On The Sensitivity Conjecture. Avishay Tal Institute for Advanced Study ICALP 2016

Transcription

1 On The Sensitivity Conjecture Avishay Tal Institute for Advanced Study ICALP 2016

2

3 Sensitivity of Boolean Functions Let ff: 0,1 nn {0,1} be a Boolean function. The sensitivity of ff at xx 0,1 nn : ss(ff, xx) = # of neighbors of xx in Hamming cube with different ff-value ( color ) than ff(xx). The sensitivity of ff: ss ff = max xx ss(ff,xx) s(f,x)=2 s(f,x)=0 s(f)=2

4 Block Sensitivity of Boolean Functions Let ff: 0,1 nn {0,1} be a Boolean function The block sensitivity (bs) of ff at xx 0,1 nn : bbbb(ff, xx) = maximal number of disjoint blocks BB 1, BB 2, nn with ff xx ff xx BB ii. The block sensitivity (bs) of ff: bbbb ff = max bbbb(ff, xx) xx Obs: ff: ss ff bbbb(ff) bs(f,x)=2 bs(f,x)=1 bs(f)=2

5 The Sensitivity Conjecture Conjecture [Nisan-Szegedy 92]: a constant cc: Boolean functions ff, bbbb ff ss ff cc. More boldly, ff:bbbb ff ss ff 22 Known upper bounds [Simon, KK, ABGMSZ, APV] ss ff bbbb ff 2 ss ff Known lower bounds [Rubinstein, Virza, AS] ff: bbbb ff Ω(ss ff 22 )

6 Other Complexity Measures D(f) R(f) Q(f) C(f) deg(f) Deterministic decision tree complexity Probabilistic decision tree complexity Quantum decision tree complexity Certificate complexity Real degree deg (f) L approximate degree bs(f) Block sensitivity [Blum-Impagliazzo, Tardos, Hartmanis, Nisan, Nisan-Szegedy, BBCMW ]: all parameters are polynomially related. The sensitivity conjecture is equivalent to MM ff ss ff OO(1) for any complexity measure M above.

7 Kenyon-Kutin s Approach The l-block sensitivity of ff at xx 0,1 nn : bbss l (f,x) = maximal number of disjoint blocks BB 1, BB 2, nn, each of size at most l, with ff xx ff xx BB ii. Observation: WLOG l ss(ff). Theorem [Kenyon-Kutin 04]: 2 l ss(ff) 1. bbss l ff 4 l ss ff bbss l 1(ff) 2. bbss l ff e ee l 1! ss ff l For l = ss(ff) this gives bbbb ff = bbss l (ff) OO ee Can we improve these inequalities? ss ff

8 Kenyon-Kutin s Approach Small l l = 2: [KK, T]: bbss 2 ff ee 2 ss ff 2 Nearly Tight [Rubinstein, Virza, AS]: ff: bbss 2 ff 2 3 ss ff 2 l = 3: [KK]: bbss 3 ff OO(ss ff 3 ) [Rubinstein, Virza, AS]: ff: bbss 3 ff 2 3 What is the right answer? ss ff 2

9 Understanding bbss 3 (ff) is important! Improving upper / lower bounds for bbss 33 vs. ss better bounds on bbbb vs. ss Claim[Folklore]: If gg: bbss 3 gg ss gg 2+εε, then inf. many ff s with bbss ff ss ff 2+εε. Thm1: If ff: bbss 33 ff < ss ff 2.999, then ff: bbss ff < 2 oo ss ff.

10 New Proof for Kenyon-Kutin BB 1 BB kk BB 2 0 nn BB 3 BB 4

11 Generalization to Bounded functions Let ff: 0,1 nn [00, 11] The sens. of ff at xx 0,1 nn : ss(ff, xx) = yy xx ff xx ff yy Similarly, generalize bbbb, bbss l New proof for Kenyon-Kutin s result works for Bounded functions Reason: most points in a ball centered at xx 0 have f-value close to ff(xx 0 ). However, in this case the bounds are tight! s(f,x)=

12 The Sensitivity Conjecture is False for Bounded Functions Thm3: For all l, nn, a function ff: 0,1 nn [0,1] with 1. bbss l ff nn l 2. ss ff = OO(l nn 1/l ) In particular, for l = log(nn), an exponential separation ss ff = OO log nn and bbbb ff Ω nn/log nn.

13 Sensitivity vs. Decision Tree First super-quadratic separation between decision-tree complexity and sensitivity Separation based on a gadget ff: 0,1 42 {0,1} with ss ff = 6, DD ff = 42 composed with itself many times. ss ff kk = 6 kk DD ff kk = 42 kk ss ff kk More examples found using computer search [Ben-David 16]: cubic separations ff: DD(ff) = Ω (ss ff 3 )

14 The Sensitivity Conjecture: Why should you believe? Believe It or Not? It stood for along time without a refutation. Best lower bounds: bbbb ff = Ω(ss ff 2 ) for more than 20 years. Conj holds for families of Boolean functions: monotone, symmetric [Nis] Consequences of the conjecture were proved, e.g. any low-sensitivity function is in NC1 [GNSTW 16]. Why should you doubt? It stood for along time without a proof. Best upper bounds: bbbb ff = exp(oo(ss ff )) for more than 30 years. Don t know how many small-sensitivity functions are there? [GNSTW 16] For s = log n, this number can be between exp (nn) to exp (nn log nn ) Can t even brute-force to check if ff with bs(f)=26 and s(f)=5. Computer-search found D(f) vs s(f) separations we weren t aware of.

15 Thank You!

16 Thm3 [Case l = 3]: a function ff: 0,1 nn [0,1] with bbss 3 ff nn/3 & ss ff = OO(nn 1/3 ). Construction: Partition [n] to nn/3 blocks of size 3: 1,2,3, 4,5,6,, nn 2, nn 1, nn Let xx 0,1 nn and HH ii (xx) be the hamming weight of x on the i-th block. 0, HH ii xx = 0 nn 2/3, HH ww ii xx = ii xx = 1 nn 1/3, HH ii xx = 2 1, HH ii xx = 3 Proof: bbbb ff, 0 nn.ff xx = min (1, ii ww ii (xx)). nn/3: Easy! Key Point for ss ff =O(nn 1/3 ): If ii ww ii (xx) 2, s ff, xx = 0. Enough to consider x s with ii ww ii (xx) < 2: #{ii HH ii (xx) = 0} nn, each may contribute nn 2/3 to s(f,x) #{ii HH ii (xx) = 1} OO(nn 2/3 ), #{ii HH ii (xx) = 2,3} OO(nn 1/3 ), each may contribute nn 1/3 to s(f,x) each may contribute 1 to s(f,x)