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CSE 6321 - Solutions to Problem Set 5 CSE 6321 - Solutions to Problem Set 5 1. 2. Solution: n 1 T T 1 T T D-Q = fhA; P0;P1;:::;Pm; b; q0; q1; : : : ; qm; r1; : : : ; rm; ti j (9x 2 R ) 2 x P0x + q0 x ≤ t; 2 x Pix + qi x + ri ≤ 0; Ax = bg. We can show that the decision version of binary integer programming reduces to D-Q in polynomial time as follows: The constraint xi 2 f0; 1g of binary integer programming is equivalent to the quadratic constrain (possible in D-Q) xi(xi − 1) = 0. The rest of the constraints in binary integer programming are linear and can be represented directly in D-Q. More specifically, let hA; b; c; k0i be an instance of D-0-1-IP = fhA; b; c; k0i j (9x 2 f0; 1gn)Ax ≤ b; cT x ≥ 0 kg. We map it to the following instance of D-Q: P0 = 0, q0 = −b, t = −k , k = 0 (no linear equality 1 T T constraint Ax = b), and then use constraints of the form 2 x Pix + qi x + ri ≤ 0 to represent the constraints xi(xi − 1) ≥ 0 and xi(xi − 1) ≤ 0 for i = 1; : : : ; n and more constraints of the same form to represent the 0 constraint Ax ≤ b (using Pi = 0). Then hA; b; c; k i 2 D-0-1-IP iff the constructed instance of D-Q is in D-Q. The mapping is clearly polynomial time computable. (In an earlier statement of ps6, problem 1, I forgot to say that r1; : : : ; rm 2 Q are also part of the input.) 3. Show that if every NP-hard language is also PSPACE-hard, then PSPACE=NP. Solution: If every NP-hard language is PSPACE-hard, then SAT is PSPACE-hard, and consequently every PSPACE language is polynomial-time reducible to SAT. Then PSPACE ⊆ NP because SAT 2 NP and therefore PSPACE=NP because we know NP ⊆ PSPACE. 4. Show that TQBF restricted to formulas where the part following the quantifiers is in conjunctive normal form is still PSPACE-complete. Solution: Let TQCNF be the language of the restricted version. TQCNF is clearly in PSPACE. To show that it is PSPACE-complete we exhibit a polynomial time reduction from TQBF to TQCNF. Let be a t.q.b.f. Applying a straightforward polynomial time transformation, we can assume all the quantifiers are at the beginning and their scope is everything that follows (prenex normal form). Let Q be the string of quantifiers in , let φ be the unquantified part of the formula so that = Qφ. The obvious approach of putting φ in c.n.f. by using rules such as distributivity and de Morgan's laws does not work as it generates a formula of exponential size in the worst case. More specifically, a formula A _ (x ^ y) with a subformula A becomes (A _ x) ^ (A _ y), where A appears twice. We can avoid this by using auxiliary variables, equivalence and quantifiers, replacing A _ (x ^ y) with the equivalent formula (9z)(z ≡ A) ^ (z _ x) ^ (z _ y), where A appears only once. Here A ≡ B is a shortcut for logical equivalence, namely the formula (A¯ _ B) ^ (B¯ _ A). To complete the argument, we need to find a CNF formula that is equivalent to x ≡ A. More specifically, given a boolean formula A without quantifiers, construct a (possibly 1 quatified) CNF formula CA;x with an extra free variable x that is equivalent to x ≡ A. We see how to do this recursively: • If A is a variable y, then CA;x is (¯x _ y) ^ (¯y _ x). (expanded CNF of x ≡ y) • If A is B¯, then CA;x is (9z)CB;z ^ (z _ y) ^ (¯y _ z¯). (expanded CNF of (9z)(z ≡ B) ^ (¯z ≡ x)) • If A is B _ D, then CA;x is (9z; w)CB;z ^ CD;w ^ (w ¯ _ x) ^ (¯z _ x) ^ (¯x _ z _ w). (expanded CNF of (9z; w)(z ≡ B) ^ (w ≡ D) ^ ((z _ w) ≡ x)) • If A is B ^ D, then CA;x is (9z; w)CB;z ^ CD;w ^ (¯x _ w) ^ (¯x _ z) ^ (x _ z¯ _ w¯). (expanded CNF of (9z; w)(z ≡ B) ^ (w ≡ D) ^ ((z ^ w) ≡ x)) In view of the previous discussion, the reduction maps in polynomial time formula , equal to Qφ, to the equivalent t.q.b.f. with c.n.f. unquantified part Q(9x)x ^ Cφ,x. 2.
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