# An analogous argument shows that one can compute all

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. An analogous argument shows that one can compute all of the s i also in time O ( L ( a ) L ( b )), and in fact, in time O ( L ( b ) 2 ). 2 We should point out that the Euclidean algorithm is not the fastest known algorithm for com- puting greatest common divisors. The asymptotically fastest known algorithm for computing the greatest common divisor of two numbers of bit length at most k runs in time O ( k (log k ) 2 log log k ). One can also compute the corresponding values s and t within this time bound as well. Fast algo- rithms for greatest common divisors are not of much practical value, unless the integers involved are very large — at least several tens of thousands of bits in length. 3.4 Computing in Z n Let n > 1. For computational purposes, we may represent elements of Z n as elements of the set { 0 , . . . , n - 1 } . Addition and subtraction in Z n can be performed in time O ( L ( n )). Multiplication can be performed in time O ( L ( n ) 2 ) with an ordinary integer multiplication, followed by a division with remainder. Given a ∈ { 0 , . . . , n - 1 } , we can determine if [ a mod n ] has a multiplicative inverse in Z n , and if so, determine this inverse, in time O ( L ( n ) 2 ) by applying the extended Euclidean algorithm. More precisely, we run the extended Euclidean algorithm to determine integers d , s , and t , such that d = gcd( n, a ) and ns + at = d . If d 6 = 1, then [ a mod n ] is not invertible; otherwise, [ a mod n ] is invertible, and [ t mod n ] is its inverse. In the latter case, by part (vi) of Theorem 3.5, we know that | t | ≤ n ; we cannot have t = ± n , and so either t ∈ { 0 , . . . , n - 1 } , or t + n ∈ { 0 , . . . , n - 1 } . Another interesting problem is exponentiation modulo n : given a ∈ { 0 , . . . , n - 1 } and a non- negative integer e , compute y = a e rem n . Perhaps the most obvious way to do this is to it- eratively multiply by a modulo n , e times, requiring time O ( e L ( n ) 2 ). A much faster algorithm, the repeated-squaring algorithm , computes y = a e rem n using just O ( L ( e )) multiplications modulo n , thus taking time O ( L ( e ) L ( n ) 2 ). This method works as follows. Let e = ( b - 1 · · · b 0 ) 2 be the binary expansion of e (where b 0 is the low-order bit). For 0 i , define e i = ( b - 1 · · · b i ) 2 . Also define, for 0 i , y i = a e i rem n , so y = 1 and y 0 = y . Then we have e i = 2 e i +1 + b i (0 i < ‘ ) , 17

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and hence y i = y 2 i +1 · a b i rem n (0 i < ‘ ) . This idea yields the following algorithm: y 1 for i - 1 down to 0 do y y 2 rem n if b i = 1 then y y · a rem n output y It is clear that when this algorithm terminates, y = a e rem n , and that the running-time estimate is as claimed above. We close this chapter by observing that the Chinese Remainder Theorem (Theorem 2.6) can be made computationally effective as well. Indeed, by just using the formulas in the proof of that theorem, we see that given integers n 1 , . . . , n k , and a 1 , . . . , a k , with n i > 1, gcd( n i , n j ) = 1 for i 6 = j , and 0 a i < n i , we can compute x such that 0 x < n and x a i (mod n i ) in time O ( L ( n ) 2 ), where n = Q i n i . We leave the details of this as an easy exercise.
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• Spring '13
• MRR

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