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Can Y@@ou D%ecry*pt This? (Posted on 2003-07-07) Difficulty: 4 of 5

Thi#s me$ssag@e* cont@ai%ns$ a sec@ret m%@es#*sa@$g$e, hid@d@en wi@th%in% it. Yo*%u ma#y enj@oy t#ryi@ng t#o so@lve$ this% p$uzz#l$e, may@be not.

See The Solution Submitted by DJ    
Rating: 4.4286 (21 votes)

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Solution re(2): solution | Comment 15 of 18 |
(In reply to re: solution by Rob)

Which part do you understand the least? I'll try to re-explain.

There are 5 symbols in this puzzle (in order they appear in the number row of the keyboard): @, #, $, %, *

There are always 0-4 letters preceding a symbol. We can write the number of preceding letters and the symbol as an ordered pair: (0, @), etc.

For example, in the title "Can Y@@ou D%ecry*pt This?"
We have 4 letters (Can Y) preceding @, so you get (4, @). Then 0 letters before the next @, so (0, @). Continuing, you get (3, %) and (4, *).

The following are all the ordered pairs in the encrypted message:
(3, #) (3, $) (4, @) (1, *) (4, @) (2, %) (2, $) (4, @) (4, %) (0, @) (2, #) (0, *) (2, @) (0, $) (1, $) (4, @) (1, @) (4, @) (2, %) (2, %) (4, *) (0, %) (3, #) (4, @) (3, #) (3, @) (3, #) (3, @) (3, $) (4, %) (1, $) (3, #) (1, $) (4, @)

Now, you'll notice that there were 5 symbols, and that the numbers in the ordered pairs range from 0-4. This gives 25 possible combinations (which turn out to be 25 distinct letters). The most common ordered pair is (4, @), so this is most likely E.

This can be confirmed after reading the hint, where it states that you will find the decryption of the title "easy". So, we can logically choose to assign:
(0, @) = A
(1, @) = B
(2, @) = C
(3, @) = D
(4, @) = E
(0, #) = F
(1, #) = G

...and so on throughout the alphabet (using the symbols in the order they appear on the keyboard).

Any additional questions, let me know!
  Posted by derek on 2003-07-09 03:22:31

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