Symmetric Functions Capture General Functions
Symmetric Functions Capture General Functions
36th MFCS, 2011, EaGL Workshop 9/11/11
Richard J. Lipton
Georgia Tech
1
Kenneth W. Regan
2
University at Bualo (SUNY)
September 11, 2011
1
Research connected to blog Gödel's Lost Letter and
2
Ditto
3
Supported by NSF CAREER grant CCF-0844796
3
Atri Rudra
P = NP
Symmetric Functions Capture General Functions
Symmetric Functions Are...
Hard:
Parity 2= AC0 .
Majority is complete for TC0 .
Easy:
2 f 0; 1 gn , depend only on #1(x ).
ACC0 symm (quasi-poly many ^) (Beigel-Tarui)
Over
The
x
elementary symmetric functions
are easy even in
Zm
(Gromulsz).
Main Theorems: Senses in which every function f
complexity-equivalent to some symmetric function
is
g.
Why care? Symmetric functions have great algebraic structure.
Symmetric Functions Capture General Functions
Symmetric Functions Over Fields (And Rings R)
: Rn ! R is symmetric if for all permutations on [n ],
(x ) = f (x ).
Symmetric functions closed under +; .
N ! R and
Hence for any symmetric functions 1 ; : : : ; n : R1
0
n
N
0
polynomials f : R0
! R, the function f : R1 ! R0 is
f
f
symmetric, where
f 0 y1 ; : : : ; yN
(
Provided each
f
When does
Note: if
F
) = f (1 (~y ); : : : ; n (~y )):
i (y1 ; : : : ; yN ) is easy to compute, f 0 f .
f 0?
is a nite eld then every function from
polynomial.
Fn
to
F
is a
Symmetric Functions Capture General Functions
Fast Symmetrization
Goal: Compute f (a1 ; : : : ; an ) over R0 .
Given: Can compute f 0 (~b ) = f (1 (~b ); : : : ; n (~b )) for any b 2 R1N .
Task: Pick the i
~b 2 R1N such that
so that given any
a1
Then
So
f
f 0.
~a 2 R0n
one can eciently nd
= 1 (~b ); a2 = 2 (~b ); : : : ; an = n (~b )
( ) = f 0 (~b ):
f ~
a
Symmetric Functions Capture General Functions
Coding Via Symmetric Functions
We want
= (1 ; : : : ; n ), so that : R1N ! R0n , to be onto R0n
eciently
invertible
as well as computable.
Complexity considerations:
Size of
R1
If N
Degree
Time
and
N?
Dene
s
= 1 + logjR0 j (jR1 jN =jR0 jn ).
= n , and R0 is a eld F , then R1 can be the eld extension F
d0
of
f0
as a symmetric polynomial,vs. degree
( ) to invert , i.e. to compute
u n
1 (~a ) = ~b :
Time
and
( ) to compute .
t n
Two main constructions in paper give dierent tradeos.
d
of
f.
s
.
Symmetric Functions Capture General Functions
1. Elementary Symmetrization
The
elementary symmetric polynomials s1 ; s2 ; : : : ; sn
are dened by
(
si b1 ; : : : ; bn
)=
X Y
J [n ];jJ j=i j 2J
s1 (~
b ) = b1 + + bn ,
~
s2 (b ) = b1 b2 + + b1 bn + b2 b3 + + bn
: Rn ! R
bj :
So
sn
= b1 b2 bn .
1 bn ,
and
Form an algebra basis for all symmetric polynomials on
Idea is to dene the following, which gives degree
f 0 b1 ; : : : ; bn
(
By counting,
s
dlog2 n e
cannot
3.
d0
Rn .
= dn :
) = f (s1 (~b ); : : : ; sn (~b )):
have jR1 j = jR0 j = q , so s > 1. Theorem:
Symmetric Functions Capture General Functions
Simple Example
The 2
2 permanent polynomial ad + bc undergoes the substitutions
a 7! e + f + g + h
b 7! ef + eg + eh + fg + fh + gh
c 7! efg + efh + egh + fgh
d 7! efgh
to yield
e 2f 2g
e 2 fh 2
+
+ e 2 fg 2 + ef 2 g 2 + e 2 f 2 h + e 2 g 2 h + f 2 g 2 h
+ ef 2 h 2 + e 2 gh 2 + f 2 gh 2 + eg 2 h 2 + fg 2 h 2
+4e 2 fgh + 4ef 2 gh + 4efg 2 h + 4efgh 2
Symmetric Functions Capture General Functions
Elementary Facts
Y( + ) =
For a formal single variable
n
i =1
x
bi
x,
xn
+
X
n
si b1 ; : : : ; bn x i 1 :
(
i =1
)
(1)
Fact: All si (~b ) are computed in O (n (log n )2 ) time by using FFT
to multiply out the product on the left-hand side of (1).
(a1 ; : : : ; an ), we want ~b = (b1 ; : : : ; bn ) such
that for each i , ai = si (~
b ). Dene
For inversion, given
= ~a (x ) = x n +
Y
= (
By fact (1), our goal is to split
ai
n
i =1
ai x i 1 :
into linear factors:
i
This will make
X
x
+ bi ):
= si (~b ) for each i .
Symmetric Functions Capture General Functions
Splitting Can Be Hard to Do
The problem is that
may notindeed by the counting, generally will
notsplit into linear factors over
R1
to be an extension
Fs.
R0 .
We need
How large must
s
R0
to be a eld
F,
and
be?
Lemma (well-known)
The minimum s equals the least common multiple of the degrees of
all irreducible factors of over F .
Alas, this
s
p
n O ( n ),
can be as high as
making the extension eld
elements themselves have exponential size.
Theorem (also known)
Pr~
a
2F [log s > log2 n ] < 2
n
p
( log n ) .
Thus there are exp-few bad
~a
that make
s
larger than
n O (log n ) .
Symmetric Functions Capture General Functions
Quasi-Good Randomized Algorithm
The theorem gives various deterministic and randomized
quasi-poly(n ) time algorithms that work on all except the bad
~a
arguments.
To get correctness on
Take a random slope
all ,
m
~
we employ one more randomization.
for a line through
~a
and dene
( ) = f (a1 + m1 y ; a2 + m2 y ; : : : ; an + mn y ):
of at least 3d + 3 points on this line will contain relatively
P~a y
A set
S
few bad points.
Using
S
and polynomial interpolation, can recover
Theorem (paper has more-general form)
( )
If the symmetric function f is in time v n , then
f
RTIME dv n n O (log n ) q O (1) . 2
[ ( )+
]
( ) = P~a (0).
f ~
a
Symmetric Functions Capture General Functions
2. Second Symmetrization
Can we do better than quasi-polynomial time overhead?
Answer is yes, but degree of
f0
becomes higher:
Still needs an extension eld, but
s
d0
1 + dlogq n e.
= q 2 dn logq n .
Less algebraically simple to dene, but running time basically
cannot be beat:
Theorem
Every function f
f 0 : Fn
! Fq
: Fqn ! Fq
is equivalent to a symmetric function
~( )
with above parameters, up to O n deterministic
time complexity (plus poly q ; s pre-processing to represent Fq s ).
q
s
Note that
f0
(
)
maps from the extension eld into the original eld.
Symmetric Functions Capture General Functions
Idea: How to encode information symmetrically?
Recall the task is to pick symmetric
can eciently
nd ~
b 2 R1N
a1
so that
i
so that given any
~a 2 R0n
= 1 (~b ); a2 = 2 (~b ); : : : ; an = n (~b )
f 0 b1 ; : : : ; bn
(
) = f (1 (~b ); : : : ; n (~b )):
Idea is to encode bi = hi ; ai i. In general we have pairs hj ; a i.
know which index
function
one
such that
i (j ).
j
gives us
Then each
ai
ai ?
How do we
We need to create a Kronecker delta
can be represented symmetrically as a sum
ai
=
X ( )
n
j =1
i j aj :
Over nite elds, all this can be done with polynomials.
Symmetric Functions Capture General Functions
Proof of Second Main Theorem
Pre-process to represent
formal root
Fq s
by an irreducible polynomial with
, giving every element of the extension eld as
=
By choice of
X = (
0
`=s 1
s
s , n q 1 , so
`
`
s 1 ; : : : ; 0 ):
embed
[n ] into rst s
1 places.
k that project out the k -th place:
k () = k :
To do so, dene V to be the Vandermonde matrix whose row `,
q`
0 ` s
1, comprises the rst s powers of . Then using
Next construct polynomials
column vectors,
V
so
k
(s 1 ; : : : ; 0 ) = (q 1 ; : : : ; q 2 ; q ; );
is obtained by invering
s
V
and dotting its
right-hand side. Use polynomial closed-form for
k -th row with the
V 1 to get k .
Symmetric Functions Capture General Functions
Key Coding Lemma
Abbreviate
Fq s
to
E
and
Fq
to
F,
and let
stand for
co-ordinate, which may be an embedded value in
Lemma
[n ].
minus its 0
2[ ]
For each j
n we can construct a symmetric polynomial
n
j E
F of degree at most sq s such that for any elements
1
n
; : : : ; in E n ,
:
!
j (1 ; : : : ; n ) =
X
i 2[n ]: =j
i0 :
i
The proof picks apart
embedded value in
j
into the
F s 1.
s
1 co-ordinates
First idea is to represent the Kronecker delta
function on the embedded values, namely
otherwise.
(js 1 ; : : : ; j1 ) of its
j (i ) = 1 if i = j
and 0
Symmetric Functions Capture General Functions
Kronecker Delta and Place Picker
This formula makes
j (j ) = 1 since the fractions are identically 1:
j (us 1 ; : : : ; u1 ) =
And
j (i ) = 0 for i 6= j
Y Y
s 1
u`
`=1 2F nf j` g j`
because the numerator hits a zero. Now dene:
j (z1 ; : : : ; zn ) =
X (
n
i =1
j
s 1 (zi ); : : : ; 1 (zi )) 0 (zi ):
i0 for which the rst s
j , thus proving the lemma's equation. Moreover j
This picks out only those
proving the lemma.
:
1 co-ordinates yield
is symmetric, thus
Symmetric Functions Capture General Functions
Completing the Construction
Finally we dene
Since each
qs 1
: E n ! F by
f 0 (~
b ) = f (1 (~
b ); 2 (~
b ); : : : ; n (~
b ):
sq s , and s is chosen to make
2
has degree at most sndq .
f 0 from f , one linear scan of ~
b can identify all the
j
nq , f 0
To compute
f0
has degree at most
terms
that will contribute to the sums in the Lemma, giving the arguments of
f.
To compute
~b
( ) with arguments from the base eld F , we need to nd
j (~b ) = aj , and nd it eciently.
f ~
a
over the extension eld such that
This is done by using the embedded natural numbers, which pick out
indices, as co-ordinates:
= (is 1 ; : : : ; i1 ; ai ):
Then for all j , j (~
b ) = 0 (bj ) = aj , as needed. This is done in O (sn )
~ (n ) time overall. time treating entries as units, which gives O
bi
Symmetric Functions Capture General Functions
Innite Fields?
The elementary symmetrization works over any eld.
The second one does not, because the coding tricks require nite
elds.
Dierent coding tricks work over the reals or complex numbers, but
do not yield polynomials.
Paper gives a result over the reals.
Symmetric Functions Capture General Functions
Open Questions
Can we prove that no symmetrization by polynomials over an
innite eld gives
~ ( ) time?
O n
Can the possibility
N
>n
be used to improve either
symmetrization?
Can either symmetrization be used in a positive way to enable
more-structured analysis of, say, symmetrized permanent
polynomials?
Can the idea be used to derive more (conditional) lower bounds?
Are elds needed? What can be done over the rings
composite?
Zm
for
m