Breaking a Cæsar Cipher

Introduction

This presents two different methods of breaking a Cæsar cipher using statistics. The ciphertext is VRVRVHDPVWUHVV; the frequency counts are: 
D    1    0.0714        R    2    0.1429       W    1    0.0714
H    2    0.1429        U    1    0.0714
P    1    0.0714        V    6    0.4286
As a preliminary, we note that the unicity distance of a Cæsar cipher is roughly 1.5, so 14 characters should be more than enough to find a unique key and decryption.

Method #1: frequency counts

The most frequent letters in English are ETAONRISH. The following table summarizes the overlap, assuming the key is the number indicated: 
key   ct  letters    key   ct  letters    key    ct  letters    key   ct  letters       
0     2   RH          7    4   VHPU       13     3   RVU        20    2   HU
1     2   PU          8    3   VPW        14     3   VHW        21    1   VD
2     3   VPW         9    2   RW         15     4   HDPW       22    3   DPW
3     6   VRHDWU     10    2   RD         16     3   HDU        23    1   P
4     3   VRW        11    2   DP         17     2   VR         24    2   RP
5     1   W          12    2   DU         18     2   WU         25    3   RHD
6     1   Y                               19     2   WU
The key value 3 leaps out. Notice that had we counted differences, the same thing would happen, as the differences are the same as the counts.

Method #2: correlations

For this, we need frequency counts of the letters in plaintext, and for this purpose (since we don't have the plaintext handy) we simply assume the plaintext follows the 1-gram, independent model. Note that this is usually a poor assumption since in English the letters are not independent, but we'll try it and see what happens.

The relevant equation is:

SUM0<=t<26Ft(c)p(t-k) = 0.0714[p(3-k) + p(15-k) + p(20-k) + p(22-k)] + 0.1429[p(7-k) + p(17-k)] + 0.4286p(21-k)

and the correlations over the values of k come out to be:

key        r        key        r        key        r        key        r        
0        0.0281     7        0.0590     13       0.0550     20       0.0252     
1        0.0295     8        0.0425     14       0.0513     21       0.0500     
2        0.0639     9        0.0331     15       0.0388     22       0.0253     
3        0.0762     10       0.339      16       0.0400     23       0.0186     
4        0.0514     11       0.0229     17       0.0763     24       0.0252     
5        0.0160     12       0.0183     18       0.0330     25       0.0341     
6        0.0214                         19       0.0312
The values 3 and 17 are just about equal, so both must be tried.

Answer

The key is 3; the message is SOSOSEAMSTRESS.

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Department of Computer Science
University of California at Davis
Davis, CA 95616-8562


Page last modified on 5/2/97