Isolation Technique

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Isolation Technique

• April 16, 2001
• Jason Ku
• Tao Li

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Outline

1. Show that we can reduce NP, with high
   probability, to US. That is NP randomized
   reduces to detecting unique solutions.
2. PH ? PPP

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Isolation Lemma

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

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Definition of weight functions

• A weight function W, maps a finite set U??
• For a set S?U, W(S)?x?SW(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 Uu1, u2, u3, let F (u1), (u2),
  (u3), (u1u2)
• define W1(u1)1 W2(u1)1
• W1(u2)2 W2(u2)1
• W1(u3)3 W2(u3)2
• W1 is good for F while W2 is bad for F.

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Isolation Lemma

• Let U be a finite set
• Let F1, F2, Fm be families of non-empty subsets
  of U
• Let D U
• Let R gt 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 lt ? lt 1, be s.t. ? gt mD/R
• Then, more than (1- ?)Z weight functions are
  good for all of F1, F2, Fm.

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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.

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Lets Count

• So, for an x?U, there are at most RD-1 weight
  functions s.t. x is ambiguous.
• There are RD weight functions, m choices for F
  and D choices for x. Thus the fraction of weight
  functions that are bad for Fi is at most
  mDRD-1/RD mD/R lt ?. So the fraction of weight
  functions good for Fi is 1- ?.

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Example of Isolating Lemma

• Let Uu1, u2, u3
• D3
• Let F1(u1), (u1,u3), (u1,u2,u3), (u2)
• m1
• R 4 gt mD 3
• Z 64
• Then at least (1 ¾)64 16 weight functions are
  good for F.
• W1(u1)1 W2(u1)2 W3(u1)1 W4(u1)1
• W1(u2)2 W2(u2)3 W3(u2)2 W4(u2)3
• W1(u3)3 W2(u3)4 W3(u3)2 W4(u3)3
• 6 variations 6 variations 3 variations 3
  variations
• 18 variations, and more.

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Definition of US

• US L (? NPTM M) (?x) x ? L ? accM(x)1

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NP randomized reduces to US

• NP ? RPUS
• 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 ? RPUS
• 6) Show x ? L ? NP implies x ? L(N) ? RPUS

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Definitions I

• Let A ltx,ygt NPTML(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. ltx,ygt?A
• Encode y as follows
• y y1y2yn i 1?i ?p(n) ? yi 1
• ex y 1001 1, 4
• (1 take right branch, 0 take left branch)

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Definitions II

• Let U(x) 1, 2, , p(x)
• D U p(x)
• Let F(x) y s.t. ltx, ygt?A (collection of
  accepting paths)
• m 1
• Let Z(x) weight functions that assign weights
  no greater than 4p(x)
• R 4p(x)

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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

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Construct an oracle B?US

• Let B ltx, W, jgt W?Z(x), 1 ? j ? p2(x),
  and ? a unique y ? F s.t. W(y) j
• NPTM MB on input u
• 1) if u is not of the form ltx,W,jgt reject
• 2) else, using p(x) non-deterministic moves,
  selects y and accepts u ? ltx,ygt?A and W(y)j.

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Why B?US

• If u?B, there is a unique path y ? F s.t. W(y)j.
  Thus machine MB will only accept once.
• If u?B, there are either zero, or more than 1 y?F
  s.t. W(y)j. Thus machine MB will have either
  zero, or more than 1 accepting path.

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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 ? 4p2(x), ask oracle B
  if ltx, W, jgt ? B. If yes, accept. If no,
  reject.

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N?RPUS and x?L ?high probability x?N

• For every x??,
• - If x?L, MB on ltx, W, jgt accepts with
  probability ¾, so N accepts with probability ¾.
• - If x?L, MB on ltx, W, jgt rejects with
  probability 1, so N rejects with probability 1.
• So,
• - x?L ?high probability x?N
• - Since x?L implies ?(N, x) ¾ gt .5
  acceptance, and x?L implies ?(N, x) 1
  rejecting, N?RP

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Definition of ?P and P

• ?P L (? NPTM M) (?x) x?L ? accM(x) is odd
• P f (? NPTM M) (?x) f(x) accM(x)

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Todas Theorem PH ? PPP

• Three major parts to prove it
• (ValiantVazrani) NP ? BPP?P
• Theorem 4.5
• Lemma 4.13
• PP?P ? PPP, hence BPP?P ? PPP

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(ValiantVazrani) 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, w1,,p(x)----1,
  ,4p(x),
• B ??P

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(Valiant Vazrani) Cont.

• For a BPP?P algorithm, consider
• On input x
• Randomly pick w
• for k1 to 4p2(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

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Note

• Valiant Vazrani Theorem is relativizable.
• In other words, we have
• NPA BPP?PA
• for every oracle A

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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 BPPBPP ? BPP, ?P?P ? ?P

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Step 1 for Thm. 4.5

• Induction hypothesis
• Since
• NPA BPP?PA for every oracle A,
• Hence,

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Step 2 Swapping

• By lemma 4.9
• ?PBPPA ? BPP?PA
• Hence

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Step 3 Collapse

• Proposition 4.6 BPPBPPA BPPA
• Proposition 4.8 ?P?P ?P
• Hence

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Todas Theorem

• Proposition 4.15
• PPP PP
• Todas Theorem PP is Hard for the polynomial
  Hierarchy
• PH ? PPP PP

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Proof for BPP?P ? PP

• Let A ? BPP?P, where A is accepted by MB and let
  f be the P function for B. Let nk 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

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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

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Proof Cont. 2

• g(x,y) is odd if and only if MB (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 2nk )
• g(x,y) is even, then g(x,y) 0(mod 2nk )
• Define h(x)

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Proof Cont.3

• The value h(x) (mod 2nk )represents the number
  of ys such that MB (x) accepts along path y
• Our PP 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

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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

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Proof Cont. 5

• Call the above machine as N
• Claim MB accepts x along y if and only if
  accN(x,y) g(x,y) is odd

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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 2nk )
• f(x) is even than g(x) 0 (mod 2nk )

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Fact 2

• Let f(x,y) be a P function, then
• Let M be the machine such that accM(x,y)f(x,y).
  Consider the following machine M
• on input x
• compute xk
• guess y of length xk
• run M(x,y)
• g(x) accM ( x,y)

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Discussions

• UL/Poly NL/Poly
• ? UL NL
• ? UP NP
• NPPSPACE UPPSPACE PSPACE
• There is an oracle relative to which NPltgtUP.

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Conclusions

• Weve shown, by use of the isolation lemma, that
  NP ? RPUS ? BPP?P.
• This was the base case of an inductive proof to
  show PH ? BPP?P.
• From there we extended to Todas theorem PH ?
  PPP PP.