Embed Size (px)
Citation preview
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!
