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November 23, 2024, 11:26:26 PM |
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Recent Posts2
Voynich Manuscript / Re: Not just a random thought - the calendar« Last post by Diane on April 29, 2015, 04:20:28 AM »I've read rather more now, and except that Panofsky usually knew what he was about, I'd dismiss the Kabbalistic link. Thing is, he did.
Against this we have the opinion of Thorndike (no less) that the ms was "an anonymous manuscript of dubious value" - as far as his own study of western science and magic was concerned. Metaphysics and religion are not in either category, though, are they? More to the point, perhaps, is that so many who pride themselves on being rational are irrationally averse to studying the history of such subjects. "Fear of the perceptual" as one wise man one said, "is the last superstition of rational men." 3
Voynich Manuscript / Re: Gallows« Last post by Aaron on April 01, 2015, 09:01:45 AM »WOW! That looks like the exact same handwriting on the Voynich!
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Just About Anything / Ciphers in Literature (Esmeralda)« Last post by Knox on February 05, 2015, 01:12:52 PM »The sample of a cipher in a novel* is unworkable but the idea has possibilities. I used that idea to create a short cryptogram. (1) Each letter of the plaintext is substituted by 4 digits, as explained below in (4). S E N D M O R E M O N E Y 7849 7560 6488 8369 4568 2651 4977 6146 6542 5530 6743 6752 4479 (2) Re-format for transposition 78 49 75 60 64 88 83 69 45 68 26 51 49 77 61 46 65 42 55 30 67 43 67 52 44 79 00 00 00 00 00 00 Columnar transposition (there was no added transposition in the book) G O L D F I S H 3..7..6..1..2..5..8..4 ----------------------- 78 49 75 60 64 88 83 69 45 68 26 51 49 77 61 46 65 42 55 30 67 43 67 52 44 79 00 00 00 00 00 00 Write columns 1-8 to a line 60 51 30 00 64 49 67 00 78 45 65 44 69 46 52 00 88 77 43 00 75 26 55 00 49 68 42 79 83 61 67 00 (3) Reformat for transmission (add one more 0 to fill out the last group) 60513 00064 49670 07845 65446 94652 00887 74300 75265 50049 68427 98361 67000 (4) How it works. The difference (positive or negative) between the pairs in each 4-digit block maps to a numerical substitute of a plaintext character. The numbers can be created ad hoc or selected from a spreadsheet. Excel method: Char Sub First Number Second Number 1 1 RANDBETWEEN(11,99) (First Number-Substitute) 2 2 RANDBETWEEN(12,99) (First Number-Substitute) 3 3 RANDBETWEEN(13,99) (First Number-Substitute) 4 4 RANDBETWEEN(14,99) (First Number-Substitute) 5 5 RANDBETWEEN(15,99) (First Number-Substitute) 6 6 RANDBETWEEN(16,99) (First Number-Substitute) 7 7 RANDBETWEEN(17,99) (First Number-Substitute) 8 8 RANDBETWEEN(18,99) (First Number-Substitute) 9 9 RANDBETWEEN(19,99) (First Number-Substitute) 0 10 RANDBETWEEN(20,99) (First Number-Substitute) A 11 RANDBETWEEN(21,99) (First Number-Substitute) B 12 RANDBETWEEN(22,99) (First Number-Substitute) C 13 RANDBETWEEN(23,99) (First Number-Substitute) ... The second pair of numbers is the reverse of columns 3 & 4 Z can be substituted by any of 108 4-digit numbers. 9963, 9862, 9761, 9660, 9559, 9458, 9357, 9256, 9155, 9054, 8953, 8852, 8751, 8650, 8549, 8448, 8347, 8246, 8145, 8044, 7943, 7842, 7741, 7640, 7539, 7438, 7337, 7236, 7135, 7034, 6933, 6832, 6731, 6630, 6529, 6428, 6327, 6226, 6125, 6024, 5923, 5822, 5721, 5620, 5519, 5418, 5317, 5216, 5115, 5014, 4913, 4812, 4711, 4610 and the reverse pairs. Y has one more substitute pair than X. X has one more substitute pair than Y, and so on. This can be made more difficult but I only want to present it as something different. *"The Cruise of the Esmeralda" by Harry Collingwood 6
Other Mysteries / Re: The Cipher of Simeone Levi« Last post by Aaron on October 30, 2014, 07:59:37 PM »Interesting that they managed to get most of that except for the one sentence. Usually I just see things stick to a single cipher throughout.
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Other Mysteries / The Cipher of Simeone Levi« Last post by Knox on October 30, 2014, 06:29:44 PM »A clever cipher and an elegant solution -- except for one sentence.
The Ciphered Autobiography of a 19th Century Egyptologist by Emanuele Viterbo, AT&T Research http://www.ecse.monash.edu.au/staff/eviterbo/simeone/simlevi.pdf Accessed 30 October 2014 via Klaus Schmeh’s List of Encrypted Books http://scienceblogs.de/klausis-krypto-kolumne/klaus-schmehs-list-of-encrypted-books/ E. Viterbo: “The Ciphered Autobiography of an 19th Century Egyptologistâ€Â, CRYPTOLOGIA, vol. XXII, n. 3, pp. 231-243, July 1998 Abstract: Simeone Levi was an Italian Egyptologist who lived in Turin during the second half of the 19th century. His major work is an eight volume hieroglyphic dictionary. He left a ciphered autobiography which remained a mistery [sic.] for about three generations in his family. We present here the procedure used to decipher the text and the structure of the code invented by Simeone Levi. *[13 pages; the PDF has pp. 193-205] Quotes from page 7, marked 199: "Simeone quotes a sentence in a mysterious foreign language: [symbols] [transliteration:] Ma pecuglie ri sifafanu cecumucela o queanru ytagno fu a pecutagno pecu nicagenu. I consulted various linguists in Ancient Egyptian, Sanskrit, Coptic and Hebrew, but the language of this sentence still remains a mystery. I leave it as a challenge for the interested readers." --- In context. English translation and undeciphered sentence. Autobiography of Simeone Levi [1843-1913] Why does telling about one's own sorrows soothe one's torments? I believe the effect is produced by the same thing which causes a continuous noise to end up inducing sleep. Ma pecuglie ri sifafanu cecumucela o queanru ytagno fu a pecutagno pecu nicagenu. Nevertheless, people prefer telling about their enjoyed happiness rather than about their suffered sorrows, because most of the times people envie [sic.] other people's victories and mock other people's sufferings, and seldom does someone else show contentment for our good and feel sympathy for our pains. But I will tell about the good and the bad, my merits and my faults, when I was right and when I was wrong, my triumphs and my failures, my good deeds and my bad deeds. --- MA PECUGLIE RI SIFAFANU CECUMUCELA O QUEANRU YTAGNO FU A PECUTAGNO PECU NICAGENU .2 .......8 .2 .......8 ........10 . ......7 .....6 .2 . ........9 ...4 .......8 If the following are combined, all words have an even number of letters. O QUEANRU 8 letters A PECUTAGNO 10 letters But could be 1:1 pre-ciphered Italian or Latin. 8
Other Mysteries / Re: STENDEC« Last post by Knox on March 14, 2014, 02:28:00 PM »Additional information, including (if I remember correctly) the mention that two other planes by the same airline disappeared.
BBC Horizon: Vanished, The Plane That Disappeared (1999) http://www.youtube.com/watch?v=Zy8Ga6OmNx8 9
Code of the Week / Re: Double Columnar Transposition« Last post by Phil_The_Rodent on December 04, 2013, 01:38:13 AM »For the record, I went back to the original post (as I had quite forgotten the general details about these) and solved 3 of them by hand today (2, 3 & 5), and had done #1 a few months back. As they were all done at the same time, and the transpositions are generally clean, they should all be doable.
For anyone else wanting to give them ago, I found this tool: http://tholman.com/other/transposition/ which replaces my paper strips well enough. Still some work to do in notepad once the first tranposition is solved... 10
Code of the Week / Re: Double Columnar Transposition« Last post by Phil_The_Rodent on December 03, 2013, 12:42:10 PM »Hi George! Welcome to the forum.
Here are the answers to your questions: 1) No 2) No What I can tell you is that the codes should all form neat rectangles. That means: a) There will be padding, or dropped letters in the body of the cipher. There will be some deviation from clean text (or, if you dropped the resulting text into MS Word, a few words will have a red underline), but I don't think anything was badly mangled. Also, every text contains at least one name, which also may not be in your dictionary. b) You do not have to attempt to solve for all numbers, just the numbers that will multiply out properly. c) Passwords were randomly generated and would not be the same for both directions. In regards to your program and an expansion of (b): The first ciphertext is 144 characters. 144 will cleanly be divided into a table of the following proportions: 1 x 144, 2 x 72, 3 x 48, 4 x 36, 6 x 24, 8 x 18, 9 x 16, 12 x 12, 16 x 9, 18 x 8, 24 x 6, 36 x 4, 48 x 3, 72 x 2, 144 x 1 I would not have used x 1 permutations, so those can be tossed. This means you have a maximum of 13 possible tables sizes, down from 625. |
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