Farley Lai, 00764474
Theory of Computation, Homework 5 7.6 i.) union Without loss of generality, let M 1 and M 2 be the TMs that decide two languages L1 and L2 in P. Then we construct a TM M that decides the union of L1 and L2 in polynomial time. M = ”on input w 1. Run M 1 on w . Accept if it accepts. 2. Run M 2 on w . Accept if it accepts. 3. Otherwise, reject.” Since stage1 and stage2 both run in polynomial time, TM M can decide the union of L1 and L2 in polynomial time. Hence, P is closed under union. ii.) concatenation Without loss of generality, let M 1 and M 2 be the TMs that decide two languages L1 and L2 in P. Then we construct a TM M that decides the concatenation of L1 and L2 in polynomial time. M = ”on input w 1. For each possible split of w = w1 w2 a. Run M 1 on w1 and run M 2 on w2 . Accept if both accept. 2. Otherwise, reject.” Assume the jwj = n . There are O(n) possible splits of w at most. Besides, the sum of the time complexities of M 1 and M 2 is also polynomial. Thus, step1 run in polynomial time and TM M can decide the concatenation of L1 and L2 in polynomial time. Hence, P is closed under concatenation. iii.) complement Without loss of generality, let M be the TM that decides a language L in P. Then we construct a TM M 0 that decides the complement of L in polynomial time. M 0 = ”on input w 1. Run M on w 2. Accept if it rejects. Otherwise, reject if it accepts.” Since step1 run in polynomial time, TM M 0 can decide the complement of L in polynomial time. Hence, P is closed under complement.
7.7 i.) union Without loss of generality, let N 1 and N 2 be the NTMs that decide two languages L1 and L2 in NP. Then we construct an NTM N that decides the union of L1 and L2 in polynomial time. N = ”on input w 1. Run N 1 on w nondeterministically. Accept if it accepts. 2. Run N 2 on w nondeterministically. Accept if it accepts. 3. Otherwise, reject.” Since stage1 and stage2 both run in polynomial time nondeterministically, NTM N can decide the union of L1 and L2 in polynomial time. Hence, N P is closed under union. ii.) concatenation Without loss of generality, let N 1 and N 2 be the NTMs that decide two languages L1 and L2 in NP. Then we construct an NTM N that decides the concatenation of L1 and L2 in polynomial time. N = ”on input w 1. For each possible split of w = w1 w2 a. Run N 1 on w1 nondeterministically and run N 2 on w2 nondeterministically. Accept if both accept. 2. Otherwise, reject.” Assume the jwj = n . There are O(n) possible splits of w at most. Besides, the sum of the time complexities of N 1 and N 2 is also polynomial. Thus, step1 run in polynomial time and NTM N can decide the concatenation of L1 and L2 in polynomial time. Hence, N P is closed under concatenation. 7.12 Since ab is exponential in the length of b , we need to break down ab and apply the modulo operation several times so that the resulting size never goes exponential. First, we treat b in its binary representation. There are at most log2 b bits. So log log Q2b b 2 i Q2b bi2 i ab mod p = ( a i ) mod p = ( (a mod p)) mod p i
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In the worst case, we assume bi = 1 for all i . However,
abi+12 mod p = (abi2 )2 mod p = (abi2 mod p)2 mod p: That is, each term can be computed by squaring the remainder of the previous term mod p . Hence, for each modulo operation, the computation is O(p 2 ) which is in polynomial time. Additionally, for each multiplication of neighboring terms, we also apply the modulo operation in O(p 2 ): Since there are at most log2 b terms, ab Ñ c (mod p) is O(2log2 b p 2 ): Therefore, M ODEXP 2 P : i+1
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7.16 1.) In the proof of theorem 7.56, the SUBSETSUM problem instance constructed contains numbers of large magnitude presented in decimal notation. However, compared with binary/decimal encoding, unary encoding will grow the input size exponentially. Therefore, the reduction to generate the corresponding input for UNARYSSUM will require a number of steps that is exponential in the size of the input, taking more than polynomial time. The proof fails. 2.) SUBSET À SUM is in P because the following polynomial time algorithm based on dynamic programming solve the problem. Input: (S; t) , where S = f x1 ; x2 ; :::; xkg Variables: Integers i; j; l[0::k][0::t] , all are initialized to 0 l[0][0]← 1 ; f or i = 1 to k do f or j = 0 to t À xi do if l[i À 1 ][j] = 1 then l[i][j + xi] ← 1; od od if l[k][t] = 1 then accept else reject; 7.17 Assume that P = N P : Let B 2 P but B = = ¶ and B = = ÆÃ such that there exist a string waccept 2 B and a string wreject 2 = B . To show B in P = N P is NPcomplete, let A be an arbitrary language in P = N P . Therefore, A can be decided by a polynomial time decider M A . Next, we show a polynomial time reduction M from A to B as follows. M = ”On input w : 1. Run M A on w . 2. If M A accepts then output waccept . If M A rejected then output wreject .” Obviously, there exists a polynomial time reduction from A to B . Hence B is NPcomplete.
7.20 a.) SP AT H is in P because the following TM M can decide it in polynomial time. M = “On input ⟨G; a; b; k⟩ where G is a graph with n nodes, two which are a and b: 1. Mark node a 0 2. For each i from 0 to n : a. If an edge (s; t) is found connecting s marked i to an unmarked node i , mark node t with i + 1 . 3. If b is marked with a value of at most b , accept. Otherwise, reject.” b.) LP AT H 2 N P because a path from a to b can be guessed by an NTM nondeterministically and checked if its length is at least k in polynomial time. Next, we show U HAMP AT H Ôp LP AT H by constructing a TM F that computes the reduction f . F = “On input ⟨G; a; b⟩ of U HAMP AT H , where G is a graph with two nodes a and b: 1. Let k be the number of nodes of G. 2. Output ⟨G; a; b; k⟩ as an instance of LP AT H ” If ⟨G; a; b⟩ 2 U HAMP AT H , then G has a Hamiltonian path of length k from a to b . That is, ⟨G; a; b; k⟩ 2 LP AT H . On the other hand, if ⟨G; a; b; k⟩ 2 LP AT H , then G must contain a simple path of length k from a to b . However, since G has only k nodes, the path must be Hamiltonian. There, ⟨G; a; b⟩ 2 U HAMP AT H: Hence, U HAMP AT H Ôp LP AT H: According to the above, LP AT H is NPComplete.
7.21 i.) DOUBLE À SAT 2 N P On input of a Boolean formula, an NTM could guess two different Boolean assignments nondeterministically and accept if both of the assignments satisfy the formula. Hence, DOUBLE À SAT is in N P : ii.) DOUBLE À SAT is NPhard SAT can be reduced to DOUBLE À SAT by constructing a TM F that computes the reduction f as follows. F = ”On input of a Boolean formula (¶) with variables x1 ; x2 ; :::; xk 1. Let ¶0 = ¶ ^ (x _ x) by introducing a new variable x 2. Output (¶0) with variables x1 ; x2 ; :::; xk; x ” If ¶ 2 SAT , its satisfying assignment plus x = true or x = true can serve as the two satisfying assignments of ¶0: On the other hand, if ¶0 2 DOUBLE À SAT is satisfiable, ¶ can be satisfied by the assignment where the new variables x and x are removed. Therefore, DOUBLE À SAT is NPhard. According to i. and ii., DOUBLE À SAT is NPComplete. 7.24 a.) In a clause of an = = assignment, there is one literal set true or two literals set true. Thus, the negation of a the clause will still induce a clause with one literal set true or two literals set true. Since a negated clause is also a clause of an = = assignment, the negation of an = = assignment is too an = = assignment. b.) Let ¶ be a 3cnfformula of an instance of 3 SAT and ¶0 be a 3cnfformula of an instance of = = SAT : The reduction of replacing a clause ci (y1 _ y2 _ y3 ) of ¶ with two clauses (y1 _ y2 _ zi) ^ (zi_ y3 _ b) of ¶0 can be computed in polynomial time for sure. Next, we show ¶ is satisfiable iff ¶0 has the corresponding = = assignment. If ¶ is satisfiable, either y1 = ture and y2 = ture , or only y3 = true . In the former case, set zi false. In the latter case, set zi true. In any case, b is set false. On the other hand, if ¶0 has an = = assignment, y1 ; y2 and y3 cannot all be set false since this may cause one of the clauses in the = = assignment to be false. Thus, the corresponding clause (y1 _ y2 _ y3 ) must be true. A satisfying assignment of ¶ is then derived. c.) Obviously, = = SAT is in NP because an NTM can guess an = = assignment and check if it satisfies the formula nondeterministically in polynomial time. According to the reduction in b, = = SAT is concluded to be NPComplete.
7.28 i.) SET À SP LIT T ING 2 N P On input of (S; C) , where S is a finite set and C = fC 1 ; C 2 ; :::; C kg is a collection of subsets of S , for some k > 0 , an NTM could guess a coloring of the elements nondeterministically and accept if no C i has all its elements colored with the same color. Hence, SET À SP LIT T ING is in N P : ii.) SET À SP LIT T ING is NPhard 3 SAT can be reduced to SET À SP LIT T ING by constructing a TM F that computes the reduction f as follows. F = ”On input of a 3CNF Boolean formula (¶) with l clauses and variables x1 ; x1 ; x2 ; x2 ; :::; xk; xk 1. Set S = f x1 ; x1 ; x2 ; x2 ; :::; xk; xk; zg 2. For each clause ci in ¶ , add a subset C i of S including z and the literals in ci as the elements. 3. add a subset C xi for each pair of xi and xi ” Clearly, C = f C 1; C 2; :::; C l; C x1; C x2; :::; C xk; g: If ¶ 2 SAT , its satisfying assignment can serve as a coloring of (S; C) by coloring all the true literals red and all the false literals as well as z blue such that each subset C i has at least one blue element z and one red literal which has a true assignment. For subset C xi , the pair of literals xi and xi must be colored exactly red for one and blue for the other. Obviously, (S; C) 2 SET À SP LIT T ING: On the other hand, if (S; C) 2 SET À SP LIT T ING has a coloring, ¶ can be satisfied by the assignment where red literals are set true while blue literals are set false. Therefore, SET À SP LIT T ING is NPhard. According to i. and ii., SET À SP LIT T ING is NPComplete.
