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The Density Attack for primes

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(1)

The Density Attack for primes

Witnesses to compositeness

of n

All numbers < n

(2)

The Density Attack for primes

1: Pick k ∈ {1, . . . , n} randomly;

2: if k| n and k ̸= n then

3: return “n is composite”;

4: else

5: return “n is (probably) a prime”;

6: end if

(3)

The Density Attack for primes (continued)

• It works, but does it work well?

• The ratio of numbers ≤ n relatively prime to n (the white area) 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.

(4)

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(20.5 log2n).

(5)

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 = ak mod nk,

has a unique solution modulo n for the unknown x.

(6)

Fermat’s “Little” Theorem

a

Lemma 55 For all 0 < a < p, ap−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.

aPierre de Fermat (1601–1665).

(7)

The Proof (concluded)

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

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

• As aΦ(p) = Φ(p), ap−1(p − 1)! = (p − 1)! mod p.

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

(8)

The Fermat-Euler Theorem

a

Corollary 56 For all a ∈ Φ(n), aϕ(n) = 1 mod n.

• The proof is similar to that of Lemma 55 (p. 437).

• 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- ber 24, 2004.

(9)

The Proof (concluded)

a

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

m∈Φ(n) m.

• Multiply all the numbers in aΦ(n) to yield aϕ(n)

m∈Φ(n) m.

• As aΦ(n) = Φ(n),

m∈Φ(n)

m = aϕ(n)

 ∏

m∈Φ(n)

m

 mod n.

• Finally, aϕ(n) = 1 mod n because n ̸ |

m∈Φ(n) m.

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

(10)

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.

(11)

Exponents

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

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

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

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

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

m = mqk+a = ma = 1 mod p, a contradiction.

Lemma 57 Any nonzero polynomial of degree k has at most k distinct roots modulo p.

(12)

Exponents and Primitive Roots

• From Fermat’s “little” theorem, all exponents divide p − 1.

• A primitive root of p is thus a number with exponent p − 1.

• Let R(k) denote the total number of residues in Φ(p) = {1, 2, . . . , p − 1} that have exponent k.

• We already knew that R(k) = 0 for k ̸ |(p − 1).

• So

k|(p−1)

R(k) = p − 1 as every number has an exponent.

(13)

Size of R(k)

• Any a ∈ Φ(p) of exponent k satisfies xk = 1 mod p.

• Hence there are at most k residues of exponent k, i.e., R(k) ≤ k, by Lemma 57 (p. 442).

• Let s be a residue of exponent k.

• 1, s, s2, . . . , sk−1 are distinct modulo p.

– Otherwise, si = sj mod p with i < j.

– Then sj−i = 1 mod p with j − i < k, a contradiction.

• As all these k distinct numbers satisfy xk = 1 mod p, they comprise all solutions of xk = 1 mod p.

(14)

Size of R(k) (continued)

• But do all of them have exponent k (i.e., R(k) = k)?

• And if not (i.e., R(k) < k), how many of them do?

• Pick s.

• Suppose ℓ < k and ℓ ̸∈ Φ(k) with gcd(ℓ, k) = d > 1.

• Then

(s)k/d = (sk)ℓ/d = 1 mod p.

• Therefore, s has exponent at most k/d < k.

• We conclude that

R(k) ≤ ϕ(k).

(15)

Size of R(k) (concluded)

• Because all p − 1 residues have an exponent, p − 1 =

k|(p−1)

R(k)

k|(p−1)

ϕ(k) = p − 1

by Lemma 54 (p. 430).

• Hence

R(k) =



ϕ(k) when k|(p − 1) 0 otherwise

• In particular, R(p − 1) = ϕ(p − 1) > 0, and p has at least one primitive root.

• This proves one direction of Theorem 49 (p. 416).

(16)

A Few Calculations

• Let p = 13.

• From p. 439, we know ϕ(p − 1) = 4.

• Hence R(12) = 4.

• Indeed, there are 4 primitive roots of p.

• As

Φ(p − 1) = {1, 5, 7, 11}, the primitive roots are

g1, g5, g7, g11 for any primitive root g.

(17)

The Other Direction of Theorem 49 (p. 416)

• We show p is a prime if there is a number r such that 1. rp−1 = 1 mod p, and

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

• Suppose p is not a prime.

• We proceed to show that no primitive roots exist.

• Suppose rp−1 = 1 mod p (note gcd(r, p) = 1).

• We will show that the 2nd condition must be violated.

(18)

The Proof (continued)

• So we proceed to show r(p−1)/q = 1 mod p for some prime divisor q of p − 1.

• rϕ(p) = 1 mod p by the Fermat-Euler theorem (p. 439).

• Because p is not a prime, ϕ(p) < p − 1.

• Let k be the smallest integer such that rk = 1 mod p.

• With the 1st condition, it is easy to show that k | (p − 1) (similar to p. 442).

• Note that k | ϕ(p) (p. 442).

• As k ≤ ϕ(p), k < p − 1.

(19)

The Proof (concluded)

• Let q be a prime divisor of (p − 1)/k > 1.

• Then k|(p − 1)/q.

• By the definition of k,

r(p−1)/q = 1 mod p.

• But this violates the 2nd condition.

(20)

Function Problems

• Decision problems are yes/no problems (sat, tsp (d), etc.).

• Function problems require a solution (a satisfying truth assignment, a best tsp tour, etc.).

• Optimization problems are clearly function problems.

• What is the relation between function and decision problems?

• Which one is harder?

(21)

Function Problems Cannot Be Easier than Decision Problems

• If we know how to generate a solution, we can solve the corresponding decision problem.

– If you can find a satisfying truth assignment efficiently, then sat is in P.

– If you can find the best tsp tour efficiently, then tsp (d) is in P.

• But decision problems can be as hard as the corresponding function problems.

(22)

fsat

• fsat is this function problem:

– Let ϕ(x1, x2, . . . , xn) be a boolean expression.

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

– Otherwise, return “no.”

• We next show that if sat ∈ P, then fsat has a polynomial-time algorithm.

• sat is a subroutine (black box) that returns “yes” or

“no” on the satisfiability of the input.

(23)

An Algorithm for fsat Using sat

1: t := ϵ; {Truth assignment.}

2: if ϕ ∈ sat then

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

4: if ϕ[ xi = true ] ∈ sat then 5: t := t ∪ { xi = true};

6: ϕ := ϕ[ xi = true ];

7: else

8: t := t ∪ { xi = false};

9: ϕ := ϕ[ xi = false ];

10: end if 11: end for 12: return t;

13: else

14: return “no”;

15: end if

(24)

Analysis

• If sat can be solved in polynomial time, so can fsat.

– There are ≤ n + 1 calls to the algorithm for sat.a – Boolean expressions shorter than ϕ are used in each

call to the algorithm for sat.

• Hence sat and fsat are equally hard (or easy).

• Note that this reduction from fsat to sat is not a Karp reduction (recall p. 237).

• Instead, it calls sat multiple times as a subroutine and moves on sat’s outputs.

aContributed by Ms. Eva Ou (R93922132) on November 24, 2004.

(25)

tsp and tsp (d) Revisited

• We are given n cities 1, 2, . . . , n and integer distances dij = dji between any two cities i and j.

• tsp (d) asks if there is a tour with a total distance at most B.

• tsp asks for a tour with the shortest total distance.

– The shortest total distance is at most

i,j dij.

∗ Recall that the input string contains d11, . . . , dnn.

∗ Thus the shortest total distance is less than 2| x | in magnitude, where x is the input (why?).

• We next show that if tsp (d) ∈ P, then tsp has a polynomial-time algorithm.

(26)

An Algorithm for tsp Using tsp (d)

1: Perform a binary search over interval [ 0, 2| x | ] by calling tsp (d) to obtain the shortest distance, C;

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

3: Call tsp (d) with B = C and dij = C + 1;

4: if “no” then

5: Restore dij to old value; {Edge [ i, j ] is critical.}

6: end if

7: end for

8: return the tour with edges whose dij ≤ C;

(27)

Analysis

• An edge that is not on any optimal tour will be eliminated, with its dij set to C + 1.

• An edge which is not on all remaining optimal tours will also be eliminated.

• So the algorithm ends with n edges which are not eliminated (why?).

• There are O(| x | + n2) calls to the algorithm for tsp (d).

• Each call has an input length of O( x |).

• So if tsp (d) can be solved in polynomial time, so can tsp.

• Hence tsp (d) and tsp are equally hard (or easy).

(28)

Randomized Computation

(29)

I know that half my advertising works, I just don’t know which half.

— John Wanamaker I know that half my advertising is a waste of money, I just don’t know which half!

— McGraw-Hill ad.

(30)

Randomized Algorithms

a

• Randomized algorithms flip unbiased coins.

• There are important problems for which there are no known efficient deterministic algorithms but for which very efficient randomized algorithms exist.

– Extraction of square roots, for instance.

• There are problems where randomization is necessary.

– Secure protocols.

• Randomized version can be more efficient.

– Parallel algorithm for maximal independent set.b

aRabin (1976); Solovay and Strassen (1977).

b“Maximal” (a local maximum) not “maximum” (a global maximum).

(31)

“Four Most Important Randomized Algorithms”

a

1. Primality testing.b

2. Graph connectivity using random walks.c 3. Polynomial identity testing.d

4. Algorithms for approximate counting.e

aTrevisan (2006).

bRabin (1976); Solovay and Strassen (1977).

cAleliunas, Karp, Lipton, Lov´asz, and Rackoff (1979).

dSchwartz (1980); Zippel (1979).

eSinclair and Jerrum (1989).

(32)

Bipartite Perfect Matching

• We are given a bipartite graph G = (U, V, E).

– U = {u1, u2, . . . , un}.

– V = {v1, v2, . . . , vn}.

– E ⊆ U × V .

• We are asked if there is a perfect matching.

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

for all i ∈ {1, 2, . . . , n}.

(33)

A Perfect Matching in a Bipartite Graph

X

X

X

X

X

Y

Y

Y

Y

Y

(34)

Symbolic Determinants

• We are given a bipartite graph G.

• Construct the n × n matrix AG whose (i, j)th entry AGij is a symbolic variable xij if (ui, vj) ∈ E and 0 otherwise.

(35)

Symbolic Determinants (continued)

• The matrix for the bipartite graph G on p. 464 is

AG =









0 0 x13 x14 0

0 x22 0 0 0

x31 0 0 0 x35

x41 0 x43 x44 0

x51 0 0 0 x55









. (6)

(36)

Symbolic Determinants (concluded)

• The determinant of AG is det(AG) = ∑

π

sgn(π)

n i=1

AGi,π(i). (7)

– π ranges over all permutations of n elements.

– sgn(π) is 1 if π is the product of an even number of transpositions and −1 otherwise.

– Equivalently, sgn(π) = 1 if the number of (i, j)s such that i < j and π(i) > π(j) is even.a

• det(AG) contains n! terms, many of which may be 0s.

aContributed by Mr. Hwan-Jeu Yu (D95922028) on May 1, 2008.

(37)

Determinant and Bipartite Perfect Matching

• In

π sgn(π)n

i=1 AGi,π(i), note the following:

– Each summand corresponds to a possible perfect matching π.

– All of these summandsn

i=1 AGi,π(i) are distinct monomials and will not cancel.

• det(AG) is essentially an exhaustive enumeration.

Proposition 58 (Edmonds (1967)) G has a perfect matching if and only if det(AG) is not identically zero.

(38)

Perfect Matching and Determinant (p. 464)

X

X

X

X

X

Y

Y

Y

Y

Y

(39)

Perfect Matching and Determinant (concluded)

• The matrix is (p. 466)

AG =









0 0 x13 x14 0

0 x22 0 0 0

x31 0 0 0 x35

x41 0 x43 x44 0

x51 0 0 0 x55









.

• det(AG) = −x14x22x35x43x51 + x13x22x35x44x51 + x14x22x31x43x55 − x13x22x31x44x55.

• Each nonzero term denotes a perfect matching.

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