 # –Def: A language L is in BPP c,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|),

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–Def: A language L is in BPP c,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), More on Randomized Class

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. X i : indicator random variable for the event that M accepts w.

By the definition of BPP c,s we have: w  L  E[X i ]  c(n), w  L  E[X i ]  s(n).

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

Using Chernoff bounds, choose  with Setting

So ■

P/poly and circuit complexity –Def: P/poly={ L |  A  P, a sequence of strings {S i } i  N and a constant k s.t. |S i |=O(i k ) 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 f n that is 1 iff its input (of length n) is in L, has circuit complexity O(n k ).

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, S n ~ poly size. Construct a poly time TM taking x and S |x| and simulate S |x| on input x.

 : Assume L  P/poly. If M decides L in TIME(O(n k )), then we can construct a circuit c k of size O(n 2k ) that simulates M running on input strings of length n. Hardwired in each machine will be the advice strings S n, which is constant for each input size n and which grows polynomial in n. ■

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] 

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. ■

Thm: BPP   2 P(Sipser,Lautemann) Pf: Suppose L  BPP. –Goal: Show that there is a  2 P 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.

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|=2 Q(n). –Define F s (y)=y  s, s  R, y  R. F s (y) is random if s is chosen uniformly. Imagine a new machine, A ’ (x,y,S), where S is a sequence random bits (s 1,s 2,…,s k ), y  R, x is the input to test if x  L A A ’ is a deterministic TM s.t.: A ’ (x,y,S)=1   s i  S, A accepts x with y  s i on its random tape.

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

If x  L A, and a specific y is chosen at random

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

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.

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, Pr h  H [h(x)=w]=1/|T| –2)  x  y  S,  w,z  T, Pr h  H [(h(x)=w) ∧ (h(y)=z)]=1/|T| 2

eg. Let S={0,1} n, T={0,1} k and for x  {0,1} n, let h M,b (x)=Mx+b, where M is a k  n Boolean matrix, b is a column vector is {0,1} k. –H={ h M,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.

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

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

Prop:  x  y  {0,1} n - and  w,z  {0,1} k, we have Pr M [Mx=w ∧ My=z]=1/2 2k. Pf: If x and y are e 1 =(1,0,…,0) and e 2 =(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=e 1, and Ay=e 2. ∵ rank(A)=n,  MA is random if M is chosen randomly. So, the truth of the proposition is clear.

M x (a 1,…,a n )  (b 1,…,b n )=c fixed Pr[a 1 b 1 +a 2 b 2 +…+a n b n =c]=? a 1,a 2,…,a n  {0,1} are selected randomly, c  {0,1}.

Prop: Let S  {0,1} n, with 2 k-2  |S|  2 k-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: –

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

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 2 k-2  |S|<2 k-1. By the previous prop.,  k has a probability  1/8 of having exactly one satisfying assignment.

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  RP UP. Thm: If UP  RP, then NP=RP.

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