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Arithmetic Algorithms, Part 1 DPV Chapter 1 Jim Royer EECS January 18, 2019 Royer Arithmetic Algorithms, Part 1 1/ 15

Multiplication à la Français function multiply(a, b) // input: two n-bit integers a and b with b 0 // output: a b if b = 0 then return 0 c multiply(a, b/2 ) if b is even then return (2 c) else return (a + 2 c) Correctness A proof by induction on b. Base Case: b = 0. Then multiply(a, b) = 0, which is correct. Induction Step: b > 0. (IH = Induction Hypothesis) IH: multiply(a, b ) = a b for b = 0,..., b 1. By the IH, c = a b/2 Case: b is even. Then: Case: b is odd. Then: (2 c) = 2 (a (b/2)) = a (2 (b/2)). = a b. (a + 2 c) = a + 2 (a b/2 ) = a (2 b/2 + 1) = a b. Royer Arithmetic Algorithms, Part 1 2/ 15

Multiplication à la Français, Continued function multiply(a, b) // input: two n-bit integers a and b with b 0 // output: a b if b = 0 then return 0 c multiply(a, b/2 ) if b is even then return (2 c) else return (a + 2 c) Run-time analysis n recursive calls (b drops by 1-bit in each call). O(n) cost of each step on the recursion. n O(n) = O(n 2 ). (Why?) Royer Arithmetic Algorithms, Part 1 3/ 15

Division Correctness Case a = 0:... Case a even and > 0:... Case a odd:... function divide(a,b) // input: two n-bit integers a and b with a 0 and b > 0 // output: (q, r) where a = q b + r and 0 r < b if a = 0 then return (0, 0) (q, r ) divide( a/2, b) q 2 q r 2 r if a is odd then r r + 1 if r b then r r b; q q + 1 return (q, r) Run-time analysis: Homework problem. On the board. On the board. Exercise for the reader. Royer Arithmetic Algorithms, Part 1 4/ 15

2019-01-18 Arithmetic Algorithms, Part 1 Division Division function divide(a,b) // input: two n-bit integers a and b with a 0 and b > 0 // output: (q, r) where a = q b + r and 0 r < b if a = 0 then return (0, 0) (q, r ) divide( a/2, b) q 2 q r 2 r if a is odd then r r + 1 if r b then r r b; q q + 1 return (q, r) Correctness Case a = 0:... On the board. Case a even and > 0:... On the board. Case a odd:... Exercise for the reader. Run-time analysis: Homework problem. Case a = 0. Then q = r = 0 and a = 0 = 0 b + 0 = q b + r and 0 = r b. Case a > 0 and a is even. Then q = 2q and r = 2r where (q, r ) = divide( a/2, b). IH: For a { 0,..., a 1 }, (q, r ) = divide(a, b) is such that a = q b + r and 0 r < b. Since a/2 < a, the IH applies with a = a/2. Hence, a/2 = q b + r and 0 r < b. Since 2 a/2 = a, a = 2 a/2 = 2q b + 2r and 0 2r < 2b SUBCASE: 2r < b: Then q = 2q and r = 2r and we are done. SUBCASE: 2r b: Then q = 2q + 1 and r = 2r b and we are done.

Modular Arithmetic Definition Suppose a, b, N N. (i) a b def a divides b, i.e., b = k a for some k N. (ii) a b (mod N) def N (a b) a b = k N for some integer k. The substitution rule Suppose a a (mod N) and b b (mod N). Then a + b a + b (mod N) and a b a b (mod N). Modular addition, subtraction, and multiplication Suppose N is n bits long and 0 a, b < N. Then computing (a + b) mod N and (a b) mod N can be done in Θ(n) time. (a b) mod N can be done in Θ(n 2 ) time. Royer Arithmetic Algorithms, Part 1 5/ 15

Modular Exponentiation Exponentiation via repeated squaring 1, if b = 0; a b = (a b/2 ) 2, if b > 0 and even; a (a b/2 ) 2, if b is odd. function modexp(a, b, N) // input: a, b, and N :: three n-bit integers // with 0 a, b and 1 < N // output: a b mod N if b = 0 then return 1 c modexp(a, b/2, N) if b is even then return c 2 mod N else return (a c 2 ) mod N Example: x 1000 via 15 multiplies x 1000 = (x 500 ) 2 x 500 = (x 250 ) 2 x 250 = x (x 125 ) 2 x 125 = x (x 62 ) 2 x 62 = (x 31 ) 2 x 31 = x (x 15 ) 2 x 15 = x (x 7 ) 2 x 7 = x (x 3 ) 2 x 3 = x (x) 2 Royer Arithmetic Algorithms, Part 1 6/ 15

Modular Exponentiation, Continued function modexp(a, b, N) // input: a, b, and N :: three n-bit integers with 0 a, b and 1 < N // output: a b mod N if b = 0 then return 1 c modexp(a, b/2, N) if b is even then return c 2 mod N else return (a c 2 ) mod N Correctness: Easy. Runtime: Let n = the number of bits in max(a, b, N). At most n-many recursive calls. Why? In each call, two or three n-bit numbers are multiplied at cost Θ(n 2 ). n Θ(n 2 ) = Θ(n 3 ). Why? Royer Arithmetic Algorithms, Part 1 7/ 15

Euclid s algorithm for greatest common divisor Definition The greatest common divisor of a and b N is the largest d N such that d divides both a and b. I.E.: gcd(a, b) = max { d d a & d b }. Example 1035 = 3 2 5 23 & 759 = 3 11 23. gcd(1035, 759) = 3 23 = 69. For a > 0, gcd(0, a) = a. gcd(0, 0) = 0 by convention. Euclid s Rule Suppose a, b N +. Then gcd(a, b) = gcd(b, a mod b). Proof on next page Royer Arithmetic Algorithms, Part 1 8/ 15

Euclid s Rule: Suppose a, b N +. Then gcd(a, b) = gcd(b, a mod b). Proof. Recall: gcd(u, v) = def max({ d d u & d v }). Claim 1. If d a & d b, then ( x, y Z) [ d (x a + y b) ]. [Proof on Board] Observe: (a) a = a b b + 1 (a mod b) (b) a mod b = 1 a + ( a b ) b By (a) & Claim 1, gcd(b, a mod b) a. Since gcd(b, a mod b) b, we have: gcd(b, a mod b) gcd(a, b). (Why?) By (b) & Claim 1, gcd(a, b) (a mod b). Since gcd(a, b) b, we have: gcd(a, b) gcd(b, a mod b). (Why?) gcd(a, b) = gcd(b, a mod b). Royer Arithmetic Algorithms, Part 1 9/ 15

Euclid s algorithm, continued Euclid s Rule Suppose a, b N +. Then gcd(a, b) = gcd(b, a mod b). function Euclid(a, b) // Input: integers a and b with a b 0. // Output: the g.c.d. of a and b. if b = 0 then return a else return Euclid(b, a mod b). Correctness. Easy. Royer Arithmetic Algorithms, Part 1 10/ 15

Euclid s algorithm, Runtime analysis function Euclid(a, b) // Input: integers a and b with a b 0. Output: the g.c.d. of a and b. if b = 0 then return a else return Euclid(b, a mod b). Lemma Suppose a b > 0. Then (a mod b) < a/2. Proof. Case: b a/2. Then: (a mod b) < b a/2. Case: b > a/2. Then: (a mod b) = (a b) (a a/2) = a/2. Since Euclid(a, b) = Euclid(b, a mod b) = Euclid(a mod b, b mod (a mod b)) (generally), every two steps the a and b values are at least halved. On n-bit numbers, Euclid stops after 2n recursions. On n-bit numbers, mod (i.e., a division) costs O(n 2 ) 2n O(n 2 ) = O(n 3 ). Royer Arithmetic Algorithms, Part 1 11/ 15

The extended Euclid algorithm Lemma Suppose d a & d b & d = xa + yb for some x, y Z. Then d = gcd(a, b). Proof. Royer Arithmetic Algorithms, Part 1 12/ 15

The extended Euclid algorithm Lemma Suppose d a & d b & d = xa + yb for some x, y Z. Then d = gcd(a, b). Proof. Since d a and d b, then d gcd(a, b). Royer Arithmetic Algorithms, Part 1 12/ 15

The extended Euclid algorithm Lemma Suppose d a & d b & d = xa + yb for some x, y Z. Then d = gcd(a, b). Proof. Since d a and d b, then d gcd(a, b). Since gcd(a, b) a & gcd(a, b) b, Royer Arithmetic Algorithms, Part 1 12/ 15

The extended Euclid algorithm Lemma Suppose d a & d b & d = xa + yb for some x, y Z. Then d = gcd(a, b). Proof. Since d a and d b, then d gcd(a, b). Since gcd(a, b) a & gcd(a, b) b, then gcd(a, b) (xa + yb), i.e., gcd(a, b) d. Therefore, gcd(a, b) d. Therefore, d = gcd(a, b). function extended-euclid(a, b) // Input: integers a and b with a b 0. // Output: (x, y, d) where d = gcd(a, b) and d = xa + yb. if b = 0 then return (1, 0, a). (x, y, d) = extended-euclid(b, a mod b) return (y, x a/b y, d) Royer Arithmetic Algorithms, Part 1 12/ 15

The extended Euclid algorithm: Base case function extended-euclid(a, b) // Input: integers a and b with a b 0. // Output: (x, y, d) where d = gcd(a, b) and d = xa + yb. if b = 0 then return (1, 0, a). (x, y, d) = extended-euclid(b, a mod b) return (y, x a/b y, d) Proof of correctness, base case. Base case: b = 0. gcd(a, b) = a & a = 1 a + 0 b. So (1, 0, a) is right. Royer Arithmetic Algorithms, Part 1 13/ 15

The extended Euclid algorithm: Induction Step function extended-euclid(a, b) // Input: integers a and b with a b 0. // Output: (x, y, d) where d = gcd(a, b) and d = xa + yb. if b = 0 then return (1, 0, a). (x, y, d) = extended-euclid(b, a mod b) return (y, x a/b y, d) Proof of correctness, induction step. Suppose b > 0. IH: extended-euclid(a, b ) is correct for all a and each b = 0,..., b 1. Let (x, y, d) = extended-euclid(b, a mod b). Note: a mod b < b. Royer Arithmetic Algorithms, Part 1 14/ 15

Modular division Definition x is the multiplicative inverse of a mod N when a x 1 (mod N). The inverse might not exist! E.g., 2 1 mod 6 does not exist. Theorem (Modular Division Theorem) Suppose N > 2 and a { 1,..., N 1 }. (a) a has an inverse mod N gcd(a, N) = 1. (b) When a 1 mod N exists, (a 1 mod N) = (x mod N), where (x, y, 1) = extended-euclid(a, N) so that 1 = a x + N y. Royer Arithmetic Algorithms, Part 1 15/ 15