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

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9.3. RECOGNIZING AND BREAKING POLYGRAPHIC CIPHERS 175<br />

Repetition <strong>St</strong>art Positions Distance Factors<br />

ELX 10 22 2 × 11<br />

XEL 12 60 2 × 2 × 3 × 5<br />

RQLXFKELRK 55 112 2 × 2 × 2 × 2 × 7<br />

LRKQF 62 88 2 × 2 × 2 × 11<br />

RLFER 28 56 2 × 2 × 2 × 7<br />

QXKLK 117 56 2 × 2 × 2 × 7<br />

LKF 120 44 2 × 2 × 11<br />

Figure 9.4: Repetitions in the unknown cipher.<br />

ones. And the large number <strong>of</strong> repetitions <strong>of</strong> length 2 at distances divisible by<br />

2 would lead us to digraphic ciphers anyway. So digraphic it is.<br />

Since we’ve recognized it, how do we break it To attack a digraphic cipher<br />

we start as we always did – with a frequency count – except we count pairs <strong>of</strong><br />

letters, not singletons. The only pairs appearing more than twice are EL and<br />

LX, 5 times each, and KO, QF and RK, 3 times each.<br />

Now what We will grant ourselves the knowledge that it is a Hill cipher.<br />

In many ways Hill ciphers are a fancy version <strong>of</strong> the Decimation and Linear<br />

Ciphers from Chapters 3 and 4. They do do a better job <strong>of</strong> mixing up the<br />

frequencies than those ciphers did, but, just as correctly guessing two letters <strong>of</strong><br />

a ciphertext enciphered with a Linear Cipher (and some math) leads to a decryption,<br />

correctly guessing two bigrams (and some math) leads to a decryption<br />

<strong>of</strong> a Hill cipher. (However, it is generally much harder to guess bigrams than<br />

letters.) Here, the pairs EL and LX must be very common bigrams. If we know<br />

which, we can solve for the enciphering matrix.<br />

For ease, we will make a (miraculously correct) guess, that LX=th and<br />

EL=at. This is, actually, fairly reasonable as th is the most common ( bigraph ) a b<br />

and at is the eleventh most common. So there is some matrix with<br />

c d<br />

( ( ) ( ( ( ) ( a b L t a b E a<br />

= and<br />

= . Substituting the values for<br />

c d)<br />

X h)<br />

c d)<br />

L t)<br />

( ) ( ) ( ) ( ( ) ( a b 12 20 a b 5 1<br />

the letters, and we have<br />

= and<br />

= .<br />

c d 24 8 c d)<br />

12 20)<br />

Multiplying out gives two sets <strong>of</strong> equations<br />

12a + 24b = 20 and 5c + 12d = 20<br />

12c + 24d = 8 5a + 12b = 1.<br />

These( two sets ) <strong>of</strong> two equations in two unknowns may now be solved to see<br />

9 5<br />

that<br />

is the original enciphering matrix. From here, the quote may<br />

10 17

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