Doranchak, somewhere you discussed at length the probability of finding two plaintext pivots in a 17 x 20 rectangle, and then the probability of encoding with a 63 symbol homophonic key and still getting pivots. I think that is very important and would like to put a link to that discussion in my Transposition and Phenomenon thread.
Wouldn’t the probability of the plaintext aligning into pivots, no transposition involved, be exactly the same as the probability of pre – route transposition plaintext being in exactly the right positions to align themselves into a pivot after route transposition ( one big inscription rectangle, one big transcription rectangle )? Wouldn’t the probability be roughly the same?
Wouldn’t the probability of the plaintext aligning into pivots, no transposition involved, be exactly the same as the probability of pre – route transposition plaintext being in exactly the right positions to align themselves into a pivot after route transposition ( one big inscription rectangle, one big transcription rectangle )? Wouldn’t the probability be roughly the same?
Here’s his reference page (with links to ZKS discussions).
http://zodiackillerciphers.com/wiki/index.php?title=Pivots
And I agree that the probability is the same.
Wouldn’t the probability of the plaintext aligning into pivots, no transposition involved, be exactly the same as the probability of pre – route transposition plaintext being in exactly the right positions to align themselves into a pivot after route transposition ( one big inscription rectangle, one big transcription rectangle )? Wouldn’t the probability be roughly the same?
Here’s his reference page (with links to ZKS discussions).
http://zodiackillerciphers.com/wiki/index.php?title=Pivots
And I agree that the probability is the same.
Thank you!
Jarlve, here is one that you might like. Inscribe into rectangles 4 x 15. Lift the plaintext from the inscription rectangles top down left right. Then transcribe into 17 x 20 rectangle.
75% of the period 1 bigrams become period 15, while 25% of the period 1 bigram become reverse period 44. Half of the period 2 bigrams become period 29, and half of the period 2 bigrams become period 30.
I’m taking a closer look at the pivot phenomenon. 4 different 90 degree pivots are possible in a grid. And they can occur in any length. If you look at the pivot table below then the pivots that appear in the 340 are type 4 and have a length of 3. Unequal lengths of the horizontal and vertical part of the pivots are also a possibility as well as gaps.
1ABC CBA2 A A B B C C C C B B A A 3ABC CBA4
Pivot probability versus 340:
Pivot type 4, length 2: 0.100755% for 1 pivot to occur per randomization
Pivot type 4, length 3: 0.001649% for 1 pivot to occur per randomization
We would expect to see 2 to 3 samples per 1 million randomizations that have 2 pivots of the same type. One idea is then to explore different transposition combinations to see the probabilities they yield.
I tried it on the 408 and in general the chances improve when adding periodical operations. I found that combining period 13 transposition along with removal of 2 characters yields a pivot in about the same region as the 340. And the chance for one pivot to occur increases to 0.08495575221238938. Probably because the pivots are unmirrored period 39 repeats and 13 is a factor.
The x-axis is removal of the first character 1 to 340 and the y-axis is removal of the second character 1 to 339. See the resemblance with where I found increases in AZdecrypt scores while adding characters earlier.
Here is one such variation of the 408. I was not able to preserve the original ASCII overlay for this sample. It has a type 4 length 3 pivot at row 15, column 15.
(M",7;8BO4".F1$<* 8>QF(QER(NT4/8QHL 3C4,'6(U#@9S:K=J- !-#,UA%M%E3-.',81 -!O="@5<7K4?I5N48 *?F-0T+K-T-*.NMF. -F&F5)EK)H$SP<@,2 =VSH,>T+$"#78M0:@ G5"%U!8K/N$<P9G<@ A4>S-ST-!RS9:I5E@ *:L6A46I01,JC@QRU E'&"AMN>-7-'D#1@) *?*T)LJ(L"JDB'M<B +U$+<&M#MT)"%!DTG SVU(FF,IV&31MD?L7 F#?4,R0J$I(+>!)M5 =GO-TBD/02DF,D98* (L+%G?8$P:E%H=:U? /(KLQ0PB+8/8:2$AB MOHSKI7K!KTI.H' 1 2 3 4 5 6 7 8 9 10 3 11 12 13 14 15 16 7 17 18 12 1 18 19 20 1 21 22 10 23 7 18 24 25 26 27 10 4 28 29 1 30 31 32 33 34 35 36 37 38 39 40 39 31 4 30 41 42 2 42 19 26 39 11 28 4 7 13 39 40 9 37 3 32 43 15 5 36 10 44 45 43 21 10 7 16 44 12 39 46 22 47 36 39 22 39 16 11 21 2 12 11 39 12 48 12 43 49 19 36 49 24 14 34 50 15 32 4 51 37 52 34 24 4 17 22 47 14 3 31 5 7 2 46 35 32 53 43 3 42 30 40 7 36 23 21 14 15 50 33 53 15 32 41 10 17 34 39 34 22 39 40 20 34 33 35 45 43 19 32 16 35 25 29 41 10 29 45 46 13 4 38 27 32 18 20 30 19 28 48 3 41 2 21 17 39 5 39 28 54 31 13 32 49 16 44 16 22 49 25 38 1 25 3 38 54 8 28 2 15 8 47 30 14 47 15 48 2 31 2 22 49 3 42 40 54 22 53 34 52 30 1 12 12 4 45 52 48 26 13 2 54 44 25 5 12 31 44 10 4 20 46 38 14 45 1 47 17 40 49 2 43 37 53 9 39 22 8 54 23 46 51 54 12 4 54 33 7 16 1 25 47 42 53 44 7 14 50 35 19 42 24 37 35 30 44 23 1 36 25 18 46 50 8 47 7 23 7 35 51 14 41 8 2 9 24 34 36 45 5 36 40 36 22 45 11 24 28
I found that combining period 13 transposition along with removal of 2 characters yields a pivot in about the same region as the 340. And the chance for one pivot to occur increases to 0.08495575221238938. Probably because the pivots are unmirrored period 39 repeats and 13 is a factor.
The x-axis is removal of the first character 1 to 340 and the y-axis is removal of the second character 1 to 339. See the resemblance with where I found increases in AZdecrypt scores while adding characters earlier.
It’s just the same pivot that appeared in all variations so the chance number inflated. And the resemblance is unworthy of note, it is just that the pivot appeared in about the same position.
Jarlve, here is one that you might like. Inscribe into rectangles 4 x 15. Lift the plaintext from the inscription rectangles top down left right. Then transcribe into 17 x 20 rectangle.
75% of the period 1 bigrams become period 15, while 25% of the period 1 bigram become reverse period 44. Half of the period 2 bigrams become period 29, and half of the period 2 bigrams become period 30.
Hey smokie,
I tried your scheme but was unable to preserve many bigrams after encoding. Then reduced it to 2 rectangles with a few variations but it just does not come out right. It is so hard to reconcile period 15 with period 29. With mixing of periods, the shift from one period to another is usually gradual.
It is so hard to reconcile period 15 with period 29.
Best chance for a transposition that has a lot of period 15 and 29 repeats would be so that period 1 becomes period 15 and period 2 becomes period 29. Or maybe vice versa. But I have not been able to find one yet.
It could be a matter of interpretation. draft the message into a rectangle with different dimensions, count the cells in another direction, and they are not period 15 and 29. Maybe there is a transposition that could easily explain a different interpretation of the two periods. But then you wouldn’t have pivots.
On my to do list is to try to make a three rectangle transposition where the plaintext is transcribed into the final 17 x 20 rectangle vertically. What we are currently interpreting as period 29 repeats derived from period 1. Forget about trying to include the period 15 repeats, just see if converting period 1 to interpreted period 29 could create a lot of pivots.
Check out this, starting at the 9th post down and definitely go the next page. Multiple consecutive incomplete rectangles can sometimes make for very interesting results. Sometimes period 1 becomes period 1 again or period 2 becomes period 2 again in certain areas of a 4th rectangle, whether horizontal or vertical. That could explain the pivots maybe, but I haven’t found the right combination yet.
viewtopic.php?f=81&t=2617&start=1380
You might enjoy making matrices from different combinations of rectangle sizes and shapes, whether complete or incomplete, and shade the cells conditioned on original position. One idea would be to generate thousands of these, not encoded, and compare to the cells of the 340 that are "true" for period 15 / 19 and period 29 / 39. See if you can find a close match, or something that creates pivots.
Another idea: Make matrices and then mirror them. Watch what happens to spikes when that happens because mirroring is reversing individual rows, not the entire message.
Another idea: Find a transposition that creates period 19 and period 5 repeats. Reading left right top bottom, there is also a spike at period 5. Several of these inhabit the last couple of columns on the right and first couple of columns on the left. When the message is mirrored, they become period 29.
At least without transposition, the pivots are astronomically improbable, and with transposition maybe a little more probable.
?
Best chance for a transposition that has a lot of period 15 and 29 repeats would be so that period 1 becomes period 15 and period 2 becomes period 29. Or maybe vice versa. But I have not been able to find one yet.
I think that your idea with transcribing around a column is in the right direction. Because when you add or remove a row in the 340, the period does not change, but it does with columns. There are other pointers that something with columns is going on, right-shift the entire cipher by one column and bigrams increase to 45. doranchak’s column period 2 increases bigrams to 44. Add one column at the right side of the cipher and period 5 bigrams increase substantially. I figure that these features could be linked to the a misalignment that was introduced vertically.
It could be a matter of interpretation. draft the message into a rectangle with different dimensions, count the cells in another direction, and they are not period 15 and 29. Maybe there is a transposition that could easily explain a different interpretation of the two periods. But then you wouldn’t have pivots.
If you wish I could so some sort of scan for that, to your specifications.
I’ve just scanned the full range of, directional * dimension * directional, and here are the highest returning bigram results:
Mirror, Dimension(14,25), Diagonal(UTP,3): 42
Mirror, Dimension(14,25), Diagonal(UTP,6): 42
Flip, Dimension(16,22), Diagonal(UTP,1): 42
Flip, Dimension(16,22), Diagonal(UTP,8): 42
Combinations processed: 265776/265776 Measurements: - Summed: 5510230 - Average: 20.7326094154476 - Lowest: 7 (Mirror, Dimension(98,4), Diagonal(UTP,3)) - Highest: 42 (Mirror, Dimension(14,25), Diagonal(UTP,3))
On my to do list is to try to make a three rectangle transposition where the plaintext is transcribed into the final 17 x 20 rectangle vertically. What we are currently interpreting as period 29 repeats derived from period 1. Forget about trying to include the period 15 repeats, just see if converting period 1 to interpreted period 29 could create a lot of pivots.
Check out this, starting at the 9th post down and definitely go the next page. Multiple consecutive incomplete rectangles can sometimes make for very interesting results. Sometimes period 1 becomes period 1 again or period 2 becomes period 2 again in certain areas of a 4th rectangle, whether horizontal or vertical. That could explain the pivots maybe, but I haven’t found the right combination yet.
Well, it seems that multiple inscription rectangles are hard to reconcile with bigram repeats in general. So it is something that does not really catch my interest at the moment.
You might enjoy making matrices from different combinations of rectangle sizes and shapes, whether complete or incomplete, and shade the cells conditioned on original position. One idea would be to generate thousands of these, not encoded, and compare to the cells of the 340 that are "true" for period 15 / 19 and period 29 / 39. See if you can find a close match, or something that creates pivots.
I get the general idea and was thinking to do that somewhere next year.
Another idea: Make matrices and then mirror them. Watch what happens to spikes when that happens because mirroring is reversing individual rows, not the entire message.
Another idea: Find a transposition that creates period 19 and period 5 repeats. Reading left right top bottom, there is also a spike at period 5. Several of these inhabit the last couple of columns on the right and first couple of columns on the left. When the message is mirrored, they become period 29.
It’s funny but we have the same ideas floating around in our head. Correlating interesting transposition schemes with the 340 is my usual modus of operandi. Further automatization of looking for particular features is on my to do list.
At least without transposition, the pivots are astronomically improbable, and with transposition maybe a little more probable.
Again. I’ve thought the same thing. Though I don’t think they randomly occured from transposition. They must be a result from transposition misalignment. That misalignment caused period 29 repeats to spike and that gave birth to the pivots. In this case, it would still be a very rare phenomena but I don’t see any other way. The 340 is not supernatural.
Here’s another plot, x-axis is a single removed character from position 1 to 340 and y-axis is a single added character from position 1 to 340. The color brightness is the AZdecrypt score. Highest return scored 21100. The 340 cipher was mirrored prior to any operation and the final operation was period 15 untranspose. The diagonal line is an artifact where the add and remove character overlap.
Remove character(47), Add character(199), Period(UTP,15) SUEHEADOURSALIFON SOHALTSCHOISSCHAS INANOTCARDIACTTHE MICANTANDARAGEONS TREGUILTOBOUTINGH OOTITNEERDETASAGU ESHIHAREATHSTRICS THISLOFRAINSURETH EGASARTOFTHESOFOU NDENTOFCERNINREAS FIRESTERTINDANDBE HELIGNSTHISREASFR THOONTHATERMISTHI NGHTHCITRENTSINTH ETHEREAFIRSTERIAN THSANDHDBOTHCREDS SATSCADRENSHENATH ICANDBORNTOFFARCH HAROLHEAROLEITAIN ERETOOUNEATURENES
If you wish I could so some sort of scan for that, to your specifications.
Thanks a lot for the offer. For now I am fine, and I had to check out for the most part, for a little while.
I understand that moving column 17 to the column 1 position increases the period 15 / 19 bigram repeats. But I am not convinced that it means anything. Doranchak has a lot of cryptograms to compare with. I would suggest finding out if moving one column in any of those cryptograms would increase the bigram repeats as much to see if it rare or not.
This might be a bit offtopic:
I have tried some transposition stuff which may not have been covered by now:
Imagine that the plaintext was transposed to chunks in which the order of the letters were shuffled. Example:
Plaintext: AVADAKEDAVRA Chunk size 4, rearrangement order "3421" Plaintext: 1234 1234 1234 AVAD AKED AVRA Ciphertext: 4312 4312 4312 DAAV DEAK ARAV
I ran tests for all permutations of chunk sizes 2, 3, 4, 5, 6, 7, 8 and 9 with AZDecrypt (409112 test ciphers at all). Sometimes I got a pretty good scoring. The highest one was 20976 from a cipher with chunk size 9 and rearrangement pattern "7, 3, 4, 8, 1, 9, 5, 2, 6".
May you check the result with your statistic tools please? It would be interesting to know if this result is just some noise again:
AZdecrypt 0.992 (Practical Cryptography 5-grams) Progressive, Index of coincidence Score: 20975.64 Ioc: 0.07417056 Entropy: 3.867192 Chi-square: 39.93682 Characters: 333 Letters: 18 aoflinduseitiscis twanedfornamhaets taterfinanddoodad itposedonhalloser reriesinorschosta tedgesahesfinisno tewherlanthouetan tstoespricenteror medlorichangestew asaparlortomanaed astherecouncieame dshopingsshorthel atehoredactstheen consthenalgacurri ghtcalliaidugfors ohanatandestantto nithieraresiasmfo raftnfrindiandint osicsafeminanifhe adicoeddes Multiplicity: 0.1891892 Characters: 333 Symbols: 63 dRaVHPbEcgfLheNTI GlBjDbWk+OqnoBYrK mpGM+WsjUtZSkubqb HLJxcwbvOodVzRKM+ +g+yF2hPu+I17Re5d 3Fb4we-oFcahOCKjk Qgl7D+VUtm8uXgGpj G2LkgIJ+yN9OLY+u+ nMZzR+h0#Bt4gKrFA dcUJ-+Vu+rvn-OUFb BI57D+Y1REO0TgBnM bK7RJTt4Icou+3#Mz BmF7k+FSBNrIG8wFt NRjcm7gtdV4p0X++f 4or1BVzCUsZE4ax+I u7Bt-mpObMKQdjmLR tfG#Tg+B+FICqcnWu +qWLtW+CjSHqPbTO5 veh1IBWFnCOBtyaoY -SHNuMZbDe
I have tried some transposition stuff which may not have been covered by now:
You basicly mean rearranging the columns in a set of dimensions. Me and daikon have done some work on it, nothing extensive.
May you check the result with your statistic tools please? It would be interesting to know if this result is just some noise again:
You’ll have to decide for yourself.
One thing you can do is look at the ioc and/or entropy. With the practical cryptography ngrams and typical AZdecrypt settings the ioc will typically inflate a bit if it can’t find a legit solution. Solves that have an ioc under 0.07 and/or entropy of over 4 are more interesting. Another thing is to look at repeating fragments, more fit plaintext will have longer repeating fragments.
Here are the higher ngram repeats of a 340 character part of the 408:
7-gram frequencies > 1: -------------------------------------------------- BECAUSE: 3 KILLING: 2 EBECAUS: 2 THEMOAT: 2 8-gram frequencies > 1: -------------------------------------------------- EBECAUSE: 2
And here is your solve:
4-gram frequencies > 1: -------------------------------------------------- sthe: 3 stat: 2 tate: 2 inan: 2 tant: 2 5-gram frequencies > 1: -------------------------------------------------- state: 2
You can also inspect the plaintext in AZdecrypt 1.0 with stats -> plaintext direction 1b and 2. And compare scores with other plaintext.
This might be a bit offtopic:
I have tried some transposition stuff which may not have been covered by now:
Imagine that the plaintext was transposed to chunks in which the order of the letters were shuffled. Example:
Plaintext: AVADAKEDAVRA Chunk size 4, rearrangement order "3421" Plaintext: 1234 1234 1234 AVAD AKED AVRA Ciphertext: 4312 4312 4312 DAAV DEAK ARAVI ran tests for all permutations of chunk sizes 2, 3, 4, 5, 6, 7, 8 and 9 with AZDecrypt (409112 test ciphers at all). Sometimes I got a pretty good scoring. The highest one was 20976 from a cipher with chunk size 9 and rearrangement pattern "7, 3, 4, 8, 1, 9, 5, 2, 6".
May you check the result with your statistic tools please? It would be interesting to know if this result is just some noise again:
AZdecrypt 0.992 (Practical Cryptography 5-grams) Progressive, Index of coincidence Score: 20975.64 Ioc: 0.07417056 Entropy: 3.867192 Chi-square: 39.93682 Characters: 333 Letters: 18 aoflinduseitiscis twanedfornamhaets taterfinanddoodad itposedonhalloser reriesinorschosta tedgesahesfinisno tewherlanthouetan tstoespricenteror medlorichangestew asaparlortomanaed astherecouncieame dshopingsshorthel atehoredactstheen consthenalgacurri ghtcalliaidugfors ohanatandestantto nithieraresiasmfo raftnfrindiandint osicsafeminanifhe adicoeddes Multiplicity: 0.1891892 Characters: 333 Symbols: 63 dRaVHPbEcgfLheNTI GlBjDbWk+OqnoBYrK mpGM+WsjUtZSkubqb HLJxcwbvOodVzRKM+ +g+yF2hPu+I17Re5d 3Fb4we-oFcahOCKjk Qgl7D+VUtm8uXgGpj G2LkgIJ+yN9OLY+u+ nMZzR+h0#Bt4gKrFA dcUJ-+Vu+rvn-OUFb BI57D+Y1REO0TgBnM bK7RJTt4Icou+3#Mz BmF7k+FSBNrIG8wFt NRjcm7gtdV4p0X++f 4or1BVzCUsZE4ax+I u7Bt-mpObMKQdjmLR tfG#Tg+B+FICqcnWu +qWLtW+CjSHqPbTO5 veh1IBWFnCOBtyaoY -SHNuMZbDe
I agree that the message could have been transposed in chunks. However, they are not small chunks. If you make a rail fence cipher with 30 columns and 2 rows, that is a chunk of 30 which will create period 15 repeats. You could make 11 of them to fit into the 340, but you will have a very difficult time getting 340 period 15 stats because each individual chunk is independent of every other chunk. Period 15 repeats are created within chunks, but not between chunks, except by coincidence.
Likewise, a bifid works with chunks. You can create a lot of period 15 repeats by working with chunks of 30 plaintext. But you get the same problem. The period 15 repeats are created within the chunks, and not between or across them except by coincidence. So you will have a difficult time matching 340 period 15 stats.
Grille cipher anagrams in chunks, but I could not get very many period 15 repeats within chunks of 64.
However, you can get 340 stats period15 stats with multiple inscription rectangles. The bigger the rectangles, the more you get.
There may be a way to detect this. Draft the message into different numbers of columns. Divisors of 15, multipliers of divisors of 15, and multipliers of 15. And starting at 340 different positions. Make a list of period 15 bigrams within each row but not from row to row. Then count the number of repeats. The arrangement that results in the highest count of repeats may show the chunk size. For example, if the result is 75 columns, then the chunk size may be 75, a multiplier of 15. The first symbol of each chunk and the last symbol of each preceding chunk may also be a period 1 bigram that has symbols that match other period 15 bigrams in the arrangement. They may appear on a horizontal number line, at increments of the chunk size, similar to your coincidence count spike.
This could be fun, can anyone find the chunksize I used?
;6MOX?&!R9=XFUHA3 P56&N$K*4#_@Z(8Q: X#HWC[2+HI>*UK7J/ ?%^M@XE'@91;P =Y&DHA[78DZ^EFZX $(<RB$3LY?KXQ'.B: +CD#PG@1[U7H9>[T5 ^Q$Q:7XD0#RJE%&@Y CMXEL%+N=K!]F28ZN 9F>@X')"*!#&$WQ=0 @A(:X#1+B^HKT%CE0 LN@O!RYX2$=T?!0AU @X7DM&96HJ#1K>@[ V%7=L;?.TF'Z)QS5' O8-NWB>C%2F8XM$# @0XEX,=3%P[+D6>UO 97QNY!+8TFN'#?%>@ =F1XARJ#N^5@(:;'X KP_9Z=U^YN#@2:QRT XH#YCM9!@+R&%=87X 1 2 3 4 5 6 7 8 9 10 11 5 12 13 14 15 16 17 18 2 7 19 20 21 22 23 24 25 26 27 28 29 30 31 5 24 14 32 33 34 35 36 14 37 38 22 13 21 39 40 41 6 42 43 3 26 44 5 45 46 7 24 18 26 10 47 1 17 11 48 7 49 14 15 34 39 29 49 27 44 43 45 12 27 5 20 28 50 9 51 20 16 52 48 6 21 5 30 46 53 51 31 36 33 49 24 17 54 26 47 34 13 39 14 10 38 34 55 18 43 30 20 30 31 39 5 49 56 24 9 40 45 42 7 26 48 33 3 5 45 52 42 36 19 11 21 8 57 12 35 29 27 19 10 12 38 26 5 46 58 59 22 8 24 7 20 32 30 11 56 26 15 28 31 5 24 47 36 51 43 14 21 55 42 33 45 56 52 19 26 4 8 9 48 5 35 20 11 55 6 8 56 15 13 26 5 44 39 49 3 7 10 2 14 40 24 47 21 38 26 34 60 42 39 11 52 1 6 53 55 12 46 27 58 30 61 18 46 4 29 62 19 32 51 38 33 42 35 12 29 44 5 3 20 24 26 56 5 45 5 63 11 16 42 17 34 36 49 2 38 13 4 10 39 30 19 48 8 36 29 55 12 19 46 24 6 42 38 26 11 12 47 5 15 9 40 24 19 43 18 26 28 31 1 46 5 21 17 25 10 27 11 13 43 48 19 24 26 35 31 30 9 55 5 14 24 48 33 3 10 8 26 36 9 7 42 11 29 39 5
