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Cryptology - Unofficial St. Mary's College of California Web Site

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9.7. EXERCISES 181<br />

(b) The use <strong>of</strong> “alphabetical” in part (a) didn’t work. Instead, when<br />

the letters are in alphabetical order, add 4 to each. When they are<br />

not, subtract 3. Leave doubles alone. Will this work (Again try to<br />

encipher and decipher some words.)<br />

3. A slightly fancier digraphic cipher is as follows. If the letters are both<br />

vowels, replace each by the vowels before them in the alphabet (ea become<br />

AU.) If the pair is two consonants, replace each by the next consonant in<br />

the alphabet (ns becomes PT). And if the pair is is one each, swap them<br />

(ho becomes OH).<br />

(a) Encipher vowels.<br />

(b) Encipher breakers.<br />

(c) Decipher ROEAP VLA.<br />

(d) Decipher EBUOVZ.<br />

(e) We’ve learned to recognized monoalphabetic ciphers by their odd<br />

appearance – too few good etaoinshr letters, too many bad jkvwxyz<br />

letters. What sort <strong>of</strong> appearance will this cipher produce<br />

4. To make a combination Caesar-AutoKey Digraphic cipher, encipher each<br />

pair <strong>of</strong> letters by using a Caesar cipher with key k on the first letter, and<br />

an Auto Key Cipher on the second letter, using the first letter as the key.<br />

Symbolically, for each pair p 1 p 2 <strong>of</strong> plaintext letters, the ciphertext pair is<br />

(p 1 + k)%26, (p 1 + p 2 )%26. So if k = 3, then to becomes WH, as W is three<br />

letters past t, and o enciphered with key T is H.<br />

(a) Encipher combinations with key 7.<br />

(b) Decipher JCLWH R, if the key was 12.<br />

(c) Encipher primer if the key is 21.<br />

(d) Decipher KZKVE MZAKV, if the key was 6.<br />

5. In a table-based digraph cipher, like Figure 9.2, it is best if each letter<br />

appears in each row exactly once, and similarly in each column exactly<br />

once. If we had a three letter language “ABC” or four letter language<br />

“ABCD” then examples <strong>of</strong> such tables are<br />

AB BC CA<br />

CC AA BB<br />

BA CB AC<br />

and<br />

DC AD BB CA<br />

CD BC AA DB<br />

AB CA CC BD<br />

BA CB DD AC<br />

These arrangements are called Greek-Latin Squares, and exists for every<br />

size <strong>of</strong> square, except 2 × 2 and 6 × 6.<br />

(The 6×6 case is <strong>of</strong>ten referred to as the 36 Officers Problem: Suppose we<br />

have 36 military <strong>of</strong>ficers <strong>of</strong> 6 ranks from 6 different states, no two <strong>of</strong> whom

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