The Gronsfeld Cipher -- A Variant of Vigenere

This page describes a method for attacking a Gronsfeld cipher. It is based on the approach described in F. Pratt, Secret and Urgent, NY: Bobbs-Merrill, 1939.

Here's a message written in a Gronsfeld Cipher.

cjifk qywtj ioipo wovlh ncxlo peosg gxrkx baiiq caguy rxrlq klcoy vewql nhsut oiddg qdrap dnfwk owpgw gzlsk xlt

The Gronsfeld cipher is a variation of the Vigenere cipher in which a key number is used instead of a keyword, e.g., 14965. Usually the key does not contain repeated digits.

For this problem, I've simplified things as follows: we allow only the digits between 0-5 (a-e) to be used in the key. The method for attacking a Gronsfeld cipher involves the following steps:

Step 1. Write the first line of the message, and then write under each of its letter, the letters that precede it in the alphabet. Since we know that this version of Gronsfeld uses only numbers between 0-5, (a-e), we need 6 rows. I've numbered the rows and columns so that we can refer to them.

	   0  1  2  3  4  5  6  7
	0  c  j  i  f  k  q  y  w  tj ioipo wovlh ncxlo peosg  gxrkx  (Message)
	1  b  i  h  e  j  p  x  v  si hnhon vnukg mbwkn odnrf  fwqjw
	2  a  h  g  d  i  o  w  u  rh gmgnm umtjf lavjm ncmqe  evpiv
	3  z  g  f  c  h  n  v  t  qg flfml tlsie kzuil mblpd  duohu
	4  y  f  e  b  g  m  u  s  pf ekelk skrhd jythk lakoc  ctngt
	5  x  e  d  a  f  l  t  r  oe djdkj rjqgc ixsgj kzjnb  bsmfs

Step 2. Construct all reasonable trigrams using combinations of letters from the first three columns -- i.e., columns 0-2 -- taking 1 letter from each column. For example, we can get the trigram 'ahe' by picking from rows 2,2,3. We would say that the number code for 'ahe' is 223. Since this represents the first word of the message, the trigrams formed should be possible ways to start a word or phrase. In this case, 'ahe' could be the start of 'ahead.' Actually, it's not a very likely trigram, since it repeats the number 2. Make a table of the trigrams, their number codes (which represent a portion of the possible key number) and their frequencies, from Table XII in Pratt.

	Trigram		Code		Frequency (Table XII in Pratt)

	aid		215		24 ******** 
	age		234		20 ********
	aff		243		9
	ahe		224		2
	agi		230		3
	agg		232		3
	big		114		4
	chi		010		22 ******** repeated numbers
	che		024		27 ********
	cei		050 052		13 
	bed		155		2
	bee		154		32 ********
	bei		150		19 ********
	bef		153		8
	beg		152		5

Step 3. Pick the most reasonable looking trigrams from the list in step 2. In this case we've picked the following entries:

	aid		215		24 ******** 
	age		234		20 ********
	bee 		154		32 ********
	bei 		150		19 ********
	che 		024		27 ********

They are all relatively frequent trigrams. They could be used as the prefix of the first word. None of them involves a repeated digit in its number code, which rules out 'chi.'

Step 4. For each of the likely trigrams, apply the number formulas to each succeeding trigram in the message. For example, if we apply 024, to the letters in columns 1,2,3 we get the trigram, 'jgb'; if we apply it to the letters in columns 2,3,4 we get 'idg,' and so on. A partial table has been constructed below. Impossible trigrams are marked with (*). Filling in the rows for 'aid' and 'age' is left as an exercise.

Column  	1	2	3	4	5

aid 215
age 234
bee 154 	idb    	hag 	efm* 	jlu* 	pts
bei 150 	idf    	hak 	efq* 	jly* 	ptw
che 024 	jgb* 	idg	fim    	kou    qws*

Step 5. Note that in the table above, some of the trigrams for 'bee' and 'bei' are reasonable looking, but they don't combine well with the assumption that 'bee' or 'bei' form the first three letters of the message. For example, we can get 'bee--pts' by combining 'bee' with the trigram that starts in column 5, the first column that has a possible trigram, since 'efm' and 'jlu' are impossible. Similarly, we can get 'bei--ptw' by combining 'bei' and 'ptw', which also starts in column 5. Neither of these strings ('bee--pts' or 'bei--ptw') look very promising as the start of the clear message. On the other hand, combining 'che' as the prefix with the trigram that begins at column 4 ('kou'), gives the following partial string: 'che-kou.' That looks pretty promising. So let's work on it.

Step 6. Now, working with our partial solution, that begins, che-kou, replace the blank with each of the 6 letters from column 3 of the table in step 1. This gives us all possible trigrams for columns 2-3-4 that are consistent with che and kou. This list consists of:

efk, eek, edk, eck, ebk, eak

We want to eliminate 'efk,' 'edk,' and 'ebk' from this list, leaving Ôeek,Ő ÔeckŐ and Ôeak.Ő If we make these substitutions we get the following candidates for partial solutions:

	Candidate	Number Code	Comment	
	
	cheekou  	0241024 		Possibly cheek our or cheek out
	checkou  	0243024 		Possible check out or check our
	cheakou  	0245024		Not very likely

Notice that a cycle is beginning to appear that goes 024-024 and we now have two candidates 02410241 and 02430243. If we replace the 7th letter for each of these candidates we get:

	02410241 = cheekouw		Impossible
	02430243 = checkout  		********* Solution!!!! ***********

Step 7. To complete the decipherment, apply the key number 0243 to the rest of the cipher text. You can use CryptoTool to complete this if you use the keyword 'aced.'

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Caesar Cipher
Simple Substitution Cipher
Vigenere Cipher
Gronsfeld Cipher
The German Enigma Machine
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