Modulo Arithmetic GCD

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Modulo ArithmeticGCD CertificatesPrimality
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Modulo

• Division 23 / 5 4
• Modulo 23 5 3
• Invariant b(a/b) ab a
• Thus, a-(a/b)b ab
• What is -23 / 5?
• Sometimes -4, sometimes -5
• What is -23 5?
• 5(-4) 3 or 5(-5) 2

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Modular Arithmetic

• Modulo has strict math
• Expressions modified by (mod N)
• -23 7 (mod 5)
• Book also uses operator mod
• Same as C/Java/Cs but always returns a
  positive number
• stuff nonsense (mod N)
• means
• stuff mod N nonsense mod N

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Properties of mod

• Substitution
• Given ab(mod N) and xy(mod N)
• Then ax by (mod N)
• And ax by (mod N)
• It is well-behaved
• x(yz) (xy)z (mod N)
• xy yx (mod N)
• x(yz) xy xz (mod N)

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Properties of mod

• Prove (if there is time)
• (a b) N (a N b N) N

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Modular Exponentiation

• How can we compute xy mod N?

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Modular Exponentiation

• int expmod(int x, int y, int N)
• if (y 0) return 1
• int z expmod(x,y/2,N)
• if (y 1) return (zz) N
• else return x(zz) N
• Correct?
• Speed?
• Can we do better?

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Greatest Common Divisor

• Given two positive integers, x and y, there is a
  unique largest integer z such that
• xz 0 and yz 0
• That z is called the Greatest Common Divisor
  (GCD) of x and y

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Algorithm for the GCD

• How can we find gcd(x,y)?
• We know x and y have unique prime factors
• We know the GCD is the product of their common
  prime factors
• Prime factorization is slow

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Algorithm for the GCD

• Observation
• (ax) mod (bx) (a mod b) x
• That is, modulo does not remove common factors
• Core of Eulcid's algorithm
• gcd(a,b) return (b0) ? a gcd(b,ab)
• Correct? Speed? Can we do better?
• How fast is ?

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GCD and certificates

• Is 52 the GCD of 4524 and 30160?

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GCD and certificates

• Lemma
• if
• ad0 and
• bd0 and
• daxby for some x,y,
• then dgcd(x,y)
• Proof (let ggcd(a,b))
• Since d is a common divisor, dg
• Since g divides a and b, it must divide axby,
  which is d, too
• Hence gd, which means g d

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GCD and certificates

• Check 1508 74524 -130160 is the GCD of
  4524 and 30160
• Check requires
• 2 multiplies O(n2), 1 add O(n), and 2 modulos
  O(n2)
• Total O(n2)
• Finding GCD requires Euclid's algorithm O(n3)
• 7 and -1 form a certificate that 1508
  gcd(4524,30160)

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Finding a certificate

• Easy case
• gcd(a,0) a 1a anything0
• Inductive step
• Assume gcd(b,ab) xb y(ab)
• gcd(a,b) gcd(b,ab) (Euclid)
• xb y(ab)
• xb y(a-(a/b)b)
• ya (x-(a/b)y)b

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Extended Eulcid GCD

• int gcd2(int a, int b)
• int ans 1,anything,a
• if (b ! 0)
• int tmp gcd2(b,ab)
• ans2 tmp2
• ans0 tmp1
• ans1 tmp0 (a/b)tmp1
• return ans

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Extended Euclid Example

• gcd2(25,11)
• gcd(11, 2511 3)
• gcd(3, 113 2)
• gcd(2, 32 1)
• gcd(1, 21 0)
• base case, use 0 return 1, 0, 1
• return 0, 1-(2/1)0 1, 1
• return 1, 0-(3/2)1 -1, 1
• return -1, 1-(11/3)-1 4, 1
• return 4, -1-(25/11)4 -9, 1

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Why certificates?

• Certificates (easier to check than to find) are
  neat, but...
• Also basis of modular division
• x inv(a) mod N iff ax 1 mod N
• inv(a) exists modulo N if and only if gcd(a,N)
  1 then we have gcd2(a,N) inv(a),junk,1
• a div b ainv(b) mod N
• Hence, inv(11) -9 (mod 25)

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Primality Testing
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Fermat's Little Theorem

• If p is prime, then
• ap-1 1 (mod p) for all 0 lt a lt p
• Example 5
• 14 1 1 (mod 5)
• 24 16 1 (mod 5)
• 34 81 1 (mod 5)
• 44 256 1 (mod 5)

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Huh?

• It's about relative primality

• 12 8 2
• 22 8 4
• 32 8 6
• 42 8 0
• 52 8 2
• 62 8 4
• 72 8 6

• 13 8 3
• 23 8 6
• 33 8 1
• 43 8 4
• 53 8 7
• 63 8 2
• 73 8 5

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Product of set

• Given S1, 2, ..., p-1
• If a and p are relatively prime,
• a, 2a, ... (p-1)a S (mod p)
• Take the product of each side
• ap-1 (p-1)! (p-1)! (mod p)
• if p is prime, (p-1)! has an inverse mod p, so
• ap-1 1 (mod p)
• which proves Fermat's little theorem

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How well does it work?

• Lemma if ap-1 ! 1 (mod p), then either
• p is a Carmichael Number, or
• ap-1 ! 1 (mod p) for at least half of all
  possible a values
• Extremely few Carmichael numbers, and book
  discusses a workaround for them
• Hence, at worst 1/2 chance of error per a we test

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One-sided error reduction

• No false negatives (if Fermat says it is
  composite, it is composite)
• At worst 1/2 chance of false positive
• Pick k different random a then under 1/2k chance
  of false positive

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Random Prime Number

• Lagrange proved there are roughly x/ln(x) primes
  less than x
• Meaning n-bit number has 1/(nln(2)) chance of
  being prime
• Hence
• Pick a random number
• Run a primality test
• Repeat if composite

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Industrial-grade primes

• For most work it's OK to just check a2 (maybe 3
  5 too)
• To see why,
• About 1,000,000,000 primes less than
  25,000,000,000
• About 20,000 composites less than 25,000,000,000
  fool Fermat with a2
• under 200 of them are Carmichael numbers
• This difference increases as numbers get larger