Suppose you compute using the algorithm for modular exponentiation discussed in lecture and shown below: modExp (b, n, m): x = 1 01 P = b mod m for i = 0: k-1: 11 return x 121 Recall that, the algorithm assumes we know or can find the binary expansion of n so that n = (ak-1, ak-2,..., ao)2. If the initial values are x and p are 1 and 11 respective the possible values that x will attain at some iteration of the algorithm. 451 221 if a_i = 1: 226 391 x = (xp) mod m P = (pp) mod m 11644 mod 645
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