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Exploring Approximate Majority and Its Computational Complexity

This document discusses the relationship between BPP (Probabilistic Polynomial Time) and circuit complexity, particularly in the context of Approximate Majority problems. It highlights key results such as the necessity of quadratic slowdown in depth-3 circuits with bottom fan-in constraints and examines the behavior of BPTime and S2Time under various conditions. The relevance of lower bounds for circuit models is emphasized, alongside results concerning the existence of polynomial-size circuits for Approximate Majority on specific input sizes.

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Exploring Approximate Majority and Its Computational Complexity

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  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))

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