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6.4. FINAL MONOALPHABETIC TRICKS 99
and penalize those that touch it.
C D F J K Q S T X total total2 total3
C 0 2 1 0 1 3 0 0 7 5 5
D 0 0 0 0 3 0 5 2 10 4 0
F 2 0 8 1 5 2 1 1 20 10 8
J 1 0 8 0 1 0 0 6 16 14 2
K 0 0 1 0 5 0 2 7 15 5 -9
Q 1 3 5 1 5 6 1 0 22 vowel
S 3 0 2 0 0 6 6 1 18 6 4
T 0 5 1 0 2 1 6 4 19 17 9
X 0 2 1 6 7 0 1 4 20 20 vowel
Step 4: Repeat twice more to find the other two aeio vowels.
C D F J K Q S T X total total2 total3 total4 total5
C 0 2 1 0 1 3 0 0 7 5 5 5 1
D 0 0 0 0 3 0 5 2 10 4 0 -10 -10
F 2 0 8 1 5 2 1 1 20 10 8 6 vowel
J 1 0 8 0 1 0 0 6 16 14 2 0 -16
K 0 0 1 0 5 0 2 7 15 5 -9 -13 -15
Q 1 3 5 1 5 6 1 0 22 vowel
S 3 0 2 0 0 6 6 1 18 6 4 -8 -12
T 0 5 1 0 2 1 6 4 19 17 9 vowel
X 0 2 1 6 7 0 1 4 20 20 vowel
So, according to this method, QXT and F are the vowels.
Now no method is foolproof. This one heavily depends on the letter that
contacts the common letters the most being a vowel, and if the first “vowel”
designation is wrong, this method will quickly lead into a morass. Here, Q has
the largest initial total, but only by 2. Both F and S have high totals, even
though they are consonants. If either had occured a couple more times, we’d
have had trouble.
Similarly, no method should be applied without thought. At the end of Step
4, C had a total of 5 and F a total of 6. Which is more likely to be a vowel
Vowels tend to dislike one another, and C touches QTX a total of once, while F
touches each of them for a total of 14 times. So, even though its total is a bit
smaller, it is more likely that C is a vowel.
6.4 Final Monoalphabetic Tricks
The examples in this chapter hopefully have convinced you that even a good
monoalphabetic substitution is not a very good cipher method. How, then,
should we make a better cipher The name mono gives us a hint: we must
change the one-for-one letter replacement and cause multiple letters to represent