 Download Presentation On approximate majority and probabilistic time On approximate majority and probabilistic time - PowerPoint PPT Presentation

Download Presentation On approximate majority and probabilistic time
An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

1. On approximate majority andprobabilistic time Emanuele Viola Institute for advanced study January 2007

2. BPP vs. POLY-TIME HIERARCHY • Probabilistic Polynomial Time (BPP): for every x, Pr [ M(x) errs ]· 1/3 • Strong belief: BPP = P [NW,BFNW,IW,…] Still open: BPP µ NP ? • Theorem [SG,L; ‘83]: BPP µS2 P • Recall NP = S1P!9 y M(x,y) S2 P !9 y 8 z M(x,y,z)

3. The problem we study • More precisely [SG,L] give BPTime(t) µS2Time( t2) • Question[This Talk]: Is quadratic slow-down necessary? • Motivation: Lower bounds Know NTime ≠ Time on some models [P+,F,…] Technique: speed-up computation with quantifiers To prove NTime ≠ BPTime cannot afford Time( t2) [DvM]

4. Approximate Majority • Input: R = 101111011011101011 • Task: Tell Pri [ Ri = 1] ¸ 2/3 from Pri [ Ri = 1] · 1/3 Approximate: Do not care if Pri [ Ri = 1] ~ 1/2 • Model: Depth-3 circuit V Æ Æ Æ Æ Æ Æ Depth V V V V V V V V R = 101111011011101011

5. The connection [FSS] |R| = 2t Ri = M(x;i) M(x;u) 2 BPTime(t) R = 11011011101011 Compute M(x): Tell Pru[M(x) = 1] ¸ 2/3 Compute Appr-Maj from Pru[M(x) = 1] · 1/3 BPTime(t) µ S2 Time(t’) = 9 8Time(t’) Running time t’ Bottom fan-in f = t’ / t • run M at most t’/t times V Æ Æ Æ Æ Æ Æ V V V V V V V V L f L 101111011011101011

6. Our Negative Result • Theorem[V] : Small depth-3 circuits for Approximate Majority on N bits have bottom fan-in W(log N) • Corollary: Quadratic slow-down necessary for relativizing techniques: BPTime A(t) µS2Time A(t1.99) • Proof of Corollary: BPTime(t) µS2 Time (t’) ) [FSS] Appr-Maj on N = 2t bits 2 depth-3, bottom fan-in t’ / t. By Theorem: t’ / t = (t). Q.E.D.

7. Quasilinear-time simulation? • Question: BPTime(t) µS3 Time(t ¢ polylog t) ? Related: Appr-Maj 2 depth-3 poly-size ? • arbitrary bottom fan-in • Previous results & problems: [SG,L] Appr-Maj 2 depth-3 size Nlog N [A] Appr-Maj 2 depth-3 size poly(N) nonuniform [A] Appr-Maj 2 depth-O(1) size poly(N)

8. Our Positive Results • Theorem[V] : There are uniform depth-3 poly(N)-size circuits for Approximate Majority on N bits • Uniform version of Ajtai’s result • Theorem[DvM,V]: BPTime (t) µS3Time (t ¢ log5 t)

9. Summary

10. Rest of slides • Proof of bottom fan-in lower bound • Other result 3Time (t) µ BPTime (t1+o(1)) on restricted models

11. Our negative result • Theorem[V]: 2N-size depth-3 circuits for Approximate Majority on N bits have bottom fan-in W(log N) • Switching lemmas fail Cannot use [H] for Approximate-Majority [SBI] ) bottom fan-in ¸ (log N)1/2 • Independently: [R] improves [SBI] alternative proof of theorem • Note: No 2W(N) bound for depth-3 w/ bottom fan-in w(1)

12. Our Negative Result • Theorem[V]: 2N-size depth-3 circuits for Approximate Majority on N bits have bottom fan-in f = W(log N) • Recall: Tells R 2 YES := { R : Pri [ Ri = 1] ¸ 2/3 } from R 2 NO := { R : Pri [ Ri = 1] · 1/3 } V Æ Æ Æ Æ Æ Æ V V V V V V V V L f L R = 101111011011101011 |R| = N

13. Proof V C1 C2 C3 LL Cs • Circuit is OR of s depth-2 circuits • By definition of OR : R 2 YES ) some Ci (R) = 1 R 2 NO ) all Ci (R) = 0 • By averaging, fix C = Ci s.t. PrR 2 YES [C (R) = 1 ] ¸ 1/s 8 R 2 NO ) C (R) = 0 • Claim: Impossible if C has bottom fan-in ·  log N

14. CNF Claim Æ • Depth-2 circuit ) CNF (x1Vx2V:x3 ) Æ (:x4) Æ (x5Vx3) bottom fan-in )clause size • Claim:All CNF C with clauses of size ¢log N Either PrR 2 YES [C (R) = 1 ] · 1 / 2N or there is R 2 NO : C(R) = 1 • Note: Claim ) Theorem V V V V V x1 x2 x3 … xN

15. Either PrR 2 YES [C(R)=1]·1/2Nor 9 R 2 NO : C(R) = 1 Proof Outline • Definition: S µ {x1,x2,…,xN} is a covering if every clause has a variable in S E.g.: S = {x3,x4} C = (x1Vx2V:x3 ) Æ (:x4) Æ (x5Vx3) • Proof idea: Consider smallest covering S Case |S| BIG : PrR 2 YES [C (R) = 1 ] · 1 / 2N Case |S| tiny : Fix few variables and repeat

16. Either PrR 2 YES [C(R)=1]·1/2Nor 9 R 2 NO : C(R) = 1 Case |S| BIG • |S| ¸ N) have N /(¢log N) disjoint clauses Gi • Can find Gi greedily • PrR 2 YES[C(R) = 1]· Pr [8 i, Gi(R) = 1 ] = Õi Pr[ Gi(R) = 1] (independence) ·Õi (1 – 1/3log N ) = Õi(1 – 1/NO()) = (1 – 1/NO()) |S|· e-N(1)

17. x3Ã 0 x4Ã 1 Either PrR 2 YES [C(R)=1]·1/2Nor 9 R 2 NO : C(R) = 1 Case |S| tiny • |S| < N) Fix variables in S • Maximize PrR 2 YES [C(R)=1] • Note: S covering ) clauses shrink Example (x1Vx2Vx3 ) Æ (:x3) Æ (x5V:x4) (x1Vx2 ) Æ (x5) • Repeat Consider smallest covering S’, etc.

18. Either PrR 2 YES [C(R)=1]·1/2Nor 9 R 2 NO : C(R) = 1 Finish up • Recall: Repeat ) shrink clauses So repeat at most ¢log N times • When you stop: Either smallest covering size ¸ N Or C = 1 Fixed · (¢log N) N¿ N vars. Set rest to 0 ) R 2 NO : C(R) = 1 Q.E.D.

19. Rest of slides • Proof of bottom fan-in lower bound • Other result 3Time (t) µ BPTime (t1+o(1)) on restricted models

20. The model Time1 Input (x1Vx2V:x3 ) Æ (:x4) Æ (x5Vx3) Æ (x1Vx6V:x3 ) Random access 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 Sequential access work tape

21. Time Lower Bound for SAT • Theorem [MS,vMR]: NTime (n) µ Time1 (n1.22) • Note: ``combinatorics’’ does not work • Palindromes 2 Time1 (n1+o(1)) • Proof by contradiction: SupposeNTime(n) µ Time1 (n1.22) µO(1) Time(n0.1) (speed-up with quantifiers) µ NTime(n0.9) (collapse by assumption) Contradicts NTime hierarchy Q.E.D.

22. The model BPTime1 Input (x1Vx2V:x3 ) Æ (:x4) Æ (x5Vx3) Æ (x1Vx6V:x3 ) Random access 0 1 Sequential access work tape with coins

23. Our BPTime Lower Bound for 3 • Theorem [V]:3Time (n) µ BPTime1 (n1+o(1)) • Proof by contradiction (Inspired by [DvM]): Suppose 3Time(n) µ BPTime1 (n1+o(1)) µ BPn0.9 Time1(n1+o(1)) (derandomize [INW]) µ3 Time(n.99) ([V] + [MS]) Contradicts 3 Time hierarchy Q.E.D. • Note: Quadratic slow-down ) won’t work for 2

24. Seen so far • Theorem[SG,L]: BPTime(t) µS2Time( t2 ) • Related to Approximate Majority • Theorem [V]: 3Time (n) µ BPTime1 (n1+o(1))