Just to be sure: You need transposition variations that are based on P19 in any way, for example a 2D transposition like the Magic Square? So are you looking for transpositions like the following?
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 301 302 303 304 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 305 306 307 308 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 309 310 311 312 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 313 314 315 316 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120 121 123 125 127 129 131 133 135 137 139 141 143 145 147 149 317 318 319 320 122 124 126 128 130 132 134 136 138 140 142 144 146 148 150 151 153 155 157 159 161 163 165 167 169 171 173 175 177 179 321 322 323 324 152 154 156 158 160 162 164 166 168 170 172 174 176 178 180 181 183 185 187 189 191 193 195 197 199 201 203 205 207 209 325 326 327 328 182 184 186 188 190 192 194 196 198 200 202 204 206 208 210 211 213 215 217 219 221 223 225 227 229 231 233 235 237 239 329 330 331 332 212 214 216 218 220 222 224 226 228 230 232 234 236 238 240 241 243 245 247 249 251 253 255 257 259 261 263 265 267 269 333 334 335 336 242 244 246 248 250 252 254 256 258 260 262 264 266 268 270 271 273 275 277 279 281 283 285 287 289 291 293 295 297 299 337 338 339 340 272 274 276 278 280 282 284 286 288 290 292 294 296 298 300
This is a zig-zag pattern based on P19 with the remaining fields LRTB filled up. The following picture shows it a bit more clearly:
P19 is therefore used as a "pattern". The interesting thing about this matrix is that there are 31 bigrams on P40 (double cipher height). Furthermore there are 6 trigrams and 2 quadgrams on P16.
I have created a lot of such matrices, but unfortunately not properly documented and archived. But I can repeat this if you need such transpositions.
EDIT: 2 Quadgrams, not 4
Another idea:
Would it make sense to solve several transposition variations with the row bound fragments solver and then compare the results with each other using the Levenshtein Distance? That way you could find out if different variations correlate with each other.
https://en.wikipedia.org/wiki/Levenshtein_distance
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My reasoning is that some of the period 19 transposition variations with higher number of bigram/trigram repeats may have the fragments in a better order by chance.
I need Z340 p15/19 transposition variations of which you think the hypothetical fragments could be in a good order by chance. But no more than say 5 variations per person.
This is a zig-zag pattern based on P19 with the remaining fields LRTB filled up.
Good idea, there’s so much left to explore.
P19 is therefore used as a "pattern".
Indeed. You’ve basically summarized the entire thread.
Would it make sense to solve several transposition variations with the row bound fragments solver and then compare the results with each other using the Levenshtein Distance? That way you could find out if different variations correlate with each other.
Can you give a layman explanation of the Levenshtein distance?
I will work on a message. Hopefully it will turn out pretty good. See you soon.
Sure, but as the fragment hill-climber works now, it will not be able to get the solution if the average fragment size is less than 5 or 6.
Would it make sense to solve several transposition variations with the row bound fragments solver and then compare the results with each other using the Levenshtein Distance? That way you could find out if different variations correlate with each other.
Can you give a layman explanation of the Levenshtein distance?
I’m not an expert in the subject, I just use the algorithm. Wikipedia describes its core as follows:
Informally, the Levenshtein distance between two words is the minimum number of single-character edits (insertions, deletions or substitutions) required to change one word into the other.
If you compare two identical strings, the Levenshtein Distance (LS) is 0. If you compare "ZODIAC" with "ZODIAK", then LS is 1. Comparing "ZODIAC" with "ZODICA" results in LS = 2. "ZODIAC" with "ZODAIK" scores 3. And "ZODIAC" with "ODIACZ" results in 2.
Here you can try it out: https://planetcalc.com/1721/
I also use LS to find repetitive but incomplete patterns in a text. Here is some pseudo-code:
Say you want to search for patterns of length 6 in z340. Then proceed as follows (pseudo-code):
for pos = 1 to 334
pattern1 = z340(0, 6)
pattern2 = z340(pos, 6)
ls = calculate_ls(pattern1, pattern2)
if (ls < 3) // arbitrary threshold value
print("Pattern found")
// Repeat that for all positions
My question now is whether this can also be used for the row bound fragment solver. Let two transposition variations be solved and then compare all fragments of solution 1 with all fragments of solution 2. If two fragments have a low LS, both transpositions probably correlate at exactly this point.
I found an implementation in free basic:
https://www.freebasic.net/forum/viewtopic.php?t=9656
I understand now. Thanks for mentioning the Levenshtein distance. At first glance it seems like a wonderful tool and I may find use for it.
My question now is whether this can also be used for the row bound fragment solver. Let two transposition variations be solved and then compare all fragments of solution 1 with all fragments of solution 2. If two fragments have a low LS, both transpositions probably correlate at exactly this point.
I think it would be simpler to compare the key since each fragment does not have an individual key.
I also use LS to find repetitive but incomplete patterns in a text. Here is some pseudo-code:
Say you want to search for patterns of length 6 in z340. Then proceed as follows (pseudo-code):
Nice, what does it say for the 340? Versus suffles for example? Just wondering if you done anything like that.
I wanted to do the test I mentioned but don’t feel motivated enough at the moment so for now I will keep it quiet.
I wanted to do the test I mentioned but don’t feel motivated enough at the moment so for now I will keep it quiet.
Okay. As soon as you need transposition variations for the test, let me know. I’ll be happy to create some for you.
Nice, what does it say for the 340? Versus suffles for example? Just wondering if you done anything like that.
So far only sporadically. Doranchak did that once with my transposition idea with the 48 bigrams and 8 trigrams. But I don’t know if the Levensthein Distance was used there, too:
I just started a test with this cipher and looked for patterns of length 6, and a maximum LS of 2. In the 48_8_phenomenon cipher there are five places where such patterns occur (partly nested). Here the distribution for 100000 shuffles:
[0, 93180] [1, 3845] [2, 2020] [3, 771] [4, 141] [5, 26] [6, 15] [7, 2]
Out of 100000 shuffels only 26 shuffels have the same values as in 48_8_phenomenon. At first sight, I would say that this is significant. Especially in view of the fact that this cipher was created with a comparatively simple transposition. But I am not sure yet.
Here are some results for the first 340 letters of z408:
Searching for patterns of lenth 6 with a max LS of 2:
z408 has 11 positions with matching patterns. Results for shuffled versions:
[0, 94535] [1, 3116] [2, 1616] [3, 591] [4, 110] [5, 17] [6, 13] [7, 2]
It is clear that a high bigram number increases the probability of such patterns. So a false positive with a high bigram count can also lead to a false positive in the pattern search. Here one would have to test further. Let me see if I find the time for it. I find this very exciting.
Here is another transposition idea:
First mark one P19 pass in orange. For the wrap-around you change the color and mark the second pass:
Now you align the lines by shifting them. The respective colors should be one below the other:
In the last step you change the reading direction (rotate + mirror):
Basically the cipher is a kind of P19 without wrap around.
You get 44 bigrams and 3 trigrams on P1. Also the quite noticeable pattern in the lower part. AZDecrypt’s transposition solver achieves fairly high values (so far at restart 10 -> 21185). Unfortunately only a PC-cycle score of 488. Currently the row bound fragment solver runs with the shown cipher. Let’s see how the result looks like.
Jarlve, two more questions:
– If z340 was transposed without wrap around, could the wrap around in period operations be a big problem? Not really, is it? At most, some fragments would be jumbled up. That shouldn’t make much difference to the solver.
– For tests with the row bound fragment solver, does it make sense to create transpositions based on "Transpose P18" instead of "Untranspose P19"? The difference is marginal, but could be critical. Or am I wrong here?
I wish everyone a happy new year!
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Here is another transposition idea:
You can of course come up with a googol different variations, especially when considering multiple steps. I would just look at 1-step period 19 stuff in a 17 by 20 grid, it is more than hard enough.
You get 44 bigrams and 3 trigrams on P1. Also the quite noticeable pattern in the lower part. AZDecrypt’s transposition solver achieves fairly high values (so far at restart 10 -> 21185). Unfortunately only a PC-cycle score of 488. Currently the row bound fragment solver runs with the shown cipher. Let’s see how the result looks like.
The solvers should not get it.
If z340 was transposed without wrap around, could the wrap around in period operations be a big problem? Not really, is it? At most, some fragments would be jumbled up. That shouldn’t make much difference to the solver.
If the fragments are properly divided over the period lines then there is no problem.
For tests with the row bound fragment solver, does it make sense to create transpositions based on "Transpose P18" instead of "Untranspose P19"? The difference is marginal, but could be critical. Or am I wrong here?
"Tranpose P18" extracts the period lines in a different order than "Untranspose P19" and that order could have the fragments in a better order by chance, usually reflected by having more bigrams/trigrams.
Happy new year.
You can of course come up with a googol different variations, especially when considering multiple steps. I would just look at 1-step period 19 stuff in a 17 by 20 grid, it is more than hard enough.
I need Z340 p15/19 transposition variations of which you think the hypothetical fragments could be in a good order by chance. But no more than say 5 variations per person.
Okay, so maybe I didn’t understand the principle after all. I thought that different P19 variations should be created to identify fragments of plain text. Just by chance, because with a bit of luck they are in a better order. But my cipher shown above meets these criteria, doesn’t it? The single steps do nothing else than to create a variation of P19.
Here is a direct comparison:
z340, untranspose P19:
H+MwICV9K+ckgEmBqaEB+o7emL-RR+4apIbMRjUVFFuzu5JN0VMqIDakZxPaLhc7IImUQLFHbOGb+gInU1o7tZG+eNclWO3DjMVE7FV7g+hbdDj+4jG++ITB4-RqWeVWq+ekg0dD4t1+tWnSPYLR57J+JYMj+ITCvuemkKb+pZ+Bm8+t+u1ZInugwK2ROb+wrmLWBOfoHyRQkgb03RBxftywLKJzdlOFvTBcXu#PYATpSMF8+BnMFGfBCOOIGqb+cgytFKuFoKnSBKgBbuOUNrBO#Nsj+Ho8hrv1-tYArgzd4OFT-+Nsd2oXu#-nSpzbc5CbtchhG+4Utr7CdWjt
My example cipher (shifted one to the left, for better comparison:
H+MwICV9K-+ckgEmBqCaEB+o7emLh-RR+4apIbtMRjUVFFuznu5JN0VMq+IDakZxPaLjMc7IImUQL+FHbOGb+gIG+U1o7tZGRqeNclWO3Dkg0VE7FV7gtWhbdDj+4j7J++ITB4-ITCWeVWq+eb+pZdD4t1+t+unSPYLR5wK2RJYMj+rmLWvuemkKyRQkg+Bm8+Bxft1ZInugzdlOFOb+wcXu#PBOfoHSMF8+Bb03RGfBCOywLKJb+cgytvTBuFoKnSYATpBbuOUNrnMF#Nsj+HOIGqrv1-tYAFKgzd4OFTBKgNsd2oXu#BOnSpzbc5o8hbtchhG+4rUtr7CdWj-+
As you can see, the principle of P19 remains, but the fragments are a bit scrambled. I thought that was the point. To find variations that explain P19 that we haven’t been able to solve yet.
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