My guess is that Jarlve’s solver will punish plaintexts that are overly repetitive, due to them having higher IoC than "normal" messages.

Jarlve ~

If there was a plaintext written LRTB in a homophonic 63 cryptogram, and it was shorter than 340 but written over and over again, how would that affect the performance of your solver? Is there a message shortness or longness that makes a difference between solve ability or not? I am not saying this is what I think the Z340 is, but just wondering about length, repetition, and solve ability. If the plaintext length was say, 34 and over and over again, could it solve? What about 68, or 136, etc.?

Imagine a plain text with a length of 17 with a multiplicity of 0.5. It solves into gibberish, now, will repeating the plain text over and over again make it easier to solve? No, because the solver also scores the same n-grams over and again. So, more or less, the difficulty is equal to the multiplicity of the single plain text string. But if the multiplicity of the resulting cipher is higher than the multiplicity of the single plain text string, then the difficulty is equal to the multiplicity of the cipher instead.

My guess is that length 34 or greater will usually solve. Something to take in mind is that any cipher with a short string repeated over and over again will have a boatload of repeats. And as you say Kasiski and other Vigenère tests will have it show up.

My guess is that Jarlve’s solver will punish plaintexts that are overly repetitive, due to them having higher IoC than "normal" messages.

SInce 1.15 AZdecrypt internally uses entropy instead of IOC.

thisisthezodiacspeaking

IOC: 0.05138339920948617
Entropy: 3.621175542919471 <—

thisisthezodiacspeaking
thisisthezodiacspeaking
thisisthezodiacspeaking
thisisthezodiacspeaking
thisisthezodiacspeaking

IOC: 0.08466819221967964
Entropy: 3.621175542919471 <—

Entropy remains the same. If the frequency distribution is "narrow" the entropy will be lower and that is now punished, but not overly so.

O.k., thanks. Just curious and thought that you may have already experimented with this.

Largo. The matrix wasn’t intended to make P19, but perhaps future ones will.

I know, but I was just curious which nGram peaks the matrix would produce and whether P19 might play a random role.

I found a way to make pivots with a cipher. I added a new spreadsheet to my transposition utility which takes a matrix and uses a mod function to repeat the same short portion over and over again in the matrix. Just changes a transposition matrix so that a short message can be repeated.

Then, I added my pivot detector, and can adjust the length of the message by adding 1 over and over again to see what lengths, if any, create a lot of pivots. Here is the original matrix at period 20.

001 018 035 052 069 086 103 120 137 154 171 188 205 222 239 256 273
290 307 324 002 019 036 053 070 087 104 121 138 155 172 189 206 223
240 257 274 291 308 325 003 020 037 054 071 088 105 122 139 156 173
190 207 224 241 258 275 292 309 326 004 021 038 055 072 089 106 123
140 157 174 191 208 225 242 259 276 293 310 327 005 022 039 056 073
090 107 124 141 158 175 192 209 226 243 260 277 294 311 328 006 023
040 057 074 091 108 125 142 159 176 193 210 227 244 261 278 295 312
329 007 024 041 058 075 092 109 126 143 160 177 194 211 228 245 262
279 296 313 330 008 025 042 059 076 093 110 127 144 161 178 195 212
229 246 263 280 297 314 331 009 026 043 060 077 094 111 128 145 162
179 196 213 230 247 264 281 298 315 332 010 027 044 061 078 095 112
129 146 163 180 197 214 231 248 265 282 299 316 333 011 028 045 062
079 096 113 130 147 164 181 198 215 232 249 266 283 300 317 334 012
029 046 063 080 097 114 131 148 165 182 199 216 233 250 267 284 301
318 335 013 030 047 064 081 098 115 132 149 166 183 200 217 234 251
268 285 302 319 336 014 031 048 065 082 099 116 133 150 167 184 201
218 235 252 269 286 303 320 337 015 032 049 066 083 100 117 134 151
168 185 202 219 236 253 270 287 304 321 338 016 033 050 067 084 101
118 135 152 169 186 203 220 237 254 271 288 305 322 339 017 034 051
068 085 102 119 136 153 170 187 204 221 238 255 272 289 306 323 340

I found out that at period 20 which I like to experiment with because it fills a 17 x 20 grid perfectly, if the message length is 11, 33 or 67, you get a lot of pivots. At any other length up to 170, no pivots at all. Here is the matrix with message length of 67.

01 18 35 52 02 19 36 53 03 20 37 54 04 21 38 55 05
22 39 56 02 19 36 53 03 20 37 54 04 21 38 55 05 22
39 56 06 23 40 57 03 20 37 54 04 21 38 55 05 22 39
56 06 23 40 57 07 24 41 58 04 21 38 55 05 22 39 56
06 23 40 57 07 24 41 58 08 25 42 59 05 22 39 56 06
23 40 57 07 24 41 58 08 25 42 59 09 26 43 60 06 23
40 57 07 24 41 58 08 25 42 59 09 26 43 60 10 27 44
61 07 24 41 58 08 25 42 59 09 26 43 60 10 27 44 61
11 28 45 62 08 25 42 59 09 26 43 60 10 27 44 61 11
28 45 62 12 29 46 63 09 26 43 60 10 27 44 61 11 28
45 62 12 29 46 63 13 30 47 64 10 27 44 61 11 28 45
62 12 29 46 63 13 30 47 64 14 31 48 65 11 28 45 62
12 29 46 63 13 30 47 64 14 31 48 65 15 32 49 66 12
29 46 63 13 30 47 64 14 31 48 65 15 32 49 66 16 33
50 67 13 30 47 64 14 31 48 65 15 32 49 66 16 33 50
67 17 34 51 01 14 31 48 65 15 32 49 66 16 33 50 67
17 34 51 01 18 35 52 02 15 32 49 66 16 33 50 67 17
34 51 01 18 35 52 02 19 36 53 03 16 33 50 67 17 34
51 01 18 35 52 02 19 36 53 03 20 37 54 04 17 34 51
01 18 35 52 02 19 36 53 03 20 37 54 04 21 38 55 05

Here is the what the pivot detector looks like with message length 67.

Here is the plaintext 67 long that repeats itself a lot of times.

O F T A N D P I T I F U L L O O K I N G I T W A S S I M P L Y A S H A M E T H E W A Y T H I N G S H A D T O H A P P E N B Y T H E T I M

The plaintext written over and over again at period 20 into a 17×20 grid.

O I A D F N M T T G E O A I T H N
T H A F N M T T G E O A I T H N T
H A D W E P T G E O A I T H N T H
A D W E P P A W P A I T H N T H A
D W E P P A W P I S A E N T H A D
W E P P A W P I S A E T S Y N D W
E P P A W P I S A E T S Y N I I T
B P A W P I S A E T S Y N I I T B
F M H Y I S A E T S Y N I I T B F
M H Y U P I T T S Y N I I T B F M
H Y U P I T L L N H I I T B F M H
Y U P I T L L N H L Y G E F M H Y
U P I T L L N H L Y G E O A S T U
P I T L L N H L Y G E O A S T O S
H I L L N H L Y G E O A S T O S H
I K H A O L Y G E O A S T O S H I
K H A O I A D F O A S T O S H I K
H A O I A D F N M T T O S H I K H
A O I A D F N M T T G E O A K H A
O I A D F N M T T G E O A I T H N

The key ( symbol 00 is a polyphone to simulate the + ).

E 00 01 02 03 04 05 06 07
T 08 09 10 11 12 13 14 00
A 15 16 17 18 19 20 00
O 21 22 23 24 25 00
I 26 27 28 29 30 00
N 31 32 33 34
S 35 36 37 38
R
H 39 40 41
D 42 43 44
L 45 46 47
C
U 48 49
F 50 51
M 52 53
W 54 55
G 56 57
P 58 59
Y 60
B 61
V
K 62
X
J
Q
Z

Randomization matrix.

15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15
35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35
35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35
35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35
35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35
35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35
35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35
35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35
35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35
35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35
35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35 35

And the cryptogram, which has two pivots that resemble the 340 pivots. Ha ha!

21 00 15 42 50 31 52 08 08 56 00 22 16 27 10 40 32
11 40 17 51 33 53 12 13 57 01 23 18 28 14 41 34 00
39 19 43 54 01 58 08 56 03 21 20 29 09 40 31 00 41
00 44 55 04 59 58 15 54 59 20 30 11 39 32 12 40 17
42 55 03 58 59 18 54 58 00 35 19 06 33 13 41 18 43
55 07 59 58 00 54 59 26 36 15 00 14 35 60 34 43 54
01 58 59 16 54 58 27 38 17 02 00 35 60 31 28 29 08
61 59 18 55 59 30 36 19 06 09 37 60 32 00 26 10 61
50 52 39 60 30 38 20 04 14 35 60 33 28 29 12 61 51
53 40 60 48 59 26 13 14 36 60 31 28 26 10 61 50 52
41 60 49 58 27 12 46 46 31 40 27 30 09 61 51 53 40
60 48 58 26 08 47 45 32 41 46 60 57 03 50 52 39 60
48 58 00 10 45 45 33 40 46 60 56 00 25 00 37 11 48
59 29 13 45 45 34 41 46 60 57 07 00 15 38 00 25 38
39 27 47 47 31 40 46 60 56 02 22 19 35 00 23 37 41
28 62 39 17 23 47 60 57 04 25 16 38 08 00 36 40 29
62 39 19 21 30 20 42 51 22 00 36 14 23 37 40 28 62
41 15 24 26 16 43 50 32 53 10 11 25 38 41 27 62 39
17 21 30 00 44 51 33 52 12 13 56 02 21 19 62 41 20
22 29 15 42 50 34 52 14 00 56 03 22 15 30 09 41 33

But the P16, 32 and 48 unigram repeats are way more than in the 340 and I haven’t checked the other stats and also wonder if a message of a certain length that does not make pivots would if nulls or skips are in certain positions, or vice versa.

It should solve at P20.

The bigram repeat chart shows a pattern unlike the 340 and the coincidence count spike is way, way higher.

But, my hypothesis is that there is a pattern of letters written over and over again, at least in the region of the pivots. Or something like that because of the cluster of P16, 32 and 48 unigram repeats in the area of the pivots. Especially the one on the right.

I have a new, adjusted hypothesis. Here is what might create pivots in the middle of the message and P29 or P39. He used a cylinder with a paper taped to it, and vertically wrote the same word or short phrase over and over again in certain columns at certain intervals. Then, he wrote vertically an actual scytale message, skipping over the word or short phrase. Then the plaintext was transcribed into the 17×20 grid, and the letters of the word or short phrase lined up diagonally and that made pivots and P29 or P39.

I got a pretty good partial solve in just a few seconds with AZD. Thanks!

21	00	15	42	50	31	52	08	08	56	00	22	16	27	10	40	32
11	40	17	51	33	53	12	13	57	01	23	18	28	14	41	34	00
39	19	43	54	01	58	08	56	03	21	20	29	09	40	31	00	41
00	44	55	04	59	58	15	54	59	20	30	11	39	32	12	40	17
42	55	03	58	59	18	54	58	00	35	19	06	33	13	41	18	43
55	07	59	58	00	54	59	26	36	15	00	14	35	60	34	43	54
01	58	59	16	54	58	27	38	17	02	00	35	60	31	28	29	08
61	59	18	55	59	30	36	19	06	09	37	60	32	00	26	10	61
50	52	39	60	30	38	20	04	14	35	60	33	28	29	12	61	51
53	40	60	48	59	26	13	14	36	60	31	28	26	10	61	50	52
41	60	49	58	27	12	46	46	31	40	27	30	09	61	51	53	40
60	48	58	26	08	47	45	32	41	46	60	57	03	50	52	39	60
48	58	00	10	45	45	33	40	46	60	56	00	25	00	37	11	48
59	29	13	45	45	34	41	46	60	57	07	00	15	38	00	25	38
39	27	47	47	31	40	46	60	56	02	22	19	35	00	23	37	41
28	62	39	17	23	47	60	57	04	25	16	38	08	00	36	40	29
62	39	19	21	30	20	42	51	22	00	36	14	23	37	40	28	62
41	15	24	26	16	43	50	32	53	10	11	25	38	41	27	62	39
17	21	30	00	44	51	33	52	12	13	56	02	21	19	62	41	20
22	29	15	42	50	34	52	14	00	56	03	22	15	30	09	41	33

Iterations: 303/1000
Score: 23363.37 Ioc: 0.06491
Ngrams: 8875690302 PC-cycles: 2807

Columnar(20*17,TP,C:1)
Offset row order(11*31,Y:25)
Offset row order(134*3,Y:1)

IORTANORTANDPINIR
OLLOOKENGITWASSIM
PLYESHAMETHEWAYTH
INGSHADTOHAPPENBY
THENIORTANDPINFRO
LLOOKINGITWASSIMP
LYASHAMETHEWAYTHE
NGSHADTOHAPPENBYT
HATFORTANDPITFROL
LOEKINGFEWESSIMPL
YASHEMENHEWAYTHIN
GSHADNOHEPPANBYTH
EEIRREANDPIEERILL
EOKINGIEWASSIMPLY
ASHAMETHEWAYTHING
SHADTOHAPPENBYTHE
EFORTANDPENFROLLO
OKINGITWASSIMPLYE
SHAMATHEWAYTHINGS
HADTOHAPPENBYTHET

O F T A N D P I T I F U L L O O K I N G I T W A S S I M P L Y A S H A M E T H E W A Y T H I N G S H A D T O H A P P E N B Y T H E T I M
ORT AND PINIR OL LOOKENG IT WAS SIMPLY E SHAME THE WAY THINGS HAD TO HAPPEN BY THE NIO

At P19, if the repeated message was 53 long, it would make pivots. Or, if there was a word or short phrase repeated at intervals of 53, there would be pivots and also P19.

Or far simpler explanation, and simple is good. A word that is 8 long, repeated over and over again in the middle of the message. Would make P19, P39 and pivots!

The numbers are word positions, just imagine an 8 letter word there. Match up 4 and 7 = P19. Match up 1 and 8 = P39. Match up 3 and 2 = P39. And the diagonals P16, P32 and P48 unigram repeats will make pivots.

But, my hypothesis is that there is a pattern of letters written over and over again, at least in the region of the pivots. Or something like that because of the cluster of P16, 32 and 48 unigram repeats in the area of the pivots. Especially the one on the right.

I also have considered this but did not get very far.

If you made a scytale message 19 x 18 and wrote vertically, and there was a section that was repeated over and over again at this interval, it would make pivots.

The message below, with only the repeated phrase in the creative pattern.

After recast into 17 x 20, it makes pivots.

Maybe we are looking at a very creative scytale message. If the phrase was just the right length I think it would make P39 but I will work on that later today or tomorrow.

hi smokie.. i have tried lots of times to get a scytale to wrap and form in a pivot with no success. i can see your line of thinking but i am not sure he would have done that.
cheers

hi smokie.. i have tried lots of times to get a scytale to wrap and form in a pivot with no success. i can see your line of thinking but i am not sure he would have done that.
cheers

That gives me an idea. Test the message by designated areas for cycle scores. Designate an area, and calculate the cycle score with all of the symbols in the message where only those symbols are in the area. For example, designate all of the cells in a diagonal row. Then calculate the cycle scores for the message after deleting all of the symbols in the message that do not appear in that diagonal row. Do that with a lot of areas of the same size and shape to see if there is one area or shape that has a really high cycle score. See if that can be used to identify repetitive patterns in the plaintext.

For the idea two posts above, cast the message into 19 x 18, then slide three columns down one position. Then slide the next three columns down two positions. Do that with different combinations of three columns. Test horizontal rows to see if you get higher cycle scores.

Bigrams: 41
– Normalized: 0.1547169811320755
Bigram ioc: 98
– Normalized: 0.0008552826796529996
Deep bigrams: 14407
N-grams: 49
Asymmetry: 2446

Bigram map:
——————————————————–
10 28 19 10
25 4 30 50 10
28 13 17 5 15 19 53
27 62 34 5 19 6
16 47 7 23 51 14 20 9 27
5 19 7 25 21 19 53
21 19 5 19 15 19 11 14 20
55 3 30 50

18 35 59 40 63 55 19 6 22 16
20 23 29 42
37 51 58 19 20 37 51 18 35 21 19
22 16 23 11 5
19 19 20 58 19 20 22 16
7 25 19 40
29 42
17 5 55 3 19 53 11 5
16 47 7 23 51 55 19 40
29 42 59 40 63 9 27 62 34 28 13
20 23 11 14 19 10 25 4

similar to what you are doing smokie.. this gives a high bigram count. simply odds evens columns then each column moved up one. then read top to bottom left to right.. the over all solve is not great but the first two words are DON CARDON which would be a neat way to start of your next code knowing Don Harden had solved the first one so easy.

Score: 20612.56 Ioc: 0.07013 Multiplicity: 0.18529
N-grams: 329 PC-cycles: 366

DONCARDONSIDEEDAN
DSPRINTIMINOREAND
ONTHEATTHESHEILLB
EATINSCHEADLETSTO
ADGUNSTLORMANYMAS
LYSHEAURIESINCECA
NIEHSHHEDTHEETORS
BYIFITTAGOFLANETH
RENATOITSAMCAPINE
ACULAWARBEANDOORE
RHASFAIRSOUNGHAMO
UNTLERONTRACIERAL
LYANDHPPYHCROSSTH
ENTEROLERICELANDP
URSELEEEALTINIONT
INDEDOUTANTTOFANE
THBYYESINTHTRFEIN
OUDIENDGUSSTOBEAT
ASOURSAWARMATIOND
RSTOOLAYEDDEDATRI