## Introduction

The Monome-Dinome cipher (Mo-Di) is an example of an alphanumeric substitution cipher. This cipher is similar to the Nihilist Substitution cipher in that a unique number represents each letter. In the case of the Nihilist, a two-digit number represents each letter. In the case of the Mo-Di, one third of the letters are represented by a single digit number (a Monome) and the remainder by a two digits number (a Dinome).

This system has two advantages over the Nihilist. First, representation of the commonest letters by a single digit can significantly decrease the size of a message. Secondly, it is far more difficult to strip away an additive key from a Mo-Di than from a Nihilist. (In fact, the ACA uses an additive key with the Nihilist cipher but not with the Mo-Di for this very reason.)

Historically, it should be noted that this systems simplicity and resistance to additive removal made it a favorite among Soviet espionage agents in the 1940’s and 1950’s.

## Example

Message encipherment is performed with a 24-letter alphabet arranged in a 3 x 8 rectangle. Two letter sets, usually I/J and W/X, by necessity share an entry. Of the ten digits 0-9, two are used to identify the rectangle rows, and eight are used to identify rectangle columns. The initial row of the rectangle will not have a decimal identification. The column digit alone will identify letters in this row. The letters from the remaining two rows will be identified by a two-digit number consisting of the row number and the column number. The row digits and the order of the column digits can be chosen in any manner that is desired. The alphabet is then usually placed into the rectangle in the usual manner based on a keyword.

Consider the following example

**KeyWord:**BLUEBIRD**Plain Text:**The early bird gets the worm.

2 4 6 8 0 1 3 5 - B L U E I/J R D A 7 C F G H K M N O 9 P Q S T V W/X Y Z

PT: t h e e a r l y b i r d g e t s t h e w o r m CT: 98 78 8 8 5 1 4 93 2 0 1 3 76 8 98 96 98 78 8 91 75 1 71

The final cipher text is the regrouped into five character groups to remove all letter separations.

CT: 98788 85149 32013 76898 96987 88917 5171

### Cryptanalysis

The cryptanalysis of this cipher mainly involves identifying the two row digits. Once these digits are identified, the cipher can be separated into the proper one and two digit numbers. At this point, the cipher is essentially a Patristocrat and can be solved by the methods appropriate for that cipher.

Identification of the row numbers can be accomplished by making use of the simple property that those two digits can never be in contact with one another. For instance, in the above example, the strings 77, 99, 79, or 97 can not appear in the cipher text. A contact analysis of the cipher text will allow the Cryptanalyst to identify potential row digits.

Further, on average approximately two third of the letters used in the plain text will have to be encoded as a number starting with one of the row digits. This will usually result in these two digits having a higher frequency than their column digit counterparts. A frequency analysis will help the Cryptanalyst to choose the proper row digits from the candidates supplied by the contact analysis.

### Sample Analysis

The example problem we will work on is JA05 E-10:

#### E-10. Monome-dinome. Looking in the mirror. (pictures) BOATTAIL

18913 18284 34787 06313 20215 21319 53718 69513 13458 03116 31848 08793 66314 93658 03118 76313 43136 29513 65803 11631 34782 52973 51868 71987 34860 34828 73446 31531 34956 39298 48718 63987 34848 01863 78058 6131

Note that the word ‘pictures’ is supplied as a crib.

The analysis starts with a contact chart. This provides the following:

R\C | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | Tot |
---|---|---|---|---|---|---|---|---|---|---|---|

0 | 0 | 1 | 1 | 4 | 0 | 1 | 1 | 0 | 1 | 0 | 9 |

1 | 0 | 3 | 0 | 11 | 1 | 2 | 2 | 0 | 8 | 2 | 29 |

2 | 1 | 2 | 0 | 0 | 0 | 1 | 0 | 0 | 2 | 3 | 9 |

3 | 0 | 15 | 1 | 0 | 9 | 1 | 4 | 2 | 0 | 2 | 34 |

4 | 0 | 0 | 0 | 2 | 1 | 1 | 1 | 2 | 6 | 2 | 15 |

5 | 0 | 3 | 2 | 2 | 0 | 0 | 1 | 0 | 4 | 0 | 12 |

6 | 1 | 1 | 1 | 9 | 0 | 2 | 1 | 0 | 1 | 1 | 17 |

7 | 1 | 3 | 0 | 4 | 0 | 0 | 1 | 0 | 3 | 1 | 13 |

8 | 6 | 0 | 3 | 0 | 4 | 0 | 6 | 8 | 0 | 1 | 28 |

9 | 0 | 1 | 1 | 2 | 0 | 4 | 0 | 1 | 3 | 0 | 12 |

This chart shows the number of times a specific number sequence (In the form RC) occurs in the cipher text. For instance, notice that the sequence ’03’ occurs 4 times in the text but the sequence ’30’ never occurs in the text.

Looking at the entries on the central diagonal of this chart allows us to check for repeated digits (e.g., 33 or 44). Note that the digits 0, 2, 3, 5, 7, 8, and 9 do not contact themselves (i.e., the analysis reveals 0 contacts). These are all candidates for the row digits. Only digits 1, 4, and 6 are ruled out by the presence of repeats in the text.

The total column at the right end of the contact chart shows the total number of times each digit is present in the text. Of the candidate row digits, 3 (with 34 appearances) and 8 with (28 appearances) are the most common. Since these digits are the most frequent, they are the most probable pair of row digits. Determining whether the sequences ’38 ‘or ’83’ appear in the cipher text can test this identification. Since neither sequence appears, the likelihood that 3 and 8 are the correct row digits is increased.

As a working assumption, we will start with 3 and 8 as row digits. When the cipher text is properly separated, the following cipher text results:

0 1 2 4 5 6 7 9 - * * * * * * * * 3 * * * * * * * * 8 * * * * * * * *

1 89 1 31 82 84 34 7 87 0 6 31 32 0 2 1 5 2 1 31 * * * * * * * * * * * * * * * * * * * * 9 5 37 1 86 9 5 1 31 34 5 80 31 1 6 31 84 80 87 9 * * * * * * * * * * * * * * * * * * * * 36 6 31 4 9 36 5 80 31 1 87 6 31 34 31 36 2 9 5 1 * * * * * * * * * * * * * * * * * * * * 36 5 80 31 1 6 31 34 7 82 5 2 9 7 35 1 86 87 1 9 * * * * * * * * * * * * * * * * * * * * 87 34 86 0 34 82 87 34 4 6 31 5 31 34 9 5 6 39 2 9 * * * * * * * * * * * * * * * * * * * * 84 87 1 86 39 87 34 84 80 1 86 37 80 5 86 1 31 * * * * * * * * * * * * * * * * *

The cipher is now been converted into the equivalent of a Patristocrat. The solution will start by attempting to place the crib ‘pictures’ into the cipher text. This word has the two high frequency letters ‘e’ and ‘t’ in close proximity to one another. If we can determine which cipher text numbers are equivalent to these two letters, we should be able to place the crib. A frequency analysis of the cipher provides the following:

0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | Tot | |
---|---|---|---|---|---|---|---|---|---|---|---|

– | 3 | 15 | 5 | 0 | 2 | 11 | 7 | 3 | 0 | 9 | 55 |

3 | 0 | 15 | 1 | 0 | 9 | 1 | 4 | 2 | 0 | 2 | 34 |

8 | 6 | 0 | 3 | 0 | 4 | 0 | 6 | 8 | 0 | 1 | 28 |

Since the single digit numbers are based on a keyword, which is likely to be composed of high frequency letters, we search in this row first. Based on the frequencies of these numbers, a first guess could be that e=1 and t=5. If this guess is later found to be incorrect, the high frequency number ’31’ can be added to possible numbers and a new ‘t-e’ pair tested.

Assuming e = 1 and t= 5 produces:

0 | 1 | 2 | 4 | 5 | 6 | 7 | 9 | |
---|---|---|---|---|---|---|---|---|

– | * | e | * | * | t | * | * | * |

3 | * | * | * | * | * | * | * | * |

8 | * | * | * | * | * | * | * | * |

1 89 1 31 82 84 34 7 87 0 6 31 32 0 2 1 5 2 1 31 e * e * * * * * * * * * * * * e t * e * 9 5 37 1 86 9 5 1 31 34 5 80 31 1 6 31 84 80 87 9 * t * e * * t e * * t * * e * * * * * * 36 6 31 4 9 36 5 80 31 1 87 6 31 34 31 36 2 9 5 1 * * * * * * t * * e * * * * * * * * t e 36 5 80 31 1 6 31 34 7 82 5 2 9 7 35 1 86 87 1 9 * t * * e * * * * * t * * * * e * * e * 87 34 86 0 34 82 87 34 4 6 31 5 31 34 9 5 6 39 2 9 * * * * * * * * * * * t * * * t * * * * 84 87 1 86 39 87 34 84 80 1 86 37 80 5 86 1 31 * * e * * * * * * e * * * t * e *

The only location where the word ‘picture’ can be fit into this partial text is in the third line. Fitting the crib at this location would provide p=4, i=9, c=36, u=80, and r=31. Substituting these values provides:

0 | 1 | 2 | 4 | 5 | 6 | 7 | 9 | |
---|---|---|---|---|---|---|---|---|

– | * | e | * | p | t | * | * | i |

3 | * | r | * | * | * | c | * | * |

8 | u | * | * | * | * | * | * | * |

1 89 1 31 82 84 34 7 87 0 6 31 32 0 2 1 5 2 1 31 e * e r * * * * * * * r * * * e t * e r 9 5 37 1 86 9 5 1 31 34 5 80 31 1 6 31 84 80 87 9 i t * e * i t e r * t u r e * r * u * i 36 6 31 4 9 36 5 80 31 1 87 6 31 34 31 36 2 9 5 1 c * r p i c t u r e * * r * r c * i t e 36 5 80 31 1 6 31 34 7 82 5 2 9 7 35 1 86 87 1 9 c t u r e * r * * * t * i * * e * * e i 87 34 86 0 34 82 87 34 4 6 31 5 31 34 9 5 6 39 2 9 * * * * * * * * p * r t r * i t * * * i 84 87 1 86 39 87 34 84 80 1 86 37 80 5 86 1 31 * * e * * * * * u e * * u t * e r

It is now obvious that the first word of the plaintext is most likely ‘every’. Further, the word ‘architecture’ appears to be placed at the break between lines 3 and 4. These guesses provide v=89, y=82, a=34, and h=2. Substituting we get:

0 | 1 | 2 | 4 | 5 | 6 | 7 | 9 | |
---|---|---|---|---|---|---|---|---|

– | * | e | h | p | t | * | * | i |

3 | * | r | * | a | * | c | * | * |

8 | u | * | y | * | * | * | * | v |

1 89 1 31 82 84 34 7 87 0 6 31 32 0 2 1 5 2 1 31 e v e r y * a * * * * r * * h e t h e r 9 5 37 1 86 9 5 1 31 34 5 80 31 1 6 31 84 80 87 9 i t * e * i t e r a t u r e * r * u * i 36 6 31 4 9 36 5 80 31 1 87 6 31 34 31 36 2 9 5 1 c * r p i c t u r e * * r a r c h i t e 36 5 80 31 1 6 31 34 7 82 5 2 9 7 35 1 86 87 1 9 c t u r e * r a * y t h i * * e * * e i 87 34 86 0 34 82 87 34 4 6 31 5 31 34 9 5 6 39 2 9 * a * * a y * a p * r t r a i t * * h i 84 87 1 86 39 87 34 84 80 1 86 37 80 5 86 1 31 * * e * * * a * u e * * u t * e r

Numerous words (e.g., whether, anything, portrait) are now apparent in the partial plain text. The completion of this CON is left as an exercise for the student.