Many of the different period 19 and 15 transpositions that we have covered simply follow a different order/distribution over what I call the "period lines" in a 17 by 20 matrix. For both period 19 and 15 there are 55 of these lines (think starting points). In the following image, a transposition matrix, the green cells are the 55 starting points of the "period 19 lines", the orange cells starting with number 73 depict one such period line. To get "period 15 lines" one can simply mirror the image and get the idea.
If we expect the Z340 to have a period 19 or 15 transposition that properly follows these period lines (no nulls/skips, diagonal transposition and etc…) then these period lines can be extracted and fed into AZdecrypt’s "Substitution + row bound" solver (solver idea by smokie treats) and the cipher should solve.
To test that I apply a normal period 19 transposition to the first 340 characters of the Z408 and extract the period 19 lines and feed it to the solver:
Score: 24419.13 IOC: 0.06533 Multiplicity: 0.15882 Repeats: 568 PC-cycles: 1468 Seconds: 1.30 Average n-gram size: 4.35087 E (0) F (0) NK (10) TH (11) MAN (34) TUE (27) ILLS (76) ETHE (88) EXPER (176) ETTER (185) OURROC (270) RLTHEB (287) IATHAEW (238) LBEREBO (408) CEANDALL (736) LLEDWILL (662) LAVESIWIL (711) OUMYNAME (585) OUWILLTRY (891) INGPEOPL (791) TISSOMUCH (935) REFUNTHA (650) ILDGAMEIN (798) TBECAUSE (875) OATDANGER (773) OFALLTOK (617) INGGIVESM (820) HRILLING (705) ITISEVENB (841) GETTINGY (793) FFWITHAGI (845) PARTOFIT (730) NIDIEIWIL (653) INPARADI (629) HEIHAVEKI (814) ECOMEMYS (617) LNOTGIVEY (829) BECAUSEY (845) ILIKEKILL (779) BECAUSEI (869) UNITIAMO (508) ILLINGW (553) EFORRES (506) ISTHEM (373) ANAMAL (280) OMETH (188) MOATT (132) ENCE (81) THAN (83) KSO (31) EST (34) HE (11) RN (10) T (0) B (0)
As you can see it properly solved the cipher though the fragment order is lost.
To test this is on the Z340 you can extract the period 19 and 15 lines yourself with AZdecrypt using the following steps:
1) Open the Z340 in the Input window. Must be in symbolic format!
2a) For period 19: go to the Format menu and pick "Offset rows bottom-to-top" (do this twice).
2b) For period 15: (mirror the cipher first!) go to the Format menu and pick "Offset rows top-to-bottom" (do this twice).
3) Go to the Format menu and pick "Square with spaces".
4) Go to the Functions menu and pick "Transposition", in the Transposition window pick "Columnar 1" and click on Untranspose.
5) Go to the Format menu and pick "Convert spaces to rows" and then remove the empty row at the top of the cipher if need be.
6) Pick the "Substitution + row bound" solver and click on Solve.
I did this for you are here are the results with the Reddit 6-grams loaded first:
Z340 period 19 and 15 lines:
> M |D FH +kN +dp R)Wk cW<S |TC7z c+ztZ y.LWBO B31c_8 lXz6PYA G1BCOO| zF*K<SBK 6N:(+H*; 29^4OFT-+ <Sf9pl/C Ucy5C^W(c +l#2E.B) -RR+4>f|p z/JNbVM) dl5||.UqL <Ut*5cZG (MVE5FV52 G++|TB4- #2b^D4ct+ 5J+JYM(+ p+fZ+B.;+ 8KjROp+8 _Rq#2pb&R 9^%OF7TB SMF;+B<MF p+l2_cFK BpzOUNyBO y7t-cYAy N:^j*Xz6- pclddG+4 H+M8|CV@K EB+*5k.L R(UVFFz9 >#Z3P>L pOGp+2| l%WO&D ^D(+4( VW)+k PYLR/ k.#K |<z2 1*H LKJ Tf G) 2 d Score: 22063.25 IOC: 0.06581 Multiplicity: 0.18529 Repeats: 250 PC-cycles: 292 Seconds: 41.25 Average n-gram size: 4.35087 E (0) T (0) OW (10) SP (10) TOA (31) TFN (23) LFVO (52) EVIN (72) OBRTS (80) ETSRA (135) DDIVEI (274) EUPERH (242) OUSESAN (505) SPERIIO (261) SSAYINEY (518) EABITPAI (539) AKEHISBUT (817) INDKNOWR (578) MEDGREVIE (624) TOCANDEF (604) ULLTHEDON (864) SWHATITF (711) FOGOODMYI (639) IMRAGEAS (519) ITINGSIGA (742) STTOBEHU (648) CATEWHERT (765) GHTHATIT (773) NTDATEDIT (844) HYCLINTH (579) RLYCANTEL (841) KEXISTBE (617) NTSITEITS (791) NTOARESY (570) ENSIMADEI (741) DTRUEAND (649) ABECAUSEU (957) NEOFFSTH (576) PTTHORITY (767) NETAGODI (491) LIMISSSK (504) ECAUSEI (633) NISNTAO (411) OXVIEW (254) EWITHI (373) IVFTO (95) SAILW (119) ODCY (49) OISA (72) PAP (29) IYH (21) BD (8) SF (10) A (0) F (0)
The difference between the period 19 and 15 lines is that the order of the lines are in reverse which does not matter for the solver. But we still need to reverse the order of each individual row to get the other direction:
1) Take the period 19 lines as from the steps above and go to Format and pick "Square with spaces".
2) Go to Functions and pick "Tranposition" and in the Transposition menu pick "Mirror" and then click Transpose.
3) Go to Format and pick "Remove spaces".
I also did this and here is the result:
> M D| HF Nk+ pd+ kW)R S<Wc z7CT| Ztz+c OBWL.y 8_c13B AYP6zXl |OOCB1G KBS<K*Fz ;*H+(:N6 +-TFO4^92 C/lp9fS< c(W^C5ycU )B.E2#l+ p|f>4+RR- )MVbNJ/z LqU.||5ld GZc5*tU< 25VF5EVM( -4BT|++G +tc4D^b2# +(MYJ+J5 +;.B+Zf+p 8+pORjK8 R&bp2#qR_ BT7FO%^9 FM<B+;FMS KFc_2l+p OByNUOzpB yAYc-t7y -6zX*j^:N 4+Gddlcp K@VC|8M+H L.k5*+BE 9zFFVU(R L>P3Z#> |2+pGOp D&OW%l (4+(D^ k+)WV /RLYP K#.k 2z<| H*1 JKL fT )G 2 d Score: 22179.49 IOC: 0.06250 Multiplicity: 0.18529 Repeats: 206 PC-cycles: 432 Seconds: 56.68 Average n-gram size: 4.35087 I (0) P (0) ML (9) YS (10) AFS (25) TFS (24) FWGT (40) LLWR (63) URDIL (123) WEUSR (95) OAWKIN (193) YARMSA (295) VEDBUTE (465) LOODAMO (416) EALLETSU (570) HTYSONAB (531) SHISONECT (742) DGETCALL (654) ROWEDINRB (686) GAINTHES (786) TLAINSTTH (821) GPANARGU (552) KABILLIEF (688) OWRITEBL (612) TIASINAPO (698) HNAILSSO (566) SERNMENTH (835) SOPERSRI (549) SHIASWAST (657) YSTOTHEY (706) TENTTHATA (967) AIRSOMEC (625) SPLASHSPL (650) ESRATEST (621) OANABOUTA (893) NVERHERN (579) HBUTTHENA (967) NSOFFERT (656) EBADLYPSY (617) KIFITSAN (687) CUSSABOT (561) KIDSWHI (471) LTSTOOT (440) MEOWME (253) ONSOME (376) FSGWA (92) GTKED (94) EHIF (60) TULL (63) YTM (24) REK (28) AI (10) GO (10) T (0) F (0)
Both the period 19 and 15 lines score around 22000 which is much less than the 24419 score of my Z408 test example. Bottom line: if the Z340 is a homophonic substitution cipher + period 19 or 15 transposition, then the transposition for some reason is not properly divided over the aforementioned period lines.
I agree 100% and think that the problem to solve is how to find the correct fragment breakdown. If a lot of the fragments are the same length, that would help.
I agree 100% and think that the problem to solve is how to find the correct fragment breakdown. If a lot of the fragments are the same length, that would help.
Indeed. It is something we must approach carefully because there are allot of options.
Bottom line: if the Z340 is a homophonic substitution cipher + period 19 or 15 transposition, then the transposition for some reason is not properly divided over the aforementioned period lines.
While the suspected period line misalignment could be caused by nulls/skips or diagonal transposition I personally find a "free" period 19 transposition in a 17 by 20 grid more appealing. The following images show such a period 19 transposition in a 17 by 20 grid:
This is the start of it. Column 1, 3 and 5 are used first while wrapping around to the top if needed. Then follow column 2, 4 and 6, then column 7, 9 and 11, etc…
Transposition matrix complete, the fragments are no longer than 3 characters each.
A free period 19 transposition in a 17 by 20 grid typically has 170 period 19 pairs. In the worst case these 170 pairs could follow a random distribution but that would destroy at least half of the bigram information and all of the trigram information so something heavy like that is unlikely.
I experimented with a concept last March, and definitely not saying that this is what we should do, but just used it as a tool to think about the issues. I called it matrix unfolding.
Start with a transposition matrix. This one is P20.
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
129 146 163 180 197 214 231 265 282 299 316 333 011 028 045 062 079
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
085 102 119 136 153 170 187 204 221 238 255 272 289 306 323 340 248
Untranspose at period 20 to get an array 340 x 1, one big fragment and score. The true answer is somewhere between 1 x 340 and 340 x 1.
Only the first 40 is shown because of screen width.
001 002 003 004 005 006 007 008 009 010 028 029 030 031 032 033 034 018 019 020 021 022 023 024 025 026 027 045 046 047 048 049 050 051 035 036 037 038 039 040
Generate random number between 1 and 340, and divide the fragment by that number. Score. Below is the 8th iteration, which found a good segment length.
001 002 003 004 005 006 007 008
009 010 028 029 030 031 032 033
034 018 019 020 021 022 023 024
025 026 027 045 046 047 048 049
050 051 035 036 037 038 039 040
041 042 043 044 062 063 064 065
066 067 068 052 053 054 055 056
057 058 059 060 061 079 080 081
082 083 084 085 069 070 071 072
073 074 075 076 077 078 096 097
098 099 100 101 102 086 087 088
089 090 091 092 093 094 095 113
114 115 116 117 118 119 103 104
105 106 107 108 109 110 111 112
130 131 132 133 134 135 136 120
121 122 123 124 125 126 127 128
129 147 148 149 150 151 152 153
137 138 139 140 141 142 143 144
145 146 164 165 166 167 168 169
170 154 155 156 157 158 159 160
161 162 163 181 182 183 184 185
186 187 171 172 173 174 175 176
177 178 179 180 198 199 200 201
202 203 204 188 189 190 191 192
193 194 195 196 197 215 216 217
218 219 220 221 205 206 207 208
209 210 211 212 213 214 232 233
234 235 236 237 238 222 223 224
225 226 227 228 229 230 231 249
250 251 252 253 254 255 239 240
241 242 243 244 245 246 247 265
266 267 268 269 270 271 272 256
257 258 259 260 261 262 263 264
282 283 284 285 286 287 288 289
273 274 275 276 277 278 279 280
281 299 300 301 302 303 304 305
306 290 291 292 293 294 295 296
297 298 316 317 318 319 320 321
322 323 307 308 309 310 311 312
313 314 315 333 334 335 336 337
338 339 340 324 325 326 327 328
329
330 331 332 011 012
013 014
015 016 017 248
Now randomly choose the start position of one fragment, and the end position of another, later fragment. Join them all together, generate random number between 1 and the length of the joined fragments, and divide.
150th iteration.
001 002 003 004 005 006 007 008
009 010 028 029 030
031 032 033 034 018
019 020 021 022 023
024 025 026 027 045
046 047 048 049 050
051 035 036 037 038
039 040 041 042 043
044 062 063 064 065
066 067 068 052 053
054 055 056
057 058 059 060 061 079 080 081
082 083 084 085 069 070 071 072
073 074 075 076 077 078 096 097
098 099 100 101 102 086 087 088
089 090 091 092 093 094 095 113
114 115 116 117 118 119 103 104
105 106 107 108 109 110 111 112
130 131 132 133 134 135 136 120
121 122 123 124 125 126 127 128 129
147 148 149 150 151 152 153 137 138
139 140 141 142 143 144 145 146 164
165 166 167 168 169 170 154 155 156
157 158 159 160 161 162 163 181 182
183 184 185 186 187 171 172 173 174
175 176 177 178 179 180 198 199 200
201 202 203
204 188 189
190 191 192
193 194 195
196 197 215
216 217 218
219 220 221
205 206 207
208 209 210
211 212 213
214 232 233
234 235 236
237 238 222
223 224 225
226 227 228
229 230 231
249 250 251
252 253 254
255 239 240
241 242 243
244 245 246
247 265 266
267 268 269
270 271 272
256 257 258
259 260 261
262 263 264
282 283 284
285 286 287
288 289 273
274 275 276 277 278 279 280 281 299
300 301 302 303 304 305 306 290 291
292 293 294 295 296 297 298 316 317
318 319 320 321 322 323 307 308 309
310 311 312 313 314 315 333 334
335 336 337 338 339 340 324 325
326 327 328 329 330 331 332 011
012 013 014 015 016 017 248
1000th iteration.
001 002 003 004 005 006 007 008 009
010 028 029 030 031 032 033 034 018
019 020 021 022 023 024 025 026 027
045 046 047 048 049 050 051 035 036
037 038 039 040 041 042 043 044 062
063 064 065 066 067 068 052 053 054
055 056 057 058 059 060 061
079 080 081 082 083 084 085
069 070 071 072 073 074 075 076 077 078
096 097 098 099 100 101 102 086 087 088
089 090 091 092 093 094 095 113
114 115 116 117 118 119 103 104
105 106 107 108 109 110 111 112
130 131 132 133 134 135 136 120
121 122 123 124 125 126 127 128 129
147 148 149 150 151 152 153 137 138
139 140 141 142 143 144 145 146 164
165 166 167 168 169 170 154 155 156
157 158 159 160 161 162 163 181
182 183 184 185 186 187 171 172
173 174 175 176 177 178 179 180
198 199 200 201 202 203 204 188
189 190 191 192 193 194 195 196
197 215 216
217 218 219 220 221 205
206 207 208 209 210 211 212 213 214
232 233 234 235 236 237 238 222 223
224 225 226 227 228 229 230 231
249 250 251 252 253 254 255 239
240 241 242 243 244 245 246 247
265 266 267 268
269 270 271 272
256 257 258 259 260 261 262 263 264
282 283 284 285 286 287 288 289 273
274 275 276 277 278 279 280 281 299
300 301 302 303 304 305 306 290 291
292 293 294 295 296 297 298 316 317
318 319 320 321 322 323
307 308 309 310 311 312 313 314 315
333 334 335 336 337 338 339 340 324
325 326 327 328 329 330 331 332 011
012 013 014 015 016 017 248
I experimented with a concept last March, and definitely not saying that this is what we should do, but just used it as a tool to think about the issues. I called it matrix unfolding.
…
Untranspose at period 20 to get an array 340 x 1, one big fragment and score. The true answer is somewhere between 1 x 340 and 340 x 1.
Fantastic, great idea, I will start working on it right away. I am looking to generalize this to a "row bound" fragment hill climber. Start with 340 by 1 and allow for example 10 fragments, then 11 fragments and so on…
Can you explain it one iteration at a time? Not sure how your algorithm works.
I don’t know if this will work for a message, but here is how the matrix unfolder works.
Step 1. First fragment is untransposed fragment 340 long ( only first 40 shown here for clarity ). Score by counting how many sequential numbers occur starting at the far left. Stop counting when arriving at the first non-sequential number. The first fragment scores 9 shown in red.
001 002 003 004 005 006 007 008 009 010 028 029 030 031 032 033 034 018 019 020 021 022 023 024 025 026 027 045 046 047 048 049 050 051 035 036 037 038 039 040
Step 2. The first fragment has start position 1 and end position 340. Generate random number between 1 and 340, and divide the fragment by that number.
Iteration 2, random number is 21. Now there are 17 fragments, each of which are scored. Now the score is 70. It found a sweet fragment length. Note the leftover symbols at the end. Now each fragment has a start position and end position. Since there are 17 fragments, generate a random number between 1 and 17. The random number is 9, so the beginning of the 9th fragment will be the beginning of a new fragment, or new set of fragments. Now generate a random number between 10 and 17, which this time is 10. The end of the 10th fragment is the end of a new fragment, or new set of fragments. See below, the beginning is red start position 211, and end is blue end position 252.
001 002 003 004 005 006 007 008 009 010 028 029 030 031 032 033 034 018 019 020 021
022 023 024 025 026 027 045 046 047 048 049 050 051 035 036 037 038 039 040 041 042
043 044 062 063 064 065 066 067 068 052 053 054 055 056 057 058 059 060 061 079 080
081 082 083 084 085 069 070 071 072 073 074 075 076 077 078 096 097 098 099 100 101
102 086 087 088 089 090 091 092 093 094 095 113 114 115 116 117 118 119 103 104 105
106 107 108 109 110 111 112 130 131 132 133 134 135 136 120 121 122 123 124 125 126
127 128 129 147 148 149 150 151 152 153 137 138 139 140 141 142 143 144 145 146 164
165 166 167 168 169 170 154 155 156 157 158 159 160 161 162 163 181 182 183 184 185
186 187 171 172 173 174 175 176 177 178 179 180 198 199 200 201 202 203 204 188 189
190 191 192 193 194 195 196 197 215 216 217 218 219 220 221 205 206 207 208 209 210
211 212 213 214 232 233 234 235 236 237 238 222 223 224 225 226 227 228 229 230 231
249 250 251 252 253 254 255 239 240 241 242 243 244 245 246 247 265 266 267 268 269
270 271 272 256 257 258 259 260 261 262 263 264 282 283 284 285 286 287 288 289 273
274 275 276 277 278 279 280 281 299 300 301 302 303 304 305 306 290 291 292 293 294
295 296 297 298 316 317 318 319 320 321 322 323 307 308 309 310 311 312 313 314 315
333 334 335 336 337 338 339 340 324 325 326 327 328 329 330 331 332 011 012 013 014
015 016 017 248
Join these two fragments into one big fragment, the length is 42.
001 002 003 004 005 006 007 008 009 010 028 029 030 031 032 033 034 018 019 020 021
022 023 024 025 026 027 045 046 047 048 049 050 051 035 036 037 038 039 040 041 042
043 044 062 063 064 065 066 067 068 052 053 054 055 056 057 058 059 060 061 079 080
081 082 083 084 085 069 070 071 072 073 074 075 076 077 078 096 097 098 099 100 101
102 086 087 088 089 090 091 092 093 094 095 113 114 115 116 117 118 119 103 104 105
106 107 108 109 110 111 112 130 131 132 133 134 135 136 120 121 122 123 124 125 126
127 128 129 147 148 149 150 151 152 153 137 138 139 140 141 142 143 144 145 146 164
165 166 167 168 169 170 154 155 156 157 158 159 160 161 162 163 181 182 183 184 185
186 187 171 172 173 174 175 176 177 178 179 180 198 199 200 201 202 203 204 188 189 190 191 192 193 194 195 196 197 215 216 217 218 219 220 221 205 206 207 208 209 210
211 212 213 214 232 233 234 235 236 237 238 222 223 224 225 226 227 228 229 230 231
249 250 251 252 253 254 255 239 240 241 242 243 244 245 246 247 265 266 267 268 269
270 271 272 256 257 258 259 260 261 262 263 264 282 283 284 285 286 287 288 289 273
274 275 276 277 278 279 280 281 299 300 301 302 303 304 305 306 290 291 292 293 294
295 296 297 298 316 317 318 319 320 321 322 323 307 308 309 310 311 312 313 314 315
333 334 335 336 337 338 339 340 324 325 326 327 328 329 330 331 332 011 012 013 014
015 016 017 248
Generate a random number between 1 and 42. Divide the joined fragment by that number. This time the number is 11.
Fragments 9 and 10 were joined, then divided into new fragments length of 11. It scores a little higher.
001 002 003 004 005 006 007 008 009 010 028 029 030 031 032 033 034 018 019 020 021
022 023 024 025 026 027 045 046 047 048 049 050 051 035 036 037 038 039 040 041 042
043 044 062 063 064 065 066 067 068 052 053 054 055 056 057 058 059 060 061 079 080
081 082 083 084 085 069 070 071 072 073 074 075 076 077 078 096 097 098 099 100 101
102 086 087 088 089 090 091 092 093 094 095 113 114 115 116 117 118 119 103 104 105
106 107 108 109 110 111 112 130 131 132 133 134 135 136 120 121 122 123 124 125 126
127 128 129 147 148 149 150 151 152 153 137 138 139 140 141 142 143 144 145 146 164
165 166 167 168 169 170 154 155 156 157 158 159 160 161 162 163 181 182 183 184 185
186 187 171 172 173 174 175 176 177 178 179 180 198 199 200 201 202 203 204 188 189
190 191 192 193 194 195 196 197 215 216 217 218 219 220 221 205 206 207 208 209 210
211 212 213 214 232 233 234 235 236 237 238
222 223 224 225 226 227 228 229 230 231 249
250 251 252 253 254 255 239 240 241 242 243
244 245 246 247 265 266 267 268 269
270 271 272 256 257 258 259 260 261 262 263 264 282 283 284 285 286 287 288 289 273
274 275 276 277 278 279 280 281 299 300 301 302 303 304 305 306 290 291 292 293 294
295 296 297 298 316 317 318 319 320 321 322 323 307 308 309 310 311 312 313 314 315
333 334 335 336 337 338 339 340 324 325 326 327 328 329 330 331 332 011 012 013 014
015 016 017 248
And keep doing that over and over. The leftover symbols after dividing have a chance of joining into a new fragment, and the number of fragments can fluctuate up and down. A group of fragments can be joined, and if the random divisor is 1, then it would stay joined.
The idea is that if the message has a lot of fragments at the same length, they will eventually be found and score really high.
The issue that the matrix unfolder made me think of is whether the algorithm should have a tendency to work on lower scoring fragments instead of higher scoring fragments.
Thanks smokie, that is a neat little algorithm! I will try it.
Here’s another example of a free period 19 transposition in a 17 by 20 grid. Start at a random column offset and fill up all odd columns and wrap around to the top if needed. Then do the same for even columns:
I transposed the first 340 characters of the Z408 with the this matrix and AZdecrypt’s "Substitution + simple transposition" solver was unable to get the solution:
^cBBS89XpLeARGL^p ISR+@%kc%9O)E9+YM rK)qR_VOOdPDVlU%j =#9EWWPd+PRe/RW^H kR6L#_YV@Le5=/Z_I 5TF6tNT^GK)GMp@/ UX+YZH9_Q8Ur!NY8X =RBJtrfI#(YLIIqVF 6EBU%QZR#BPDMk!Y eJ9KWtWAJtlBPz5l7 qNMkHH9%efdOqVOU) SpTAJ5z=K%8UrPEWR A(kt)l9ScI%Ye^k ^=U+/A#ZOA(E)I@6p F7LS%VX9Bq8PR/6M+ qNHDMrkTq+LWq#N9q X%GkVzqlcTE@dqGBt IV8%D)BG!YZ5H9/_B TFYp5ZNSPZ8qKRMAP #9Q#Eecf(YDe5_qUU 133 313 125 306 117 299 1 292 173 181 165 334 157 327 149 320 141 142 321 134 314 126 307 10 300 2 189 174 182 166 335 158 328 150 151 329 143 322 135 315 19 308 11 197 3 190 175 183 167 336 159 160 337 152 330 144 323 28 316 20 205 12 198 4 191 176 184 168 169 185 161 338 153 331 37 324 29 213 21 206 13 199 5 192 177 178 193 170 186 162 339 46 332 38 221 30 214 22 207 14 200 6 7 201 179 194 171 187 55 340 47 229 39 222 31 215 23 208 15 16 209 8 202 180 195 64 188 56 237 48 230 40 223 32 216 24 25 217 17 210 9 203 73 196 65 245 57 238 49 231 41 224 33 34 225 26 218 18 211 82 204 74 253 66 246 58 239 50 232 42 43 233 35 226 27 219 91 212 83 261 75 254 67 247 59 240 51 52 241 44 234 36 227 100 220 92 269 84 262 76 255 68 248 60 61 249 53 242 45 235 109 228 101 277 93 270 85 263 77 256 69 70 257 62 250 54 243 118 236 110 285 102 278 94 271 86 264 78 79 265 71 258 63 251 127 244 119 293 111 286 103 279 95 272 87 88 273 80 266 72 259 136 252 128 301 120 294 112 287 104 280 96 97 281 89 274 81 267 145 260 137 309 129 302 121 295 113 288 105 106 289 98 282 90 275 154 268 146 317 138 310 130 303 122 296 114 115 297 107 290 99 283 163 276 155 325 147 318 139 311 131 304 123 124 305 116 298 108 291 172 284 164 333 156 326 148 319 140 312 132
Here’s another example of a free period 19 transposition in a 17 by 20 grid. Start at a random column offset and fill up all odd columns and wrap around to the top if needed. Then do the same for even columns:
I did a lot of similar tests a few months ago. When I experimented with the Matrix tool, I thought of a lot of ways to create a P19 cipher. What you are showing right now could also be called columns odd/even + 2x diagonal transposition. Simpler: the even columns contain a diagonal transposition and the odd columns also contain a diagonal transposition.
This reminds me a bit of this test cipher:
viewtopic.php?f=81&t=3196&p=69008#p69008
I’ve added this transposition to my solver for the next AZdecrypt release. It has solved your cipher and no results on the 340 so far. It can pick any set of dimensions and then make a horizontal or vertical split at any offset of which each part could have its own transposition (none, mirrored, flipped, columnars, diagonals).
Shouldn’t your example above be solvable by AZDecrypt if you combine column order period 2 (odd/even) and a "split" diagonal transposition? It doesn’t work for me, even if I do the first step manually.
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I did a lot of similar tests a few months ago. When I experimented with the Matrix tool, I thought of a lot of ways to create a P19 cipher. What you are showing right now could also be called columns odd/even + 2x diagonal transposition. Simpler: the even columns contain a diagonal transposition and the odd columns also contain a diagonal transposition.
Shouldn’t your example above be solvable by AZDecrypt if you combine column order period 2 (odd/even) and a "split" diagonal transposition? It doesn’t work for me, even if I do the first step manually.
Yes, good thinking, but since I picked random columns to start it would probably still need to fix the column offset for each part of the split:
I experimented with a concept last March, and definitely not saying that this is what we should do, but just used it as a tool to think about the issues. I called it matrix unfolding.
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Untranspose at period 20 to get an array 340 x 1, one big fragment and score. The true answer is somewhere between 1 x 340 and 340 x 1.Fantastic, great idea, I will start working on it right away. I am looking to generalize this to a "row bound" fragment hill climber. Start with 340 by 1 and allow for example 10 fragments, then 11 fragments and so on…
Next week I will release AZdecrypt 1.17 with the updates beijinghouse and I worked on. This will include a "row bound fragments" solver (idea by smokie treats). My testing shows that it can solve ciphers which have an average fragment size of 5 or higher and perhaps 6 and higher is a more safe limit.
I have not done much testing on the Z340 yet but plan to do a large test. 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. So, I would like to run through allot of these variations hoping to get lucky. Considering running through about 25 to 50 variations + shuffles of the Z340 that have the same number of bigram/trigram repeats as the period 19 variations acting as a control group.
For every variation I will also run through the reversed character order. Will include the basic period 15 and 19, the diagonal, "special bigram" and Skytale variations, Largo’s 48 bigram and 8 trigram repeats, etc… Mr lowe, smokie, Largo and anyone else feel free to submit some, please! In doing this we can also build up a small library of potential period 19 untranspositions for other tests.
looking forward to the new AZ ..
this is odds evens column shift up one. its wwhat brings period19 together easy. lots of variations make it tuff though.
i hope you hit the jackpot soon Jarlve.
16 14 32 12 30 1 10 28 22 31 8 26 33 19 10 6 24
26 23 52 25 4 22 39 45 9 4 13 2 20 19 30 50 10
28 13 17 5 36 6 17 17 15 19 53 15 33 34 43 48 55
36 27 62 34 13 31 41 5 19 6 16 46 36 51 31 11 29
40 16 47 7 24 23 51 43 14 20 9 27 13 3 54 44 31
49 3 23 5 19 44 7 25 21 19 53 21 50 41 19 41 27
37 21 19 5 23 15 5 19 16 11 15 19 19 11 14 20 53
55 3 21 38 8 51 51 40 47 29 38 48 30 50 36 39 15
1 19 37 44 11 56 8 60 31 40 54 41 18 61 8 37 33
18 35 7 49 30 59 40 63 55 19 6 22 16 2 28 20 33
20 5 40 23 38 18 34 20 23 29 42 32 47 5 6 54 56
42 37 51 58 19 20 29 37 51 63 18 35 21 19 1 30 58
46 3 57 22 16 5 61 52 3 15 12 20 56 23 23 11 5
19 32 39 19 20 28 58 19 20 45 12 36 46 44 22 16 61
7 25 53 36 48 19 36 19 40 48 39 21 37 8 2 50 51
8 50 16 36 26 29 42 17 6 50 11 11 28 38 57 13 19
17 5 55 3 3 19 53 4 32 11 5 51 1 38 36 34 50
56 7 26 21 36 37 16 47 7 53 23 51 14 55 19 40 51
30 31 29 42 20 31 6 59 40 63 9 27 62 34 28 13 26
20 23 11 14 56 43 40 3 33 26 10 19 10 18 11 25 4
looking forward to the new AZ ..
this is odds evens column shift up one. its wwhat brings period19 together easy. lots of variations make it tuff though.
i hope you hit the jackpot soon Jarlve.
Thanks Mr lowe. I am skeptic but trying does not hurt.
Hi, Jarlve,
that’s good news. I am very excited about the new version. In the next few days I will certainly find time to collect, comment and post some of my transposition ideas. Since I have a reasonably up-to-date computer, I can run tests if you like. Then I would only need to know the necessary settings for AZDecrypt.
I will work on a message. Hopefully it will turn out pretty good. See you soon.
