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Isolation Technique April 16, 2001 Jason Ku Tao Li.

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Presentation on theme: "Isolation Technique April 16, 2001 Jason Ku Tao Li."— Presentation transcript:

1 Isolation Technique April 16, 2001 Jason Ku Tao Li

2 Outline 1)Show that we can reduce NP, with high probability, to US. That is: NP randomized reduces to detecting unique solutions. 2)PH  P PP

3 Isolation Lemma 1)Definitions 2)Isolation Lemma 3)Example of using Isolation Lemma

4 Definition of weight functions A weight function W, maps a finite set U  + For a set S  U, W(S)=  x  S W(x) Let F be a family of non-empty subsets of U. A weight function W is “good for” F if there is a unique minimum weight set in F, and “bad for” F otherwise. Ex: let U={u 1, u 2, u 3 }, let F ={(u 1 ), (u 2 ), (u 3 ), (u 1 u 2 )} define W 1 (u 1 )=1W 2 (u 1 )=1 W 1 (u 2 )=2 W 2 (u 2 )=1 W 1 (u 3 )=3W 2 (u 3 )=2 W 1 is good for F while W 2 is bad for F.

5 Isolation Lemma Let U be a finite set Let F 1, F 2, …F m be families of non-empty subsets of U Let D = ||U|| Let R > mD Let Z be a set of weight functions s.t. the weight of any individual element of U is at most R Let , 0 mD/R Then, more than (1-  )||Z|| weight functions are good for all of F 1, F 2, …F m.

6 Proof of Isolation Lemma Proof sketch: By definition, a weight function W is bad if there are at least 2 different minimum weight sets in F. Let S1 and S2 be 2 different sets with the same minimum weights, then  x  S1 s.t. x  S2. Call x ambiguous. If we know the weights of all other elements in U, either x is unambiguous, or there is one specific weight for x that makes x ambiguous.

7 Lets Count So, for an x  U, there are at most R D-1 weight functions s.t. x is ambiguous. There are R D weight functions, m choices for F and D choices for x. Thus the fraction of weight functions that are bad for F i is at most mDR D-1 /R D = mD/R < . So the fraction of weight functions good for F i is 1- .

8 Example of Isolating Lemma Let U={u1, u2, u3} D=3 Let F 1 ={(u1), (u1,u3), (u1,u2,u3), (u2)} m=1 R = 4 > mD = 3 ||Z|| = 64 Then at least (1 – ¾)64 = 16 weight functions are good for F. W1(u1)=1W2(u1)=2W3(u1)=1W4(u1)=1 W1(u2)=2W2(u2)=3W3(u2)=2W4(u2)=3 W1(u3)=3W2(u3)=4W3(u3)=2W4(u3)=3 6 variations6 variations3 variations3 variations 18 variations, and more.

9 Definition of US US = {L | (  NPTM M) (  x) x  L  #acc M(x) =1}

10 NP randomized reduces to US NP  RP US Proof Map: 1) Definitions I, II 2) Apply Isolation Lemma to get a probability 3) Construct an oracle B  US 4) Construct a machine N that uses oracle B 5) show N  RP US 6) Show x  L  NP implies x  L(N)  RP US

11 Definitions I Let A = { | NPTM L(x) on path y accepts} for L  NP,  a polynomial p, s.t.  x  *, x  L   at least 1 y, |y| = p(|x|), s.t.  A Encode y as follows: y = y 1 y 2 …y n = {i | 1  |i|  p(n)  y i = 1} ex: y = 1001 = {1, 4} (1 take right branch, 0 take left branch)

12 Definitions II Let U(x) = {1, 2, …, p(|x|)} D = ||U|| = p(|x|) Let F(x) = y s.t.  A (collection of accepting paths) m = 1 Let Z(x) = weight functions that assign weights no greater than 4p(|x|) R = 4p(|x|)

13 Applying Isolation Lemma By the Isolation Lemma: if x  L, 3/4 of weights functions assigns F a unique minimum weight set if x  L, there are no accepting paths y  F so zero weight functions are assigns F a unique minimum weight set

14 Construct an oracle B  US Let B = { | W  Z(x), 1  j  p 2 (|x|), and  a unique y  F s.t. W(y) = j} NPTM M B on input u: 1)if u is not of the form reject 2)else, using p(|x|) non-deterministic moves, selects y and accepts u   A and W(y)=j.

15 Why B  US If u  B, there is a unique path y  F s.t. W(y)=j. Thus machine M B will only accept once. If u  B, there are either zero, or more than 1 y  F s.t. W(y)=j. Thus machine M B will have either zero, or more than 1 accepting path.

16 Construct an RP machine with oracle B NPTM N on input x: 1)Create random weight functions W properly bounded. 2)For each j, 1  j  4p 2 (|x|), ask oracle B if  B. If yes, accept. If no, reject.

17 N  RP US and x  L  high probability x  N For every x  *, - If x  L, M B on accepts with probability ¾, so N accepts with probability ¾. - If x  L, M B on rejects with probability 1, so N rejects with probability 1. So, - x  L  high probability x  N - Since x  L implies  (N, x) = ¾ >.5 acceptance, and x  L implies  (N, x) = 1 rejecting, N  RP

18 Definition of  P and #P  P = {L | (  NPTM M) (  x) x  L  #acc M(x) is odd} #P ={f | (  NPTM M) (  x) f(x) = #acc M(x) }

19 Toda’s Theorem PH  P PP Three major parts to prove it: (Valiant&Vazrani) NP  BPP  P Theorem 4.5 Lemma 4.13 PP  P  P PP, hence BPP  P  P PP

20 (Valiant&Vazrani) NP  BPP  P Proof: Let A  NP, A = L(M) and M runs in time p(|x|). Let B={(x,w,k): M has an odd # of accepting paths on input x having weight k}, w:{1,…,p(|x|)}----{1,…,4p(|x|)}, B  P

21 (Valiant &Vazrani) Cont. For a BPP  P algorithm, consider On input x Randomly pick w for k:=1 to 4p 2 (|x|) ask if (x,w,k) is in B if so, then halt and accept end-for if control reaches here, then halt and reject

22 Note Valiant &Vazrani Theorem is relativizable. In other words, we have NP A = BPP  P A for every oracle A

23 Theorem 4.5 PH  BPP  P We prove it by induction Three steps for induction step: Apply Valiant & Vazrani to the base machine Swap BPP and  P in the middle Collapse BPP BPP  BPP,  P  P   P

24 Step 1 for Thm. 4.5 Induction hypothesis: Since NP A = BPP  P A for every oracle A, Hence,

25 Step 2: Swapping By lemma 4.9  P BPP A  BPP  P A Hence

26 Step 3: Collapse Proposition 4.6: BPP BPP A = BPP A Proposition 4.8:  P  P =  P Hence

27 Toda’s Theorem Proposition 4.15 P PP = P #P Toda’s Theorem: PP is Hard for the polynomial Hierarchy PH  P PP = P #P

28 Proof for BPP  P  P #P Let A  BPP  P, where A is accepted by M B and let f be the #P function for B. Let n k be the running time of M. Assume first that M makes only one query along any path. Then let g(x,y) be a #P function that is defined to be the number of accepting paths of the following machine:

29 Proof cont. 1 On input x,y run M(x) along path y when a query “w is in B?” is made then flip a coin c in {0,1} and use this as the oracle answer and continue simulating M(x) if the simulation accepts, then generate f(w)+(1-c) paths and accept

30 Proof Cont. 2 g(x,y) is odd if and only if M B (x) accepts along y For g(x,y), consider a #P function g’(x,y) such that : g(x,y) is odd, then g’(x,y) =1(mod 2 ^ n k ) g(x,y) is even, then g’(x,y) = 0(mod 2 ^ n k ) Define h(x)=

31 Proof Cont.3 The value h(x) (mod 2^n k )represents the number of y’s such that M B (x) accepts along path y Our P #P algorithm: on input x, using the oracle h(x), decides if the following holds: if so, x is accepted, and if not x is rejected

32 Proof Cont. 4 If M makes more than one query, modify g(x,y) as follows: on input x,y repeat run M(x) along with path y when a query “w is in B?’’ is made then flip a coin c in {0,1} and generate f(w)+(1-c) paths use c as the oracle answer and continue simulating M(x) until no more queries are asked; if the simulation of M(x) along path y accepts with this sequence of guessed oracle queries then accept else reject

33 Proof Cont. 5 Call the above machine as N Claim : M B accepts x along y if and only if #acc N (x,y) = g(x,y) is odd

34 Fact 1 Let k in N, f in #P, then there exists g in #P such that f(x) is odd then g(x) = 1 (mod 2^n k ) f(x) is even than g(x) = 0 (mod 2^n k )

35 Fact 2 Let f(x,y) be a #P function, then Let M be the machine such that #acc M (x,y)=f(x,y). Consider the following machine M’: on input x compute |x| k guess y of length |x| k run M(x,y) g(x)= #acc M’ ( x,y)

36 Discussions UL/Poly = NL/Poly ? UL= NL ? UP = NP NP PSPACE = UP PSPACE = PSPACE There is an oracle relative to which NP<>UP.

37 Conclusions We’ve shown, by use of the isolation lemma, that NP  RP US  BPP  P. This was the base case of an inductive proof to show PH  BPP  P. From there we extended to Toda’s theorem: PH  P PP = P #P.

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