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

On the degree of symmetric functions on

the Boolean cubeJoint work with Amir

Shpilka

The basic question of complexity

: 0,1 0,1n

f

The basic question of complexity

: 0,1 0,1n

f

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

The basic question of complexity

: 0,1 0,1n

f

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…

Polynomials as computers

: 0,1 0,1n

f

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.

Polynomials as computers

: 0,1 0,1n

f

1, , np x x

0,1n

x p x f x

Our model of computation – Polynomials.

Polynomials as computers

: 0,1 0,1n

f

1, , np x x

0,1n

x p x f x

Our model of computation – Polynomials.

Complexity of is deg degf f p

Tight lower boundNisan and Szegedy (94) proved

assuming f depend on all n variables.

2deg log log logf n O n

Tight lower boundNisan and Szegedy (94) proved

assuming f depend on all n variables.

Can we get stronger lower bounds on more restricted natural classes of functions?

2deg log log logf n O n

Symmetric Boolean functions

1 1, , , ,n n nS f x x f x x

Symmetric Boolean functions

Von zur Gathen and Roche (97) proved

assuming f is non-constant.

1 1, , , ,n n nS f x x f x x

0.525deg f n O n

Symmetric Boolean functions

1 1, , n nf x x F x x

Symmetric Boolean functions

1 1, , n nf x x F x x

: 0,1 0,1 : symmetric : 0,1, , 0,1n

f f F n

Symmetric Boolean functions

: 0,1, , 0,1,2, ,f n c

0 1 2 3 4 5 6 . . . n012

.

.

.

c

Symmetric Boolean functions

: 0,1, , 0,1,2, ,f n c

0 1 2 3 4 5 6 . . . n012

.

.

.

c

Symmetric Boolean functions

: 0,1, , 0,1,2, ,f n c

deg fWhat can be said about ?

0 1 2 3 4 5 6 . . . n012

.

.

.

c

Symmetric functions

: 0,1, , 0,1,2, ,f n c

deg fWhat can be said about ?

For c=1 we got

For c=n the function has degree 1.

deg f n o n

f k k

Symmetric functions

: 0,1, , 0,1,2, ,f n c

deg fWhat can be said about ?

For c=1 we got

For c=n the function has degree 1.

How does the degree behaves?

deg f n o n

f k k

Symmetric functions

Von zur Gathen and Roche noted that

1deg

1

nf

c

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.

1deg

1

nf

c

/ 2c n

Relative degree

: 0,1, , 0,1,2, ,f n c

Define

1min deg : as abovecD n f fn

Relative degree

: 0,1, , 0,1,2, ,f n c

Define

is monotone decreasing in c.

1min deg : as abovecD n f fn

cD n

Relative degree

: 0,1, , 0,1,2, ,f n c

Define

is monotone decreasing in c.

has a crazy behavior in n.

1min deg : as abovecD n f fn

cD n

cD n

Relative degree

: 0,1, , 0,1,2, ,f n c

Define

is monotone decreasing in c.

has a crazy behavior in n.

1min deg : as abovecD n f fn

cD n

cD n

1

1cD nc

6 stages of first-time research

Stage 1

6 stages of first-time research

Stage 2

6 stages of first-time research

Stage 3

6 stages of first-time research

Stage 4

6 stages of first-time research

Stage 5

6 stages of first-time research

Stage 6

6 stages of first-time research

Stage

1…

Our main result

1

91

22nD n o

Main theorem

This proves a threshold behavior at c=n.

Main theorem

This proves a threshold behavior at c=n.

Yet another theorem

Our main result

1

91

22nD n o

: 0,1,..., , 1f n C C O

2deg

3f n o n

Proof strategy – reducing c

2n o n p n Lemma 1. For any n there exist a prime p such that and

1 4

11

2nD n D p o

Proof strategy – reducing c

2n o n p n Lemma 1. For any n there exist a prime p such that and

Together with the trivial bound , we already get a threshold behavior

1 4

11

2nD n D p o

4 1/ 5D p

1

11

10nD n o

Proof strategy – reducing nn mLemma 2. For every c,m,n such that , it

holds that

c cD n D m

Dream

version

Proof strategy – reducing nn mLemma 2. For every c,m,n such that , it

holds that

1c cD n D m o

Dream

version

Proof strategy – reducing nn mLemma 2. For every c,m,n such that , it

holds that

11c c

mD n D m o

m

Dream

version

Proof strategy – reducing n2mn cLemma 2. For every c,m,n such that , it

holds that

11c c

mD n D m o

m

Proof of the main theorem

A computer search found that .By Lemma 2

By Lemma 1

4 21 6 / 7D

4

21 6 91 1

21 1 7 11D n o o

1 4

1 1 9 91 1 1

2 2 11 22nD n D p o o o

Periodicity and degree

Low degree Strong periodical structure

Dream

version

Periodicity and degree

Low degree Strong periodical structure

Strong periodical structure High degree

Dream

version

Periodicity and degree

Low degree Strong periodical structure

Strong periodical structure High degree

Hence no function has “to low” degree.

Dream

version

Periodicity and degree

Low degree Strong periodical structure

Strong periodical structure High degree

Not the same sense of periodical structure…

Low degree implies strong periodical structure

pf p j f j

Lemma 3. Let with . Let be a prime number. Then for all such that it holds that

: 0,1, , 0,1, ,f n c

deg f d d p n 0 j d

0 1 2 3 . . . d . . . p

01

.

.

c

n

p j n

Low degree implies strong periodical structure

pf p j f j

Lemma 3. Let with . Let be a prime number. Then for all such that it holds that

: 0,1, , 0,1, ,f n c

deg f d d p n 0 j d

0 1 2 3 . . . d . . . p q

01

.

.

c

n

p j n

Low degree implies strong periodical structure

pf p j f j

Lemma 3. Let with . Let be a prime number. Then for all such that it holds that

: 0,1, , 0,1, ,f n c

deg f d d p n 0 j d

0 1 2 3 . . . d . . . p q r

01

.

.

c

n

p j n

Strong periodical structure implies high degree

0 :TP f k n T f k f k T

Definition. Let and define

: 0,1, , 0,1, ,f n c 1T

Strong periodical structure implies high degree

0 :TP f k n T f k f k T

Definition. Let and define

: 0,1, , 0,1, ,f n c

, 10 T

Lemma 4. Let . Then for all

If then

If then or

Strong periodical structure implies high degree

: 0,1, , 0,1, ,f n c

0 :TP f k n T f k f k T

Definition. Let and define

: 0,1, , 0,1, ,f n c

0, 1T

0

0

1T

deg Tf P f

deg Tf P f deg 1f

Proof of Lemma 1

2n o n p n Lemma 1. For any n there exist a prime p such that and

1 4

11

2nD n D p o

Proof of Lemma 1

0 1 2 . . . n012

.

.

.

n-1

f

Proof of Lemma 1

0 1 2 . . . p . . . 2p n012

.

.

.

n-1

o(n)

f

Proof of Lemma 1

0 1 2 . . . p . . . 2p n012

.

.

.

n-1

o(n)

f

Proof of Lemma 1

0 1 2 . . . p . . . 2p n012

.

.

.

n-1

o(n)

We might as well assume that

non-constantf

f

Proof of Lemma 1

0 1 2 . . . p . . . 2p n012

.

.

.

n-1

o(n)

We might as well assume that

non-constant deg f pf

f

Proof of Lemma 1

Define

2 0,1,...,f p k f k

g k k pp

Proof of Lemma 1

Define

From Lemma 3

2 0,1,...,f p k f k

g k k pp

0,1,...,pf p k f k k p

Proof of Lemma 1

Define

From Lemma 3

and also

2 0,1,...,f p k f k

g k k pp

0,1,...,pf p k f k k p

, 0,1, 2,..., 1 2f p k f k n n p o p

Proof of Lemma 1

0,1,...,pf p k f k k p

From Lemma 3

Hence : 0,1, , 0,1, 2,3, 4g p

Proof of Lemma 1

Case 1: g is a non-constant

and we are done.

2f p k f k

g kp

1 4 4deg deg deg2n

nn D n f f g p D p o n D p

Proof of Lemma 1

Case 2: g is a constant G

Hence , by Lemma 4

2f p k f k

g kp

21 deg deg

2G p

n p

nn D n f f P f p o n

2 0,1,...,f p k f k G p k p

Proof of Lemma 1

Case 2: g is a constant G

Hence , by Lemma 4

or is linear.

2f p k f k

g kp

2 0,1,...,f p k f k G p k p

f

21 deg deg

2G p

n p

nn D n f f P f p o n

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.

2f p k f k

g kp

Rf k f n k

f

deg deg Rf fRf

Open Questions Main question - Better understand .

Improve the lower bounds to non-linear, if possible.

cD n

Thank you!