Cryptograms: The Solutions
The following are the solutions for the puzzles on the page
Cryptogram # 1
(Quote from a famous mathematician.)
fq pqv yqtta cdqsv aqwt fkhhkewnvkgu
kp ocvjgocvkeu. k ecp cuuwtg aqw okpg atg uvknn itgcvgt. - cndgtv
Do not worry about your difficulties
in mathematics. I can assure you mine are still greater. - Albert
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
(Quote about intelligence.)
3325863186 2432 881621163412 3216 24313124331933248826
1932 3216878621163412 53243325 89863232 248833868989242686884286
198834 87163186 3286883286 33251988 5386 25195786. - 341688 258631168934
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
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:
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.