Solutions Manual for an Introduction to Cryptography with by Wade Trappe, Lawrence C. Washington

By Wade Trappe, Lawrence C. Washington

The accompanying strategies guide to an advent to Cryptography with Coding idea (2nd variation) by means of Wade Trappe, Lawrence C. Washington (Pearson).

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Additional resources for Solutions Manual for an Introduction to Cryptography with Coding Theory (2nd Edition)

Example text

Bw−j , B = bw−j+1 . . bw P and c = 2j P for some value of j. If a is even, step 3 does nothing, so the output still has the desired form. If a is odd, then the last bit bw−j of a is 1. The new a is a = b1 . . bw−j−1 0. Also, the new B is bw−j bw−j+1 . . bw P + 2j P = (1bw−j bw−j+1 . . bw )P = bw−j bw−j+1 . . bw P . The new C is still 2j P . If the new a = 0, then j = w−1, so B = xP . Therefore step 5 outputs xP , as desired. Otherwise, step 4 sends us to step 2, which outputs a = b1 . .

0), or has more than t nonzero entries, in which case it has at most t 0’s. Then it has distance at most t from (1, 1, . . , 1). Therefore, the spheres of radius t around (0, 0, . . , 0) and (1, 1, . . , 1) cover the space of all n-tuples. Since d = 2t + 1 for this code, it is perfect (see the bottom of page 307). (b) We know that a perfect code satisfies the Hamming bound with equality. This means that 2n , 2=M = t n j=0 t n j=0 j j n−1 which implies that =2 . 16. (a) Let C1 consist of all vectors of the form (a, a, .

6. (a) P + P = (5, 16). Now compute 3P = 2P + P . The slope is (9 − 16)/(10 − 5) = −7/5. But gcd(5, 35) = 5, so we have the factorization 35 = 5 · 7. (b) We have 2yy ′ ≡ 3x2 +5, which yields 56y ′ ≡ 8, so y ′ = 1/7. But gcd(7, 35) = 7, so we have the factorization 35 = 5 · 7. 7. The tangent line at (2,0) has vertical slope, so 2P = ∞. This is infinity mod all factors of n, so no factor is singled out. The gcd that we hope will give us a factor, will give us the factor n of n. 8. Choose an elliptic curve E mod some large prime p, and choose a random 40 41 point Q on E.

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