cse 311: foundations of computing Spring 2015 Lecture 13: Primes, - - PowerPoint PPT Presentation

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cse 311: foundations of computing Spring 2015 Lecture 13: Primes, - - PowerPoint PPT Presentation

Jun 02, 2023 259 likes β€’487 views

cse 311: foundations of computing Spring 2015 Lecture 13: Primes, GCDs, modular inverses review: repeated squaring Since a mod m a (mod m) for any a we have a 2 mod m = (a mod m) 2 mod m and a 4 mod m = (a 2 mod m) 2 mod m

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cse 311: foundations of computing Spring 2015 Lecture 13: Primes, GCDs, modular inverses

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review: repeated squaring

Since

a mod m ≑ a (mod m) for any a we have a2 mod m = (a mod m)2 mod m and a4 mod m = (a2 mod m)2 mod m and a8 mod m = (a4 mod m)2 mod m and a16 mod m = (a8 mod m)2 mod m and a32 mod m = (a16 mod m)2 mod m

Can compute 𝑏𝑙 mod 𝑛 for 𝑙 = 2𝑗 in only 𝑗 steps

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review: general algorithm

ModPow(a, k, m) should compute 𝑏𝑙 mod 𝑛. If 𝑙 == 0 then return 1 If (𝑙 mod 2 == 0) then return ModPow(𝑏2 mod 𝑛, 𝑙/2, 𝑛) else return (𝑏 Γ— ModPow(𝑏, 𝑙 βˆ’ 1, 𝑛)) mod 𝑛

𝑙 = 81453 = 10011111000101101 2 = 216 + 213 + 212 + 211 + 210 + 29 + 25 + 23 + 22 + 20

Total # of arithmetic operations ~ 4 Γ— 16 = 64

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primality An integer p greater than 1 is called prime if the only positive factors of p are 1 and p. A positive integer that is greater than 1 and is not prime is called composite.

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An integer p greater than 1 is called prime if the only positive factors of p are 1 and p. A positive integer that is greater than 1 and is not prime is called composite.

pr prima imali lity

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fundame ament nt al al theore rem m of

  • f ari

rith thmetic ic

Every positive integer greater than 1 has a unique prime factorization

48 = 2 β€’ 2 β€’ 2 β€’ 2 β€’ 3 591 = 3 β€’ 197 45,523 = 45,523 321,950 = 2 β€’ 5 β€’ 5 β€’ 47 β€’ 137 1,234,567,890 = 2 β€’ 3 β€’ 3 β€’ 5 β€’ 3,607 β€’ 3,803

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f ac actoriza

  • rization

ion

If π‘œ is composite, it has a factor of size at most π‘œ.

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eucl clid id’s s theor

  • rem

There are an infinite number of primes. Proof by contradiction:

Suppose that there are only a finite number of primes: π‘ž1, π‘ž2, … , π‘žπ‘œ

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f amous

  • us algori

gorithmic thmic problems

  • blems
  • Primality Testing

– Given an integer π‘œ, determine if π‘œ is prime – Fermat’s little theorem test: If π‘ž is prime and 𝑏 β‰  0, then π‘π‘žβˆ’1 ≑ 1 (mod π‘ž)

  • Factoring

– Given an integer π‘œ, determine the prime factorization of π‘œ

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f ac actoring

  • ring

Factor the following 232 digit number [RSA768]:

123018668453011775513049495838496272077285 356959533479219732245215172640050726365751 874520219978646938995647494277406384592519 255732630345373154826850791702612214291346 167042921431160222124047927473779408066535 1419597459856902143413

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123018668453011775513049495838496272077285356959533479219 732245215172640050726365751874520219978646938995647494277 406384592519255732630345373154826850791702612214291346167 042921431160222124047927473779408066535141959745985690214 3413 334780716989568987860441698482126908177047949837 137685689124313889828837938780022876147116525317 43087737814467999489 367460436667995904282446337996279526322791581643 430876426760322838157396665112792333734171433968 10270092798736308917

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grea eate test st com

  • mmo

mon n di divisor isor

GCD(a, b): Largest integer 𝑒 such that 𝑒 ∣ 𝑏 and 𝑒 ∣ 𝑐

– GCD(100, 125) = – GCD(17, 49) = – GCD(11, 66) = – GCD(13, 0) = – GCD(180, 252) =

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gcd d and and f ac actoring

  • ring

a = 23 β€’ 3 β€’ 52 β€’ 7 β€’ 11 = 46,200 b = 2 β€’ 32 β€’ 53 β€’ 7 β€’ 13 = 204,750 GCD(a, b) = 2min(3,1) β€’ 3min(1,2) β€’ 5min(2,3) β€’ 7min(1,1) β€’ 11min(1,0) β€’ 13min(0,1)

Factoring is expensive! Can we compute GCD(a,b) without factoring?

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usef eful ul GCD D f ac act

If 𝑏 and 𝑐 are positive integers, then gcd 𝑏, 𝑐 = gcd(𝑐, 𝑏 mod 𝑐) Proof: By definition 𝑏 = 𝑏 div 𝑐 β€’ 𝑐 + (𝑏 mod 𝑐) If 𝑒 ∣ 𝑏 and 𝑒 ∣ 𝑐 then 𝑒 ∣ 𝑏 mod 𝑐 . If 𝑒 ∣ 𝑐 and 𝑒 ∣ 𝑏 mod 𝑐 then 𝑒 ∣ 𝑏.

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eucl clid id’s s al algor

  • rithm

thm

GCD(660,126) Repeatedly use the GCD fact to reduce numbers until you get GCD 𝑦, 0 = 𝑦.

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GCD(x, y) = GCD(y, x mod y)

int GCD(int a, int b){ /* a >= b, b > 0 */ int tmp; while (b > 0) { tmp = a % b; a = b; b = tmp; } return a; }

Example: GCD(660, 126)

eucl clid id’s s al algor

  • rithm

thm

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Bezout

  • ut’s

s theorem

  • rem

If a and b are positive integers, then there exist integers s and t such that gcd(a,b) = sa + tb

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ex exten tended ded eu eucli clidea dean n algori

  • rithm

thm

  • Can use Euclid’s Algorithm to find 𝑑, 𝑒 such that

gcd 𝑏, 𝑐 = 𝑑𝑏 + 𝑒𝑐

  • e.g. gcd(35,27):

35 = 1 β€’ 27 + 8 35 - 1 β€’ 27 = 8 27= 3 β€’ 8 + 3 27- 3 β€’ 8 = 3 8 = 2 β€’ 3 + 2 8 - 2 β€’ 3 = 2 3 = 1 β€’ 2 + 1 3 - 1 β€’ 2 = 1 2 = 2 β€’ 1 + 0

  • Substitute back from the bottom

1 = 3 - 1 β€’ 2 = 3 – 1 (8 - 2 β€’ 3) = (-1) β€’ 8 + 3 β€’ 3 = (-1) β€’ 8 + 3 (27- 3 β€’ 8 ) = 3 β€’ 27 + (-10) β€’ 8 =

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mu mul tiplic iplicative e inv nver erse e mod 𝑛

Suppose GCD 𝑏, 𝑛 = 1 By BΓ©zout’s Theorem, there exist integers 𝑑 and 𝑒 such that 𝑑𝑏 + 𝑒𝑛 = 1. 𝑑 mod 𝑛 is the multiplicative inverse of 𝑏: 1 = 𝑑𝑏 + 𝑒𝑛 mod 𝑛 = 𝑑𝑏 mod 𝑛

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sol

  • lving

ing mo modu dula lar r equa uatio ions ns

Solving 𝑏𝑦 ≑ 𝑐 (mod 𝑛) for unknown 𝑦 when gcd 𝑏, 𝑛 = 1.

1. Find 𝑑 such that 𝑑𝑏 + 𝑒𝑛 = 1 2. Compute π‘βˆ’1 = 𝑑 mod 𝑛, the multiplicative inverse of 𝑏 modulo 𝑛 3. Set 𝑦 = π‘βˆ’1 β‹… 𝑐 mod 𝑛

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example ample

Solve: 7𝑦 ≑ 1 (mod 26)