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 Difference is a factor of log210 ≈ 3.32 but the individual operations are harder

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

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!!!

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

Master Theorem on D+C recurrences T(n) = aT(n/b) + cne Let L = logba Recurrence has the solution: T(n) = Θ(ne) if e > L T(n) = Θ(ne log 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 so Θ(n1.58…) [Case 3]

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) 33 can be computed with 2 multiplications So 2+1+2 = 5 multiplications in all!

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

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) February 28, 2007 Harvard Bits

III. Modular arithmetic 1 2 3 4 5 6 7 6 + 5 = 3 (mod 8) February 28, 2007

Math Quiz 2 x 6 = mod 11 2 x 6 x 5 23 mod 7 2300 1 5 1 1 = (23)100 = 1100 = 1 February 28, 2007

(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) February 28, 2007 Harvard Bits

Fast Modular Exponentiation Problem: Given q and p and n, find y < p such that qn = y (mod p) Method 1: multiply q by itself n-1 times Requires n-1 multiplications Method 2: use successive squaring Requires about log2n multiplications 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. February 28, 2007 Harvard Bits

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