Summyry of "Wrapping", "Patterns from hints" and "Road to nowhere" Wrapping Tuesday, 09-Nov-1999 03:50:23 193.158.174.47 writes: My 'theory' about the wrapping seems to stand the light of day : The tuts say : 2. If the letters of the plaintext digraph are in the same horizontal row, the letter to the right of each plain letter is the cipher equivalent. When one of the paired plain letters is at the right end of the row, the first letter at the left end of the row becomes the cipher letter. (Same for the columns) The Seeker *********************** Patterns from hints Tuesday, 09-Nov-1999 08:21:51 208.214.188.4 writes: There are not a lot of patterns in the hint, unfortunately..! Red Penguin Frenzy: RE DP EN GI UN FR EN ZY ^^ ^^ ?R ED PE NG IU NF RE NZ Y? Beware Ice Weasels: BE WA RE IC EW EA SE LS ?B EW AR EI CE WE AS EL S? ^^ ^^ Stop: ST OP ?S TO P? Norsk Hydro: NO RS KH YD RO ?N OR SK HY DR O? Rjukan: RJ UK AN ?R JU KA N? Norway: NO RW AY ?N OR WA Y? But of course it would be surprising to see literal names in such a message.. I guess they'd use codewords.. ST/OP would have been nice if it had been encoded two times on even boundaries, but apparently we're out of luck... My guess is stop appears two times in the message, one starting even and one odd. The message should finished with END. We have FU UF BG as last ciphers.. This could be: FU UF BG E? ?E ND Or, if the message doesn't end evenly: FU UF BG NE EN DX In any cases, FU/UF aren't anywhere else in the message... If FU/UF = NE/EN, it would makes senses with REDPENGUINFRENZY starting on an odd boundary since we can't find the EN/NE couples.. -Tx ****************** Revised indexes Tuesday, 09-Nov-1999 08:38:16 208.214.188.9 writes: If we take into account we can't find the patterns from both phrases in the cipher, here's the revised indexes (0-based, again) of where they can be encoded: Red Penguin Frenzy (39 possibilities): ?R ED PE NG IU NF RE NZ Y? 9, 13, 15, 19, 27, 33, 37, 41, 43, 45, 47, 49, 51, 53, 69, 73, 79, 81, 83, 91, 93, 95, 97, 99, 101, 105, 107, 109, 117, 121, 125, 127, 133, 137, 139, 141, 145, 147, 149 Beware Ice Weasels (30 possibilities): BE WA RE IC EW EA SE LS 4, 14, 16, 20, 26, 28, 52, 54, 64, 76, 78, 82, 84, 86, 90, 102, 114, 116, 120, 122, 124, 126, 128, 132, 138, 146, 150, 154, 156, 158 -Tx *********************** Re: Revised indexes Tuesday, 09-Nov-1999 08:58:30 194.78.232.85 writes: Hi Tx, I had some times during lunch to think a bit about it. I think we should focus on the second phrase (beware ice weasels). We know it MUST appear, and it gives the minimum possibilities (btw, I got a little difference in my possible index. I don't have 76, but well 96 ... who's wrong?) Also, if we look at that phrase wa can notice some interesting stuff : we have lots of duet with 'E' : BE, RE, WE(EW), AE(EA), SE. This is 'E' + 5 (five) letters. So, either two (or more) of those 5 letters appear on the same row or col, or 1 (or more) appear on the same row (or col) as the 'E'. I think this should be explored more deeper, because i suspect 'something special' happens when 2 letters are in the same row (col) and appears in different duet (ie R&S on the same row , RE and SE should give ?X and ?X). Hope this is clear enough, I'm not as good as Tx to explain things :( Back to work for a few hours now :( I'll work on this later :-), Laurent. laurent ********************* Warning with the 'special' cases (degenerated rectangles) Tuesday, 09-Nov-1999 10:59:37 194.138.158.162 writes: Hi everybody! Hope you people keep in mind that if both letters forming a pair are in the same row or column, then the diagonalization feature is corrupted, or so I guess. Taking these special cases into account,I suspect that the weasels phrase could show up on even as well as odd positions. Bye. Don Quijote ****************** Re: Warning with the 'special' cases (degenerated rectangles) Tuesday, 09-Nov-1999 13:26:10 208.214.188.10 writes: Hmm I don't think so.. In all cases (same line/col or not), a two characters couple will always encore in the same two characters cipher.. if AB encodes as CD then BA will encode as DC.. that's the type of pattern we were looking for.. couples formed of the same letters in the same or the reverse order.. -Tx ***************** Re: Re: Revised indexes Tuesday, 09-Nov-1999 13:49:04 208.214.188.9 writes: For 76 and 96, my mistake.. They were in the original email but apparently I can't be trusted to type things by hand ;) Both 76 and 96 are valid... There are no identical letters starting at 76: L B A E E W T A O R L E Q I A C D E F W H E S A F S Z E W L N S Have to get back to 'real' work ;) -Tx *************** Re: my mistake too, 76 is valid ... Tuesday, 09-Nov-1999 14:11:59 216.224.154.150 writes: not sure what I am doing but I looked for several patterns: 1 BE WA RE IC EW EA SE LS skipping the 'b' and beginning with 'e' I counted over to the next 'e' that e was at 4 places next 'e' is 3 places next 'e' is 2 next is 3 first e being my zero counting to right 0-4-3-2-3 I then started with each letter from the list: (manually cause I don't know how to write programs) 1. The message, separated into pairs : VY TE SY ED LU TE RV LF NV UH DW AR DL CF FB SD EW NP XK IC FT RE OL KA LZ YL SL TO BK EV LY AR MK RB OD NA LD YP LA ET OL QA DF HS FZ WN AI DS MU RU OL HR YL LO TW FY LD IC VL US VS SF ZY LU NF FX LK TG BC DO BF AL EW RP FY WL HU LD AR LI TF LA BF FZ CY FU UF BG XX XX ....and could not find any where where any of the letters above produced a matching letter (of any kind) at 0-4-3-2-3 I then did the same for the red penguin frenzy the en in penguin and en in frenzy are 6 letters apart. I found no where where two combinations of letters 6 spaces apart matched... I then looked for three letters in a row that might match for STO (for stop) ??? nada; nothing;zip are they encripting the encription ??? (Am I at least on the right path here guys???) ************************************** 208.214.188.10 writes: That's the right idea.. But you have to work with couples.. encoding AB and AC won't always give you a cipher couple with the same starting letter... -Tx ****************************** Re: Yup..okay -Tx Tuesday, 09-Nov-1999 15:04:24 216.224.154.150 writes: okay; I'm a little behind...I was doing the first project and wonderin what the sam hell u guys were goin on about...slowly catching up on all the threads...saw this one from Seeker: http://www.ftech.net/~monark/crypto/crypt/playfair.htm I was wondering what happened to your 'j' in your list yesterday...but continuously showing my ignorance with dumb questions depresses me...heheh and I am now reading it... a question please. the example for cryptogram seems to have left out the R in gram...this is because its already been used once in the word??? CRYPTOGAM BDEFHIKLN QSUVWXZ-- pkay am going back to read..I just noted that u were ON right now so wanted to hurry and ask questions....any other reading I can do to catch up and learn or did The Seeker posts the URls I need? thanks jeff *********************** another approach (failed ..up to now) + 4 Jeff Tuesday, 09-Nov-1999 15:14:51 193.158.174.50 writes: Another approach : Still hanging around with "END" (at the end). There are three possibilities : UF BG EN DX ********* . F BG E ND ********* BG XX EN D ********* Took every possibility and replaced the according letters, for example in nr. 1 U == E and vice versa F == N and vice versa B == D and vice versa I KNOW, that this is NOT 100% correct and brings some wrong letters, but just tried to get a 'feeling' for the words. Could get no 'feeling' up to now, could find no scheme up to now. :( Nevertheless I tend to say, that E==F (and F is the second-most) Later more Jeff : Correct with the "R", every letter appears only once ! The Seeker *************************** welcome on board, jeff + some thought Tuesday, 09-Nov-1999 15:27:01 195.238.22.213 writes: Hi jeff, nice to see you onboard :) Yep, you start writing your 'password' in the cells (startint top-left) and you skip all letters that's already in the grid. Why ? Well, we have 25 cells (5x5), and our alphabet have 25 letters (yes, 25, J is replaced by I) so, we can only put each letters 1 times. btw, wife will come home soon, so i'll have to quit :( ... unless she is in good mood, lol :) Well, here are my thought of the evening : - if you have a duet like ?E (or E?), then the crypted duet will only contain letters in the same row or col than the E. This is due to the 'torus' form of our grid (you can swap 2 rows or cols, the encryption will remain the same ! ). Anyway, to resume, each time you know a duet with 'E' (for example), then the 2 letters of the correspoding due can only comes from a set of 8 letters (the one in the row and col of the 'E'). if it's not clear i'll try to write something 'α la Tx' :) with this, I could reduce to about 21 possibilities, but i still have to check them ... later, laurent. ************************* Sets of 8... Tuesday, 09-Nov-1999 15:52:19 208.214.188.4 writes: Gee, am I becoming some sort of standard ? That's a lot of pressure LOL :) Good point.. I hadn't looked really at the forward process before but damn that's a fact.. a letter will supply 8 letters as possible cipher... hmm.. too bad that's not reversible.. Don't know if it'll help anything, but I've compiled a list of which letters are paired to each letter of the alphabet: A: IKLNQR B: CFGKR C: FIY D: EFLOSW E: DRTVW F: BCDLTUXYZ G: BT H: RSU I: ACL K: ABLX L: ADFIKOSUVWYZ M: KU N: AFPVW O: DLT P: NY Q: A R: ABEHPUV S: DFHLUVY T: EFGOW U: FHLMRS V: LENRSY W: DELNT X: FK Y: CFLPSZ Z: FLY That means that if L isn't paired in same col/line cases, it's still used by 75% of the rest of the board ! So my guess is the 8 letters in the lines and columns where L is in the matrix must be pretty often used letters... Wondering if there's any way to reconstruct the matrix ? I bet we'll have to do both: decrypt and construct the matrix so they feed each other.... Btw, as an informal poll: does anyone here have trouble reading Perl code ? That's all I use to work on the cipher... works great :) -Tx ****************************** Re: Sets of 8... Tuesday, 09-Nov-1999 17:00:03 216.224.154.150 writes: BE WA RE IC EW EA SE LS B E W A R I C S L X D F G H K M N O P Q T U V Y Z is this correct then? jeff **************************** Re: OOPS FORGOT Tuesday, 09-Nov-1999 17:09:04 216.224.154.150 writes: forgot this part B I D M T E C F N U W S G O V U W S G O A L H P Y R X K Q Z so now u search by coloum and by row to match whats in the message? I don't get this....Im going to take a break jeff ****************************** Re: Re: OOPS okay I think i get it Tuesday, 09-Nov-1999 18:12:03 216.224.154.150 writes: ... we have knowns of BE WA RE IC EW EA SE LS and we have the tabel: VY TE SY ED LU TE RV LF NV UH DW AR DL CF FB SD EW NP XK IC FT RE OL KA LZ YL SL TO BK EV LY AR MK RB OD NA LD YP LA ET OL QA DF HS FZ WN AI DS MU RU OL HR YL LO TW FY LD IC VL US VS SF ZY LU NF FX LK TG BC DO BF AL EW RP FY WL HU LD AR LI TF LA BF FZ CY FU UF BG XX XX so now we have to develope the KEYWORD they used to do this? (example KEYWORD PALMERSTON) Oooooooooh boy; jeff ******************************** Re: Re: Re: OOPS okay I think i get it Tuesday, 09-Nov-1999 18:42:11 195.238.20.201 writes: Yes jeff, you got it :) the goal is to find a 'plausible' position for the plain text you know (it may be 'beware...' or STOP or END or whatever else) and for each of these possible position to try to reconstruct a working 'grid'. The first goal is, IMHO, to first minimize the number of possible position, cause the only way I see to reconstruct the grid is by hand ... going to bed now ... see you tommorow ... ooops, no, no more tomorow, but later this morning :) Regards, Laurent. laurent ****************************** My road ... to nowhere :( Tuesday, 09-Nov-1999 18:36:58 195.238.20.201 writes: Hi guys, well, this is a resume of the road i am following. Although it should lead to somewhere, I still didn't moved :( ... so maybe someone could check this over and even try it. Maybe (probably) I am wrong somewhere : - I work with the 'phrase' 'BEWAREICEWEASELS'. - I run it to every position (from 0 to 160) and keep only the EVEN positions where a plain letter doesn't correspond to the same letter in the cipher. ex : BCDOBFALEWRPFYWL->is refused (B,W are at the same place) BEWAREICEWEASELS This leave about 31 possible position (Tx published them somewhere already). - Then for each pair of my plain text where appear a 'E' (BE RE EW EA SE), i look at the corresponding cipher letters (both one). I can't get more than 8 different letters. This leave about