On the Sensitivity Conjecture for Disjunctive Normal Forms. Karthik C. S. (Weizmann Institute of Science)

Transcription

1 On the Sensitivity Conjecture for Disjunctive Normal Forms Karthik C. S. (Weizmann Institute of Science) Joint work with Sébastien Tavenas (Université de Savoie-Mont-Blanc).

2 Boolean Function f : {0, 1} n {0, 1} 1

3 Boolean Function f : {0, 1} n {0, 1} n =

4 Boolean Function f : {0, 1} n {0, 1} x = n x = n 1 x = (n/2) + 1 x = n/2 x = (n/2) 1 x = 1 x = 0 1

5 Sensitivity s(f, x) = s(f ) = n f (x) f (x e i ) i=1 max s(f, x) x {0,1} n 2

6 Sensitivity s(f, x) = s(f ) = n f (x) f (x e i ) i=1 max s(f, x) x {0,1} n x = 3 f (x) = x mod 2 x = 2 x = 1 x =

7 Sensitivity s(f, x) = s(f ) = n f (x) f (x e i ) i=1 max s(f, x) x {0,1} n x = 3 f (x) = x mod 2 s(f ) = n x = 2 x = 1 x =

8 Sensitivity s(f, x) = s(f ) = n f (x) f (x e i ) i=1 max s(f, x) x {0,1} n 2

9 Sensitivity s(f, x) = s(f ) = n f (x) f (x e i ) i=1 max s(f, x) x {0,1} n f (x) = n i=1 x i x = 3 x = 2 x = 1 x =

10 Sensitivity s(f, x) = s(f ) = n f (x) f (x e i ) i=1 max s(f, x) x {0,1} n f (x) = n i=1 s(f ) = n x i x = 3 x = 2 x = 1 x =

11 Block Sensitivity bs(f, x, B = {B 1,..., B k }) = bs(f ) = max x {0,1} n max B k f (x) f (x e Bi ) i=1 bs(f, x, B) 3

12 Block Sensitivity bs(f, x, B = {B 1,..., B k }) = bs(f ) = max x {0,1} n max B k f (x) f (x e Bi ) i=1 bs(f, x, B) x = 3 f (x) = x mod 2 x = 2 x = 1 x =

13 Block Sensitivity bs(f, x, B = {B 1,..., B k }) = bs(f ) = max x {0,1} n max B k f (x) f (x e Bi ) i=1 bs(f, x, B) x = 3 f (x) = x mod 2 bs(f ) = n x = 2 x = 1 x =

14 Block Sensitivity bs(f, x, B = {B 1,..., B k }) = bs(f ) = max x {0,1} n max B k f (x) f (x e Bi ) i=1 bs(f, x, B) 3

15 Block Sensitivity bs(f, x, B = {B 1,..., B k }) = bs(f ) = max x {0,1} n max B k f (x) f (x e Bi ) i=1 bs(f, x, B) x = 3 f (x) = n i=1 x i x = 2 x = 1 x =

16 Block Sensitivity bs(f, x, B = {B 1,..., B k }) = bs(f ) = max x {0,1} n max B k f (x) f (x e Bi ) i=1 bs(f, x, B) x = 3 f (x) = n i=1 bs(f ) = n x i x = 2 x = 1 x =

17 Block Sensitivity bs(f, x, B = {B 1,..., B k }) = bs(f ) = max x {0,1} n max B k f (x) f (x e Bi ) i=1 bs(f, x, B) bs(f ) s(f ) 3

18 Sensitivity Conjecture: Statement Conjecture (Nisan and Szegedy, 1994) There exist constants c and δ such that for all Boolean functions we have: bs c s δ. 4

19 Sensitivity Conjecture: Equivalent forms Theorem The following statements are equivalent: 1. bs poly(s). 2. deg, C, DT poly(s). 3. There exist constants c and δ such that for every induced subgraph G of Q n with V (G) 2 n 1 we have: max{ (G), (Q n G)} c n δ. 5

20 Sensitivity Conjecture: Progress Simon 83 Kenyon & Kutin 04 bs 4 s s ( ) e bs 2π e s s Ambainis et. al 14 bs 2 s 1 s (s 1) Ambainis, Prusis, & Vihrovs 16 bs 2 ( ) s 1 s 1 3 He, Li, & Sun 16 bs ( o(1)) 2 s 1 s 6

21 Sensitivity Conjecture: Known Separation Rubinstein 95 bs = 1 2 s2 Virza 11 Ambainis & Sun 11 bs = 1 2 s s bs = 2 3 s2 1 3 s 7

22 Sensitivity Conjecture: Special Cases Turán 84 Graph Property and Symmetric functions Nisan 91 Chakraborty 09 Gao, Mao, Sun, & Zuo 12 Bafna, Lokam, Tavenas, & Velingker 16 Monotone functions Min-term Transitive Bipartite Graph Property Constant depth Regular Read-k Formulas 8

23 Block Property Wlog we assume bs(f ) = bs(f, 0 n ) and f (0 n ) = 0. 9

24 Block Property Wlog we assume bs(f ) = bs(f, 0 n ) and f (0 n ) = 0. A 1 A 1 A i A i A d A d Block Property: f is said to admit block property if i, j [d ], i j, we have A i A j =. 9

25 Block Property Wlog we assume bs(f ) = bs(f, 0 n ) and f (0 n ) = 0. A 1 A 1 A i A i A d A d Block Property: f is said to admit block property if i, j [d ], i j, we have A i A j =. Motivation: Captures best known separation examples. 9

26 Our Results Theorem For every Boolean function f admitting the Block Property we have: bs(f ) 4s(f ) 2. 10

27 Our Results Theorem For every Boolean function f admitting the Block Property we have: bs(f ) = d. s(f ) d /2. s(f ) d /2d. 11

28 Block structure: bs(f ) = d A 1 A 1 A i A i A d A d Note that A i s are pairwise disjoint. This implies bs(f, 0 n, A = {A 1,..., A d, A d +1}) d. 12

29 Block structure: bs(f ) = d A 1 A 1 A i A i A d A d Note that A i s are pairwise disjoint. This implies bs(f, 0 n, A = {A 1,..., A d, A d +1}) d. bs(f ) d follows from disjointness of blocks and assumption on bs(f, 0 n ). 12

30 Local argument: s(f ) d /2 Fix i = argmax i d i. Let x = e Ai. x x = A i 13

31 Local argument: s(f ) d /2 Fix i = argmax i d i. Let x = e Ai. x x = A i x = A i 1 13

32 Local argument: s(f ) d /2 Fix i = argmax i d i. Let x = e Ai. x = A i + 1 x x = A i x = A i 1 13

33 Local argument: s(f ) d /2 Fix i = argmax i d i. Let x = e Ai. x = A i + 1 x x = A i x = A i 1 13

34 Global argument: s(f ) d /2d Consider the following graph G: V (G): AND gates. ( i, j ) E(G) iff A i A j or A i A j. 14

35 Global argument: s(f ) d /2d Consider the following graph G: V (G): AND gates. ( i, j ) E(G) iff A i A j or A i A j. Claim: G has an independent set E of size d /2d. 14

36 Global argument: s(f ) d /2d Consider the following graph G: V (G): AND gates. ( i, j ) E(G) iff A i A j or A i A j. Claim: G has an independent set E of size d /2d. Let A i A i obtained by removing one element of A i. Consider input x, x = e A i E i. 14

37 Global argument: s(f ) d /2d Consider the following graph G: V (G): AND gates. ( i, j ) E(G) iff A i A j or A i A j. Claim: G has an independent set E of size d /2d. Let A i A i obtained by removing one element of A i. Consider input x, x = e A i E i. We have, s(f, x) d /2d. 14

38 Our Results Theorem For every Boolean function f admitting the Block Property we have: bs(f ) 4s(f ) 2. 15

39 t Block Property A Boolean function is said to admit the t-block property if x X, we have: {A i x A i } t. 16

40 t Block Property A Boolean function is said to admit the t-block property if x X, we have: {A i x A i } t. Theorem Let f be a Boolean function admitting the t-block property, with t d, for some ε > 0. Then, we have the following: d 1+ε bs(f ) t (3s(f )) 1+ 1 ε. 16

41 A Computational Question Sensitivity Problem (S): Given a circuit C : {0, 1} n {0, 1}, x {0, 1} n, and blocks B 1,..., B k, the sensitivity problem is to find y {0, 1} n such that: s(c, y) bs(c, x, B 1,..., B k )/c. 17

42 A Computational Question Sensitivity Problem (S): Given a circuit C : {0, 1} n {0, 1}, x {0, 1} n, and blocks B 1,..., B k, the sensitivity problem is to find y {0, 1} n such that: s(c, y) bs(c, x, B 1,..., B k )/c. S is in Search-NP. 17

43 A Computational Question Sensitivity Problem (S): Given a circuit C : {0, 1} n {0, 1}, x {0, 1} n, and blocks B 1,..., B k, the sensitivity problem is to find y {0, 1} n such that: s(c, y) bs(c, x, B 1,..., B k )/c. S is in Search-NP. If (strong) Sensitivity conjecture is true then S is in TFNP. 17

44 A Computational Question Sensitivity Problem (S): Given a circuit C : {0, 1} n {0, 1}, x {0, 1} n, and blocks B 1,..., B k, the sensitivity problem is to find y {0, 1} n such that: s(c, y) bs(c, x, B 1,..., B k )/c. S is in Search-NP. If (strong) Sensitivity conjecture is true then S is in TFNP. Is S in P? 17

45 End of the Talk Thank you! 18