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I am trying to find the plain text for the following cipher text using a frequency analysis

vr pvst yqlp mq nvf

But for the letters above this is really difficult, as if I use the alphabetic substitution technique, where I assume that v corresponds to e as this occurs most in the english alphabet. What about p, this also occurs three times, does this mean it corresponds to e. Same for r,s,t,y,l,m,n,j,f which occur only once. Is there an algorithm which looks at such duplicated frequencies, or do I guess:

v and p corresponds to e q corresponds to t r to f correspond to a

?

even using the bigram method for something like this is difficult as a bigram of pv is found, if p = t and v = h that means p doesn't correspond to t anymore?

Thanks in advance!

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  • $\begingroup$ Cross-posted on Cryptography.SE. crypto.stackexchange.com/q/12330/351 Please don't cross-post -- that is forbidden under site rules. $\endgroup$
    – D.W. ♦
    Commented Dec 16, 2013 at 1:53

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I'm by no means an expert but it seems to me that your sample is far too short to be amenable to frequency analysis or, indeed, to cryptanalysis of any kind. Assuming a substitution cypher, you have only the information that eleven letters are used, and characters 1, 4 and 14; 3 and 10; and 8 and 11 are the same. There are fifteen unused characters so there's a huge amount of freedom. You only have a little more information than a one-time pad would give you.

In a long sample of text, it is likely but is not guaranteed that the most frequent character is 'e'; in a fifteen character sample, it's perfectly possible that there are no e's at all. ("Look at my odd words" is a sixteen-letter example.) Frequency analysis helps you figure out the substitutions but it doesn't guarantee to give you the answer with no more work.

Assuming the spaces are significant, your message could be "My emus love no imp". Or it could be something completely different.

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  • $\begingroup$ how did you work out the message? $\endgroup$
    – joker
    Commented Dec 15, 2013 at 10:34
  • $\begingroup$ @joker I just wrote the cyphertext on a piece of paper and started guessing letters underneath. I went through a couple of iterations before I found one that made reasonable sense as an English sentence. And, just now, I changed it because I found one that read better. $\endgroup$
    – David Richerby
    Commented Dec 15, 2013 at 11:13
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    $\begingroup$ @joker I'm sorry for not being very helpful in my previous comment. To figure out a sentence that matches the code, I started by guessing the two-letter word at the start, since the first letter of that appears twice more. I then tried to guess the first letter of the second word, bearing in mind that the last letter of the third word has to be the same. At that point, I had XY ZX-- ---Z -- -X- and you either figure out something to put in the rest of the gaps, or you don't so you try something else. "At Ears Cove, go mad." "Ay, Dave sold to Ian." $\endgroup$
    – David Richerby
    Commented Dec 15, 2013 at 12:08

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