3 into a simple algorithm as follows while b 6 0 do

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We can easily turn the scheme described in Theorem 3.3 into a simple algorithm as follows: while b 6 = 0 do Compute q, r such that a = bq + r , with 0 r < b ( a, b ) ( b, r ) output a By Theorem 3.3, this algorithm, known as Euclid’s algorithm , outputs the greatest common divisor of a and b . Theorem 3.4 Euclid’s algorithm runs in time O ( L ( a ) L ( b )) . Proof. The running time is O ( τ ), where τ = i =1 L ( r i ) L ( q i ). We have τ ≤ L ( b ) X i L ( q i ) ≤ L ( b ) X i (log 2 q i + 1) = L ( b )( + log 2 ( Y i q i )) . Note that a = r 0 r 1 q 1 r 2 q 2 q 1 ≥ · · · ≥ r q · · · q 1 q · · · q 1 . We also have log b/ log φ + 1. Combining this with the above, we have τ ≤ L ( b )(log b/ log φ + 1 + log 2 a ) = O ( L ( a ) L ( b )) , which proves the theorem. 2 Let d = gcd( a, b ). We know that there exist integers s and t such that as + bt = d . The extended Euclidean algorithm , which we now describe, allows us to compute s and t . 15
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Theorem 3.5 Let a , b , r 0 , r 1 , . . . , r +1 , and q 1 , . . . , q be as in Theorem 3.3. Define integers s 0 , s 1 , . . . , s +1 and t 0 , t 1 , . . . , t +1 as follows: s 0 := 1 , t 0 := 0 , s 1 := 0 , t 1 := 1 , and for 1 i , s i +1 := s i - 1 - s i q i , t i +1 := t i - 1 - t i q i . Then (i) for 0 i + 1 , we have s i a + t i b = r i ; in particular, s a + t b = gcd( a, b ) ; (ii) for 0 i , we have s i t i +1 - t i s i +1 = ( - 1) i ; (iii) for 0 i + 1 , we have gcd( s i , t i ) = 1 ; (iv) we have | s +1 | ≤ b and | t +1 | ≤ a ; (v) for 0 i , we have | t i | ≤ | t i +1 | , and for 1 i , we have | s i | ≤ | s i +1 | ; (vi) for 0 i + 1 , we have | s i | ≤ b and | t i | ≤ a . Proof. (i) and (ii) are easily proved by induction on i (exercise). (iii) follows directly from (ii). To prove (iv), note that s +1 a + t +1 b = r +1 = 0. We have t +1 6 = 0, since otherwise, both s +1 and t +1 would be zero, contradicting (ii). This implies that t +1 /s +1 = - a/b , and then (iv) follows from the fact (iii) that gcd( s +1 , t +1 ) = 1. For (v), one proves the statement for the t i by induction, but with the stronger hypothesis that t i t i +1 0 (i.e., the sign alternates) and | t i | ≤ | t i +1 | for 0 i (exercise). One argues similarly for the statement for the s i . (vi) follows immediately from (iv) and (v). 2 Example 3.2 We continue with Example 3.1. The numbers s i and t i are easily computed from the q i : i 0 1 2 3 4 r i 100 35 30 5 0 q i 2 1 6 s i 1 0 1 -1 7 t i 0 1 -2 3 -20 2 We can easily turn the scheme described in Theorem 3.5 into a simple algorithm, as follows: s 1 , t 0 s 0 0 , t 0 1 while b 6 = 0 do Compute q, r such that a = bq + r , with 0 r < b ( s, t, s 0 , t 0 ) ( s 0 , t 0 , s - s 0 q, t - t 0 q ) ( a, b ) ( b, r ) output a, s, t 16
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This algorithm, known as the extended Euclidean algorithm , computes the greatest common divisor d of a and b , together with s and t such that as + bt = d . Theorem 3.6 The extended Euclidean algorithm runs in time O ( L ( a ) L ( b )) . Proof. It suffices to analyze the cost of computing the sequences { s i } and { t i } . Consider first the cost of computing all of the t i , which is O ( τ ), where τ = i =1 L ( t i ) L ( q i ). By Theorem 3.5 part (vi), and arguing as in the proof of Theorem 3.4, we have τ = L ( q 1 ) + X i =2 L ( t i ) L ( q i ) ≤ L ( q 1 ) + L ( a )( - 1 + log 2 ( Y i =2 q i )) = O ( L ( a ) L ( b )) , using the fact that Q i =2 q i b . An analogous argument shows that one can compute all of the
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