Download presentation

Presentation is loading. Please wait.

1
–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

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

3
By the definition of BPP c,s we have: w L E[X i ] c(n), w L E[X i ] s(n).

4
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), –

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 {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 ).

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, S n ~ 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(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. ■

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 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.

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|=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.

14
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)

15
If x L A, and a specific y is chosen at random

16
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.

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, 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

19
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.

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 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.

23
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}.

24
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: –

25
–

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 2 k-2 |S|<2 k-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 RP UP. Thm: If UP RP, then NP=RP.

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google