Perfect Matching for General Graphs

Perfect Matching for General Graphs

• Page 438 is about bipartite perfect matching

• Now we are given a graph G = (V, E).

– V = {v₁, v₂, . . . , v_2n}.

• We are asked if there is a perfect matching.

– A permutation π of {1, 2, . . . , 2n} such that (v_i, v_π(i)) ∈ E

for all v_i ∈ V .

The Tutte Matrix

• Given a graph G = (V, E), construct the 2n × 2n Tutte matrix T^G such that

T_ij^G =









x_ij if (v_i, v_j) ∈ E and i < j,

−x_ij if (v_i, v_j) ∈ E and i > j, 0 othersie.

• The Tutte matrix is a skew-symmetric symbolic matrix.

• Similar to Proposition 58 (p. 442):

Proposition 60 G has a perfect matching if and only if det(T^G) is not identically zero.

William Thomas Tutte (1917–2002)

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

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 flips, at most half of them lead to the wrong answer.

The Markov Inequality

Lemma 61 Let x be a random variable taking nonnegative integer values. Then for any k > 0,

prob[ x ≥ kE[ x ] ] ≤ 1/k.

• Let p_i denote the probability that x = i.

E[ x ] = X

i

ip_i

= X

i<kE[ x ]

ip_i + X

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

• 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.

• By running this algorithm m times, we reduce the error probability to ≤ k^−m.^a

An Application of Markov’s Inequality (concluded)

• 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.

fsat for k-sat Formulas (p. 425)

• 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 · · · + ²)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).

Random Walk Works for 2sat

Theorem 62 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.

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

aPapadimitriou (1991).

The Proof

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

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

• If T 6= ˆT or T is not equal to any other satisfying truth assignment, then we need to flip 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.

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

The Proof (continued)

• 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)

• As we are only interested in upper bounds, we solve x(0) = 0

x(n) = x(n − 1) + 1

x(i) = x(i − 1) + x(i + 1)

2 + 1, 0 < i < n

• This is one-dimensional random walk with a reflecting and an absorbing barrier.

The Proof (continued)

• Add the equations up to obtain

x(1) + x(2) + · · · + x(n)

= x(0) + x(1) + 2x(2) + · · · + 2x(n − 2) + x(n − 1) + x(n) 2

+n + x(n − 1).

• Simplify to yield

x(1) + x(n) − x(n − 1)

2 = n.

• As x(n) − x(n − 1) = 1, we have x(1) = 2n − 1.

The Proof (continued)

• Iteratively, we obtain

x(2) = 4n − 4, ...

x(i) = 2in − i².

• The worst case happens when i = n, in which case x(n) = n².

The Proof (concluded)

• We therefore reach the conclusion that t(i) ≤ x(i) ≤ x(n) = n².

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

• The algorithm picks a running time 2n².

• This amounts to invoking the Markov inequality (p. 460) with k = 2, with the consequence of having a probability of 0.5.

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

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

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^n/2) steps, where n = log₂ N .

The Density Attack for primes

1: Pick k ∈ {2, . . . , N − 1} randomly; {Assume N > 2.}

2: if k | N then

3: return “N is composite”;

4: else

5: return “N is a prime”;

6: end if

Analysis

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

• The probability of success is

< 1 − φ(N )

N = 1 − (P − 1)(Q − 1)

P Q = P + Q − 1 P Q .

• In the case where P ≈ Q, this probability becomes

< 1

P + 1

Q ≈ 2

√N .

• This probability is exponentially small.

aSee also p. 407.

The Fermat Test for Primality

Fermat’s “little” theorem on p. 409 suggests the following primality test for any given number p:

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

2: if a^{N −1} 6= 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)

• Unfortunately, there are composite numbers called Carmichael 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 .

• 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.

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 56 (p. 414).

– 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 one is true is trivial.^a

aNo efficient deterministic root-finding algorithms are known yet.

Euler’s Test

Lemma 63 (Euler) Let p be an odd prime and a 6= 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 6= 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 mod p or r^(p−1)/2 = −1 mod p.

• But as r is a primitive root, r^(p−1)/2 6= 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 = h

r^(p−1)/2 i_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}.

– Because there are (p − 1)/2 such a’s and each a has two distinct square roots.

• 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 = h

r^(p−1)/2

i_2j+1

= (−1)^2j+1 = −1 mod p.

The Legendre Symbol^a and Quadratic Residuacity Test

• By Lemma 63 (p. 481) a^(p−1)/2 mod p = ±1 for a 6= 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 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).

Gauss’s Lemma

Lemma 64 (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.

• 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)

• Consider the set R⁰ of residues that result from R if we 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.

• All residues in R⁰ are now at most (p − 1)/2.

• In fact, R⁰ = {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, R⁰ = {±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 R⁰.

• 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 65 (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 R⁰ in the previous proof in mod2.

• On one hand, this is just P_(p−1)/2

i=1 i mod 2.

• On the other hand, the sum equals

(p−1)/2X

i=1

µ

qi − p

¹qi p

º¶

+ mp mod 2

=



q

(p−1)/2X

i=1

i − p

(p−1)/2X

i=1

¹qi p

º

 + mp mod 2.

– Signs are irrelevant under mod2.

– m is as in Lemma 64 (p. 487).

The Proof (continued)

• Ignore odd multipliers to make the sum equal





(p−1)/2X

i=1

i −

(p−1)/2X

i=1

¹qi p

º

 + m mod 2.

• Equate the above with P_(p−1)/2

i=1 i mod 2 to obtain

m =

(p−1)/2X

i=1

¹qi p

º

mod 2.

The Proof (concluded)

• P_(p−1)/2

i=1 b^qi_p c is the number of integral points under the line

y = (q/p) x for 1 ≤ x ≤ (p − 1)/2.

• Gauss’s lemma (p. 487) says (q|p) = (−1)^m.

• Repeat the proof with p and q reversed.

• So (p|q) = (−1)^m0, where m⁰ 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+m0.

• But m + m⁰ is the total number of integral points in the

p−1 q−1 p−1 q−1

Eisenstein’s Rectangle

(p,q)

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

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) = Yk i=1

(a | p_i).

– Note that the Jacobi symbol equals ±1.

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

• Define (a | 1) = 1.

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 64 on p. 487).

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 64 (p. 487) and some parity arguments.

Calculation of (2200|999)

Similar to the Euclidean algorithm and does not require factorization.

(202|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 66 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 67 If (M |N ) = M^{(N −1)/2} mod N for all M ∈ Φ(N ), then N is 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 (R88526067) pointed out that the textbook’s

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 66 (p. 499), there exists a primitive root r modulo p^a.

• From the assumption, M^{N −1} =

h

M^{(N −1)/2} i₂

= (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 66 (p. 499), there exists a primitive root r modulo p^a.

• From the assumption, M^{N −1} =

h

M^{(N −1)/2} i₂

= (M |N )² = 1 mod N for all M ∈ Φ(N ).

The Proof (continued)

• In particular,

M^{N −1} = 1 mod p^a (7)

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. (7), 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 .

The Number of Witnesses to Compositeness

Theorem 68 (Solovay and Strassen (1977)) If N is an odd composite, then (M |N ) 6= M^{(N −1)/2} mod N for at least half of M ∈ Φ(N ).

• By Lemma 67 (p. 500) there is at least one a ∈ Φ(N ) such that (a|N ) 6= a^{(N −1)/2} mod N .

• Let B = {b₁, b₂, . . . , b_k} ⊆ Φ(N ) be the set of all distinct residues such that (b_i|N ) = b^{(N −1)/2}_i mod N .

• Let aB = {ab_i mod N : i = 1, 2, . . . , k}.

The Proof (concluded)

• |aB| = k.

– ab_i = ab_j mod N implies N |a(b_i − b_j), which is

impossible because gcd(a, N ) = 1 and N > |b_i − b_j|.

• aB ∩ B = ∅ because

(ab_i)^{(N −1)/2} = a^{(N −1)/2}b^{(N −1)/2}_i 6= (a|N )(b_i|N ) = (ab_i|N ).

• Combining the above two results, we know

| B |

φ(N ) ≤ | B |

| B ∪ aB | = 0.5.

1: if N is even but N 6= 2 then

2: return “N is composite”;

3: else if N = 2 then

4: return “N is a prime”;

5: end if

6: Pick M ∈ {2, 3, . . . , N − 1} randomly;

7: if gcd(M, N ) > 1 then

8: return “N is a composite”;

9: else

10: if (M |N ) 6= M^{(N −1)/2} mod N then

11: return “N is composite”;

12: else

13: return “N is a prime”;

14: end if

15: end if

Analysis

• The algorithm certainly runs in polynomial time.

• There are no false positives (for compositeness).

– When the algorithm says the number is composite, it is always correct.

• The probability of a false negative is at most one half.

– If the input is composite, then the probability that the algorithm says the number is a prime is ≤ 0.5.

• So it is a Monte Carlo algorithm for compositeness.

The Improved Density Attack for compositeness

All numbers < N

Witnesses to compositeness of

N via Jacobi Witnesses to

compositeness of N via common

factor