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PHLIV VFIPH XVA

      To find n, write the cipher text on a long strip of paper, and copy it onto a second strip of paper beneath the first strip. For a displacement of 1, move the bottom strip to the left by one letter and count the number of character coincidences. Repeat this process for several displacements until the maximum displacement is obtained. The shift shown below corresponds to a displacement of 3:

V V H Z K U H R G F H G K D ellipsis ellipsis
V V H Z K U H R G F H G K D K I T ellipsis ellipsis
Displacement Number of coincidences
1 4
2 4
3 9
4 12
5 5
6 2
7 7
8 7

      From our results, the maximum number of occurrences appears for a displacement of 4. Since we know the maximum displacement occurs for a scalar multiple of the period, the period is likely either 2 or 4.

      Remark

      In applying the second principle, we are using a probabilistic argument. That is, in the above example, we cannot be certain that the period is either 2 or 4; however, we can say with high probability that it is likely to be either 2 or 4. If we were unable to decipher the text with a keyword length of 2 or 4, we would then try with the next highest number of coincidences, which occurs for displacement 3.

      Finding the keyword

      We will first try the case where the period is 4, and we will determine the character frequencies for the {1 Superscript s t Baseline comma 5 Superscript t h Baseline comma 9 Superscript t h Baseline comma ellipsis} letters, {2 Superscript n d Baseline comma 6 Superscript t h Baseline comma 1 0 Superscript t h Baseline comma ellipsis} letters, and so on. Taking the 1 Superscript s t Baseline comma 5 Superscript t h Baseline comma 9 Superscript t h Baseline comma ellipsis letters, we get

VKGKT EVPUG JVCCP GDGPD DMGKG PVPA

      from which we obtain the following table of frequencies:

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A B C D E F G H I J K L M
1 0 2 3 1 0