integer programming

integer programming

• integer programming asks whether a system of linear inequalities with integer coefficients has an integer

solution.

• In contrast, linear programming asks whether a

system of linear inequalities with integer coefficients has a rational solution.

integer programming Is NP-Complete

• set covering can be expressed by the inequalities Ax ≥ 1, _n

i=1 x_i ≤ B, 0 ≤ x_i ≤ 1, where – x_i is one if and only if S_i is in the cover.

– A is the matrix whose columns are the bit vectors of the sets S1, S2, . . ..

– 1 is the vector of 1s.

– The operations in Ax are standard matrix operations.

• This shows integer programming is NP-hard.

• Many NP-complete problems can be expressed as an integer programming problem.

Christos Papadimitriou (1949–)

Easier or Harder?

• Adding restrictions on the allowable problem instances will not make a problem harder.

– We are now solving a subset of problem instances or special cases.

– The independent set proof (p. 364) and the knapsack proof (p. 417): equally hard.

– circuit value to monotone circuit value (p. 317): equally hard.

– sat to 2sat (p. 344): easier.

aThanks to a lively class discussion on October 29, 2003.

Easier or Harder? (concluded)

• Adding restrictions on the allowable solutions (the solution space) may make a problem harder, equally hard, or easier.

• It is problem dependent.

– min cut to bisection width (p. 392): harder.

– linear programming to integer programming (p. 434): harder.

– sat to naesat (equally hard by p. 357) and max cut to max bisection (p. 390): equally hard.

– 3-coloring to 2-coloring (p. 401): easier.

coNP and Function Problems

coNP

• NP is the class of problems that have succinct certificates (recall Proposition 38 on p. 329).

• By definition, coNP is the class of problems whose complement is in NP.

• coNP is therefore the class of problems that have succinct disqualifications:

– A “no” instance of a problem in coNP possesses a short proof of its being a “no” instance.

– Only “no” instances have such proofs.

coNP (continued)

• Suppose L is a coNP problem.

• There exists a polynomial-time nondeterministic algorithm M such that:

– If x ∈ L, then M(x) = “yes” for all computation paths.

– If x ∈ L, then M(x) = “no” for some computation path.

• Note that if we swap “yes” and “no” of M, the new algorithm M^ decides ¯L ∈ NP in the classic sense (p.

103).

\HV [ ∉ /

\HV QR

\HV QR

\HV [ ∈ /

\HV

\HV

\HV

\HV

coNP (continued)

• So there are 3 major approaches to proving L ∈ coNP.

1. Prove ¯L ∈ NP.

2. Prove that only “no” instances possess short proofs.

3. Write an algorithm for it directly.

coNP (concluded)

• Clearly P ⊆ coNP.

• It is not known if

P = NP ∩ coNP.

– Contrast this with

R = RE ∩ coRE (see Proposition 14 on p. 169).

Some coNP Problems

• validity ∈ coNP.

– If φ is not valid, it can be disqualified very succinctly:

a truth assignment that does not satisfy it.

• sat complement ∈ coNP.

– sat complement is the complement of sat.

– The disqualification is a truth assignment that satisfies it.

• hamiltonian path complement ∈ coNP.

– The disqualification is a Hamiltonian path.

Some coNP Problems (concluded)

• optimal tsp (d) ∈ coNP.

– optimal tsp (d) asks if the optimal tour has a total distance of B, where B is an input.^a

– The disqualification is a tour with a length < B.

aDefined by Mr. Che-Wei Chang (R95922093) on September 27, 2006.

A Nondeterministic Algorithm for sat complement (See also p. 113)

φ is a boolean formula with n variables.

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

2: Guess x_i ∈ {0, 1}; {Nondeterministic choice.}

3: end for

4: {Verification:}

5: if φ(x1, x2, . . . , xn) = 1 then

6: “no”;

7: else

8: “yes”;

9: end if

Analysis

• The algorithm decides language {φ : φ is unsatisfiable}.

– The computation tree is a complete binary tree of depth n.

– Every computation path corresponds to a particular truth assignment out of 2ⁿ.

– φ is unsatisfiable if and only if every truth assignment falsifies φ.

– But every truth assignment falsifies φ if and only if every computation path results in “yes.”

An Alternative Characterization of coNP

Proposition 50 Let L ⊆ Σ^∗ be a language. Then L ∈ coNP if and only if there is a polynomially decidable and

polynomially balanced relation R such that L = {x : ∀y (x, y) ∈ R}.

(As on p. 328, we assume | y | ≤ | x |^k for some k.)

• ¯L = {x : ∃y (x, y) ∈ ¬R}.

• Because ¬R remains polynomially balanced, ¯L ∈ NP by Proposition 38 (p. 329).

• Hence L ∈ coNP by definition.

coNP-Completeness

Proposition 51 L is NP-complete if and only if its complement ¯L = Σ^∗ − L is coNP-complete.

Proof (⇒; the ⇐ part is symmetric)

• Let ¯L^ be any coNP language.

• Hence L^ ∈ NP.

• Let R be the reduction from L^ to L.

• So x ∈ L^ if and only if R(x) ∈ L.

• By the law of transposition, x ∈ L^ if and only if R(x) ∈ L.

coNP Completeness (concluded)

• So x ∈ ¯L^ if and only if R(x) ∈ ¯L.

• The same R is a reduction from ¯L^ to ¯L.

• This shows ¯L is coNP-hard.

• But ¯L ∈ coNP.

• This shows ¯L is coNP-complete.

Some coNP-Complete Problems

• sat complement is coNP-complete.

• validity is coNP-complete.

– φ is valid if and only if ¬φ is not satisfiable.

– The reduction from sat complement to validity is hence easy.

• hamiltonian path complement is coNP-complete.

Possible Relations between P, NP, coNP

1. P = NP = coNP.

2. NP = coNP but P = NP.

3. NP = coNP and P = NP.

• This is the current “consensus.”^a

aCarl Gauss (1777–1855), “I could easily lay down a multitude of such propositions, which one could neither prove nor dispose of.”

The Primality Problem

• An integer p is prime if p > 1 and all positive numbers other than 1 and p itself cannot divide it.

• primes asks if an integer N is a prime number.

• Dividing N by 2, 3, . . . ,√

N is not efficient.

– The length of N is only log N, but √

N = 2^{0.5 log N}. – It is an exponential-time algorithm.

• A polynomial-time algorithm for primes was not found until 2002 by Agrawal, Kayal, and Saxena!

• The running time is ˜O(log^7.5 N).

1: _{if n = a}^b for some a, b > 1 then

2: return “composite”;

3: _{end if}

4: for r = 2, 3, . . . , n − 1 do

5: if gcd(n, r) > 1 then

6: return “composite”;

7: _{end if}

8: if r is a prime then

9: Let q be the largest prime factor of r − 1;

10: _{if q ≥ 4}^√r log n and n^(r−1)/q = 1 mod r then

11: break; {Exit the for-loop.}

12: _{end if} 13: _{end if}

14: end for{r − 1 has a prime factor q ≥ 4√

r log n.}

15: for a = 1, 2, . . . , 2√

r log n do

16: _{if (x − a)}ⁿ _{= (x}ⁿ − a) mod (x^r − 1) in Zn[x ] then

17: return “composite”;

18: _{end if} 19: _{end for}

The Primality Problem (concluded)

• Later, we will focus on efficient “randomized” algorithms for primes (used in Mathematica, e.g.).

• NP ∩ coNP is the class of problems that have succinct certificates and succinct disqualifications.

– Each “yes” instance has a succinct certificate.

– Each “no” instance has a succinct disqualification.

– No instances have both.

• We will see that primes ∈ NP ∩ coNP.

– In fact, primes ∈ P as mentioned earlier.

Primitive Roots in Finite Fields

Theorem 52 (Lucas and Lehmer (1927)) ^a A number p > 1 is a prime if and only if there is a number 1 < r < p such that

1. r^p−1 = 1 mod p, and

2. r^(p−1)/q = 1 mod p for all prime divisors q of p − 1.

• This r is called the primitive root or generator.

• We will prove the theorem later.^b

aFran¸cois Edouard Anatole Lucas (1842–1891); Derrick Henry Lehmer (1905–1991).

bSee pp. 469ff.

Derrick Lehmer

(1905–1991)

a

Pratt’s Theorem

Theorem 53 (Pratt (1975)) primes ∈ NP ∩ coNP.

• primes is in coNP because a succinct disqualification is a proper divisor.

– A proper divisor of a number n means n is not a prime.

• Now suppose p is a prime.

• p’s certificate includes the r in Theorem 52 (p. 457).

• Use recursive doubling to check if r^p−1 = 1 mod p in time polynomial in the length of the input, log₂ p.

– r, r², r⁴, . . . mod p, a total of ∼ log p steps.

The Proof (concluded)

• We also need all prime divisors of p − 1: q1, q2, . . . , q_k. – Whether r, q1, . . . , q_k are easy to find is irrelevant.

– There may be multiple choices for r.

• Checking r^(p−1)/qi = 1 mod p is also easy.

• Checking q1, q2, . . . , qk are all the divisors of p − 1 is easy.

• We still need certificates for the primality of the qi’s.

• The complete certificate is recursive and tree-like:

C(p) = (r; q1, C(q1), q2, C(q2), . . . , q_k, C(q_k)). (4)

• We next prove that C(p) is succinct.

The Succinctness of the Certificate

Lemma 54 The length of C(p) is at most quadratic at 5 log²₂ p.

• This claim holds when p = 2 or p = 3.

• In general, p − 1 has k ≤ log₂ p prime divisors q¹ = 2, q², . . . , qk.

– Reason:

2^k ≤

k i=1

q_i ≤ p − 1.

• Note also that, as q1 = 2,

k

q_i ≤ p − 1

2 . (5)

The Proof (continued)

• C(p) requires:

– 2 parentheses;

– 2k < 2 log2 p separators (at most 2 log2 p bits);

– r (at most log₂ p bits);

– q1 = 2 and its certificate 1 (at most 5 bits);

– q2, . . . , q_k (at most 2 log₂ p bits);^a – C(q2), . . . , C(qk).

aWhy?

The Proof (concluded)

• C(p) is succinct because, by induction,

|C(p)| ≤ 5 log₂ p + 5 + 5

k i=2

log²₂ q_i

≤ 5 log₂ p + 5 + 5

 _k



i=2

log₂ q_i

²

≤ 5 log₂ p + 5 + 5 log²₂ p − 1

2 by inequality (5)

< 5 log2 p + 5 + 5[ (log2 p) − 1 ]²

= 5 log²₂ p + 10 − 5 log₂ p ≤ 5 log²₂ p for p ≥ 4.

A Certificate for 23

• Note that 5 is a primitive root modulo 23 and 23 − 1 = 22 = 2 × 11.^b

• So

C(23) = (5; 2, C(2), 11, C(11)).

• Note that 2 is a primitive root modulo 11 and 11 − 1 = 10 = 2 × 5.

• So

C(11) = (2; 2, C(2), 5, C(5)).

aThanks to a lively discussion on April 24, 2008.

bOther primitive roots are 7, 10, 11, 14, 15, 17, 19, 20, 21.

A Certificate for 23 (concluded)

• Note that 2 is a primitive root modulo 5 and 5 − 1 = 4 = 2².

• So

C(5) = (2; 2, C(2)).

• In summary,

C(23) = (5; 2, C(2), 11, (2; 2, C(2), 5, (2; 2, C(2)))).

– In Mathematica, PrimeQCertificate[23] yields {23, 5, {2, {11, 2, {2, {5, 2, {2}}}}}}

Turning the Proof into an Algorithm

• How to turn the proof into a polynomial-time nondeterministic algorithm?

• First, guess a log₂ p-bit number r.

• Then guess up to log₂ p log₂ p-bit numbers q1, q2, . . . , q_k.

• Then recursively do the same thing for each of the q_i to form a certificate (4) on p. 460.

• Finally check if the two conditions of Theorem 52 (p.

457) hold throughout the tree.

aContributed by Mr. Kai-Yuan Hou (B99201038, R03922014) on November 24, 2015.

Basic Modular Arithmetics

• Let m, n ∈ Z⁺.

• m | n means m divides n; m is n’s divisor.

• We call the numbers 0, 1, . . . , n − 1 the residue modulo n.

• The greatest common divisor of m and n is denoted gcd(m, n).

• The r in Theorem 52 (p. 457) is a primitive root of p.

• We now prove the existence of primitive roots and then Theorem 52 (p. 457).

a

Basic Modular Arithmetics (concluded)

• We use

a ≡ b mod n if n | (a − b).

– So 25 ≡ 38 mod 13.

• We use

a = b mod n

if b is the remainder of a divided by n.

– So 25 = 12 mod 13.

Euler’s

Totient or Phi Function

• Let

Φ(n) = {m : 1 ≤ m < n, gcd(m, n) = 1}

be the set of all positive integers less than n that are prime to n.^b

– Φ(12) = {1, 5, 7, 11}.

• Define Euler’s function of n to be φ(n) = | Φ(n) |.

• φ(p) = p − 1 for prime p, and φ(1) = 1 by convention.

• Euler’s function is not expected to be easy to compute without knowing n’s factorization.

aLeonhard Euler (1707–1783).

Two Properties of Euler’s Function

The inclusion-exclusion principle^a can be used to prove the following.

Lemma 55 φ(n) = n 

p|n(1 − ¹_p).

• If n = p^e₁¹p^e₂² · · · p^e_^ is the prime factorization of n, then φ(n) = n

 i=1



1 − 1 p_i

.

Corollary 56 φ(mn) = φ(m) φ(n) if gcd(m, n) = 1.

aConsult any textbooks on discrete mathematics.

A Key Lemma

Lemma 57 

m|n φ(m) = n.

• Let n = _

i=1 p^k_iⁱ be the prime factorization of n and consider

 i=1

[φ(1) + φ(pi) + · · · + φ(p^k_iⁱ) ]. (6)

• Equation (6) equals n because φ(p^k_i ) = p^k_i − p^k−1_i by Lemma 55 (p. 471) so φ(1) + φ(p_i) + · · · + φ(p^k_iⁱ) = p^k_iⁱ.

• Expand Eq. (6) to yield

n = 

k^≤k ,...,k^≤k

 i=1

φ(p^k_i^i).

The Proof (concluded)

• By Corollary 56 (p. 471),

 i=1

φ(p^k_i^i) = φ

 _



i=1

p^k_i^i

 .

• So Eq. (6) becomes

n = 

k^₁≤k1,...,k_^≤k_

φ

 _



i=1

p^k_i^i

 .

• Each _

i=1 p^k_i^i is a unique divisor of n = _

i=1 p^k_iⁱ.

• Equation (6) becomes

φ(m).

Leonhard Euler (1707–1783)

The Density Attack for primes

Witnesses to compositeness

of n

All numbers < n

The Density Attack for primes

1: Pick k ∈ {1, . . . , n} randomly;

2: if k | n and k = 1 and k = n then

3: return “n is composite”;

4: else

5: return “n is (probably) a prime”;

6: end if

The Density Attack for primes (continued)

• It works, but does it work well?

• The ratio of numbers ≤ n relatively prime to n (the white ring) is

φ(n) n .

• When n = pq, where p and q are distinct primes, φ(n)

n = pq − p − q + 1

pq > 1 − 1

q − 1 p.

The Density Attack for primes (concluded)

• So the ratio of numbers ≤ n not relatively prime to n (the grey area) is < (1/q) + (1/p).

– The “density attack” has probability about 2/√

n of factoring n = pq when p ∼ q = O(√

n ).

– The “density attack” to factor n = pq hence takes Ω(√

n) steps on average when p ∼ q = O(√ n ).

– This running time is exponential: Ω(2^{0.5 log2n}).

The Chinese Remainder Theorem

• Let n = n1n2 · · · nk, where ni are pairwise relatively prime.

• For any integers a1, a2, . . . , ak, the set of simultaneous equations

x = a1 mod n1, x = a2 mod n2,

...

x = a_k mod n_k,

has a unique solution modulo n for the unknown x.

Fermat’s “Little” Theorem

Lemma 58 For all 0 < a < p, a^p−1 = 1 mod p.

• Recall Φ(p) = {1, 2, . . . , p − 1}.

• Consider aΦ(p) = {am mod p : m ∈ Φ(p)}.

• aΦ(p) = Φ(p).

– aΦ(p) ⊆ Φ(p) as a remainder must be between 1 and p − 1.

– Suppose am ≡ am^ mod p for m > m^, where m, m^ ∈ Φ(p).

– That means a(m − m^) = 0 mod p, and p divides a or m − m^, which is impossible.

The Proof (concluded)

• Multiply all the numbers in Φ(p) to yield (p − 1)!.

• Multiply all the numbers in aΦ(p) to yield a^p−1(p − 1)!.

• As aΦ(p) = Φ(p), we have

a^p−1(p − 1)! ≡ (p − 1)! mod p.

• Finally, a^p−1 = 1 mod p because p  |(p − 1)!.

The Fermat-Euler Theorem

Corollary 59 For all a ∈ Φ(n), a^φ(n) = 1 mod n.

• The proof is similar to that of Lemma 58 (p. 480).

• Consider aΦ(n) = {am mod n : m ∈ Φ(n)}.

• aΦ(n) = Φ(n).

– aΦ(n) ⊆ Φ(n) as a remainder must be between 0 and n − 1 and relatively prime to n.

– Suppose am ≡ am^ mod n for m^ < m < n, where m, m^ ∈ Φ(n).

– That means a(m − m^) = 0 mod n, and n divides a or m − m^, which is impossible.

aProof by Mr. Wei-Cheng Cheng (R93922108, D95922011) on Novem-

The Proof (concluded)

• Multiply all the numbers in Φ(n) to yield 

m∈Φ(n) m.

• Multiply all the numbers in aΦ(n) to yield a^φ(n) 

m∈Φ(n) m.

• As aΦ(n) = Φ(n),



m∈Φ(n)

m ≡ a^φ(n)

⎛

⎝ 

m∈Φ(n)

m

⎞

⎠ mod n.

• Finally, a^φ(n) = 1 mod n because n  | 

m∈Φ(n) m.

aSome typographical errors corrected by Mr. Jung-Ying Chen (D95723006) on November 18, 2008.

An Example

• As 12 = 2² × 3,

φ(12) = 12 ×



1 − 1 2



1 − 1 3

= 4.

• In fact, Φ(12) = {1, 5, 7, 11}.

• For example,

5⁴ = 625 = 1 mod 12.

Exponents

• The exponent of m ∈ Φ(p) is the least k ∈ Z⁺ such that m^k = 1 mod p.

• Every residue s ∈ Φ(p) has an exponent.

– 1, s, s², s³, . . . eventually repeats itself modulo p, say sⁱ ≡ s^j mod p, which means s^j−i = 1 mod p.

• If the exponent of m is k and m^ = 1 mod p, then k | .

– Otherwise,  = qk + a for 0 < a < k, and

m^ = m^qk+a ≡ m^a ≡ 1 mod p, a contradiction.

Lemma 60 Any nonzero polynomial of degree k has at most k distinct roots modulo p.