How To Test If a Polynomial Is Identically Zero? •

How To Test If a Polynomial Is Identically Zero?

• det(A^G) is a polynomial in n² variables.

• There are exponentially many terms in det(A^G).

• Expanding the determinant polynomial is not feasible.

– Too many terms.

• If det(A^G) ≡ 0, then it remains zero if we substitute arbitrary integers for the variables x₁₁, . . . , x_nn.

• But what is the likelihood of obtaining a zero when det(A^G) ̸≡ 0?

Number of Roots of a Polynomial

Lemma 59 (Schwartz (1980)) Let p(x₁, x₂, . . . , x_m) ̸≡ 0 be a polynomial in m variables each of degree at most d. Let M ∈ Z⁺. Then the number of m-tuples

(x₁, x₂, . . . , x_m) ∈ {0, 1, . . . , M − 1}^m such that p(x₁, x₂, . . . , x_m) = 0 is

≤ mdM^m−1.

• By induction on m (consult the textbook).

Density Attack

• The density of roots in the domain is at most mdM^m−1

M^m = md

M . (8)

• So suppose p(x¹, x₂, . . . , x_m) ̸≡ 0.

• Then a random

(x₁, x₂, . . . , x_m) ∈ { 0, 1, . . . , M − 1 }^m has a probability of ≤ md/M of being a root of p.

• Note that M is under our control!

– One can raise M to lower the error probability, e.g.

Density Attack (concluded)

Here is a sampling algorithm to test if p(x₁, x₂, . . . , x_m) ̸≡ 0.

1: Choose i₁, . . . , i_m from {0, 1, . . . , M − 1} randomly;

2: if p(i₁, i₂, . . . , i_m) ̸= 0 then

3: return “p is not identically zero”;

4: else

5: return “p is (probably) identically zero”;

6: end if

A Randomized Bipartite Perfect Matching Algorithm

We now return to the original problem of bipartite perfect matching.

1: Choose n² integers i₁₁, . . . , i_nn from {0, 1, . . . , 2n² − 1}

randomly; {So M = 2n².}

2: Calculate det(A^G(i₁₁, . . . , i_nn)) by Gaussian elimination;

3: if det(A^G(i₁₁, . . . , i_nn)) ̸= 0 then

4: return “G has a perfect matching”;

5: else

6: return “G has no perfect matchings”;

7: end if

aLov´asz (1979). According to Paul Erd˝os, Lov´asz wrote his first sig- nificant paper “at the ripe old age of 17.”

Analysis

• If G has no perfect matchings, the algorithm will always be correct as det(A^G(i₁₁, . . . , i_nn)) = 0.

• Suppose G has a perfect matching.

– The algorithm will answer incorrectly with

probability at most md/M = 0.5 with m = n², d = 1 and M = 2n² in Eq. (8) on p. 473.

– Run the algorithm independently k times.

– Output “G has no perfect matchings” if and only if all say no.

– The error probability is now reduced to at most 2^−k.

L´ oszl´ o Lov´ asz (1948–)

Remarks

• Note that we are calculating

prob[ algorithm answers “no”| G has no perfect matchings ], prob[ algorithm answers “yes”| G has a perfect matching ].

• We are not calculating^b

prob[ G has no perfect matchings| algorithm answers “no” ], prob[ G has a perfect matching| algorithm answers “yes” ].

aThanks to a lively class discussion on May 1, 2008.

bNumerical Recipes in C (1988), “[As] we already remarked, statistics is not a branch of mathematics!”

But How Large Can det(A

(i

, . . . , i

)) Be?

• It is at most

n! (

2n²)n

.

• Stirling’s formula says n! ∼ √

2πn (n/e)ⁿ.

• Hence

log₂ det(A^G(i₁₁, . . . , i_nn)) = O(n log₂ n) bits are sufficient for representing the determinant.

• We skip the details about how to make sure that all intermediate results are of polynomial sizes.

An Intriguing Question

• Is there an (i¹¹, . . . , i_nn) that will always give correct answers for the algorithm on p. 475?

• A theorem on p. 571 shows that such an (i¹¹, . . . , i_nn) exists!

– Whether it can be found efficiently is another matter.

• Once (i¹¹, . . . , i_nn) is available, the algorithm can be made deterministic.

aThanks to a lively class discussion on November 24, 2004.

Randomization vs. Nondeterminism

• What are the differences between randomized algorithms and nondeterministic algorithms?

• One can think of a randomized algorithm as a

nondeterministic algorithm but with a probability associated with every guess/branch.

• So each computation path of a randomized algorithm has a probability associated with it.

aContributed by Mr. Olivier Valery (D01922033) and Mr. Hasan Al- hasan (D01922034) on November 27, 2012.

Monte Carlo Algorithms

• The randomized bipartite perfect matching algorithm is called a Monte Carlo algorithm in the sense that

– If the algorithm finds that a matching exists, it is always correct (no false positives).

– If the algorithm answers in the negative, then it may make an error (false negatives).

aMetropolis and Ulam (1949).

Monte Carlo Algorithms (concluded)

• The algorithm makes a false negative with probability

≤ 0.5.^a

– Note this probability refers to^b

prob[ algorithm answers “no”| G has a perfect matching ] not

prob[ G has a perfect matching| algorithm answers “no” ].

• This probability is not over the space of all graphs or determinants, but over the algorithm’s own coin flips.

– It holds for any bipartite graph.

aEquivalently, among the coin flip sequences, at most half of them lead to the wrong answer.

bIn general, prob[ algorithm answers “no”| input is a “yes” instance ].

The Markov Inequality

Lemma 60 Let x be a random variable taking nonnegative integer values. Then for any k > 0,

prob[ x ≥ kE[ x ] ] ≤ 1/k.

• Let pⁱ denote the probability that x = i.

E[ x ] = ∑

i

ip_i

= ∑

i<kE[ x ]

ip_i + ∑

i≥kE[ x ]

ip_i

≥ kE[ x ] × prob[x ≥ kE[ x ]].

aAndrei Andreyevich Markov (1856–1922).

Andrei Andreyevich Markov (1856–1922)

An Application of Markov’s Inequality

• Suppose algorithm C runs in expected time T (n) and always gives the right answer.

• Consider an algorithm that runs C for time kT (n) and rejects the input if C does not stop within the time bound.

• By Markov’s inequality, this new algorithm runs in time kT (n) and gives the wrong answer with probability

≤ 1/k.

An Application of Markov’s Inequality (concluded)

• By running this algorithm m times (the total running time is mkT (n)), we reduce the error probability to

≤ k^−m.^a

• Suppose, instead, we run the algorithm for the same running time mkT (n) once and rejects the input if it does not stop within the time bound.

• By Markov’s inequality, this new algorithm gives the wrong answer with probability ≤ 1/(mk).

• This is much worse than the previous algorithm’s error probability of ≤ k^−m for the same amount of time.

aWith the same input. Thanks to a question on December 7, 2010.

fsat for k-sat Formulas (p. 453)

• Let ϕ(x¹, x₂, . . . , x_n) be a k-sat formula.

• If ϕ is satisfiable, then return a satisfying truth assignment.

• Otherwise, return “no.”

• We next propose a randomized algorithm for this problem.

A Random Walk Algorithm for ϕ in CNF Form

1: Start with an arbitrary truth assignment T ;

2: for i = 1, 2, . . . , r do

3: if T |= ϕ then

4: return “ϕ is satisfiable with T ”;

5: else

6: Let c be an unsatisfied clause in ϕ under T ; {All of its literals are false under T .}

7: Pick any x of these literals at random;

8: Modify T to make x true;

9: end if

10: end for

11: return “ϕ is unsatisfiable”;

3sat vs. 2sat Again

• Note that if ϕ is unsatisfiable, the algorithm will not refute it.

• The random walk algorithm needs expected exponential time for 3sat.

– In fact, it runs in expected O((1.333· · · + ϵ)ⁿ) time with r = 3n,^a much better than O(2ⁿ).^b

• We will show immediately that it works well for 2sat.

• The state of the art as of 2006 is expected O(1.322ⁿ) time for 3sat and expected O(1.474ⁿ) time for 4sat.^c

aUse this setting per run of the algorithm.

bSch¨oning (1999).

cKwama and Tamaki (2004); Rolf (2006).

Random Walk Works for 2sat

Theorem 61 Suppose the random walk algorithm with r = 2n² is applied to any satisfiable 2sat problem with n variables. Then a satisfying truth assignment will be

discovered with probability at least 0.5.

• Let ˆT be a truth assignment such that ˆT |= ϕ.

• Assume our starting T differs from ˆT in i values.

– Their Hamming distance is i.

– Recall T is arbitrary.

aPapadimitriou (1991).

The Proof

• Let t(i) denote the expected number of repetitions of the flipping step^a until a satisfying truth assignment is

found.

• It can be shown that t(i) is finite.

• t(0) = 0 because it means that T = ˆT and hence T |= ϕ.

• If T ̸= ˆT or any other satisfying truth assignment, then we need to flip the coin at least once.

• We flip a coin to pick among the 2 literals of a clause not satisfied by the present T .

• At least one of the 2 literals is true under ˆT because ˆT satisfies all clauses.

aThat is, Statement 7.

The Proof (continued)

• So we have at least 0.5 chance of moving closer to ˆT .

• Thus

t(i) ≤ t(i − 1) + t(i + 1)

2 + 1

for 0 < i < n.

– Inequality is used because, for example, T may differ from ˆT in both literals.

• It must also hold that

t(n) ≤ t(n − 1) + 1 because at i = n, we can only decrease i.

The Proof (continued)

• Now, put the necessary relations together:

t(0) = 0, (9)

t(i) ≤ t(i − 1) + t(i + 1)

2 + 1, 0 < i < n, (10)

t(n) ≤ t(n − 1) + 1. (11)

• Technically, this is a one-dimensional random walk with an absorbing barrier at i = 0 and a reflecting barrier at i = n (if we replace “≤” with “=”).^a

aThe proof in the textbook does exactly that. But a student pointed out difficulties with this proof technique on December 8, 2004. So our proof here uses the original inequalities.

The Proof (continued)

• Add up the relations for

2t(1), 2t(2), 2t(3), . . . , 2t(n − 1), t(n) to obtain^a 2t(1) + 2t(2) + · · · + 2t(n − 1) + t(n)

≤ t(0) + t(1) + 2t(2) + · · · + 2t(n − 2) + 2t(n − 1) + t(n) +2(n − 1) + 1.

• Simplify it to yield

t(1) ≤ 2n − 1. (12)

aAdding up the relations for t(1), t(2), t(3), . . . , t(n−1) will also work, thanks to Mr. Yen-Wu Ti (D91922010).

The Proof (continued)

• Add up the relations for 2t(2), 2t(3), . . . , 2t(n − 1), t(n) to obtain

2t(2) + · · · + 2t(n − 1) + t(n)

≤ t(1) + t(2) + 2t(3) + · · · + 2t(n − 2) + 2t(n − 1) + t(n) +2(n − 2) + 1.

• Simplify it to yield

t(2) ≤ t(1) + 2n − 3 ≤ 2n − 1 + 2n − 3 = 4n − 4 by Eq. (12) on p. 495.

The Proof (continued)

• Continuing the process, we shall obtain t(i) ≤ 2in − i².

• The worst upper bound happens when i = n, in which case

t(n) ≤ n².

• We conclude that

t(i) ≤ t(n) ≤ n² for 0 ≤ i ≤ n.

The Proof (concluded)

• So the expected number of steps is at most n².

• The algorithm picks r = 2n².

– This amounts to invoking the Markov inequality

(p. 484) with k = 2, resulting in a probability of 0.5.^a

• The proof does not yield a polynomial bound for 3sat.^b

aRecall p. 486.

bContributed by Mr. Cheng-Yu Lee (R95922035) on November 8, 2006.

Christos Papadimitriou (1949–)

Boosting the Performance

• We can pick r = 2mn² to have an error probability of

≤ 1 2m by Markov’s inequality.

• Alternatively, with the same running time, we can run the “r = 2n²” algorithm m times.

• The error probability is now reduced to

≤ 2^−m.

Primality Tests

• primes asks if a number N is a prime.

• The classic algorithm tests if k | N for k = 2, 3, . . . ,√ N .

• But it runs in Ω(2^{(log2 N )/2}) steps.

Primality Tests (concluded)

• Suppose N = P Q is a product of 2 distinct primes.

• The probability of success of the density attack (p. 434) is

≈ 2

√N when P ≈ Q.

• This probability is exponentially small in terms of the input length log₂ N .

The Fermat Test for Primality

Fermat’s “little” theorem (p. 437) suggests the following primality test for any given number N :

1: Pick a number a randomly from {1, 2, . . . , N − 1};

2: if a^N−1 ̸= 1 mod N then

3: return “N is composite”;

4: else

5: return “N is a prime”;

6: end if

The Fermat Test for Primality (concluded)

• Carmichael numbers are composite numbers that will pass the Fermat test for all a ∈ {1, 2, . . . , N − 1}.^a

– The Fermat test will return “N is a prime” for all Carmichael numbers N .

• Unfortunately, there are infinitely many Carmichael numbers.^b

• In fact, the number of Carmichael numbers less than N exceeds N^2/7 for N large enough.

• So the Fermat test is an incorrect algorithm for primes.

aCarmichael (1910).

bAlford, Granville, and Pomerance (1992).

Square Roots Modulo a Prime

• Equation x² = a mod p has at most two (distinct) roots by Lemma 57 (p. 442).

– The roots are called square roots.

– Numbers a with square roots and gcd(a, p) = 1 are called quadratic residues.

∗ They are

1² mod p, 2² mod p, . . . , (p − 1)² mod p.

• We shall show that a number either has two roots or has none, and testing which is the case is trivial.^a

aBut no efficient deterministic general-purpose square-root-extracting algorithms are known yet.

Euler’s Test

Lemma 62 (Euler) Let p be an odd prime and a ̸= 0 mod p.

1. If

a^(p−1)/2 = 1 mod p, then x² = a mod p has two roots.

2. If

a^(p−1)/2 ̸= 1 mod p, then

a^(p−1)/2 = −1 mod p and x² = a mod p has no roots.

The Proof (continued)

• Let r be a primitive root of p.

• By Fermat’s “little” theorem, r^(p−1)/2 is a square root of 1.

• So

r^(p−1)/2 = 1 or −1 mod p.

• But as r is a primitive root, r^(p−1)/2 ̸= 1 mod p.

• Hence

r^(p−1)/2 = −1 mod p.

The Proof (continued)

• Let a = r^k mod p for some k.

• Then

1 = a^(p−1)/2 = r^k(p−1)/2 = [

r^(p−1)/2 ]k

= (−1)^k mod p.

• So k must be even.

• Suppose a = r^2j for some 1 ≤ j ≤ (p − 1)/2.

• Then a^(p−1)/2 = r^j(p−1) = 1 mod p, and a’s two distinct roots are r^j,−r^j(= r^j+(p−1)/2 mod p).

– If r^j = −r^j mod p, then 2r^j = 0 mod p, which implies r^j = 0 mod p, a contradiction.

The Proof (continued)

• As 1 ≤ j ≤ (p − 1)/2, there are (p − 1)/2 such a’s.

• Each such a has 2 distinct square roots.

• The square roots of all the a’s are distinct.

– The square roots of different a’s must be different.

• Hence the set of square roots is {1, 2, . . . , p − 1}.

• As a result, a = r^2j, 1 ≤ j ≤ (p − 1)/2, exhaust all the quadratic residues.

The Proof (concluded)

• If a = r^2j+1, then it has no roots because all the square roots have been taken.

• Now,

a^(p−1)/2 = [

r^(p−1)/2

]2j+1

= (−1)^2j+1 = −1 mod p.

The Legendre Symbol^a and Quadratic Residuacity Test

• By Lemma 62 (p. 506) a^(p−1)/2 mod p = ±1 for a ̸= 0 mod p.

• For odd prime p, define the Legendre symbol (a | p) as

(a| p) =









0 if p| a,

1 if a is a quadratic residue modulo p,

−1 if a is a quadratic nonresidue modulo p.

• Euler’s test (p. 506) implies

a^(p−1)/2 = (a| p) mod p for any odd prime p and any integer a.

• Note that (ab|p) = (a|p)(b|p).

aAndrien-Marie Legendre (1752–1833).

Gauss’s Lemma

Lemma 63 (Gauss) Let p and q be two odd primes. Then (q|p) = (−1)^m, where m is the number of residues in

R = { iq mod p : 1 ≤ i ≤ (p − 1)/2 } that are greater than (p − 1)/2.

• All residues in R are distinct.

– If iq = jq mod p, then p|(j − i) q or p|q.

– But neither is possible.

• No two elements of R add up to p.

– If iq + jq = 0 mod p, then p|(i + j) or p|q.

– But neither is possible.

The Proof (continued)

• Replace each of the m elements a ∈ R such that a > (p − 1)/2 by p − a.

– This is equivalent to performing −a mod p.

• Call the resulting set of residues R^′.

• All numbers in R^′ are at most (p − 1)/2.

• In fact, R^′ = {1, 2, . . . , (p − 1)/2} (see illustration next page).

– Otherwise, two elements of R would add up to p, which has been shown to be impossible.

5 1 2 3 4

6 5

1 2 3 4

6

p = 7 and q = 5.

The Proof (concluded)

• Alternatively, R^′ = {±iq mod p : 1 ≤ i ≤ (p − 1)/2}, where exactly m of the elements have the minus sign.

• Take the product of all elements in the two representations of R^′.

• So

[(p − 1)/2]! = (−1)^mq^(p−1)/2[(p − 1)/2]! mod p.

• Because gcd([(p − 1)/2]!, p) = 1, the above implies 1 = (−1)^mq^(p−1)/2 mod p.

Legendre’s Law of Quadratic Reciprocity

• Let p and q be two odd primes.

• The next result says their Legendre symbols are distinct if and only if both numbers are 3 mod 4.

Lemma 64 (Legendre (1785), Gauss)

(p|q)(q|p) = (−1)^{p−12 q−12} .

aFirst stated by Euler in 1751. Legendre (1785) did not give a correct proof. Gauss proved the theorem when he was 19. He gave at least 6 different proofs during his life. The 152nd proof appeared in 1963.

The Proof (continued)

• Sum the elements of R^′ in the previous proof in mod2.

• On one hand, this is just ∑(p−1)/2

i=1 i mod 2.

• On the other hand, the sum equals

mp +

(p∑−1)/2 i=1

(

iq − p

⌊iq p

⌋)

mod 2

= mp +



q

(p−1)/2∑

i=1

i − p

(p−1)/2∑

i=1

⌊iq p

⌋ mod 2.

– m of the iq mod p are replaced by p − iq mod p.

– But signs are irrelevant under mod2.

– m is as in Lemma 63 (p. 512).

The Proof (continued)

• Ignore odd multipliers to make the sum equal

m +





(p∑−1)/2 i=1

i −

(p−1)/2∑

i=1

⌊iq p

⌋ mod 2.

• Equate the above with ∑(p−1)/2

i=1 i mod 2 to obtain m =

(p∑−1)/2 i=1

⌊iq p

⌋

mod 2.

The Proof (concluded)

• ∑(p−1)/2

i=1 ⌊^iq_p ⌋ is the number of integral points below the line

y = (q/p) x for 1 ≤ x ≤ (p − 1)/2.

• Gauss’s lemma (p. 512) says (q|p) = (−1)^m.

• Repeat the proof with p and q reversed.

• Then (p|q) = (−1)^m′, where m^′ is the number of integral points above the line y = (q/p) x for 1 ≤ y ≤ (q − 1)/2.

• As a result, (p|q)(q|p) = (−1)^m+m′.

• But m + m^′ is the total number of integral points in the [1, ^p−1₂ ] × [1, ^q−1₂ ] rectangle, which is ^p−1₂ ^q−1₂ .

Eisenstein’s Rectangle

(p,q)

(p - 1)/2 (q - 1)/2

Above, p = 11 and q = 7.

The Jacobi Symbol

• The Legendre symbol only works for odd prime moduli.

• The Jacobi symbol (a | m) extends it to cases where m is not prime.

• Let m = p¹p₂ · · · p^k be the prime factorization of m.

• When m > 1 is odd and gcd(a, m) = 1, then

(a|m) =

∏k i=1

(a| pⁱ).

– Note that the Jacobi symbol equals ±1.

– It reduces to the Legendre symbol when m is a prime.

• Define (a | 1) = 1.

aCarl Jacobi (1804–1851).

Properties of the Jacobi Symbol

The Jacobi symbol has the following properties, for arguments for which it is defined.

1. (ab | m) = (a | m)(b | m).

2. (a| m¹m₂) = (a| m¹)(a | m²).

3. If a = b mod m, then (a | m) = (b | m).

4. (−1 | m) = (−1)^(m−1)/2 (by Lemma 63 on p. 512).

5. (2| m) = (−1)^(m2−1)/8.^a

6. If a and m are both odd, then (a| m)(m | a) = (−1)^{(a−1)(m−1)/4}.

aBy Lemma 63 (p. 512) and some parity arguments.

Properties of the Jacobi Symbol (concluded)

• These properties allow us to calculate the Jacobi symbol without factorization.

• This situation is similar to the Euclidean algorithm.

• Note also that (a | m) = 1/(a | m) because (a | m) = ±1.^a

aContributed by Mr. Huang, Kuan-Lin (B96902079, R00922018) on December 6, 2011.

Calculation of (2200 |999)

(202|999) = (2|999)(101|999)

= (−1)^(9992−1)/8(101|999)

= (−1)¹²⁴⁷⁵⁰(101|999) = (101|999)

= (−1)(100)(998)/4

(999|101) = (−1)²⁴⁹⁵⁰(999|101)

= (999|101) = (90|101) = (−1)^(1012−1)/8(45|101)

= (−1)¹²⁷⁵(45|101) = −(45|101)

= −(−1)(44)(100)/4

(101|45) = −(101|45) = −(11|45)

= −(−1)^(10)(44)/4(45|11) = −(45|11)

= −(1|11) = −1.

A Result Generalizing Proposition 10.3 in the Textbook

Theorem 65 The group of set Φ(n) under multiplication mod n has a primitive root if and only if n is either 1, 2, 4, p^k, or 2p^k for some nonnegative integer k and and odd

prime p.

This result is essential in the proof of the next lemma.

The Jacobi Symbol and Primality Test

Lemma 66 If (M|N) = M^(N−1)/2 mod N for all M ∈ Φ(N), then N is a prime. (Assume N is odd.)

• Assume N = mp, where p is an odd prime, gcd(m, p) = 1, and m > 1 (not necessarily prime).

• Let r ∈ Φ(p) such that (r | p) = −1.

• The Chinese remainder theorem says that there is an M ∈ Φ(N) such that

M = r mod p, M = 1 mod m.

aMr. Clement Hsiao (B4506061, R88526067) pointed out that the text- book’s proof for Lemma 11.8 is incorrect in January 1999 while he was a senior.

The Proof (continued)

• By the hypothesis,

M^(N−1)/2 = (M | N) = (M | p)(M | m) = −1 mod N.

• Hence

M^(N−1)/2 = −1 mod m.

• But because M = 1 mod m,

M^(N−1)/2 = 1 mod m, a contradiction.

The Proof (continued)

• Second, assume that N = p^a, where p is an odd prime and a ≥ 2.

• By Theorem 65 (p. 525), there exists a primitive root r modulo p^a.

• From the assumption, M^N−1 =

[

M^(N−1)/2 ]2

= (M|N)² = 1 mod N for all M ∈ Φ(N).

The Proof (continued)

• As r ∈ Φ(N) (prove it), we have

r^N−1 = 1 mod N.

• As r’s exponent modulo N = p^a is ϕ(N ) = p^a−1(p − 1), p^a−1(p − 1) | (N − 1),

which implies that p| (N − 1).

• But this is impossible given that p | N.

The Proof (continued)

• Third, assume that N = mp^a, where p is an odd prime, gcd(m, p) = 1, m > 1 (not necessarily prime), and a is even.

• The proof mimics that of the second case.

• By Theorem 65 (p. 525), there exists a primitive root r modulo p^a.

• From the assumption, M^N−1 =

[

M^(N−1)/2 ]2

= (M|N)² = 1 mod N for all M ∈ Φ(N).

The Proof (continued)

• In particular,

M^N−1 = 1 mod p^a (13)

for all M ∈ Φ(N).

• The Chinese remainder theorem says that there is an M ∈ Φ(N) such that

M = r mod p^a, M = 1 mod m.

• Because M = r mod p^a and Eq. (13), r^N−1 = 1 mod p^a.

The Proof (concluded)

• As r’s exponent modulo N = p^a is ϕ(N ) = p^a−1(p − 1), p^a−1(p − 1) | (N − 1),

which implies that p| (N − 1).

• But this is impossible given that p | N.