More on Randomized Class

1. More on Randomized Class • Def: A language L is in BPPc,s ( 0s(n)c(n)1, nN) if there exists a probabilistic poly-time TM M s.t. : • 1. wL, Pr[M accepts w]  c(|w|) , • 2. wL, Pr[M(x) accepts]  s(|w|) . • Thm: (Amplification of BPP) For all choices of poly. computable functions c(n) and s(n) : {0,1}n {0,1}, such that there exists a poly. Q(n) s.t. n c(n)-s(n)  1/Q(n) and m=O(1),

2. Pf: Given a BPP machine M with c(n), s(n). We construct a BPP machine for the same language with for any m=O(1). • Define M’: 1. Run M on k times independently. 2. Accept if the number of time M accepted is  k‧(c(n)+s(n))/2 . Xi: indicator random variable for the event that M accepts w.

3. By the definition of BPPc,s we have: • wL  E[Xi]  c(n) , • wL  E[Xi]  s(n) .

4. Chernoff bound: For any k independent identically distributed random variable X1,…Xk with values in {0,1}, and with expected values E[Xi]=p, for any (0,1),

5. Using Chernoff bounds, choose  with Setting

6. So ■

7. P/poly and circuit complexity • Def: P/poly={ L | AP, a sequence of strings {Si}iN and a constant k s.t. |Si|=O(ik) and xL  (x,S|x|)A } • Def: A language L has poly circuit complexity if there exists a constant k such that for all n, the function fn that is 1 iff its input (of length n) is in L, has circuit complexity O(nk) .

8. Prop: LP/poly iff L has poly. circuit complexity. • Pf:  : If L has poly circuit complexity, then for each n, there is a circuit of size poly in n that decides membership in L for all words of length n. Encode this circuit on a string, Sn ~ poly size. Construct a poly time TM taking x and S|x| and simulate S|x| on input x.

9.  : Assume LP/poly. If M decides L in TIME(O(nk)), then we can construct a circuit ck of size O(n2k) that simulates M running on input strings of length n . Hardwired in each machine will be the advice strings Sn, which is constant for each input size n and which grows polynomial in n . ■

10. Thm: BPP  P/poly. • Pf: Let L be an arbitrary language in BPP. • By amplification of BPP, we have a TM M that decides L . Classify all possible random string R as follows: • R is bad for an input x if M(x,R) is wrong. • R is bad if there exists an input w for which R is bad. • R is good otherwise. Fix w, Pr[R is bad for w] 

11. Pr[R is bad]  Pr[R is bad for w]  Therefore, Pr[R is good] = 1-Pr[R is bad] > 0 Thus, there exists a poly size advice string for any input of length n. ■

12. Thm: BPP  2P (Sipser,Lautemann) • Pf: Suppose LBPP. • Goal: Show that there is a 2P Machine that decides L. • I.e. show that  a deterministic poly time TM M(x,y,z) s.t. • xL  y s.t. z M(x,y,z)=1 • xL  y z s.t. M(x,y,z)=0 .

13. Let A be a BPP machine that uses Q(n) random bits with c(n)= ½ and s(n)=1/3Q(n) where n is the input length and Q(n) is poly. • Let R be the set of all random string of length Q(n) used on A’s random tape. |R|=2Q(n) . • Define Fs(y)=ys, sR, yR . Fs(y) is random if s is chosen uniformly. • Imagine a new machine, A’(x,y,S), where S is a sequence random bits (s1,s2,…,sk), yR, x is the input to test if xLA • A’ is a deterministic TM s.t.: A’(x,y,S)=1  siS, A accepts x with ysion its random tape.

14. If xLA, and a specific S is chosen at random, then I.e. if xLA,then for any SRk, yR s.t. A’(x,y,S)=0 . let k2Q(n)

15. If xLA, and a specific y is chosen at random

16. Let k=2Q(n) I.e. if xLA, then  an SRk, s.t. yR, A’(x,y,S)=1 Therefore, xLA  SRk, s.t. yR, A’(x,y,S)=1 . So a 2P machine decides LA by guessing S, guessing all y and checking A’(x,y,S)=1 .

17. USAT:  is USAT if  is satisfied by exactly one truth assignment. • Suppose  is satisfiable by at most one truth assignment. We want to decide if USAT . It turns out to decide USAT is as difficult as to decide SAT.

18. Randomized reduction from SAT to USAT. M: randomized poly-time TM M s.t. • SAT  M()SAT (USAT) • SAT  Prob[M()USAT]  1/8 . • Universal Hashing: Given sets S and T, a family H of functions from S to T is a universal family of hash functions from S to T if • 1) xS, wT, PrhH[h(x)=w]=1/|T| • 2) xyS, w,zT, PrhH[(h(x)=w)∧(h(y)=z)]=1/|T|2

19. eg. Let S={0,1}n, T={0,1}k and for x{0,1}n, let hM,b(x)=Mx+b, where M is a k  n Boolean matrix, b is a column vector is {0,1}k . • H={ hM,b: for all possible M and b } . • Prop: The above H is a family of universal hash functions from {0,1}n to {0,1}k .

20. Pf: • 1) For any fixed x{0,1}n- and y{0,1}k Pr[x+y+1=1]=?

21. 2) For xy{0,1}n and w,z{0,1}k .

22. Prop:xy {0,1}n- and w,z{0,1}k, we have PrM[Mx=w∧My=z]=1/22k . • Pf: • If x and y are e1=(1,0,…,0) and e2=(0,1,0,…,0), respectively, then it is true. • Since neither x nor y is , they’re linear independent. • Thus, there exists rank n matrix A s.t. Ax=e1, and Ay=e2 . ∵ rank(A)=n, MA is random if M is chosen randomly. • So, the truth of the proposition is clear .

23. M x (a1,…,an)  (b1,…,bn)=c fixed Pr[a1b1+a2b2+…+anbn=c]=? a1,a2,…,an{0,1} are selected randomly, c{0,1} .

24. Prop: Let S{0,1}n, with 2k-2|S|2k-1 ,Then Pr[! sS s.t. Ms= 0 ]1/8, where the probability is taken over the uniform choice of M from the set of all k x n Boolean matrices. • Pf:

26. Successive restrictions: Given a CNF formula  on n variables , choose n+1 random vectors and create I for 1 i  n+1 as follows:

27. Lemma: If  is not satisfiable, then none of the i’s are satisfiable . • Lemma: If  is satisfiable, then with probability at least 1/8, at least one of the i’s has a unique satisfying assignment . • Pf: Let S be the set of satisfying assignments of : by hypothesis |S|1. Let k be such that 2k-2|S|<2k-1 . By the previous prop., k has a probability  1/8 of having exactly one satisfying assignment .

28. Thus, detecting unique solutions is an hard as NP . • UP:the class of promise problems where instances are promised to whose either zero or one solution . • Thm: NPRPUP . • Thm: If UPRP, then NP=RP .