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Arithmetic

Arithmetic. I. Fast Multiplication and the Master Theorem on Divide and Conquer. How fast can we multiply?. Adding two n -bit numbers takes O(n ) operations How many operations to multiply two n -bit numbers? Or two n -decimal-digit numbers

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Arithmetic

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  1. Arithmetic

  2. I. Fast Multiplication and the Master Theorem on Divide and Conquer

  3. How fast can we multiply? • Adding two n-bit numbers takes O(n) operations • How many operations to multiply two n-bit numbers? • Or two n-decimal-digit numbers • Difference is a factor of log210 ≈ 3.32 but the individual operations are harder

  4. Grade School Algorithm is Θ(n2) But answer is only O(n) bits: Can we do better?

  5. A Divide and Conquer Algorithm Suppose n is even, n = 2m To compute a∙b Write a = a1∙2m + a0, b = b1∙2m + b0, where a1, a0, b1, b0 are m-bit numbers (numbers < 2m) – the first and last m bits of a and b a∙b = a1b1∙22m + (a1b0+a0b1)∙2m + a0b0 = a1b1∙(22m+2m) + (a1-a0)(b0-b1)∙2m + a0b0∙(2m+1) Only 3 m-bit multiplications!!!

  6. How Fast? T(1)=1 T(n) = 3T(n/2) + cn But how to solve this?

  7. Master Theorem on D+C recurrences • T(1) = 1 • T(n) = aT(n/b) + cne • Let L = logba • Recurrence has the solution: • T(n) = Θ(ne) if e > L • T(n) = Θ(nelog n) if e = L • T(n) = Θ(nL) if e < L • Binary search: a=1, b=2, e=0, L=0 [Case 2] • Merge sort: a=2, b=2, e=1, L=1 [Case 2] • Ordinary mult: a=4, b=2, e=1, L=2 [Case 3] • Fast mult: a=3, b=2, e=1, L=lg 3 soΘ(n1.58…) [Case 3]

  8. II: Fast Exponentiation Compute 313: 313 = 3∙3∙3∙3∙3∙3∙3∙3∙3∙3∙3∙3∙3 (12 multiplications, or Θ(exponent)) 313 =36∙36∙3 (2 multiplications)36 = 33∙33 (1 multiplication)33can be computed with 2 multiplications So 2+1+2 = 5 multiplications in all!

  9. Fast Exponentiation compute ab using registers X,Y,Z,R X:= a; Y:= 1; Z:= b; REPEAT: if Z=0, then return Y R:= remdr(Z,2); Z:= quotnt(Z,2) if R=1,then Y:= X⋅Y X:= X2

  10. Powers by Repeated Squaring • Problem:compute ab • Method 1: multiply a by itself n-1 times • Requires n-1 multiplications • Method 2: use successive squaring • How many times can you divide n by 2 before it is reduced to 1? • Repeated squaring requires between log2n and 2∙log2n multiplications • Huge savings! n = 1000 => at most 20 multiplications! (since log21000 < 10) Harvard Bits

  11. 0 7 1 6 2 5 3 4 III. Modular arithmetic 6 + 5 = 3 (mod 8)

  12. Math Quiz 1 5 1 1 = (23)100 = 1100 = 1

  13. (mod p) notation • Think of the (mod p) at the end of the line as referring to everything in the equation • (23)100 = 1100 = 1 (mod 7) means “(23)100 , 1100 , and 1 are all equivalent if you divide by 7 and keep just the remainder” Often written a ≡ b (mod p) Harvard Bits

  14. Fast Modular Exponentiation • Problem: Given qand pand n,find y< psuch that qn= y (mod p) • Method 1: multiplyqby itself n-1 times • Requires n-1 multiplications • Method 2: use successive squaring • Requires about log2nmultiplications • Same idea works for multiplication modulo p • Example: If n is a 500-digit number, we can compute qn (mod p) in about 1700 (= lg 10500) steps. Harvard Bits

  15. FINIS

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