My BLT
in [Cryptography]
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From: WTShaw on 21 Feb 2010 09:21 Recognizing BLT Ciphertext. One of the most difficult to solve but highly portable entries into the category of quick, portable, and better-than-nothing pincil & paper ciphers is my BLT Cipher. The machine verson automatically adheres to a standard which might make BLT suspect as the cipher from its ciphertext but this does not in any way weaken its security if the following is true: 1) In a casual quantity, for each 26 letters of the alphabet no single letter should appear again within several locations of itself, three or more. Plus, each used cipher text triad might ever repeat itself as there are hundreds of possibilites for each plaintext letter, homographic diversity. 2) Identical plaintext should never be encrypted again with BLT as two or more known compatable ciphertexts might compromise all of the BLT advantage. 3) And, as with many other simple ciphers, another gross mistake would be pairing known plaintext with any ciphertext. If the same message needs repeating, rephrase it to destroy its otherwise identical nature. BLT was done to highlight certain valid strengths in spite of characteristics that would not be allowed in an ideal cipher. Still, if held in strict observation of non-repetition, BLT can present considerable solution difficulty. Consider how the cipher works to demonstrate its portability: The key is based on 27 characters arranged into 9 groups of 3 characters each. The most simple key generation method involves attempting to write a pangram hash from a sentence containing as many different letters of the alphabet as is convenient. While not requiring that a perfect pangram be used as the source of the key, unused letters are to be added to the end of the derived sequence. For example, the first sentence of this paragraph would equate to the following: "52) Consider how the cipher works to demonstrate its portability" <<Incomplete>> considerhwtpkmably (18 of 26 letters); MISSING ARE: fgjquvxz or, as trigraphs, "con sid er/ hwt pkm abl yfg jqu vxz" where in the Slant Sign represents where the first space occurs in the source text. Make three groups of nine characters each based on positions of letters in the smaller groupings: Pool 1=csehpayjv; Pool 2=oirwkbfqx; Pool 3=nd/tmlguz (Although the Slant Sign occurs in the lists and spaces may be encrypted, it should not be used in ciphertext as a clue that a 27 character scheme is in use.) From the pangram list, reference the location of each pangram character in the series 111 to 333. For example 111=c, 113=e, space=133, x=332, z=333, etc. For each character (or space) in plaintext, pick letters from the pools to represent its address. Try to use different combinations of characters drawn from the pools so that no triad of ciphertext characters ever repeats in a given message. Group the produced alphabetic ciphertext characters in groups of five characters to mask the three-character emphasis. In decryption, convert each 3 letters of ciphertext to numbers and then record the proper plaintext letter, easily done if you know the key which would permit deducing a given same plaintext from many possible equivalent ciphertexts. I have implemented BLT for different platforms using a variety of languages and anyone would be encouraged to do likewise. I especially like my Java Script version. OK, if you care, and few will or should, given the above, break the following message according to the old tradition of crypto which is THE final valid test of any algorithm, can you do it or just talk about how easy it would be: frvci wdgnq yeikv drcjx kybho vqjwr ioldq hewgx amdju oiwyd rvnci qxjwu noaqg mycki xnfde ymaow kdxis ryvjw eidom ancbe ydoxi rcjeq bxnhc atvxb edfao hsydj xkrga pioch fvbga xmonb ihpar nqbmc taygq opjwe scxqn ahvmx kibwy vskce odjgx pqfdm tleyi rxpdb vahri kxoqy vcjab gmiof cduma gxyjq hiocm ndtqb xghak jredn qaxfo cdqlr jamui wegrd uvofm jbgqa hxfmj tvsli yjrvh mgaby vhmed jngfx kdbwt ojiyn pvboq gyjsm ohedv cxgou yqfkj eoycg vdmfr jeixp oajhg vwpif ydgch jiyap vmqke jiohv naris xuhcr ojagy vpofi xkwjg qlrtf iocjr ayuho ejamb gxequ jagid fhenj ridax ymuif vpqlo mibxd eroia blesr pidqo vbxye fnvho jcdab qnkyv jxsdi mqfwg exmvd fachi revqa jhomx nvrqb ideuv yfhpc ewulv riync daxfb winpo vracy jqhsr cdigo bxckn mpidx qsmvn haltc mghrb xiqun hcebo pdjgs leiwo hgspj ioyrx ejnlf iyewj ronvh mprie aljxh qmgvd bifjr vphiq jydcb voqhe nlidw rfovb ymkrd halpn ygh
From: Mok-Kong Shen on 21 Feb 2010 10:07 WTShaw wrote: [snip] > or, as trigraphs, "con sid er/ hwt pkm abl yfg jqu vxz" where in the > Slant Sign represents where the first space occurs in the source > text. > > Make three groups of nine characters each based on positions of > letters in the smaller groupings: Pool 1=csehpayjv; Pool 2=oirwkbfqx; > Pool 3=nd/tmlguz [snip] > From the pangram list, reference the location of each pangram > character in the series 111 to 333. For example 111=c, 113=e, > space=133, x=332, z=333, etc. I guess (not sure) that in 113=e the 3 means that e occupies the 3rd position in Pool 1. But how does one get x=332? (x in in Pool 2!) > For each character (or space) in plaintext, pick letters from the > pools to represent its address. Try to use different combinations of > characters drawn from the pools so that no triad of ciphertext > characters ever repeats in a given message. The first sentence is rather unclear for me, the second is even more so. Could you use an example to illustrate the meaning of these? M. K. Shen
From: WTShaw on 22 Feb 2010 17:03 On Feb 21, 9:07 am, Mok-Kong Shen <mok-kong.s...(a)t-online.de> wrote: > WTShaw wrote: > > [snip] > > > or, as trigraphs, "con sid er/ hwt pkm abl yfg jqu vxz" where in the > > Slant Sign represents where the first space occurs in the source > > text. > > > Make three groups of nine characters each based on positions of > > letters in the smaller groupings: Pool 1=csehpayjv; Pool 2=oirwkbfqx; > > Pool 3=nd/tmlguz > [snip] > > From the pangram list, reference the location of each pangram > > character in the series 111 to 333. For example 111=c, 113=e, > > space=133, x=332, z=333, etc. > > I guess (not sure) that in 113=e the 3 means that e occupies the 3rd > position in Pool 1. But how does one get x=332? (x in in Pool 2!) Build a table like I start below and it will make better sense. > > > For each character (or space) in plaintext, pick letters from the > > pools to represent its address. Try to use different combinations of > > characters drawn from the pools so that no triad of ciphertext > > characters ever repeats in a given message. > > The first sentence is rather unclear for me, the second is even more > so. Could you use an example to illustrate the meaning of these? > > M. K. Shen It's a matter of counting, dividing all 27 characters into three major groups, dividing each of these major groups in groups of triads, and dividing each triad into three characters. For 332, pick any two of the characters in pool 3 and one in pool 2. for x: mtr zgb dmx all are x For "con sid er/ hwt pkm abl yfg jqu vxz," c=111 o=112 n=113 s=121 i=122 d=123 e=131 r=132 /=133 for the first nine. Pick 1 character, d=123. pick a letter from poo1, pool2, and pool3. any letter from each pool. these would work; eku aiz cwu as all mean "d." For d, there are 9X9X8 possibilities. For x, there are 8X7X9 possibilities, remembering that after one use from pool3 which also contains / there is a shinking pool for choosing the next pool3 ciphertext letter.
From: Mok-Kong Shen on 23 Feb 2010 12:36 WTShaw wrote: > It's a matter of counting, dividing all 27 characters into three major > groups, dividing each of these major groups in groups of triads, and > dividing each triad into three characters. > > For 332, pick any two of the characters in pool 3 and one in pool 2. > for x: mtr zgb dmx all are x > > For "con sid er/ hwt pkm abl yfg jqu vxz," c=111 o=112 n=113 s=121 > i=122 d=123 e=131 r=132 /=133 for the first nine. > > Pick 1 character, d=123. pick a letter from poo1, pool2, and pool3. > any letter from each pool. > these would work; eku aiz cwu as all mean "d." For d, there are 9X9X8 > possibilities. > For x, there are 8X7X9 possibilities, remembering that after one use > from pool3 which also contains / there is a shinking pool for choosing > the next pool3 ciphertext letter. Sorry for my cluelessness. You seem to define a way of mapping each letter of the alphabet to three numerals. But how does the encryption process goes after that step? Is your scheme essentially a monoalphabetic substitution? Or else a homophonic substitution (but then how, i.e. in which manner)? (Excuse me that I don't have yet much understood of what you wrote.) M. K. Shen
From: bmearns on 23 Feb 2010 14:04
On Feb 21, 9:21 am, WTShaw <lure...(a)gmail.com> wrote: [snip] > frvci wdgnq yeikv drcjx kybho vqjwr ioldq hewgx amdju oiwyd rvnci > qxjwu noaqg mycki xnfde ymaow kdxis ryvjw eidom ancbe ydoxi rcjeq > bxnhc atvxb edfao hsydj xkrga pioch fvbga xmonb ihpar nqbmc taygq > opjwe scxqn ahvmx kibwy vskce odjgx pqfdm tleyi rxpdb vahri kxoqy > vcjab gmiof cduma gxyjq hiocm ndtqb xghak jredn qaxfo cdqlr jamui > wegrd uvofm jbgqa hxfmj tvsli yjrvh mgaby vhmed jngfx kdbwt ojiyn > pvboq gyjsm ohedv cxgou yqfkj eoycg vdmfr jeixp oajhg vwpif ydgch > jiyap vmqke jiohv naris xuhcr ojagy vpofi xkwjg qlrtf iocjr ayuho > ejamb gxequ jagid fhenj ridax ymuif vpqlo mibxd eroia blesr pidqo > vbxye fnvho jcdab qnkyv jxsdi mqfwg exmvd fachi revqa jhomx nvrqb > ideuv yfhpc ewulv riync daxfb winpo vracy jqhsr cdigo bxckn mpidx > qsmvn haltc mghrb xiqun hcebo pdjgs leiwo hgspj ioyrx ejnlf iyewj > ronvh mprie aljxh qmgvd bifjr vphiq jydcb voqhe nlidw rfovb ymkrd > halpn ygh Was this intended to be gibberish? The first line reads: "pizhpjiswimjdtgkre" It's an intriguing cipher, but do you have any cryptanalysis on it? If it's actually somewhat secure, it would be a nice thing to have in a pencil-and-paper crypto toolbox, but not without seeing some in rigorous analysis. -Brian |