Zodiac Discussion Forum

Homophonic substitu…
 
Notifications
Clear all

Homophonic substitution

1,434 Posts
21 Users
0 Reactions
304.3 K Views
doranchak
(@doranchak)
Posts: 2614
Member Admin
 

And I am interested in what the books available to Zodiac would yield.

My copy of the Laffin book (Codes and Ciphers) arrived today. I will get it scanned and posted as soon as I get a chance.

http://zodiackillerciphers.com

 
Posted : January 30, 2016 7:13 am
Jarlve
(@jarlve)
Posts: 2547
Famed Member
Topic starter
 

Thanks doranchak. Can’t wait to see the skytale section.

Has anyone out there tried transpositions with more than one plaintext at a time, or differing counts of plaintext at a time, such as AMSCO?

I haven’t. daikon has a nomenclator solver, I provided him with a difficult test cipher and he solved it. It turned up nothing for the 340 excluding transposition.

viewtopic.php?f=81&t=2782

AZdecrypt

 
Posted : January 30, 2016 1:28 pm
smokie treats
(@smokie-treats)
Posts: 1626
Noble Member
 

AMSCO is a columnar transposition cipher that preserves about two thirds (?) of the period 1 repeats, and does not have a period equal to the number of rows because shuffling the columns destroys the period equal to the number of rows. Am I thinking right?

But a one direction rectangular route transposition that carries more than one contiguous symbol at a time from the original plaintext to the transposed plaintext, similar to AMSCO, would have two periods. It would preserve some of the period 1 repeats, and also create new repeats at the route period.

We might be able to detect this by comparing the list of period 1 bigrams ( not repeats ) with the list of period 19 bigrams ( not repeats ). Some of the period 19 bigrams were originally period 1 bigrams, but then separated by the route period during transposition. There would have been some with symbol pairs matching those of the preserved period 1 bigrams.

Alternatively, the symbols carried from the original plaintext to the transposed plaintext do not necessarily have to be contiguous. They could be at any period, and perhaps cause two spikes. Think about that for a minute. And then I don’t know if any cryptography books of the time had any examples of this. It would have been an original idea if not.

I am thinking start with unoriginal ideas.

Unfinished thought here, but the AMSCO cipher gives me inspiration. I tried to find out where it came from, and need to do some more research. But one of the websites mentioned that it was created by a person named A.M. Scott in the 19th century. I don’t know how accurate this information is; the website was not very thorough and the other sites don’t talk about origin.

 
Posted : January 30, 2016 3:36 pm
smokie treats
(@smokie-treats)
Posts: 1626
Noble Member
 

Here you go. After making smokie27a, it was difficult to get exact same statistics for smokie27b. So I made smokie 27b-d, which are all pretty close.

smokie27a is a homophonic period 19 scytale transposition
48 41 56 52 25 56 31 16 14 42 37 14 21 20 15 45 46
43 9 26 19 38 22 10 32 27 11 9 47 10 39 16 19 45
7 28 5 1 20 40 11 12 13 9 49 2 53 3 50 12 37
4 51 15 16 6 5 38 23 54 48 58 8 19 46 41 34 19
10 24 17 19 11 9 19 42 25 14 21 1 2 34 13 30 19
31 58 52 15 34 33 53 54 39 56 40 17 15 37 49 26 47
3 22 55 19 24 45 4 16 24 43 20 52 20 43 57 10 21
35 49 34 19 22 23 12 13 51 7 14 17 24 48 53 31 19
41 27 6 21 32 16 2 19 29 1 22 54 19 42 12 44 23
31 49 38 20 43 46 13 39 55 11 2 8 18 47 34 45 19
4 14 19 50 19 24 40 19 60 19 21 26 36 24 46 19 41
20 9 47 9 58 15 23 1 5 37 59 25 16 57 5 16 42
8 43 12 2 30 26 13 31 14 38 19 30 3 37 58 15 13
9 35 33 36 52 40 33 57 9 51 41 10 4 17 60 12 14
45 11 53 54 19 1 24 9 27 13 21 51 55 42 47 47 52
45 40 21 10 21 24 35 38 28 34 40 46 27 52 22 38 9
43 22 41 53 19 42 13 15 44 9 51 42 19 54 36 16 19
12 5 35 3 43 25 10 19 32 41 44 23 55 42 6 43 47
45 46 8 56 19 52 35 37 43 47 19 53 26 13 24 30 20
41 37 49 42 54 24 32 18 4 44 59 11 22 15 55 47 15

smokie27b is a homophonic cipher
48 25 19 37 26 38 20 52 21 12 9 27 57 13 47 53 42
19 42 43 39 21 14 17 19 31 32 41 58 15 10 23 28 35
46 42 56 40 11 54 24 16 57 19 50 50 12 33 51 48 52
13 47 37 19 21 14 49 1 38 9 43 39 52 22 15 19 53
23 16 45 50 19 10 19 58 2 51 24 25 20 21 3 40 11
41 37 12 7 42 56 34 9 8 31 26 36 5 43 38 6 19
4 46 10 5 56 55 32 27 48 55 13 46 49 23 45 1 39
30 18 46 41 35 19 23 12 11 1 47 30 2 33 34 15 60
58 3 60 52 24 4 54 31 16 9 19 42 55 21 19 12 40
19 28 37 13 45 43 19 36 17 41 46 19 32 33 19 22 1
53 23 19 59 19 38 53 27 39 2 40 9 50 3 58 54 24
16 11 25 57 13 47 21 4 9 42 44 12 37 13 10 53 24
16 29 2 36 6 15 11 10 43 19 47 59 23 16 38 24 19
47 12 1 8 21 12 10 19 22 14 34 15 11 20 16 2 18
31 3 48 22 17 47 41 36 52 24 14 42 53 21 12 45 48
13 32 14 8 54 46 26 7 33 19 35 44 44 27 15 47 8
16 11 55 22 14 20 34 43 43 36 19 39 10 23 16 19 47
25 13 11 52 41 18 42 46 20 14 54 23 25 49 54 23 46
19 6 6 28 40 19 22 15 1 9 2 37 10 31 42 41 30
16 9 3 5 42 56 55 50 44 4 45 30 32 19 38 20 6

smokie27c is a homophonic cipher
48 25 19 38 26 38 20 52 21 12 9 27 57 13 45 53 41
19 42 43 39 22 14 17 19 31 32 41 58 15 10 23 28 35
46 42 56 40 11 54 24 14 57 19 49 50 12 33 51 48 55
13 47 37 19 21 14 49 2 38 9 43 39 52 22 12 19 53
23 16 45 50 19 10 19 59 2 51 24 25 20 21 3 40 11
41 37 12 7 42 56 34 9 8 31 28 36 5 43 38 6 19
4 46 10 5 56 54 31 27 48 55 14 47 49 22 45 19 39
30 18 46 41 35 19 23 14 11 1 46 30 2 33 32 15 60
59 3 60 52 24 4 53 31 13 9 19 42 55 21 16 12 40
19 28 37 13 46 43 19 35 17 41 46 19 32 33 19 22 1
52 23 14 59 15 39 55 19 39 2 40 10 50 3 58 54 24
16 10 25 57 19 47 22 4 9 42 44 12 37 13 10 55 22
14 29 19 35 6 13 11 9 43 41 47 59 23 16 38 24 19
46 12 2 7 24 13 10 19 22 14 34 15 10 20 12 2 18
31 3 51 23 17 47 42 35 52 24 13 42 53 21 12 45 48
13 32 13 7 54 46 27 7 33 3 35 44 44 27 14 46 8
16 9 54 22 13 19 34 43 19 35 19 39 10 23 13 19 45
28 13 9 52 42 18 42 46 20 14 53 21 26 49 55 24 47
43 5 5 19 40 20 22 15 19 10 2 38 10 33 41 41 30
16 11 3 5 42 56 55 50 44 4 46 30 32 27 38 20 5

smokie27d is a homophonic cipher
51 25 19 37 26 38 20 52 21 12 9 28 57 13 46 53 41
19 42 43 39 21 16 17 19 31 32 41 58 15 10 23 25 35
46 42 56 40 11 54 24 16 57 19 49 50 12 33 51 51 52
13 47 37 19 21 14 49 2 38 9 41 39 52 22 15 19 53
23 16 45 50 19 10 19 59 2 51 24 25 20 21 3 40 11
41 39 12 7 42 56 31 9 8 31 26 36 5 43 38 6 19
4 46 10 5 56 55 31 27 48 19 13 47 49 23 47 19 38
30 18 46 41 36 19 23 14 10 1 47 30 2 32 34 15 60
58 19 60 52 24 19 53 31 16 9 54 42 55 21 19 12 40
20 28 37 15 45 43 19 36 17 41 46 19 32 33 19 22 1
52 23 14 59 15 38 53 19 39 2 40 10 49 3 58 54 21
13 11 26 57 15 47 21 4 9 42 44 12 37 14 10 53 22
14 29 1 35 6 15 11 9 43 41 45 58 23 16 38 24 15
45 12 1 7 21 13 10 19 22 14 33 15 11 20 16 2 18
31 3 50 23 17 47 19 36 52 24 16 42 53 21 12 45 48
13 32 14 8 54 46 26 7 34 4 35 44 44 27 15 47 8
13 10 55 21 13 19 31 42 19 36 19 39 10 23 12 53 46
27 13 11 52 43 18 42 47 20 14 55 24 19 49 54 22 45
42 5 6 25 40 20 22 14 19 11 2 37 9 32 19 41 30
16 10 3 5 41 56 55 50 44 3 47 30 32 26 38 20 5

 
Posted : January 30, 2016 8:13 pm
Jarlve
(@jarlve)
Posts: 2547
Famed Member
Topic starter
 

Thanks allot!

AZdecrypt

 
Posted : January 30, 2016 8:35 pm
Jarlve
(@jarlve)
Posts: 2547
Famed Member
Topic starter
 

I confused AMSCO with nomenclator.

There will be two things we need to know or explore, the amount of columns and the connection pattern (1,2,1,2) or (2,1,2,1). The number of possible connection patterns is equal to 2^columns. So if Zodiac used 19 columns 524288 variations will need to be checked. Very much in range of brute force.

AMSCO is a columnar transposition cipher that preserves about two thirds (?) of the period 1 repeats, and does not have a period equal to the number of rows because shuffling the columns destroys the period equal to the number of rows. Am I thinking right?

AMSCO is columnar transposition, but I think the letter pattern will – to some degree – randomize periodic bigram counts.

AZdecrypt

 
Posted : January 30, 2016 10:28 pm
 Soze
(@soze)
Posts: 810
Prominent Member
 

Aside from the obvious Amsco references above, the only thing I found, was the following:

Alexander Malcolm Scott was born in 1841 and died in 1892. He wrote a book called The Battle of Langside, 1568. The book is about Mary Queen of Scotts and concerns her abdication from the thrown. Mary Queen of Scotts, during her fued with England, created a polyalphabetic cipher to transport her messages.

I found this when doing a search on A. M. Scott and ciphers. May be of use or maybe not. Just thought I would fill in a possible gap on who A. M. Scott might have been.

Soze

 
Posted : January 30, 2016 11:00 pm
smokie treats
(@smokie-treats)
Posts: 1626
Noble Member
 

Soze, thank you very much for finding that information. I really appreciate it.

 
Posted : January 30, 2016 11:22 pm
 Soze
(@soze)
Posts: 810
Prominent Member
 

I had made a previous post of three questions but deleted it. I realized I was right on the first question, wrong on the second and wasn’t sure at all how valid the third question would be given my second question was wrong.

It seems knowing the number of columns as well as the order of columns is important to the Amsco cipher. I guess i have no choice but to go ahead and just ask:

What about just 3 columns? What does that say, if anything, in terms of the research being performed here?

Soze

 
Posted : January 31, 2016 2:34 am
smokie treats
(@smokie-treats)
Posts: 1626
Noble Member
 

Well I don’t really think it is AMSCO, at least because AMSCO involves columnar transposition. But I went to the University library today and found the shelf with the cryptography books. I wanted to look at old books and the chapters about transposition.

I used the guide published by Humbolt_Squid here: viewtopic.php?f=81&t=2702, and found among mostly newer books Cryptography: The Science of Secret Writing by Laurence D. Smith 1955.

It is a good book. Simple, but explains fundamentals very well. The transposition chapter discusses military applications, terminology such as inscription and transcription, geometric shapes for route ciphers, use of nulls to fill voids in irregular shapes, orthogonal routes, and use of multiple shapes for multiple transcriptions. At the end of the chapter, Smith discusses "polyliteral transposition":

"Polyliteral transposition. A variation may be made in the size of the unit transposed. One need not transpose single letters at a time. Pairs of letters, or groups of three, four or five letters may be employed in the same manner as that used in monoliteral or single letter transposition."

To me, the period 19 repeats shows that a rearrangement of columns was not used. But that Zodiac may have transposed more than once with different shaped rectangles and inadvertently caused the period 19 repeats, or he transposed groups of letters at a time from a rectangle of either width or height of 19.

 
Posted : January 31, 2016 2:47 am
(@mr-lowe)
Posts: 1197
Noble Member
 

Smokie .. Lenovo fired up.. But struggling to get solvers to load.. Will take a while.. Question who’s got the best solver..and easiest to use for a novice..
Doranchak…also eagerly awaiting the scanned book..have you had a chance to look at the book .. And if so does anything stand out..
Cheers

 
Posted : January 31, 2016 11:47 am
Jarlve
(@jarlve)
Posts: 2547
Famed Member
Topic starter
 

"Polyliteral transposition. A variation may be made in the size of the unit transposed. One need not transpose single letters at a time. Pairs of letters, or groups of three, four or five letters may be employed in the same manner as that used in monoliteral or single letter transposition."

Awesome work smokie. It’s something I had not considered yet, looking into it.

AZdecrypt

 
Posted : January 31, 2016 2:11 pm
doranchak
(@doranchak)
Posts: 2614
Member Admin
 

The book is now scanned and posted: viewtopic.php?f=81&t=2915

Haven’t had much of a chance to go through it myself yet – meanwhile, I’m hoping you guys will find something!

http://zodiackillerciphers.com

 
Posted : January 31, 2016 3:38 pm
doranchak
(@doranchak)
Posts: 2614
Member Admin
 

Question who’s got the best solver..and easiest to use for a novice..

Jarlve’s solver (AZdecrypt) has my vote for the best solver. It is the fastest and most feature-rich solver I’ve ever used!

http://zodiackillerciphers.com

 
Posted : January 31, 2016 3:42 pm
Jarlve
(@jarlve)
Posts: 2547
Famed Member
Topic starter
 

Thanks doranchak and I also still have to go through the book as well. Glimpsed over it and noticed the skytale 9 by 6 format and the vigenere grid caught my attention again in relation to the pivots.

ABCD
BCDE
CDEF
DEFG

I worked on smokie’s find, polyliteral transposition, transposing with a unit size variable. Here’s a period routine (code down below) I made which accounts for this, I have stress tested it with AZdecrypt’s manipulation solver using random inputs and it does not seem to produce any truncated output.

Imagine a unit size of 2 and a period of 5. Since 5 does not divide equally by 2 the unit size of the 3rd part is reduced to 1. The read in pattern would then look as following.

112231122311223...

Somewhere tomorrow the big test that considers any continues part in the 340 as null + period 15, 19 transposition will finally come to an end and I will let AZdecrypt loose with this new period routine considering all 4 orientations.

't=transpose/untranspose
'u=unit size
'p=period

for i=1 to p step u
	if p-i<u-1 then vu=p-i+1 else vu=u 'variable unit size if necessary
	for j=i to c step p
		for k=0 to vu-1
			if j+k>c then exit for 'do not extend beyond cipher length
			l+=1
			if t=1 then cipher_out(j+k)=d1(l) else cipher_out(l)=d1(j+k) 'transpose/untranspose
		next k
	next j
next i

Something I added to AZdecrypt this morning is the option to include "advanced" statistics, bigrams and my m_2s_cycles measurement (renamed to just Sequential). For the 340, worst case, it only reduces iteration speed by 5%. I plan to use it to possibly gauge the "moldability" of the 340’s cycles.

Multiplicity: 0.1852941 Characters: 340 Symbols: 63 Sequential: 2078.79 Bigrams: 25

AZdecrypt

 
Posted : January 31, 2016 5:32 pm
Page 60 / 96
Share: