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# Cryptograms: The Solutions

The following are the solutions for the puzzles on the page Cryptogram Puzzles.

### Cryptogram # 1

(Quote from a famous mathematician.)

fq pqv yqtta cdqsv aqwt fkhhkewnvkgu kp ocvjgocvkeu. k ecp cuuwtg aqw okpg atg uvknn itgcvgt. - cndgtv gkpuvgkp

The Solution:

Do not worry about your difficulties in mathematics. I can assure you mine are still greater. - Albert Einstein

This is an easy one, to say the least. If you read the page Cryptograms, you know that a Caesar cipher simply substitutes for each letter another letter that is a fixed number of positions away in the alphabet. In this case the shift-value is two, so a=c, b=d, c=e, etc.

If you suspect a simple shift-cipher (in this case it was a given), you don't need to use letter frequency analysis. What is referred to as a "brute force attack" will work fine. Start with a shift of one and test a piece of the ciphertext. In this case, "fq" would be "ep" (move back one position in the alphabet for each letter). It isn't a word, so you can stop there. A shift of two gives you "do," which is a word.

Looking at the alphabet in front of you (a b c d e f g h i j k l m n o p q r s t u v w x y z), you can easily move back two positions for each letter in the second word: "pqv" = "not". The rest can be deciphered in a couple minutes now that you have the key.

Interestingly, in this case, you could also have guessed that a "famous mathematician" might be Albert Einstein, and compared the name to the obvious attribution at the end of the ciphertext. Sure enough, the number of letters matches, and "gkpuvgkp" has a repeating pair of letters, just like the "ei" that repeats in "Einstein". If the shift had been greater than two, this might have been the faster way to solve the puzzle.

### Cryptogram # 2

3325863186 2432 881621163412 3216 24313124331933248826 1932 3216878621163412 53243325 89863232 248833868989242686884286 198834 87163186 3286883286 33251988 5386 25195786. - 341688 258631168934

The Solution:

There is nobody so irritating as somebody with less intelligence and more sense than we have. - Don Herold

First of all, you know that there cannot be one digit substituted for each letter, since there are only 10 digits. On the other hand, numbers of two, three or more digits could be substituted for each letter, so how do you determine how many are used? You make an educated guess.

You might notice that there is an even number of digits in each word, so you know that either two or four digits are being used for the letters. Why not more? Because there are four words with just four digits, which have to have at least one letter, right? But in fact, four one-letter words in a short quote seems less likely than four two-letter words, so two digit numbers are most probable (and if you are wrong, you start over and try it another way).

Now that you have decided the letters are represented by two-digit numbers, it will be easier to solve the cryptogram by separating out the letters. You can copy and paste the puzzle onto any word processing program to do this, or use pen and paper. I would also use dividers for words, as in this example:

33 25 86 31 86 | 24 32 | 88 16 21 16 34 12 | 32 16 | 24 31 31 24 33 19 33 24 88 26 | 19 32 | 32 16 87 86 21 16 34 12 | 53 24 33 25 | 89 86 32 32 | 24 88 33 86 89 89 24 26 86 88 42 86 | 19 88 34 | 87 16 31 86 | 32 86 88 32 86 | 33 25 19 88 | 53 86 | 25 19 57 86. - | 34 16 88 | 25 86 31 16 89 34

Now you can more easily see the numbers that represent the letters. Solving cryptograms like this usually involves "letter frequency analysis". On the page, "Code Breaking," there is a table showing the statistical averages for how often letters show up in English, but here is a simple distribution of letters from most-frequent to least frequent: e-t-a-o-i-n-s-h-r-d-l-c-u-m-w-f-g-y-p-b-v-k-j-x-q-z

If you write a list of the numbers used in the cryptogram, and then count how many times each appears, you can see that "86" is the most common:

11-0
12-2
13-0
14-0
15-0
16-8
19-5
24-7
25-5
26-2
27-0
21-2
31-5
32-8
33-6
34-5
42-1
53-2
55-0
57-1
86-13
87-2
88-8
89-4
90-0
91-0

In fact, it occurs 13 times, which is 5 more than the next closest number. It is almost certainly one of the first three letters in the frequency table, so you start with "e", since it is the most common by far. Replacing every "86" with an "e", you can look for any other related clues. The next to last word before the author's name is "53 86", for example, meaning it is a two-letter word ending in "e", probably "be", "he", "me", or "we". "53" then, is one of the four letters: b, h, m or w. Since it shows up only twice, it is most likely a "b" or "w", since these are less common than the others.

You can play with this for a minute, to see if it yields anything. For example, "53 24 33 25" (8th word) is the only other word with "53" in it, so what four letter words start with "b" or "w"? There are fewer "w" words, so start with those. "With" and what" and "when" come to mind, but we can eliminate the last because there is no "86" in the word ("e").

"With" would make the following true: 53=w, 24=i, 33=t, 25=h. Applying these to the first word of the cryptogram, you get "t-h-e-31-e". that sure looks like a word if the "31" is an "r". Furthermore, if the first word is "there", what could the two-letter second word be? Almost certainly it would be "is", giving us 24=i and 32=s. This is a speculative guessing game to some extent, and it isn't likely you took this route to a solution. But you can guess like this and test your guesses.

Otherwise you return to the frequency table and note that "16", "32" and "88" tie for the second most common numbers, appearing 8 times each. Anyone of these are likely to be any of the next five or six letters in the frequency table. You may notice that a small sample of ciphertext makes using the frequency table more difficult. In this case, for example, the letter "t" is not the second most common letter, as is normal in the English language, but sixth.

Essentially you keep testing and looking for patterns. If a given assumption doesn't work to produce a word, you drop it and try another. If you notice that the word "89 86 32 32" ends in a double letter, you might stop to think about the possibilities. An "ss" or "ll" is most likely, and you know that "86" is probably "e", so you can look at possible four-letter "ess" or "ell" words to determine the first letter.

Of course, don't limit yourself to clues within the cryptogram when looking for solutions. I did mention in the original puzzle that it was a quote about intelligence. That made it likely that the word "intelligence" might be in the message. Look for a 12-letter word with a double letter in in and you'll find one: "24 88 33 86 89 89 24 26 86 88 42 86". That gives you a lot of information to work with.

Note: All the pages on codes, ciphers and cryptograms are listed on the page: Secret Codes.

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