Randomization vs. Nondeterminism
a• 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
a• 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 (continued)
• The algorithm makes a false negative with probability
≤ 0.5.a
– Note this probability refers tob
prob[ algorithm answers “no”| G has a perfect matching ] not
prob[ G has a perfect matching| algorithm answers “no” ].
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 ].
Monte Carlo Algorithms (concluded)
• This probability 0.5 is not over the space of all graphs or determinants, but over the algorithm’s own coin flips.
– It holds for any bipartite graph.
The Markov Inequality
aLemma 61 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
ipi = ∑
i<kE[ x ]
ipi + ∑
i≥kE[ x ]
ipi
≥ ∑
i≥kE[ x ]
ipi ≥ kE[ x ] ∑
i≥kE[ x ]
pi
≥ 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.
– Here, we treat C as a black box without going into its internal code.a
• By Markov’s inequality, this new algorithm runs in time kT (n) and gives the wrong answer with probability
≤ 1/k.
aContributed by Mr. Hsien-Chun Huang (R03922103) on December 2, 2014.
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. 491)
• Let ϕ(x1, x2, . . . , xn) 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
aTheorem 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.
aPapadimitriou (1991).
The Proof
• Let t(i) denote the expected number of repetitions of the flipping stepa 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.
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, (10)
t(i) ≤ t(i − 1) + t(i + 1)
2 + 1, 0 < i < n, (11)
t(n) ≤ t(n − 1) + 1. (12)
• 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 obtaina 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. (13)
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. (13) on p. 536.
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. 525) with k = 2, resulting in a probability of 0.5.a
• The proof does not yield a polynomial bound for 3sat.b
aRecall p. 527.
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. 472) 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. 475) suggests the following primality test for any given number N :
1: Pick a number a randomly from {1, 2, . . . , N − 1};
2: if aN−1 ̸= 1 mod N then
3: return “N is composite”;
4: else
5: return “N is (probably) 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 N2/7 for N large enough.
• So the Fermat test is an incorrect algorithm for primes.
aCarmichael (1910). Lo (1994) mentions an investment strategy based on such numbers!
bAlford, Granville, and Pomerance (1992).
Square Roots Modulo a Prime
• Equation x2 = a mod p has at most two (distinct) roots by Lemma 58 (p. 480).
– 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 63 (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.
• Fermat’s “little” theorem says rp−1 = 1 mod p, so r(p−1)/2
is a square root of 1.
• In particular,
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 = rk mod p for some k.
• Then
1 = a(p−1)/2 = rk(p−1)/2 = [
r(p−1)/2 ]k
= (−1)k mod p.
• So k must be even.
• Suppose a = r2j for some 1 ≤ j ≤ (p − 1)/2.
• Then a(p−1)/2 = rj(p−1) = 1 mod p, and a’s two distinct roots are rj,−rj(= rj+(p−1)/2 mod p).
– If rj = −rj mod p, then 2rj = 0 mod p, which implies rj = 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 = r2j, 1 ≤ j ≤ (p − 1)/2, exhaust all the quadratic residues.
The Proof (concluded)
• If a = r2j+1, then it has no roots because all the square roots have been taken.
• Finally,
a(p−1)/2 = [
r(p−1)/2
]2j+1
= (−1)2j+1 = −1 mod p.
The Legendre Symbola and Quadratic Residuacity Test
• By Lemma 63 (p. 547),
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.
aAndrien-Marie Legendre (1752–1833).
The Legendre Symbol and Quadratic Residuacity Test (concluded)
• Euler’s test (p. 547) 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 distinct 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) 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,a which has been shown to be impossible.
aBecause iq ≡ −jq mod p for some i, j.
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
a• Let p and q be two distinct 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 8 different proofs during his life. The 152nd proof appeared in 1963.
A computer-generated formal proof was given in Russinoff (1990). As of 2008, there have been 4 such proofs. According to Wiedijk (2008),
“the Law of Quadratic Reciprocity is the first nontrivial theorem that a student encounters in the mathematics curriculum.”
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 64 (p. 554).
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 modulo 2 and then simplify to obtain
m =
(p∑−1)/2 i=1
⌊iq p
⌋
mod 2.
The Proof (concluded)
• ∑(p−1)/2
i=1 ⌊iqp ⌋ is the number of integral points below the line
y = (q/p) x for 1 ≤ x ≤ (p − 1)/2.
• Gauss’s lemma (p. 554) 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−1] × [1, q−1] rectangle, which is p−1 q−1.
Eisenstein’s Rectangle
(p,q)
(p - 1)/2 (q - 1)/2
Above, p = 11, q = 7, m = 7, m′ = 8.
The Jacobi Symbol
a• 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 · · · pk 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.
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. 554).
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. 554) 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 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, pk, or 2pk 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
aLemma 67 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 = pa, where p is an odd prime and a ≥ 2.
• By Theorem 66 (p. 567), there exists a primitive root r modulo pa.
• From the assumption, MN−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
rN−1 = 1 mod N.
• As r’s exponent modulo N = pa is ϕ(N ) = pa−1(p − 1), pa−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 = mpa, 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. 567), there exists a primitive root r modulo pa.
• From the assumption, MN−1 =
[
M(N−1)/2 ]2
= (M|N)2 = 1 mod N for all M ∈ Φ(N).
The Proof (continued)
• In particular,
MN−1 = 1 mod pa (14)
for all M ∈ Φ(N).
• The Chinese remainder theorem says that there is an M ∈ Φ(N) such that
M = r mod pa, M = 1 mod m.
• Because M = r mod pa and Eq. (14), rN−1 = 1 mod pa.
The Proof (concluded)
• As r’s exponent modulo N = pa is ϕ(N ) = pa−1(p − 1), pa−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) ≡ M(N−1)/2 mod N for at most half of M ∈ Φ(N).
• By Lemma 67 (p. 568) there is at least one a ∈ Φ(N) such that (a|N) ̸≡ a(N−1)/2 mod N .
• Let B = {b1, b2, . . . , bk} ⊆ Φ(N) be the set of all distinct residues such that (bi|N) ≡ b(Ni −1)/2 mod N .
• Let aB = {abi mod N : i = 1, 2, . . . , k}.
• Clearly, aB ⊆ Φ(N), too.
The Proof (concluded)
• |aB| = k.
– abi ≡ abj mod N implies N | a(bi − bj), which is
impossible because gcd(a, N ) = 1 and N > |bi − bj|.
• aB ∩ B = ∅ because
(abi)(N−1)/2 = a(N−1)/2b(Ni −1)/2 ̸= (a|N)(bi|N) = (abi|N).
• Combining the above two results, we know
| B |
ϕ(N ) ≤ | B |
| B ∪ aB | = 0.5.
1: if N is even but N ̸= 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 composite”;
9: else
10: if (M|N) ≡ M(N−1)/2 mod N then
11: return “N is (probably) a prime”;
12: else
13: return “N is composite”;
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.
– Suppose the input is composite.
– The probability that the algorithm says the number is a prime is ≤ 0.5 by Theorem 68 (p. 575).
• 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
Randomized Complexity Classes; RP
• Let N be a polynomial-time precise NTM that runs in time p(n) and has 2 nondeterministic choices at each step.
• N is a polynomial Monte Carlo Turing machine for a language L if the following conditions hold:
– If x ∈ L, then at least half of the 2p(n) computation paths of N on x halt with “yes” where n = | x |.
– If x ̸∈ L, then all computation paths halt with “no.”
• The class of all languages with polynomial Monte Carlo TMs is denoted RP (randomized polynomial time).a
aAdleman and Manders (1977).