Improved Derandomization of BPP using a hitting set generator Oded Goldreich Avi Wigderson

1. Improved Derandomization of BPP using a hitting set generatorOded GoldreichAvi Wigderson D91922010(狄彥吾),R91922045(游家牧),R91922046(林宜禎) National Taiwan University Department of Computer Science and Information Engineering 05/14/2004

2. Outline • Introduction to Hitting-set generator. • Derandomization procedure. • Patch.

3. Our Goal • Theorem : If there exists a hitting-set generator G running in time tG then BPP is belong to DTime(poly(tG(poly(n))). • If G runs in polynomial time then BPP=P. • This paper gives a simplified proof.

4. The road to P=BPP • Andreev ever gave another derandomization procedure having running time DTime(poly(tG(tG(poly(n))))) • It seems that it the same as the theorem in this paper. • On the one hand, if tG(s)=poly(s), it’s good. • On the other hand, if tG(tG(s))=2s, it’s bad.

5. Hitting-Set Generator(HSG) • An algorithm G is called a hitting-set generator for circuits if for every n, s are belong to N generates as output a set of n-bit strings G(n,s) with the following property : every circuit of size s on n input bits, which accepts at least half its inputs, accepts at least one element from the set G(n,s). • Let s=s(n), tG(s) denote the running time of G, NG(s) denote the size of output set of G. • s=s(n) is the essental complexity parameter (n ≤ s)

6. Hitting-set generator (n,s) G(n,s)={ s1, s2, … ,sNG(s) } Circuit with size at most s n-bit input Accept at least half its Inputs and at least one element from the set G(n,s)

7. One-Side Error and Two-Side Error • RP : One-Side Error • BPP : Two-Side Error • If polynomial time hitting-set generator exist then P=RP. • It is sufficient to run the algorithm on each random sequence generated by the HSG and accept the input if and only if one of these sequences makes the algorithm accept.

8. Hitting-set generator (n,s) G(n,s)={ s1, s2, … ,sNG(s) } RP algorithm x accept / reject s1 If any algorithm above accept x then we say that x is belong to L RP algorithm x accept / reject s2 . . . RP algorithm x accept / reject sNG(s)

9. Difference • The output of HSG is a set of string, s1,s2, … ,sN. • The procedure which we prove P=RP use these strings one by one. • In the procedure which we prove P=BPP, we see the random bit as matrix Q, Q(i,j)=sisj.

10. Before Derandomization • Given L is belong to BPP, we have a probabilistic polynomial-time algorithm A which on inputs of length n use 2l=poly(n) many ramdom bits and error with probability at most 2-(l+1). • A(x,r) denote the output of algorithm A on input x  {0,1}n and random-tape contents r  {0,1}2l . • p is some fixed polynomial so that the computation of A on inputs of length n can be implemented by circuits of size p(l)/l.

11. Derandomization procedure • On input ,letting A and l be as above. Invoking the hitting-set generator G obtain , where each is in . • Construct an N by N matrix, , so that .

12. Using a procedure to determine if (1) for every l

13. Prove Idea • We first show that if then Eq. (1) holds, and analogously if then • To the end we use a general technical lemma which implies that it cannot be the case that both Eq.(1) and Eq.(2) hold.

14. Proposition 5 If (resp., ) then EQ.(1) (resp., Eq.(2)) holds.

15. Proof • Let be the characteristic function of L. • Fixing the inputto algorithm A, we consider the circuit which takes an 2l-bit input r and outputs A(x,r). • By the above hypothesis ,we have

16. Thus, at most values of satisfy • And at least half the values of satisfy

17. Hence for any fixed sequence then • Consider the circuit . • Then by the above • On the other hand, the size of is

18. Thus, if there exists a hitting-set generator G. The set H=G(l, p(l)) must contain a string z so that • It follows that holds for every • The above holds for every Hence we have proved that for every l rows in M there exists a column on which the value of the matrix is . We proved that if then Eq. (1) holds.

19. A similar argument applies to sets of lcolumns in M. • Specifically, for every • Again, we conclude that for every there existsso that for every • Thus, for every l columns in M there exists a row on which the value of the matrix is

20. Equations • (1) (2).

21. Proposition 5 • Shows for x L(respectively,x L) ,Eq.(1) (res. Eq.(2) ) holds. • But It is not clear whether Eq.(1) and Eq.(2) can hold simultaneously.

22. Corollary 6 • For n-by-n Boolean matrix,with ,it is impossible for the following 2 conditions to hold simultaneously.

23. 2 conditions • 1.for arbitrary k rows there exists a column so that all the k rows have a 0-entry in this column.

24. 2 conditions 2.for arbitrary k columns there exists a row so that all the k columns have a 1-entry in this row

25. Furthermore • Assuming one of both conditions holds,we can decide which holds in (deterministic) polynomial time.

26. Proof of Corollary 6 • Suppose Eq. (1) holds. • Then,arbitrary k rows contain a 0-entry,and so cannot be the all 1`s row.

27. Proof of Corollary 6 Suppose Eq. (2) holds. • Then,arbitrary k columns contain a 1-entry,and so cannot be the all 0`s column.

28. Proof of Corollary 6 • Both the 2 equations hold will cause contradiction.

29. Proof of Corollary 6 • Let S0=[n],R=ø,and report for i=0,1,2,… • Take a row j not in R which at least | Si |/2 1`s in Si .

30. Proof of Corollary 6 • Add j to R, and let Si+1 be the part of Si that had 0`s in row j.

31. Proof of Corollary 6 • We get stuck if for any i,no row in current [n]-R has at least | Si |/2 1`s in Si • otherwise,we terminate when Si= ø

32. Proof of Corollary 6 • If we never get stuck,then we generated at most log2n k rows(and the last Si is empty) • On the other hand,if we got stuck at iteration i,let S= Si • Note that every row has at least /2 0`s in the column S.

33. Proof of Corollary 6 • Now picking greedily columns from S in sequence so as to contain the largest number of 0`s in the remaining rows . • a set T of at most k columns from S were chosen (these columns have 0`s more than 1`s,and these rows with 0-entries are taken to T)

34. Proof of Corollary 6 • Note that the above procedure for constructing R,S,and T is implementable in polynomial-time • We may output the set of rows R,and otherwise the set T of columns.

35. Proof of Theorem 2 • Proposition 5 shows that for x L either Eq.(1) or Eq.(2) holds. • Furthermore,the former holds whenever x L • By applying Corollary 6 as indicated above it follows that only one of these equation may hold. • Using the decision procedure guaranteed by this corollary, we implement these steps in our derandomized procedure,and Theorem 2 follows.