Download Polynomial Identity

Complexity
23-1
More Probabilistic
Algorithms
Complexity
Andrei Bulatov
Complexity
Polynomial Identity
As we know, Primes can be solved in polynomial time. Therefore,
the probabilistic algorithm for this problem has great practical
significance, but no “theoretical” interest.
For the following problem, no polynomial time deterministic algorithm
is known yet
Polynomial Identity
Instance: Two polynomials P and G in variables x1 , x2 ,, xn
Question: Is P identically equal to G?
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Complexity
Representation of Polynomials
If P and G are represented explicitly, then the problem makes no sense:
Fact
Two polynomials on N are identical if and only if their
coefficients are equal.
The polynomials are given implicitly
There is a polynomial time algorithm that, given a1 , a2 ,, an
computes P(a1 , a2 ,, an ) and G(a1 , a2 ,, an )
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Complexity
23-4
Example
 A11

A21

Eigenvalues of a matrix A 
 

 Ak 1
the determinant
A11  x
A21

Ak 1
A12
A22  x

Ak 2
A12
A22

Ak 2








A1k 

A2 k 
are the roots of
 

Akk 
A1k
A2 k

Akk  x
This determinant can be computed explicitly (using Gaussian elimination),
but the time complexity of this algorithm is O ( n 3 ) under the assumption
that operations on polynomials are done in one step; which is unrealistic
Complexity
Algorithm
Clearly, it is enough to check if P( x1 , x2 ,, xn )  G( x1 , x2 ,, xn )  0
Thus, we assume that G( x1 , x2 ,, xn ) is 0
Let d be the degree of P and N = 3d
On the input P of degree d
• choose randomly a1 , a2 ,, an such that 0  ai  N
• if P(a1 , a2 ,, an )  0 then accept; otherwise reject
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Complexity
Analysis
We prove that
• if P is equal 0 identically then Pr[P accepted] = 1
• if P is not equal 0 identically then Pr[P accepted]  1/3
The first part is obvious.
The second part is based on the following theorem
Theorem (The Main Theorem of Algebra)
A polynomial F(x) of degree d has at most d roots.
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Complexity
23-7
Suppose first that P is a polynomial in one variable
Then if P does not equal 0, then P(a)  0 for at least 2/3 of the
numbers a 1,2,…,3d
Let P(x,y) be in two variables
Then, for any 0  a  3d, the
polynomial P(a,y) is a
polynomial in one variable.
y
x
0
a
This polynomial has at most d
roots in the interval
1,2,…,3d
3d
Therefore, P(a,b)  0 for at least 2/3 of the numbers b 1,2,…,3d
Complexity
Random Walk
We have seen random walk algorithms for graphs
Similar algorithms are possible for other problems. For example for SAT
Given a CNF 
• take an arbitrary truth assignment T
• for i = 1 to k do
- if T is a satisfying assignment then accept
- otherwise take any unsatisfied clause (all literals in this clause
are false)
- randomly flip a literal from this clause updating T
• reject
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Complexity
Analysis
If the formula is unsatisfiable, then it cannot be accepted
A satisfiable formula can be accepted and can be rejected
How large must the length of the walk be to guarantee significant
probability of success?
Lemma
If  is a satisfying formula with n variables and k  ln 2  2n
then Pr[ rejected]  1/2.
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Complexity
23-10
Proof
If  is satisfiable, it has at least 1 satisfying assignment.
Therefore the probability that a random assignment is satisfying is at least 2n
The probability that the algorithm fails after one attempt is at most 1  2n
and after k attempts at most (1  2 n ) k
We need a k such that
1
2
k  ln(1  2 n )   ln 2
(1  2 n ) k 
It is known that | log(1  x) | x
ln 2
ln 2

2 n | ln(1  2 n ) |
Thus we may take any k  ln 2  2n
k
ln 2
| ln(1  2 n ) |
Complexity
2-SAT
Theorem
If  is a satisfiable 2-CNF with n variables and k  2n 2
then Pr[ rejected]  1/2.
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Complexity
3-SAT
For any n, we consider CNF  n consisting of the following clauses
•
( X1 )  ( X 2 )  ( X n )
• for every different 1  i, j, k  n, the clause ( X i  X j  X k )
Lemma
For any polynomial p(n), any n, and k = p(n),
Pr[  n rejected]  1  3  n.
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