Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 4 - Section 4.6 - Cryptography - Supplementary Exercises - Page 307: 10

Answer

$k = 13$ for the case k = 13, the enciphering function for the shift cipher is then the ame as the deciphering function.

Work Step by Step

Enciphering function of the shift cipher: f(p) = (p+k) mod 26 Corresponding deciphering function: f-(p) = (p-k) mod 26 Note: the trivial case k = 0 also makes the enciphering function for the shift cipher is the same as the deciphering function. Let us assume that & is not zero. We want the two functions to be the same (p + k) mod 26 = (p - k) mod 26 Subtract p from each side of the previous equation: k mod 26 = -k mod 26 Adding 26 does not have an effect on the result due to the modulo 26: k mod 26 = 26 - k mod 26 Since k is a number between 0 and 25 (including), 26-k is a number between 1 and 26. Assuming that & is not zero, we then obtain: k = 26 - k Add & to each side of the equation: 2k = 26 Divide each side of the equation by 2: k = 13 for the case k = 13, the enciphering function for the shift cipher is then the as the deciphering function.
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