, if the original problem involves finding a structure that optimizes some quantity, and its corresponding decision problem can be efficiently solved, then almost invariably the original problem can also be efficiently solved.

Example 2. Suppose the MST decision problem can also be solved in polynomial time. Show that its corresponding optimization problem can be solved in polynomial time, and that an mst can be constructed in polynomial time.

3 When working with decision problem L, we often replace the phrase “L can be solved” with that of “L can be decided”. Similarly, solvable decision problems are often referred to as being decidable.

Many-to-one Reducibility

Given a decision problem L and a problem instance x of L, we call x a positive instance of L, if an algorithm for solving L returns true on input x. Similarly, x a negative instance of L, if an algorithm for solving L returns false on input x. For example, x1 ∨ x2 is a positive instance of SAT, while x1 ∧ x1 is a negative instance of SAT.

Let L1 and L2 be two decision problems. We say that L1 is many-to-one, polynomial-time p reducible to L2, written as L1 ≤m L2, iﬀ there exists a function f : L1 → L2 that satisﬁes the following properties.

1. f(x) is polynomial-time computable in the size |x| of an L1-problem instance x.

2. f(x) is a positive instance of L2 iﬀ x is a positive instance of L1.

Theorem 1. The Clique problem is the problem of deciding if a simple graph G = (V,E) has a clique of size k, for some integer k ≥ 0; i.e. a subset C ⊆ V of k vertices that are pairwise adjacent. Show that 3SAT is polynomial-time many-to-one reducible to Clique.

Proof of Theorem 1. Let C be a collection of m clauses, where clause ci, 1 ≤ i ≤ m, has the form ci = li1 ∨ li2 ∨ li3. We proceed to construct a simple graph f(C) = G = (V,E) that has an m-clique if and only if C is satisfiable. G is defined as follows. V consists of 3m vertices, one for each literal lij, 1 ≤ i ≤ m and 1 ≤ j ≤ 3. Then (lij, lrs) ∈ E iff i 6= r and lij is not the negation of lrs (i.e. they are logically consistent).

Then G can be constructed from C in time polynomial in the size of C, since the order of G is m 3m, and the size of G is not more than 9 = O(m2). 2

Now assume, that C is satisﬁable. Given a satisfying assignment for C, let liji , 1 ≤ i ≤ m, denote a literal from clause i that is satisﬁed by this assignment. Then, since each pair of these literals is consistent, (liji , lrjr ) ∈ E for all i 6= r. In other words, l1j1 , l2j2 , . . . , lmjm is an m-clique for G.

Conversely, assume G has an m-clique. Then the clique must have the form R = {l1j1 , l2j2 , . . . , lmjm }, since no two literals from the same clause are adjacent. Moreover, by the deﬁnition of G, R is a consistent set of literals and the i th literal of R is a member of ci. Hence the assignment aR induced by R satisifes every clause of C, and thus C is satisﬁable.

4 Example 3. Show the reduction given in the previous example using

C = {(x1, x1, x2), (x1, x2, x2), (x1, x2, x2)}.

5 The Complexity Class P

A complexity class represents a set of decision problems, each of which can be solved within a certain limit placed on time and space (i.e. memory). For example, the complexity class P denotes all decision problems that can be solved in a polynomial amount of time. More speciﬁcally, decision problem L is a member of P iﬀ there exists a predicate function P (x) for which

1 if x is a positive instance of L P (x) = 0 otherwise and P (x) can be decided in an amount of time bounded by p(|x|), for some polynomial p.

Example 4. From previous lectures we may conclude that the decision-problem versions of the the following problems belong to class P: Minimum Spanning Tree, Single Source Distances, Unit Task Scheduling, Huffman Coding, Fuel Reloading, Fractional Knapsack, Find Statistic, Find Median, Minimum Distance Pair, Addition, Multiplication, Matrix Multiplication, Minimum Subsequence Sum, Edit Distance, Optimal Binary Search Tree, Matrix Chain Multiplication, DAG Longest Path, Optimal Change, All-Pairs Distances, 2SAT, Maximum Matching, Maximum Flow.

The Complexity Class NP

Decision problem L is a member of complexity class NP iﬀ there exists a polynomial-time decidable predicate P (x, y), such that x is a positive instance of L iﬀ the predicate ∃yP (x, y) evaluates to true, where the domain elements of y that are tested in P satisfy |y| ≤ p(|x|), for some polynomial p.

Although, for any given value of y, P (x, y) can be be decided in polynomial time, we see that the addition of quantiﬁer ∃y adds a potential order of complexity to the deciding of x, since it may be necessary to compute P (x, y) an exponential number of times in order to ﬁnd an appropriate witness y0, for which P (x, y0) evaluates to true. This is true since there can potentially exist an exponential number of domain values of y whose size does not exceed p(|x|), for some polynomial p, and hence it may be necessary to evaluate P (x, y) an exponential number of times before identifying a witness.

6 Example 5. We show that SAT ∈ NP. Given Boolean formula instance F (x1, . . . , xn) of SAT, let y be a variable whose domain consists of all possible assignments α over V = {x1, . . . , xn}. Then F is satisfiable iff ∃y eval(F, y) evaluates to true. Moreover, from the previous lecture, we know that eval(F, y) can be decided in time proportional to |F |. Moreover, each assignment α ∈ dom(y) satisfies |α| = n ≤ |F |. Therefore, SAT ∈ NP.

7 Example 6. An instance of Clique is i) a simple graph G = (V,E), and ii) a nonnegative integer k ≤ |V | = n. The problem is to decide if G has a clique of size at least k. Show that Clique ∈ NP.

8 Theorem 2. P ⊆ NP.

Proof of Theorem 2. Let L ∈ P be given. Then there is a polynomial-time computable predicate function P (x) that decides L. Let y be a variable for which dom(y) = {1} consists of a single element. Define predicate function P 0(x, y) as P 0(x, y) = P (x). Then P 0 is polynomial-time computable, and x is a positive instance of L iff P (x) evaluates to true, iff P 0(x, 1) evlauates to true, iff ∃yP (x, y) evlauates to true. Therefore, by definition, L ∈ NP.

Do there exist decision problems that are in NP but not in P? This would imply that the use of the ∃ operation is necessary for solving some problems with a polynomial-time predicate function. The question remains unanswered since there are decision problems in NP for which no eﬃcient polynomial-time algorithms are currently known, while at the same time no one has yet to prove that these problems do not admit eﬃcient algorithms.

9 The Complexity Class co-NP

Given a decision problem L, the complement of L, denoted L, is that decision problem for which a positive (respectively negative) instance x of L is a negative (respectively positive) instance of L.

Example 7. If L is the problem of deciding if a positive integer is prime, then deﬁne its complement L.

Example 7 Solution.

10 Example 8. Deﬁne the complement of the SAT decision problem.

Example 8 Solution.

11 A Logical deﬁnition of Co-NP

It is left as an exercise to show that if L ∈ P then L ∈ P. On the other hand, the complexity class co-NP is defined as follows. Problem L is the the class co-NP if and only if L is a member of NP. It is generally believed that NP and co-NP represent two unequal complexity classes. One reason for this is that co-NP has its own independent logical definition. Indeed, it is an exercise to show that decision problem L is a member of co-NP iff there exists a polynomial-time decidable predicate P (x, y), such that x is a positive instance of L iff the predicate ∀yP (x, y) evaluates to true, where the domain elements of y that are tested in P satisfy |y| ≤ p(|x|), for some polynomial p.

12 The Classes PH and PSPACE

The Polynomial Hierarchy generalizes the classes NP and co-NP via recursive definitions using the Pp Qp ∃ and ∀ operations. For the base cases, 0 = 0 = P. For the inductive cases, for k ≥ 1, decision Pp Qp problem L is a member of k iff there exists a k−1-computable predicate function P (x, y), such that x is a positive instance of L iff the predicate ∃yP (x, y) evaluates to true, where the domain elements of y that are tested in P satisfy |y| ≤ q(|x|), for some polynomial q. Similarly, decision Qp Pp problem L is a member of k iff there exists a k−1-computable predicate function P (x, y), such that x is a positive instance of L iff the predicate ∀yP (x, y) evaluates to true, where the domain elements of y that are tested in P satisfy |y| ≤ q(|x|), for some polynomial q.

Pp Qp From the above deﬁnitions, we see that 1 = NP, while 1 = co-NP. Moreover, class PH is deﬁned ∞ Pp Qp as PH = ∪ ( k ∪ k). k=0

Example 9. Consider the deciison problem Clique Partition, where an instance consists of i) a simple graph G = (V,E), where each vertex of G is labeled with a nonnegative integer, and ii) a nonnegative intger k. The problem is to decide if every k-clique of G can be partitioned into two sets, where the labels of the ﬁrst set sum to the labels of the second set. Prove that Qp Clique Partition ∈ 2.

We note in passing that PSPACE denotes all decision problems that can be decided using a polynomial amount of space. It is left as an exercise to prove that PH ⊆ PSPACE. Moreover, it can be shown Pp Qp that any problem in PSPACE can be logically defined in a manner similar to k and k, but now the number of alternating quantifiers is allowed to grow polynomially (as opposed to being fixed at k) in the size of input x.

13 NP-Complete Decision Problems

Now that we have an idea of what kind of decision problems belong in NP, we would like a method for demonstrating that a decision problem is in some sense hard and possibly may not be in P. Furthermore, if we think about what might constitute a difficult decision problem in a class of decision problems, the most difficult decision problem would seem to be one for which every other p decision problem in that class can be reduced to. Indeed, if A ≤m B, then A is no harder than B, since an efficient algorithm for B leads to an efficient algorithm for A. We say that such a problem is complete for that class of decision problems. This leads to the following definition. A decision problem L ∈ NP is said to be NP-complete iff

1. L ∈ NP

p 2. for every other decision problem L2 ∈ NP, L2 ≤m L.

Note that, in case L 6∈ NP, then we refer to L as being NP-hard. In particular, the optimization version of an NP-complete decision problem is usually NP-hard.

Lemma 1. Let P (x, y) be a polynomial-time computable predicate function, and let p be a polynomial. Then for each input x, there exists a Boolean formula Fx for which the following is true:

1. Fx may be computed in polynomial time,

2. the variables of Fx encode precisely those inputs y for which |y| ≤ p(|x|), and

3. for every input y0 for which |y0| ≤ p(|x|), P (x, y0) evaluates to true iﬀ eval(Fx, α) evaluates to true, where α is the assignment over the variables of Fx that encodes y0.

The proof of Lemma 1 is beyond the scope of this course, in that it requires a rigorous mathematical deﬁnition of “polynomial-time algorithm”. Such a deﬁnition is typically a subject for a course on the theory of computation.

Theorem 2. SAT is NP-complete.

Proof of Theorem 2. Let L be an instance of NP. Then there exists a polynomial-time decidable predicate P (x, y), such that x is a positive instance of L iff the predicate ∃yP (x, y) evaluates to true, where the domain elements of y that are tested in P satisfy |y| ≤ p(|x|), for some polynomial p. Now given this p and an instance x of L, by Lemma 1 there exists a polynomial-time computable function f(x) that maps x to a Boolean formula f(x) = Fx, and that has all three properties described in Lemma 1. Thus f reduces x in polynomial time to an instance Fx of SAT. Moreover, Fx is satisfiable iff there exists an assignment α over the variable of Fx for which eval(Fx, α) evaluates to true. But, by Lemma 1, this will be true iff the predicate ∃yP (x, y) evaluates to true, where the domain elements

14 of y that are tested in P are exactly those that can be encoded by an assignment α. Therefore, x is a positive instance of L iﬀ f(x) = Fx is a positive instance of SAT, and the proof is complete.

We now build on Theorem 2 to show that a host of other problems are NP-complete (to this date there are several thousand known NP-complete problems from several areas of computer science and mathematics). We do this with the help of the following lemma. Its proof is left as an exercise.

Lemma 2. Suppose decision problems L1 and L2 are in NP, and L1 is both NP-complete and polynomial-time reducible to L2. Then L2 is NP-complete.

Theorem 3. CNF-SAT is NP-complete.

Proof of Theorem 3. Let F be a Boolean formula. We map F to an instance CF of CNF-SAT. To do this, let V = {x1, . . . , xn}, and V = {x1,..., xn} denote the literals of F . Recall that each leaf of F is labeled with one of these literals. If F has no internal nodes, then it is already an instance of CNF-SAT. So suppose F has m ≥ 1 internal nodes. Introduce variable set Y = {y1, . . . , ym}, and label each internal node ni with yi, i = 1, . . . , m. Assume n1 is the root of F . Now let lj1 , . . . , lji ∈ V ∪V ∪Y denote the literal labels of the children of ni. Then, if ni is a ∧-node (respectively, ∨-node), then

Fi = lj1 ∧ · · · ∧ lji (respectively, Fi = lj1 ∨ · · · ∨ lji ). It then follows that F is satisﬁable iﬀ the formula

y1 ∧ (y1 ↔ F1) ∧ · · · ∧ (ym ↔ Fm) is satisfiable, and we leave it as an exercise to show that each subformula yi ↔ Fi can be translated to CNF in linear time with respect to the number of children of ni. Finally, letting CF denote the union of all clauses obtained from translating each yi ↔ Fi subformula, it follows that F can be translated to CF in polynomial time, and F is satisfiable iff CF is satisfiable.

15 Example 10. Demonstrate the reduction described in the proof of Theorem 3 for the formula F (x1, x2, x3) = (x1 ∧ x2) ∨ (x1 ∨ x3).

16 Theorem 4. 3SAT is NP-complete.

Proof of Theorem 2. 3SAT is in NP using an argument similar to SAT. we reduce CNF-SAT to 3SAT. To do this, let C be a collection of disjunctive clauses. If c = l1 ∨ l2 ∨ · · · ∨ lm is one such clause, then we can effectively map to f(c), a collection of 3-SAT clauses, by possibly introducing new variables. The effective mapping is defined as follows.

• Case 1. m = 1. Then f(c) = l1 ∨ l1 ∨ l1

• Case 2. m = 2. Then f(c) = l1 ∨ l2 ∨ l2 • Case 3. m = 3. Then f(c) = c

• Case 4. m = 4. Then f(c) = {(l1 ∨ l2 ∨ z), (z ∨ l3 ∨ l4)} • Case 5. m ≥ 5. Then

f(c) = {(l1 ∨ l2 ∨ z1), (z1 ∨ l3 ∨ z2),..., (zm−4 ∨ lm−2 ∨ zm−3), (zm−3 ∨ lm−1 ∨ lm)}.

Notice that f(c) can be eﬀectively computed from c in time polynomial in the size of c.

Now suppose C has a satisfying assignment a. Then a satisfies some literal l of c, for every c ∈ C. In Cases 1-4, f(c) will also be satisfied. In Case 5, the clause containing l will be satisified, and the clauses not containing l can be satisfied by assigning the z’s appropriate values (verify this!).

Similarly, if C is unsatisiable, and a is some assignment that leaves c ∈ C unsatisﬁed, then one can verify, by examining all ﬁve cases, that no assignment of the z-variables can satisfy all the clauses in f(c).

Corollary 1. Clique is NP-complete.

17 Example 12. Given c = (x1 ∨ x2 ∨ x3 ∨ x4 ∨ x5 ∨ x6 ∨ x7), compute f(c) and verify that an assignment a satisifies c iff there is some assignment of the z variables that, along with a, satisfies f(c).

18 Theorem 5. The Vertex Cover decision problem is the problem of deciding if a simple graph G = (V,E) has a vertex cover of size k ≥ 0, for some integer k, i.e. a subset of k vertices C ⊆ V for which every edge e ∈ E is incident with at least one vertex of C. Then Vertex Cover is NP-complete.

Proof of Theorem 5. The fact that Vertex Cover is in NP is left as an exercise. We show a polynomial-time reduction from 3SAT. Let C be a collection of m clauses over n variables, where clause cj, 1 ≤ j ≤ m, has the form cj = lj1 ∨ lj2 ∨ lj3. We proceed to construct a simple graph f(C) = G = (V,E) that has a vertex cover of size 2m + n iff C is satisfiable. G = (V,E) is defined as follows. V consists of the following 3m + 2n vertices.

Clause Vertices 3m vertices labeled as ljk, 1 ≤ j ≤ m and 1 ≤ k ≤ 3

Variable Vertices 2n vertices labeled as xi and xi

There are also three kinds of edges.

Clause Edges The three vertices corresponding to a clause are pairwise connected to form a triangle. In other words, for every 1 ≤ j ≤ m,

(lj1, lj2), (lj2, lj3), (lj1, lj3) ∈ E.

Variable Edges Variable vertices xi and xi are adjacent, i = 1, . . . , n.

Communication Edges A variable vertex l is adjacent to a clause vertex ljk iﬀ ljk and l have the same lable, i.e. represent the same literal. For example, if l13 = x5, then clause vertex l13 is adjacent to variable vertex x5.

Notice that G can be constructed from C in time polynomial in m and n, since |V | = 2n + 3m and |E| = 6m + n.

Now assume, that C is satisfiable. Let a = (a1, . . . , an) be a satisfying assignment for C. Without loss of generality, assume that the first literal lj1 of clause cj is satisfied by a, j = 1, . . . , m. Then vertex cover C ⊆ V has the following 2m + n vertices: i) variable vertex li, where li = xi if ai = 1, and li = xi otherwise, and ii) clause vertices lj2, lj3, j = 1, . . . , m. To see that C is a cover, notice that, by i), C covers all variable edges, and, by ii) C covers each triangle edge of each clause. Furthermore, for each clause cj, by ii), the communication edges that are incident with lj2 and lj3 are covered. Finally, the communication edge incident with lj1 is covered since lj1 is satisfied by a, and hence lj1 = li, for some i ∈ {1, . . . , n}.

Conversely, assume C is a vertex cover of size 2m + n for G. Then by the way G is constructed, C must consist of i) n variable vertices li, i = 1, . . . , n, where li = xi or li = xi, and ii) two clause vertices from each clause (WLOG, assume they are lj2 and lj3, j = 1, . . . , m). Indeed, C must have at least one variable vertex li, i = 1, . . . , n, since otherwise some variable edge would not be covered. Similarly, C must have at least two clause vertices from each clause, since otherwise some triangle

19 edge would not be covered. Moreover, since |C| = 2m + n, we see that it must have exactly one variable vertex from each of the n pairs, and exactly two clause vertices from each of the m clauses. Now, since R = {l1, . . . , ln} is a consistent set of literals, consider the variable assignment aR induced by R. For each j = 1, . . . , m, notice that lj1 6∈ C. Hence, its communication edge must be incident with some l ∈ R. In other words, lj1 = li, for some i ∈ {1, . . . , n}, and thus aR satisifes cj, for all j = 1, . . . , m. Therefore, C is satisﬁable.

Example 13. Show the reduction given in Theorem 5 using

C = {(x1, x2, x3), (x1, x2, x3), (x1, x2, x3), (x1, x2, x3)}.

20 Theorem 6. The Directed Hamilton Path Problem (DHP) is the problem of deciding if directed graph G = (V,E) possesses a directed simple path having length n − 1. Such a path is called a Hamilton path. Then DHP is NP complete.

Proof of Theorem 6. The fact that DHP is in NP is left as an exercise. We show a polynomial-time reduction from 3SAT. Let C be a collection of m ternary clauses over n variables. We proceed to construct a directed simple graph f(C) = G = (V,E) that has a Hamilton path iff C is satisfiable. G is defined as follows. G has m clause vertices c1, . . . , cm and n diamond subgraphs, one corresponding to each variable xi, 1 ≤ i ≤ n. Diamond subgraph Di consists of a top vertex ti, bottom vertex bi, left vertex li, and right vertex ri, along with directed edges

(ti, li), (ti, ri), (li, bi), (ri, bi).

In addition, there is a string of 3m + 1 vertices that connect li with ri:

si1, lci1, rci1, si2, lci2, rci2, . . . , sim, lcim, rcim, si(m+1).

The s vertices are called separators, while the lc and rc pairs correspond with each of the m clauses. This string of vertices is bidirectionally pairwise connected; i.e.

(li, si1), (si1, li), (si1, lci1), (lci1, si1),..., (si(m+1), ri), (ri, si(m+1)) ∈ E.

Finally, if xi is a literal of clause cj, then edges (rcij, cj), (cj, lcij) are added. On the other hand, if xi is a literal clause cj, then edges (lcij, cj), (cj, rcij) are added.

Finally, for 1 ≤ i ≤ n − 1 the edges (bi, ti+1) are added to connect the n diamond graphs to- gether. Note that f(C) = G = (V,E) can be constructed in time that is polynomial in m and n. Moreover, our goal is prove that C is satisﬁable iﬀ f(C) = G = (V,E) has a Hamilton Path starting at t1 and ending at bn.

For each diamond Di, i = 1, . . . , n, consider a path that begins at the top vertex ti and ends at the bottom vertex bi, and visits every vertex in the diamond. One such path is in the form of a left lightning bolt that moves from the top to the left vertex li, cuts across the chain of 3m+1 horizontal vertices from left to right vertex ri, and moves down to bi. The other path is of the form of a right lightning bolt that moves from ti down to ri, cuts across the chain of 3m + 1 horizontal vertices from right to left vertex li, and moves down to bi. Moreover, a left lightning bolt path with excursions (LBPWE) is a lightning bolt path that, when reaching vertex lcj, takes an excursion and next visits vertex cj, then returns to the diamond via rcj, provided i) cj has yet to be visited by a LBPWE within an earlier diamond, and ii) xi ∈ cj. Thus, left LBPWE’s correspond with assigning xi = 0. Finally, a right lightning bolt path with excursions (LBPWE) is a lightning bolt path that, when reaching vertex rcj, next makes an excursion to vertex cj, and returns to the diamond via lcj, provided i) cj has yet to be visited by a LBPWE within an earlier diamond, and ii) xi ∈ cj. Thus, right LBPWE’s correspond with assigning xi = 1.

Now assume C is satisﬁable via truth assignment a = (a1, . . . , an). Then let P = P1 ··· Pn be a concatenation of n paths P1,...,Pn, where Pi is a left (respectively, right) LBPWE provided ai = 0 (respectively, ai = 1), i = 1, . . . , n. Since a lightning bolt path visits every diamond vertex exactly

21 once, it remains to show that every clause vertex c will be visited exactly once. Let i be the least index for which either xi ∈ c and ai = 1, or xi ∈ c and ai = 0. Since a satisﬁes c, at least one of the two must be true. WLOG, assume xi ∈ c and ai = 1. Then Pi is a right LBPWE and will make an excursion visit to c, since i) xi ∈ c and hence (rc, c) and (c, lc) are edges connecting diamond Di to c, and ii) no other previous LBPWE has yet to visit c. Therefore P is directed Hamilton path that begins at t1 and ends at bn.

Conversely, suppose G has a directed Hamilton path P . Then this path must begin at t1 and end at bn.

Claim: when P enters a diamond D, it will not subsequently reach a vertex of any other diamond D0 until ﬁrst visiting every vertex of D. Suppose, by way of contradiction, that after entering D, P reaches a vertex in another diamond D0 before reaching the bottom of D. The only way this is possible is if P makes an excursion visit to some clause c, and, instead of returning back to D, hops to some other diamond, say D0. WLOG, assume that P made the excursion to c via edge (lc, c) (i.e. P is moving across D from left to right). Then how will P be able to return to D and visit vertex rc? Notice that the only vertices adjacent to rc are lc, c, and s, where s is a separator vertex to the immediate right of rc. Moreover, since lc and c have already been visited, the only way to reach rc would be via s. But once rc is visited, there is no other place to go, since all adjacent vertices have been visited, and the path would have to terminate at rc, which contradicts the assumption that P is a Hamilton path from t1 to bn.

From the claim it follows that P must have the form P = P1 ··· Pn, i.e. a concatenation of n paths P1,...,Pn, where each Pi is an LBPWE, i = 1, . . . , n. Now let a = (a1, . . . , an) be the truth assignment where ai = 1 iff Pi is a right LBPWE. It remains to show that a satisfies each c of C. Since P is a Hamilton path, there must be some subpath Pi for which Pi makes an excursion to c. WLOG, assume Pi is a right LBPWE. Then, since (rc, c), and (c, lc) are edges that connect c to Di, it follows that xi ∈ c and, hence, a satisfies c since ai = 1. A similar argument holds for when Pi is a left LBPWE. Therefore, C is satisfiable.

22 Example 14. Show the reduction given in Theorem 6 using

C = {(x1, x2), (x1, x2), (x1, x2)}.

23 Subset Sum Problem. Given a ﬁnite subset S ⊆ of integers, and an integer t, does there exist a subset A ⊆ S such that the elements in A sum to t?

Theorem 7. Subset Sum is NP-complete.

Proof of Theorem 7. It is an exercise to show that Subset Sum is in NP. We now show a polynomial- time reduction to Subset Sum from 3-SAT. Let C be a collection of m ternary clauses over n variables. The following table provides the reduction from C to f(C) = (S, t). The table rows correspond to the numbers comprising set S, while t is the number at the bottom.

1 2 3 4 ··· n c1 c2 ··· cm y1 1 0 0 0 ··· 0 1 0 ··· 0 z1 1 0 0 0 ··· 0 0 0 ··· 0 y2 1 0 0 ··· 0 1 0 ··· 0 z2 1 0 0 ··· 1 0 0 ··· 0 y3 1 0 ··· 0 1 1 ··· 0 z3 1 0 ··· 1 0 0 ··· 1 ...... yn 1 0 0 ··· 0 zn 1 0 0 ··· 0 g1 1 0 ··· 0 h1 1 0 ··· 0 g2 1 ··· 0 h2 1 ··· 0 . . . . gm ··· 1 hm ··· 1 t 1 1 1 1 ··· 1 3 3 ··· 3

Number yi correpsonds to variable xi, while zi corresponds to literal xi. Moreover, the fact that t’s whole part consists of all 1’s satisﬁes the requirement that, for a truth assignment A, either ai is set to 1 or ai is set to 0, but not both. Thus we see that either yi or zi must be added to the subset, but not both. Also, the fractional part of yi (respectively zi) indicates to which clauses xi (respectively xi) belongs.

Now suppose C is satisﬁable via some assignment A. Then the subset A needed to sum to t includes the following numbers. For 1 ≤ i ≤ n, if ai = 1, then add yi to A; otherwise add zi to A. To see which gj’s and hj’s to add to A, if any, consider clause cj. If cj is satisﬁed by A only once, then add both gj and hj to A; otherwise add only gj to A. The reader can check that the elements in A indeed add to t.

Now suppose that C is unsatisfiable. For the elements of a subset A to sum to t, it is necessary that either yi or zi must be added to A, but not both. Thus, assuming the elements of A sum to t, let A be the truth assignment defined by ai = 1 if yi ∈ A, and ai = 0 if zi ∈ A, for 1 ≤ i ≤ n. Then, since C is unsatisfiable, it must be the case that some clause cj is not satisfied by A, in which case, when

24 summing y and z elements of A in the j th column to the right of the decimal, we see that this sum equals 0, since none of the y’s or z’s added to A satisfy cj. Thus, even if we choose both gj and hj to add to A, the j th column to the right of the decimal will add to at most 2, which is one less than the required value of 3. Therefore, the elements in A cannot sum to t.

Example 15. Show the reduction given in Theorem 7 using

C = {(x1, x2, x3), (x1, x2, x3), (x1, x2, x3), (x1, x2, x3)}.

25 Exercises

1. Redeﬁne the Minimum Spanning Tree optimization problem as a decision problem.

2. Let A be an algorithm for solving the Minimum Spanning Tree decision problem of the previous exercise. Given a positive instance of this decision problem, show how A can used to ﬁnd an actual minimum spanning tree with the desired cost for the given instance.

3. Redeﬁne the 0-1 Knapsack optimization problem as a decision problem.

4. Let A be an algorithm for solving the 0-1 Knapsack decision problem of the previous exercise. Given a positive instance of this decision problem, show how A can used to ﬁnd an actual subset of goods with the desired proﬁt, and the desired total weight that does not exceed the knapsack capacity.

5. Suppose A is an algorithm that decides if an arbitrary Boolean formula F (x1, . . . , xn) is sat- isﬁable. Explain how you can use A to ﬁnd a satisfying assignment a for F , assuming one exists.

6. Suppose A is an algorithm that decides the Vertex Cover decision problem. For a positive instance of A, describe a procedure that uses A to determine an actual vertex cover of the desired size for the positive instance.

7. Prove that the 0-1 Knapsack decision problem is in NP. For this exercise (and all subsequent ones involving showing membership in a complexity class) i) define all variables and, for each existential (respectively, universal) variable, define its domain and the polynomial p that bounds the size of each domain element being tested, and ii) define the polynomial-time decidable predicate function P and justify that it can be computed in polynomial time.

8. A Hamilton Cycle in a simple graph is a cycle that visits every vertex in the graph at most once. Deﬁne the associated decision problem and show that this problem is in NP.

9. Two simple (undirected) graphs G = (V1,E1) and H = (V2,E2) are called isomorphic iﬀ there exists a one-to-one correspondence f : V1 → V2 between V1 and V2 such that (u, v) ∈ E1 iﬀ (f(u), f(v)) ∈ E2, for every pair of vertices u, v ∈ V1. Are the following graphs isomorphic?

G = ({1, 2, 3, 4}, {(1, 2), (2, 3), (3, 4), (1, 4)}),

and H = ({a, b, c, d}, {(a, c), (a, d), (b, c), (b, d)}). If yes, provide the one-to-one correspondence. If no, explain why not.

10. Repeat the previous problem for graphs

G = ({1, 2, 3, 4, 5}, {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 5)}),

and H = ({a, b, c, d, e}, {(a, d), (a, e), (e, d), (b, c), (a, b), (b, d), (c, e)}).

11. Let ISO be the problem of deciding if two simple graphs G and H are isomorphic. Prove that ISO ∈ NP.

26 12. Subset Sum is the decision problem that takes as input a set S of nonnegative integers, a nonnegative integer t, and decides if there is a subset A ⊂ S such that X a = t. a∈A Show that the following is a positive instance of SUBSET SUM:

S = {2, 5, 10, 13, 19, 21, 25, 29}

, t = 43. Show that Subset Sum ∈ NP. 13. Let G = (V,E) be a simple graph. Subset I ⊂ V is called an independent set of G iff, for every u, v ∈ I,(u, v) 6∈ E. In other words, I is an independent set iff every pair of vertices in I are non-adjacent. The Independent Set problem is stated as follows. Given G = (V,E) and non-negative integer k, does G have an independent set I of size k? Show that this problem is in NP. 14. Let Triangle be the problem of deciding if a simple graph G has a 3-clique. Prove that Triangle is in P. 15. Define the complement of the Triangle problem. Give an example of a positive instance of this problem.

16. A permutation on the set {1, . . . , k} is a one-to-one and onto function on this set. When p is a permutation, pt means the composition of p with itself t times. Let Perm Power is the problem of, given input (q, p, t), deciding if q = pt, where p and q are permuations (over {1, . . . , k}), and t is a nonnegative integer. Prove that Perm Power is in P, Hint: the obvious algorithm does not run in polynomial time, since the problem size is logarithmic (and not linear) with respect to the value of t.

17. Let Unary Primes be the set {1n|n is a prime number}. Prove that Unary Primes is in P.

18. Let Unary Subset Sum be the subset sum problem in which all numbers are written in unary. Show that Unary Subset Sum is in P. Explain why the NP-completeness reduction for Subset Sum does not work for Unary Subset Sum.

19. Provide a deﬁnition for the complement of the Subset Sum problem.

20. Set Partition is the decision problem that takes as input a set S of nonnegative integers and decides if there is a subset A ⊂ S such that X X a = ( s)/2. a∈A s∈S In other words, can the set be partitioned into two subsets, each of whose elements sum to the same value? Is the following a positive instance of SUBSET SUM:

S = {2, 5, 10, 13, 19, 21, 25, 29}

? If yes, provide the partition. If not, give a sound argument as to why not. Show that Set Partition is polynomial-time reducible to Subset Sum. Show that Subset Sum is polynomial- time reducible to Set Partition.

27 21. Prove that Primes is in co-NP.

22. Prove that polynomial-time reducibility is both a reﬂexive and transitive relation on the set of all decision problems.

23. An instance of LPath is a simple graph G = (V,E), two vertices a, b ∈ V , and a nonnegative integer k ≥ 0. The problem is to decide if there is a simple path of length equal to k from a to b. An instance of Hamilton Path is a simple graph G = (V,E) along with two vertices a, b ∈ V . The problem is to decide if there is a Hamilton Path path from a to b, i.e. a simple path from a to b that visits every vertex. Provide a polynomial-time reduction from Hamilton Path to LPath, and defend your answer.

24. Let Double-SAT be the problem of deciding if a Boolean formula has at least two satisfying assignments. Provide a polynomial-time reduction from SAT to Double-SAT.

25. Prove that if P = NP, then every problem in P is NP-complete.

26. Prove that if some problem L ∈ P is NP-complete, then P = NP. Hint: when proving that set A equals set B it is necessary and suﬃcient to prove that A ⊂ B and B ⊂ A.

27. Let Half Clique be the problem of deciding if a simple graph G = (V,E) has a clique of size bn/2c, where n = |V |. Provide a polynomial-time reduction from Clique to Half Clique. Defend your answer.

28. Let C be a 3-CNF formula. A 6=-assignment to C is a truth assignment that satisﬁes C, but in such a way that every clause of C has at least one literal set to true, but also has one literal set to false. Prove that, if a is a 6=-assignment, then so is its negation a, where the negation of an assignment is the assignment that is obtained by negating each assignment value of a. Let 6=SAT be the problem of deciding if a 3-CNF Formula has a 6=-assignment. Prove that 3-SAT is polynomial-time reducible to 6=SAT, by mapping each clause i of the form (l1 ∨ l2 ∨ l3) to the two clauses (l1 ∨ l2 ∨ zi) and (zi ∨ l3 ∨ b),

where zi is a newly introduced variable (one for each clause of C), and b is a single additional new variable.

29. Given a collection of m-bit binary strings s1, s2, . . . , sn, and a nonnegative integer K, the String

Sum decision problem is to decide if there are K strings si1 , si2 , . . . , siK from the collection such that

si1 + si2 + ··· + siK = 1, where the addition is bitwise Boolean-algebra addition (e.g. 1001 + 1010 = 1011), and 1 is the m-bit binary string consisting of all ones. Prove that String Sum is an NP-complete decision problem. Hint: reduce from Vertex Cover.

30. An instance of Set Splitting is a ﬁnite set S and a collection of subsets C = {C1,...,Cm} of S. The problem is to decide whether or not S can be partitioned into two sets A and B such that

a. S = A ∪ B b. A ∩ B = ∅

28 c. Ci ∩ A 6= ∅, for all i = 1, 2, . . . , m

d. Ci ∩ B 6= ∅, for all i = 1, 2, . . . , m. Prove that Set Splitting is NP-complete. Hint: reduce from 6=-SAT.

29 Exercise Hints and Answers.

1. Inputs: weighted graph G = (V, E, w) and integer k. Does G have an mst that has a cost not exceeding k?

2. Let A be an algorithm that takes as input weighted graph G = (V, E, w) and integer k, and returns true iﬀ G has an mst with cost not exceeding k. Suppose A returns true on some input (G, k). We now use A to determine an mst of G whose cost does not exceed k. For each edge e ∈ E of G, remove e from E: E ← E − e. Run A on input (G, k). If A returns true, then e is not needed to form an mst of cost not exceeding k. Thus, permanently remove e from E. Otherwise, add e back to E: E ← E + e. Proceed to the next edge. After iterating over each edge, E will consist entirely of those edges that form an mst of G with cost not exceeding k.

3. Inputs: i) Goods G = {g1, . . . , gn}, where good gi has weight wi and proﬁt pi, i = 1, . . . , n; ii) capacity M; iii) integer P . Is there a subset of goods whose weights sum to a value not exceeding M, and whose proﬁts sum to a value that equals or exceeds P ?

4. Let A be an algorithm that takes as input i) Goods G =}g1, . . . , gn}, where good gi has weight wi and profit pi, i = 1, . . . , n; ii) capacity M; iii) integer P . Suppose A returns true on some input (G, M, P ). We now use A to determine a subset of G whose weights sum to a value not exceeding M, and whose profits sum to a value that equals or exceeds P . For each good g ∈ G of G, remove g from G: G ← G − g. Run A on input (G, M, P ). If A returns true, then g is not needed to add to the knapsack in order to obtain a profit of at least P . Thus, permanently remove g from G. Otherwise, add g back to G: G ← G + g. Proceed to the next good. After iterating over each good, G will consist entirely of a set of goods whose weights sum to a value not exceeding M, and whose profits sum to a value that equals or exceeds P .

5. Define F0 = F . Suppose Fi has been defined. Let Fi+1 be the result of substiuting xi+1 with the value true in formula Fi. If Fi+1 is satisiable, then the satisfying assignment will assign xi+1 to true. Otherwise, define Fi+1 be the result of substiuting xi+1 with the value false in formula Fi. Then Fi+1 must be satisfiable (since Fi is satisfiable, and hence some satisfying assignment must assign xi+1 either true or false). Repeating this for variables xi, i = 1, . . . , n, will lead to a satisfying assignment for F .

6. Initialize the desired vertex cover to C = ∅. Initialize V0: V0 ← V and E0 ← E. Suppose Vi and Ei have been deﬁned. To deﬁne Vi+1, remove vi+1 from Vi and all edges incident with vi+1 from Ei. Run the algorithm to check if Gi = (Vi,Ei) has a cover of size K − 1. If true, then let Vi+1 = Vi (with vi+1 removed), Ei+1 = Ei (with vi+1’s edges removed), update C to C ← C + vi+1, and update K: K ← K − 1. Otherwise, Vi+1 = Vi (without vi+1 removed), and Ei+1 = Ei (without vi+1’s edges removed), and C and K remain the same. Then the desired vertex cover is C which consists of the K removed vertices.

7. Let G = {g1, . . . , gn} denote the set of goods, W = {w1, . . . , wn} the weights, P = {p1, . . . , pn} the profits, M the capacity, and k ≥ 0 the profit threshold. Let s be a variable whose domain consists of all subsets of G. Then |s| = O(n) is polynomial in the input size. Fi- nally, P ((G, W, P, M, k), s) is the predicate function that evaluates to true iff s is a subset of goods, the sum of whose weight does not exceed M, and the sum of whose profit is at least k. P ((G, W, P, M, k), s) is polynomial-time decidable since the summing of the weights and

30 profits of elements of s can be performed in time linear in n, log M, and log k (we assume that all weights have O(log M) bits, and all profits have O(log k) bits). Finally, (G, W, P, M, k) is a positive instance iff ∃sP ((G, W, P, M, k), s) evaluates to true. Therefore, 0-1 Knapsack is in NP.

8. Let G = (V,E) be an instance of Hamilton Cycle, with V = {v1, . . . , vn}. Let variable p have domain equal to the set of permutations of the numbers 1, . . . , n. Then |p| = O(n) is polynomial in the size of G. Finally, let P (G, p) denote the predicate function that evaluates to true iﬀ (vp(1), vp(2)),..., (vp(n−1), vp(n)), (vp(n), vp(1)) ∈ E. Then P (G, p) is polynomial-time decidable since checking if each of the above pairs is in E can be done in time linear in both n and m = |E|. Finally, G is a positive instance iﬀ ∃pP (G, p) evaluates to true. Therefore, Hamilton Cycle is in NP.

9. f(1) = a, f(2) = c, f(3) = b, f(4) = d.

10. f(1) = b, f(2) = e, f(3) = d, f(4) = c, f(5) = a.

11. Let (G1 = (V1,E1),G2 = (V2,E2)) be an instance of Graph Isomorphism, where both graphs have n vertices and m edges. WLOG, assume V1 = V2 = {1, . . . , n}. Let variable f have domain equal to the set of one-to-one correspondences from {1, . . . , n} to itself. Then |f| = O(n) is polynomial in the size of the input. Finally, let P ((G1,G2), f) denote the predicate function that evaluates to true iff, for all i = 1, . . . , n,(i, j) ∈ E1 iff (f(i), f(j)) ∈ E2. Then P ((G1,G2), f) is polynomial-time decidable since checking the above can be performed in time linear in both n and m by constructing hash tables for both E1 and E2. Finally, (G1,G2) is a positive instance iff ∃fP ((G1,G2), f) evaluates to true. Therefore, Graph Isomorphism is in NP. 12. Let (S, t) be an instance of Subset Sum, where |S| = n. Let s be a variable whose domain consists of all subsets of S. Then |s| = O(n) is polynomial in the input size. Finally, let P ((S, t), s) denote the predicate function that evaluates to true iff the elements of s sum to t. Then P ((S, t), s) is polynomial-time decidable since summing n numbers of S can be done in time linear in n and log t (assuming each element of S has O(log t) bits. Finally, (S, t) is a positive instance iff ∃sP ((S, t), s) evaluates to true. Therefore, Subset Sum is in NP.

13. Almost identical to the solution for Example 6.

14. For each triple of vertices (x, y, z), check if (x, y), (x, z), (y, z) ∈ E. If yes, return true. Running n time: O( ) = O(n3) 3

15. Given a simple graph G = (V,E), is it the case that G does not have a triangle? Positive instance: G = ({v1}, ∅). 16. Compute p2j , j = 1,..., blog tc, and use these powers to compute pt. Thus, pt can be computed using a O(k log t) steps (since each product requires O(k) steps) and compared with q in k additional steps.

17. √Problem input size is now n. Return true√ iﬀ it’s not the case that some integer between 2 and n divides n. This can be checked in O( n) steps.

31 18. It would take exponential time to compute the elements of set S in unary. Thus the reduction is no longer polynomial time.

19. Given a set S of integers and integer t, is it the case that there is no subset of S whose elements add to t?

p P 20.SP ≤m SS. f(S) = (S,M/2), where M = s. Clearly S is a positive instance of SP iﬀ f(S) is s∈S a positive instance of Subset Sum. Moreover, f(S) is polynomial-time computable in the size of S. p ˆ ˆ SS ≤m SP. f((S, t) = S, where S = S ∪ {J}, and J = M − 2t. Clearly, f is polynomial-time computable in the size of (S, t). Moreover, if A ⊆ S with P (a) = t, then Sˆ is a positive a∈A instance of SP, since, given Aˆ = A ∪ {J} ⊂ Sˆ, we have P (ˆa) = t + J. But 2(t + J) = M + J, aˆ∈Aˆ is the sum of all elements in Sˆ. This is true, since

2(t + J) = 2t + 2J = M + J ⇔ J = M − 2t.

Thus, if (S, t) is a positive instance, so is f((S, t)). Conversely, if Aˆ ⊂ Sˆ for which P (ˆa) = t + J, then WLOG we may assume J ∈ Sˆ (why?). aˆ∈Aˆ Then, letting A = Aˆ − {J}, we have A ⊆ S and P (a) = t + J − J = t. Thus, if f((S, t)) is a a∈A positive instance, then so is (S, t).

21. Let n be√ an instance of Primes. Let d be a variable whose domain consists of all elements {2,..., n}. Then |d| = O(log n) is polynomial in the size of n. Finally, let P (n, d) denote the predicate function that evaluates to true iﬀ d does not divide evenly into n. Then P (n, d) is polynomial-time decidable since division of two numbers having log n bits can be performed in O(log n) steps. Finally, n is a positive instance iﬀ ∀dsP n, d) evaluates to true. Therefore, Primes is in co-NP.

p p 22. Reflexive: f(x) = x. Transitive. Assume L1 ≤m L2 via f, and L2 ≤m L3 via g. Let h : L1 → L3 be defined by h(x) = g(f(x)). Then x is positive for L1 iff f(x) is positive for L2 iff h(x) = g(f(x)) is positive for L3. Moreover, h(x) is computable in time polynomial in |x|, since computing f(x) can be done in time O(p(|x|) for some polynomial p. Moreover, this means that |f(x)| ≤ p(|x|). Hence, since g(y) can be computed in time O(q(|y|), and |y| = |f(x)| ≤ p(|x|), we see that g(y) can be computed in time O(q(p(|x|)), which is bounded by a polynomial of degree k + l, where k is the degree of p, and l is the degree of q.

23. Given graph G = (V,E) and vertices a, b ∈ V , suppose |V | = n. Then a Hamilton path from a to b is a simple path of length n − 1. Thus, by deﬁnition, (G, a, b) is a positive instance of HP iﬀ f(G, a, b) = (G, a, b, n − 1) is a positive instance of LPath. Moreover, f is computable in polynomial time since we only need to add the fourth component of n − 1.

24. f(F ) = Fˆ, where Fˆ = F ∧ (z ∨ z), where z is a variable that does not appear in F . Then F is satisfiable iff Fˆ has two satisfying assignments (one that sets z = 1, the other that sets z = 0). In other words, F is a positive instance of SAT iff f(F ) is a positive instance for Double-SAT. Note also that f(F ) can be computed in polynomial time, since we simply add two nodes to F ’s tree.

32 25. Use Exercise 17.

26. Given Lˆ ∈ NP, let x be an instance of Lˆ. Compute f(x), where f is the reduction from Lˆ to L. This takes polynomial time. Moreover, deciding if f(x) is positive can also be done in polynomial time, since L ∈ P. Hence, we can decide if x is positive for Lˆ in polynomial time. Therefore, Lˆ ∈ P. 27. Given Clique instance (G = (V,E), k), there are three cases for f((G, k)). Case 1: k = |V |/2. In this case f((G, k) = G, since G has a k-clique iff it has a half-clique. Case 2: k < |V |/2. Then f((G, k)) = Gˆ, where Gˆ is the result of adding |V | − 2k more vertices to G and connecting each of them all other vertices in Gˆ. Then, in an analysis similar to the reduction of Subset Sum to Set Partition (See Exercise 20), G has a k-clique iff Gˆ has a half clique, since k + |V | − 2k = 1/2(|V | + |V | − 2k) = |V | − k, means that Gˆ will have a Half Clique iff G has a k-clique. Case 3: k > |V |/2. Then f((G, k)) = Gˆ, where Gˆ is the result of adding 2k − |V | additional isolated vertices to G. Then G has a k-clique iff Gˆ has a half-clique, since k = 1/2(|V | + (2k − |V |). Finally, notice that f is computable in polynomial time with respect to m = |E| and n = |V |, since at most n vertices and m2 edges must be added to form Gˆ.

28. Let C be a satisiﬁable instance of 3SAT having variables x1, . . . , xn. Let a be a satisfying assignment for C. We can get a satisfying 6= assignment for f(C) over the variables x1, . . . , xn, z1, . . . , zm, b, by extending a toa ˆ in the following way. First, assign b = 0. Next, consider clause ci = (l1, l2, l3) of C. Case 1: a is not consistent with both l1 and l2. Then a is consistent with l3. Then assign zi = 1. In this case,a ˆ is a 6= assignment for

(l1 ∨ l2 ∨ zi) and (zi ∨ l3 ∨ b).

Case 2: a is consistent with either l1 or l2. In this case assign zi = 0. Then again,a ˆ is a 6= assignment for (l1 ∨ l2 ∨ zi) and (zi ∨ l3 ∨ b). Thus, assigning b = 0 and the z’s in the above manner show that a positive instance of 3SAT maps to a positive instance of 6= SAT. Now assume that C is unsatisﬁable. Suppose by way of contradiction that f(C) has a 6= assignmenta ˆ. Let a be the restriction ofa ˆ to variables x1, . . . , xn of C. Then, since a does not satisfy C (remember that C is unsatisﬁable), there must be a clause ci for which a is not consistent with ci = (l1, l2, l3). Buta ˆ is a 6= assignment for

(l1 ∨ l2 ∨ zi) and (zi ∨ l3 ∨ b).

The only way this can happen is ifa ˆ assigns zi = 1 and b = 1. Now let ˆb be the complement ofa ˆ. We know from the ﬁrst part of this exercise that ˆb is also ˆ a 6= assignment for f(C). Once again, let b be the restriction of b to variables x1, . . . , xn of C. Then, since b does not satisfy C, there must be a clause cj for which b is not consistent with ‘ ‘ ‘ ˆ cj = (l1, l2, l3). But b is a 6= assignment for ‘ ‘ ‘ (l1 ∨ l2 ∨ zj) and (zj ∨ l3 ∨ b).

33 ˆ ˆ Again, the only way this can happen is if b assigns zj = 1 and b = 1. But b must assign b = 0, since it is the complement ofa ˆ, anda ˆ assigns b = 1. Thus, f(C) must not have a 6= assignment, proving that a negative instance of 3SAT maps to a negative instance of 6= SAT. Finally, notice that the reduction is clearly polynomial time, since f(C) has n + m + 1 variables and 2m clauses; both quantities being linear in m + n.

29. Given graph G = ({v1, . . . , vn}, {e1, . . . , em}), and integer k, define f((G, k)) = (S, k), where S = {s1, . . . , sn} and si is a binary string of length m, where sij = 1 iff edge ej is incident with vi. Then with a few moments thought, we can see that G will have a vertex cover of size k iff S has k strings that sum to 1m, since the latter simply encodes the fact that each edge is covered by a k-subset of the vertices.

30. Given C, an instance of 6= SAT, let x1, . . . , xn be its variables, and c1, . . . , cm its constraints. Then f(C) consists of set S = {x1, x1, . . . , xn, xn}, while C consists of sets c1, . . . , cm, along with {x1, x1},..., {xn, xn}. Then if a is a 6= assignment for C, let A be the subset of literals that are consistent with a. For example, if a = (x1 = 1, x2 = 0, x3 = 1), then A = {x1, x2, x3}. Furthermore, let B = A be the complement of all the literals of A. For example, if A = {x1, x2, x3, then B = {x1, x2, x3}. Clearly, A ∪ B = S, and both A and B intersect each of the sets of C. Indeed, since a is a 6= assignment, each clause c will contain an element of A and an element of B; while each set {xi, xi} will have either xi ∈ A and xi ∈ B, or xi ∈ B and xi ∈ A. Conversely, if there exist A and B for which S = A∪B, and A and B intersect all of the clauses and literal sets {xi, xi}, then A and B must both be consistent sets of literals. Let a be the assignment over {x1, . . . , xn} induced by A, and b the assignment induced by B. Then, b is the complement of a. Moreover, since both a and b satisfy each clause (since A and B intersect each clause), we see that a is a 6= assignment for C. Finally, notice that the reduction can be performed in polynomial time, since the number of sets C is m + n, which is linear in m + n.