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).

**Read or Download Solutions Manual for an Introduction to Cryptography with Coding Theory (2nd Edition) PDF**

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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.