This page is being published with the kind permission of of the editor of "The Cryptogram" from  November - December 1982, Vol XLVIII No 7, pages 6 and 7 published by the American Cryptogram Association (ACA).

First my comments:
This article was written by "Trio" a pseudonym used by members of the ACA who was one of three people that solved the challenge messages given in the prior issue.
The abbreviation "pt" or subscript "p" means plain text.  The abbreviation "ct" or subscript "c" means cypher text.
A double alphabet on a strip means a repeated alphabet "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz" this is done so that the center 26 letters always line up with the center 26 letters no matter what offset is used on any one strip.
The solution below is based on knowing the construction of the disks AND having some plain text (in this case "areyou").  Note the the location of "are you" in the message is unknown.
A more general solution would be based ONLY on the test masseages and knowing the construction of the disks WITHOUT any known plain text.
 

S 0 L U T I 0 N S 0 F T H E

M - 9 4 T E S T M E S S A G E S

By TRIO

Mauborgne wasted his time dreaming up ridiculous test phrases.

Without knowing the 25 mixed alphabets, I reckon ciphertext

using even the most obvious words would be unbreakable.

 
 
Herewith, given the 25 mixed alphabets JA82 Cm), the plaintexts of all 25 (why all?) of Mauborgne's test messages - and very dull reading they make:

Gap

1  14 Chlorine and oxygen have not b
2   8 Where did you meet each other
3   1 Drink this potion quickly for
4  20 Well, make me the same shape but
5  13 Cyanogen is a colorless gas in
6  25 Phenols are benzene derivati
7   2 Xylonite and artificial ivor
8  12 I went to a new theatre, the Pala
9  24 Picric acid is explosive and i
10 18 Llangollen is a town in Wales a
11  6 Yvette, are you going shopping
12 23 Orthophosphoric is the compo
13 16 Caoutchouc is closely allied
14 17 Olefiant gas, ethene or ethyle
15  3 See the terrible tank tackle a
16  7 It is a thin limpid liquid that
17 22 If it is insoluble in water it i
18 11 Silver has been known from rem
19 21 Hot concentrated sulphuric a
20  4 Small coefficient of expansi
21  9 Palladium possesses a power o
22 19 Absorbing and condensing thi
23 15 Compounds of platinum form tw
24 10 Gold occurs widely dristribut
25  5 Oxidation caused by it probab

The disks are threaded in the following order. 4, 14, 5, 6, 2, 13, 3, 24, 21, 11, 19, 8, 7, 16, 15, 17, 10, 9, 1, 12, 25, 23, 22, 20, 18. Any keyword from this sequence escapes me.

The word, Gap, in the list above refers to the interval between two letters on a disk -what Friedman calls a "generatrix." So, for example, r = T (gap 7) means that the number of letters between pt r and ct T on one or more disks is 7 (eg., disk 11: ...RFBEJALT...).

I found Riverbank No. 20 remarkably unhelpful on the subject. In Friedman's first example (p. 41), the gaps are supplied; in the next, a pt word is given, but for any particular gap the ct equivalents are far fewer than in the Mauborgne alphabets; and then he goes rambling off into the reconstruction of keywords. So, for me, it was back to first principles.

My method was simple but extremely tedious, and I am sure a computer could do it far quicker. This is my method:

1. Prepare 25 double alphabets as slides, one for each disk. Slides, of course, are long strips that do the same job as disks, which is what I shall call them - disks numbered the same as published in the JA82 Cm.

2. Start by matching the six letters of pt to the first six letters of message 1: areyoup= VFDJLQC. Set all 25 disks to "a" and list all possible gaps ap = Vc. There are 18 different gaps, and the remaining seven are duplicates. Incidentally, Parker Hitt on p. 60 of Riverbank No. 20 can only manage 8 gaps for a = V.

3. With the top row of all disks set at "r", repeat the process for "F" but this time exclude all gaps not in a = V. Repeat for e = D, y = J, o = L, and u = Q, ignoring any gap not common to all the previous pairs. With any luck, areyou = VFDJLQ for gaps 1, 9, 23, 25 only will result.

4. The next thing to discover is which disk relates to which gap. Taking gap 1, there is no choice; the disks are 21, 11, 4, 24, 1, 14. (Check this against the table on p. 4 of JA82 Cm - I could be wrong!) But with gap 9, alternatives complicate things: r = F (gap 9) appears on disks 2, 7, 12. (Better check this, too, if you don't mind - just to make absolutely sure we are on the same wavelength.)

In sets of six disks, there are one set for gap 1; nine sets for gap 9; nine sets for gap 23; and - our first bit of luck - zero sets for gap 25, since both r = F (gap 25) and o = L (gap 25) are on the same disk 24 with no alternatives avail- able. Altogether, that totals 19 sets.

It is now established that "areyou" can be enciphered as VFDJLQ in 19 different ways, using a variety of disks but only gaps 1, 9, or 23. At this point I did a few lightning calculations on the back of an envelope and came to the conclusion that "areyou" could fit the total ct in well over 7,000 different disk-gap combinations. But was I downhearted? Yes, I was!

5. The next step is to see whether any of these 19 combinations of six disks will produce any other pt apart from "are you." Using disks 21, 11, 4, 24, 1, 14 (see para. 4), set the top row to ct CGNJMZ (the start of message 2) and see whether any other row makes good pt. The best of a poor bunch is "yokhew." Pass on to ct CSTDTS (message 3) and so on down to message 25.

6. Return to message 2 but this time use disks 1, 2, 8, 17, 22, 15. This is the first combination for gap 9, which can be verified by setting the top row to ct VFDJLQ when pt "areyou" appears at row 17. Why 17? Because this time, starting with ct, if pt to ct gives gap 9, then ct to pt gives 26-9 = 17.

Repeat for the other eight combinations for gap 9 and then for the nine choices for gap 23. Regretfully, no pt emerges. The possibilities for areyou = VFDJLQ are now exhausted - 19 down, 6,981 to go.

7. Now, work down the column (areyou CGNJMZ, etc.). Occasionally, gratefully, there is no need to compare. For example, with areyou = SSEIQD (the start of message 7), e=E, which means the whole row has to be gap zero and hence pt = et - which it clearly doesn't.

To save time, (as on those TV cooking demonstrations in which the smiling chef, having rapidly mixed the ingredients, pops them into the oven with one hand and produces the finished article, cooked to a turn, with the other hand) it is now five or six days later, local time. Toiling down columns 7-12, which starts areyou = MMJBHS, areyou = UPMBNA (message 11) is reached. Checking that against disks 3, 24, 21, 11, 19, 8, gap 6 certainly gives "areyou." Therefore, set the top line to MMJBHS and look down the rows for pt.

Being a bit tired, I completely missed "neando" at gap 14. I then set up VKQCJP and, once again, overlooked "idyoum" at gap 8. So, on to SDJNJD in message 3, and this time "hispot" at gap 1 rings a bell. In no time at all, "terrib" at gap 3 emerges, followed by "centra" at gap 21 and "ingand" at gap 19. At last, a group of six disks and all 25 gaps are fixed. The rest is relatively straightforward. It's as good as solved!

I did make some mistakes: I assumed "centra" was followed by "l" so that central = DXGPLCP (gap 21). But only on disk 6 does l = P (gap 21), and the set of seven disks must become 3, 24, 21, 11, 19, 8, 6. However, that ruins the pt of message 15: terribf = LNXESCM (gap 3).

I had to try something else. Message 20, "oeffic" became "coefficient" without much trouble, fixing five more disks, and so on. That's about it. On the whole, progress was slow but sure. I hope the krewe finds my method marginally less boring than the messages themselves, and I hope there are no "gaps" in my explanation.

One interesting fact emerged from my labours: Mauborgne certainly managed "the fewest number of repetitions of pairs of letters." I checked this for gap 1 and found there are only three duplications: D=E in disks 2 and 16, U=B in disks 3 and 16, and Y=A in disks 16 and 22. I am now wondering whether a perfect gap 1 with no duplications at all could be achieved. I've done it for four disks using five letters (ABEDC, ACDEB, ADBCE, AECBD) but nothing larger Perhaps there is a connection with Latin squares or even Euler (Graeco-Latin) squares.

(Editor's note: TRIO won the prize offered by MEROKE. REEY and THE BRUIN also sent correct solutions.)

"vvvvvvvvvvvvvvvvvvvvvvwBOOK REVIEWS by MEROKE vvvvvvvvvvvvvvvvvvv"

Elman, R., The Menu Cypher, Macmillan Pub. Co., 866 Third Ave., New York, NY 10022, 1982, 216 pp., $12.95. A satire on the CIA's world of espionage which is set in 1980 dealing with Central American foreign policy in a burlesque fashion. The title is the gastronomic code used by agents; its complete glossary is provided.

Pioneers of Computing. Two sets of 10 audio cassettes, L53 each set or L6 each tape. Hugo Informatics, 331 Workingham Rd., RG6 2EB Reading, England. Oral history of computing produced by the Science Museum in London. Relevant tapes lasting about one hour each: M.H.A. Newman discusses cryptanalytic computing at Bletchley Park; T.H. Flowers tells of designing COLOSSUS with Turing at B.P.; and A.V.M. Coombs talks about building the MOSAIC defense computer to Turing's design.

ND82 7

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