If you need a reminder on how the Nihilist Substitution Cipher works click here.

To find the period you assume it is a particular period and put in blocks of 2 in columns of the period, then you do an diagraphic index of coincidence calculation on each column and take the average of all the columns.

This is an example of the difference between the expected English index of coincidence (0.0667) and the average Index of Coincidence Calculation for periods 2-40. Hence the smaller the bar the closer it is to that of English.

As you can see for this particular text it is very obvious that the period is 3 because all the of multiples of 3s are very close to English. This is because the key ‘MAN’ – period 3 is the same as ‘MANMAN’ – period 6.

Once the period has been identified place the ciphertext into blocks of 2 in columns of the correct period.

Example:

345173345643531536543672… has been found to have a period of 3

? ? ? = Key 34 51 73 34 56 43 53 15 36 54 36 72 ........

From this point on you treat each column separately as they are all encoded by a different letter. From here we use each number digraph to narrow down the possible keys. We can infer things from ciphertext for example if the second digit is 0 there was only one way it could have been created that would be the plaintext number and the key number ending in a 5.

This can be extended to create inequalities for all possible ciphertext number digraphs. This is some pseudocode to create an inequalities for both the row and column.

rowMin = 1 rowMax = 5 colMin = 1 colMax = 5 no = ciphertext number digraph IF no is smaller than 11 THEN no = no + 100 col = no % 10 IF col equals 0 THEN colMin = 5 colMax = 5 no = no - 10 ELIF col smaller than 7 THEN colMin = 1 colMax = col - 1 ELSE colMin = col - 5 colMax = 5 row = floor(no / 10) % 10 IF row equals 0 THEN rowMin = 5 rowMax = 5 ELIF row smaller than 7 THEN rowMin = 1 rowMax = row - 1 ELSE rowMin = row - 5 rowMax = 5

You apply this algorithm to all number digraphs in each column and then create an equation for the row and column of the key number. The equation will be…

rowMin <= row <= rowMax colMin <= col <= colMax

You then use these to narrow down the possibility, lets say you had the inequalities …

2 <= row <= 4 & 3 <= row <= 5 & 2 <= row <= 3

From these three inequalities you can infer that:

3 <= row <= 3 hence row = 3

So you now know that for that columns the key number must starts with a 3. You can then get the inequalities for the column and then create the full key which in this case will now be 31, 32, 33, 34 or 35.

Once the key has been found for each column subtract it away from each number in its respective column. Now if there have been no mistakes there should be less than 25 (size of polybius square with I/J being 1 character) number digraphs. Convert each unique one into a unique letter. Example: swap out all 24 for ‘A’s all 45 for ‘B’s, all 86 for ‘C’s etc.

You are now left will a simple substitution cipher, I wont go into detail on how to break it here, but I have a page here on how to break a simple substitution cipher. Tips: The most common letter in the new ciphertext will likely be ‘E’, the most common trigraph ‘THE’ and so on.