Presentation is loading. Please wait.

Presentation is loading. Please wait.

On the degree of symmetric functions on the Boolean cube Joint work with Amir Shpilka.

Similar presentations


Presentation on theme: "On the degree of symmetric functions on the Boolean cube Joint work with Amir Shpilka."— Presentation transcript:

1 On the degree of symmetric functions on the Boolean cube Joint work with Amir Shpilka

2 The basic question of complexity

3 How complex is it (how hard it is to compute f?)

4 The basic question of complexity How complex is it (how hard it is to compute f?) That depends on the computational model at hand. e.g. Turing machines, Circuits, Decision trees, etc…

5 Polynomials as computers How complex is it (how hard it is to compute f?) That depends on the computational model at hand. e.g. Turing machines, Circuits, Decision trees, etc… Our model of computation – Polynomials.

6 Polynomials as computers Our model of computation – Polynomials.

7 Polynomials as computers Our model of computation – Polynomials.

8 Tight lower bound Nisan and Szegedy (94) proved assuming f depend on all n variables.

9 Tight lower bound Nisan and Szegedy (94) proved assuming f depend on all n variables. Can we get stronger lower bounds on more restricted natural classes of functions?

10 Symmetric Boolean functions

11 Von zur Gathen and Roche (97) proved assuming f is non-constant.

12 Symmetric Boolean functions

13

14 0 1 2 3 4 5 6... n 0 1 2....... c

15 Symmetric Boolean functions 0 1 2 3 4 5 6... n 0 1 2....... c

16 Symmetric Boolean functions What can be said about ? 0 1 2 3 4 5 6... n 0 1 2....... c

17 Symmetric functions What can be said about ? For c=1 we got For c=n the function has degree 1.

18 Symmetric functions What can be said about ? For c=1 we got For c=n the function has degree 1. How does the degree behaves?

19 Symmetric functions Von zur Gathen and Roche noted that

20 Symmetric functions Von zur Gathen and Roche noted that In particular, even for this observation doesn’t exclude the existence of a parabola interpolating on some function.

21 Relative degree Define

22 Relative degree Define is monotone decreasing in c.

23 Relative degree Define is monotone decreasing in c. has a crazy behavior in n.

24 Relative degree Define is monotone decreasing in c. has a crazy behavior in n.

25 6 stages of first-time research Stage 1

26 6 stages of first-time research Stage 2

27 6 stages of first-time research Stage 3

28 6 stages of first-time research Stage 4

29 6 stages of first-time research Stage 5

30 6 stages of first-time research Stage 6

31 6 stages of first-time research Stage 1…

32 Our main result Main theorem This proves a threshold behavior at c=n.

33 Main theorem This proves a threshold behavior at c=n. Yet another theorem Our main result

34 Proof strategy – reducing c Lemma 1. For any n there exist a prime p such that and

35 Proof strategy – reducing c Lemma 1. For any n there exist a prime p such that and Together with the trivial bound, we already get a threshold behavior

36 Proof strategy – reducing n Lemma 2. For every c,m,n such that, it holds that Dream version

37 Proof strategy – reducing n Lemma 2. For every c,m,n such that, it holds that Dream version

38 Proof strategy – reducing n Lemma 2. For every c,m,n such that, it holds that Dream version

39 Proof strategy – reducing n Lemma 2. For every c,m,n such that, it holds that

40 Proof of the main theorem A computer search found that. By Lemma 2 By Lemma 1

41 Periodicity and degree Low degree Strong periodical structure Dream version

42 Periodicity and degree Low degree Strong periodical structure Strong periodical structure High degree Dream version

43 Periodicity and degree Low degree Strong periodical structure Strong periodical structure High degree Hence no function has “to low” degree. Dream version

44 Periodicity and degree Low degree Strong periodical structure Strong periodical structure High degree Not the same sense of periodical structure…

45 Low degree implies strong periodical structure Lemma 3. Let with. Let be a prime number. Then for all such that it holds that 0 1 2 3... d... p 0 1. c n

46 Low degree implies strong periodical structure Lemma 3. Let with. Let be a prime number. Then for all such that it holds that 0 1 2 3... d... p q 0 1. c n

47 Low degree implies strong periodical structure Lemma 3. Let with. Let be a prime number. Then for all such that it holds that 0 1 2 3... d... p q r 0 1. c n

48 Strong periodical structure implies high degree Definition. Let and define

49 Strong periodical structure implies high degree Definition. Let and define

50 Lemma 4. Let. Then for all If then If then or Strong periodical structure implies high degree Definition. Let and define

51 Proof of Lemma 1 Lemma 1. For any n there exist a prime p such that and

52 Proof of Lemma 1 0 1 2... n 0 1 2....... n-1

53 Proof of Lemma 1 0 1 2... p... 2p n 0 1 2....... n-1 o(n)

54 Proof of Lemma 1 0 1 2... p... 2p n 0 1 2....... n-1 o(n)

55 Proof of Lemma 1 0 1 2... p... 2p n 0 1 2....... n-1 o(n) We might as well assume that non-constant

56 Proof of Lemma 1 0 1 2... p... 2p n 0 1 2....... n-1 o(n) We might as well assume that non-constant

57 Proof of Lemma 1 Define

58 Proof of Lemma 1 Define From Lemma 3

59 Proof of Lemma 1 Define From Lemma 3 and also

60 Proof of Lemma 1 From Lemma 3 Hence

61 Proof of Lemma 1 Case 1: g is a non-constant and we are done.

62 Proof of Lemma 1 Case 2: g is a constant G Hence, by Lemma 4

63 Proof of Lemma 1 Case 2: g is a constant G Hence, by Lemma 4 or is linear.

64 Proof of Lemma 1 Case 2: If happens to be linear, apply the proof so far on. Since we are done unless it also happens that is linear. But this means f itself must be linear. Since f is not constant it means f assumes n+1 distinct values – a contradiction.

65 Open Questions Main question - Better understand. Improve the lower bounds to non-linear, if possible.

66 Thank you!


Download ppt "On the degree of symmetric functions on the Boolean cube Joint work with Amir Shpilka."

Similar presentations


Ads by Google