« Prizegiving | Main | Fialka Excel spreadsheet »
January 02, 2006
Rotor turning
If you imagine the pins of the rotors numbered 0 to 35 with 0 standing for 0, 1 for 1 and so on, 10 for A, 11 for B and so on, then rotor 1 turns so that the numbers increase while rotor 2 turns so that the numbers decrease.
Posted by Harry at January 2, 2006 08:33 PM
Comments
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Thanks for a great challenge Harry. It's been great fun. The new method of scoring with the time bands was very nice.
And of course, Well done to the Winners
Posted by: Mike at January 11, 2006 03:22 PM
i say well done harry and everyone else, challenging challenge (obviously) but extremely fun at the same time :) well done all, hope to see you at southhampton, if its going ahead :)
[It will go ahead, date to be announced. Watch this space. Harry]
Posted by: martyn compton at January 10, 2006 08:54 AM
Hi Harry, I'd just like to say thanks for a great challenge, I've really enjoyed it and I've learned so much that I can't wait til next year! I have a couple of questions: 1) I never managed to crack 8B (partly due to lack of time, and partly due to my inferior programming skills!) so when will you publish the solution? I want to know how the story ends. :) 2) Are there going to be prizes for 8A like the other part A challenges?
Posted by: Jess at January 9, 2006 07:58 AM
Did you know, if you type "Damn you Harry" into google, cipher challenge comes up as the first entry?
Posted by: girl with too much time on her hands at January 8, 2006 09:19 PM
so do the winners already know they have won? there go our hopes!
Posted by: Ellie at January 8, 2006 05:43 PM
What does 'early next week' mean?
[It means I don't know exactly when I can publish the winners' names as I have to discuss it with them and with the sponsors. Harry]
Posted by: Laura at January 8, 2006 08:42 AM
Cecil, have another go at J1 by hand. It's true that Joker A can be in one of 13 positions, and you need to know which one to proceed. But have a careful look at the deck (a pencil layout might help). Notice how, for each of the 13 positions of the joker, the cards move around. You will find that 10 of them can be eliminated ALL AT ONCE - for the same reason - in the first two letters of the crib. (It has to do with other unknown cards' possible values.) One other position fails in two letters; the 12th fails at letter 3, leaving the joker known.
Posted by: John at January 7, 2006 09:10 PM
Is "John" the same John as last year, who posted in detail his way of solving the cipher by hand which helped me work it out? If it is, a belated thanks, it was great. :) Aah, the sense of achievement I felt....What fun I had......
Posted by: cecil at January 7, 2006 07:16 PM
although I didn't try, I don't think I would have been able to do challenge J1 because the way i did the solitaire depended on the two jokers being in the part of the pack we were given.
Posted by: cecil at January 7, 2006 07:00 PM
Jason,
Bit of fun to go back to our Solitaire spreadsheets from last year. John's right about your challenge being possible by hand (or at least in Excel). After the end of the crib a couple of combinations for the next two cards give all but 2 of the next 20 letters - and it looks more or less like plaintext! Advantage of Wombat manual methods is how easy it is to recognise stuff that looks nearly plaintext. Took us about 4 hours to get to "Old pen and paper ciphers INCLUDE". And that's with 41 cards given - which shows what a neat cipher Solitaire is. Good one!
Posted by: The Wombats at January 7, 2006 09:15 AM
Just to add my thanks to all the other ones floating around in cyberspace - thank you very much Harry for a great challenge and lots of fun. I'm just looking forward to next year now!
Kati
[Thanks, Kati for all your help, Harry]
Posted by: Katriel Cohn-Gordon at January 6, 2006 11:06 PM
I'd like to say a big well done to Harry and his crew on this challenge, it was really enjoyable for me and my team (The Dewberrys), and we all feel we learnt a massive amount about ciphers, a topic we'd never actually come across before. I'll admit that unfortunately Martyn was right, we did sort of give up on the last one, for several reasons, including problems keeping in contact during the holidays and various seasonal activities people had. So, a big thumbs up to the people who managed the final challenge, and those who tried their darnest to complete it :)
Though next year we'll be too old to take part in this challenge (which is a shame, wish we'd discovered it before!), we think a forum would be great idea, to allow for greater discussion of the challenges and to possibly get to know other people involved. Also, I think it would be a great help to post after each challenge how to solve it, especially with the final challenges. We found the penultimate challenge with the Hill cipher really challenging, but at the end of the day the results we got only applied to the first eighteen characters - we wouldn't mind seeing where we went wrong in applying it to the rest of the message!
And I think I've rabbled on long enough now lol... a final question, though. Is there a deadline for sending an email about the prizegiving thing?
[Thanks for that. I would get back back to us about tickets asap - they are pretty popular. Harry]
Posted by: Vicky Bird at January 6, 2006 09:17 PM
Harry, thanks a lot for all your hard work on the challenge - it must have taken ages. I really enjoyed it and I learnt a lot about programming and different ciphers. We even dashed down off Snowdon to solve the one in the holidays. It was great fun and I hope thet there will be another one next year.
Posted by: Hannah at January 6, 2006 08:48 PM
Yes. I would be interested in the excel spreadsheet and with macros. Incidentally I have used excel now for checking the settings you have given harry but still cannot get lomonsov ridge weather report.
Posted by: david at January 6, 2006 08:34 PM
Harry
Thanks for everything! I really enjoyed it and learnt a lot. Hope it's just as good next year.
A. White Knight
Posted by: Sir Sovlalot at January 6, 2006 08:04 PM
yes, yes, yes, ciphertigers. i would love to see your excel simulator.
Posted by: lara at January 6, 2006 07:31 PM
Thanks Harry, really good challenge this year... Any more thoughts about that forum where we can discuss our crackings etc.?
Cheers again!
[Working on it! Harry]
Posted by: Josh at January 6, 2006 07:15 PM
sorry to ask again, but are the results likely to come out tonight? as i'm checking continuosly...
[Oh sorry, that's not likely. Look out for an announcement early next week. meanwhile the site remains open for comments. Harry]
Posted by: excited at January 6, 2006 06:55 PM
yes plz ciphertigers!!! sounds really cool
Posted by: Mark at January 6, 2006 05:56 PM
Would anyone be interested in an example of an Excel spreadsheet which simulates the Fialka machine and has a macro to search through the rotor settings? I've added loads of comments for those not familiar with Excel functions or macros, so it could be quite a good learning aid. I could e-mail it to Harry and he could put it up on the site somewhere if anyone thinks it would help. (Not if there's going to be an Enigma machine next year of course.)
[Sounds great, anyone interested? Harry]
Posted by: ciphertigers at January 6, 2006 05:39 PM
When will the honours board go up Harry?
[Soon, but there are some things I need to do first. Harry]
Posted by: Em at January 6, 2006 04:25 PM
thank you Harry for doing the cipher challenge. it was really fun and challenging and i learned a lot. thanks for such a great experience.
(and thanks to everyone else who i've pestered for hints for being so patient and for helping me!)
[That's OK, I enjoyed it lots too. Harry]
Posted by: lara at January 6, 2006 04:24 PM
could anyone give the first 10/15 letters of the plain text - I haven't done it yet, but I'm close. I would like to see if my program would have worked.
[Here it is. I'll hold off publishing the final solutino for a little bit longer. Harry
Lomonosov Ridge Weather]
Posted by: stephen harris at January 6, 2006 12:12 PM
i didn't manage to get anywhere, but thanks everyone for your help. once this is over, can i have the plaintext and a really detailed explanation of how you get there from anyone whose done it please?
Posted by: lara at January 5, 2006 11:27 PM
what exactly does the movement of the rotors do?
Posted by: lara at January 5, 2006 11:07 PM
well I think this was a great experience second year enjoyed this one a lot more I must say. a big thankyou to Harry for working hard on these codes without them there would be no cipher challenge. I admit defeat with this my pringles machine must have been faulty somewhere and I congratulate all who have managed to solve it.
Becci
Posted by: Becci at January 5, 2006 11:00 PM
Harry, on the reflector, do you start linking at 0 or 1?
[To be honest I can't remember! Harry]
Posted by: lara at January 5, 2006 10:30 PM
Harry, i feel so guilty saying this but you couldn't give me the wirings for the reflector could you? pleeeeeeeeeease? i printed the diagram, but my printers kind of busted so you can't see the diagram properly.
[Here they are:
15, 34, 29, 25, 22, 16, 27, 11, 26, 31, 14, 7, 35, 33, 10, 0, 5, 23, 32,
28, 30, 24, 4, 17, 21, 3, 8, 6, 19, 2, 20, 9, 18, 13, 1, 12
Good luck! Harry]
Posted by: lara at January 5, 2006 10:20 PM
i was just wondering if any credit is given to part of the text being submitted or do we need the entire text as my pringle machine isn't working very well at all
Becci
[You get a score for anything! Do as much as you can and submit it before midnight. Harry]
Posted by: Becci at January 5, 2006 09:19 PM
please please please could anyone give me the connections for the punchcard and tell me how it works? pleeeeeeeeease only 2 hours and 45 minutes to go... is anyone feeling kind? Harry?...
[The punchcard is wired as
0, 1, 18, 25, 28, 22, 6, 16, 23, 33, 26, 27, 34, 17, 29, 32, 7, 13, 2,
35, 24, 30, 5, 8, 20, 3, 10, 11, 4, 14, 21, 31, 15, 9, 12, 19
so 0 and 1 are "straight through" 2 and 18 are switched and so on. Good luck, Harry]
Posted by: lara at January 5, 2006 09:14 PM
Lara, yes it is!
Posted by: Помогать at January 5, 2006 08:48 PM
Is this cipher text right?
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
[Here is my version - I didn;t get a chance to check yours, sorry, but the exclamation mark is a bit suspicious. Harry:
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
Posted by: lara at January 5, 2006 08:43 PM
Lara,
The rotor machine works as follows:
1. Letter typed in
2. Letter goes through the punchcard (so is swapped for another letter)
3. Letter goes through rotor 1 (if you've made the pringle Enigma, it will go through starting at the letter touching the one it ends up at after the punchcard)
4. Letter goes through rotor 2
5. Letter goes through the reflector. You can stick your reflector on your Pringle can lid and follow the wires across as long as you've got it the right way round. I think somebody said somewhere that it needs to be flipped before it will work if you are sticking it directly on the lid of the pringles can.
6. Letter goes back through rotor 2 (on the Pringle machine it will start at the letter where the reflector wire it's been through comes out)
7. Letter goes back through rotor 1
8. Letter goes back through punchcard
9. Letter comes out as whatever it is changed to by the punchcard.
Katriel says that rotor 1 starts with F (15) where the punchcard 0 is, rotor 2 with Z (35) and the reflector with 9, all lined up. Remember that the rotors rotate in opposite directions (I'm not sure which way they actually go, but if you are getting nonsense then chances are they go the other way) and that the second rotor only moves when the first rotor goes from the Z position to the 0 position.
There is not a lot more I can think of to say that might help (Harry, can you? I think people need all the information now), but I hope you can crack it. You certainly deserve to as you've clearly worked very hard on the challenges. Good luck!
[I honestly think everything is known now. Thanks to all of you for your help with hints, and congratulations to everyone who tried. I received one or two great photos of Pringle Fialka/Enigma machines. Harry]
Posted by: Помогать at January 5, 2006 08:33 PM
Will the last honours board be up tomorrow Harry? I want to know who won!!!!!
[We'll put it up asap, but there are some preliminaries for us to complete first. Harry]
Posted by: Laura at January 5, 2006 07:20 PM
Harry, or anyone, what does the reflector do?
[It bounces the letters back through the rotors again to complete their encryption. Harry]
Posted by: lara at January 5, 2006 06:03 PM
i'm still so stuck. please help. can anyone tell me how you do it by hand please?
Posted by: lara at January 5, 2006 06:02 PM
Jason,
Your Solitaire cipher challenge is considerably harder than last year's 8B. Nonetheless, with the crib you gave, it is possible to solve it by hand, but slow.
After 3 letters of the crib ('ENC'), the positions of the joker and one other card are known. By the end of the crib, 5 cards are known. 3 letters later, a 'count cut' depends on an unknown card, which may point to other unknowns, with about 20 different possibilities to try. So by hand, it takes hours and hours.
If your program can crack it in 2 minutes, you deserve a round of applause.
Posted by: John at January 5, 2006 03:51 PM
just wondering harry, how do you actually encipher your messages, do you have a programme to do it? if you do can you post them for download so we can much about with them :)
[It's a mixture of perl scripts (implemented in the fantastic text editor BBEdit), off the shelf components from around the net and Maple worksheets, plus a healthy dollop of hand checking and ad hoc messing about with Excel. If I get a chance I would like to build a dedicated all singing all dancing cipher machine but that has the disadvantage of tending to make everything a bit too uniform. Harry]
Posted by: martyn compton at January 5, 2006 03:22 PM
i dont know if anyone feels the same but it seems like most people have just given up, the other team from my college didnt even try, a bit of a shame really but theyve probably got their reasons :(
[It is a shame but it's not unusual with the final challenge. It is very tough. All credit to those who have tried. Harry]
Posted by: martyn compton at January 5, 2006 03:04 PM
good tips kati, thanks :) see if they help now :s
Posted by: martyn compton at January 5, 2006 02:58 PM
Just a last hint for all those who still have not solved it:
Neither of the rotors is inverted or switched; they are in the given orientations. Their positions are:
Rotor 1: 15; Rotor 2: 25; Reflector: 9
so try those in your program. However, if they don't work, don't give up hope. Here is the testing scheme I used; I hope it helps.
1. Set your program so it traces and prints the path of every character as they go through the rotors (e.g. a-b-1-3-g - that isn't real, by the way).
2. Encrypt a test, say "helloworld", with the rotors at position 0,0,0. Then try to decrypt it. If it doesn't decrypt back to "helloworld" then something is wrong; use the trace to try to figure out where and fix it. One important thing to look at is to see whether you have implemented the stepping of the rotors correctly.
3. Once you have your test working, try running your brute force subroutine on it to see if it cracks it. If not, you will need to use a lot of print statements to see if you can find exactly why it is not cracking it. This bit requires a lot of intuition and a good bit of luck.
4. Once you have this working (i.e. you can brute force any messages that you encrypt yourself) it's time to get to work on the real code. The first thing is of course to try the positions given above. If they don't work, try adding or subtracting one to all of them, or taking their complement in 36, or both. If that still doesn't work, try changing the direction the rotors rotate - mine were coded so you had to subtract one from the position instead of add one. In other words, fiddle with the settings. You should also invert the reflector if you have taken the wirings from the Pringle guide as it was given reflected.
5. If it still doesn't work, you could try running a full brute force of all the possibilities overnight.
A simple way to test for the plaintext is to look for the string "temperature".
I hope that helps, and please feel free to ask me for clarification.
Kati
Posted by: Katriel Cohn-Gordon at January 4, 2006 10:39 PM
Knowing that the original setting is 15, 36, or -35 is only helpful if you know what position 0 is (0 is the position the photograph was taken, how convenient ;)) also, 36 = 0 mod 36, so there are only two settings you need to test, for the first rotor. You also need to know the position of the rotors, which is that they both face forwards (the photographed end faces the keyboard) and they are in the right order, (1 first, then 2). The orientation of the reflector is how it is on the pringle enigma pdf. That leaves you 1 day to get the settings from the last rotor, good luck :)
Posted by: Frank Hamand at January 4, 2006 04:55 PM
That's good advice by the Wombats - the way I tested the plaintext was to search for the word "temperature" (from the forecast) but I like the idea of searching for sections of plaintext.
By the way, don't immediately assume that because the original Fialka used one of the three rotor settings given, yours has to do the same - the way I coded my rotor meant that any settings given had to be used as their complement in 36...
Posted by: Katriel Cohn-Gordon at January 4, 2006 12:02 PM
How many teams have solved 8B, Harry? (Not long left!)
[25 completely correct entries so far! Good luck to the rest of you. Harry]
Posted by: Lizzy at January 4, 2006 07:56 AM
are you going to tell us the settings for the pringle machine or do we just have to use brute force until the end and are any of the rotors not linked by the 15, 35 or 36 set up? and which way does the reflector go because its making even more combinations still to bee tested?
Becci
[Becci, you know just about all the settings now (read around on the site) - you have the wirings, you know that the initial setting for rotor 1 is 15, 36 or -35, you know which way the punchcard goes and you know a crib which you can use to test the other settings. That doesn't leave many possibilities. Do read what other people have written. Good luck, Harry]
Posted by: Becci at January 3, 2006 07:51 PM
I'm not going to crack this unless a miracle occurs. (so i'm basically just waiting for the miracle!!!!!!!!!!!!!!!!!) what the wombats said sounds good, but how do you tell excel to recognise anything? the most advanced thing i can do involving a computer and codebreaking is find and replace on MS word. i know i'm not going to win anything, but i'd really like to crack this. please help me.
Posted by: lara at January 3, 2006 01:22 PM
Harry has already told us the punchcard goes so 0 is 0 and 1 is 1
Posted by: Ruth at January 3, 2006 11:20 AM
Harry,
Don't know if you'll let any of this through at this stage?
Lots of stuff here about programming, but Wombats can still only use Excel. Making a Pringle Machine in Excel is fairly straightforward (given all the rotor wirings) which gives the complete cipher/plaintext alphabet for any given setting.
Finding the solution still requires picking from 2 punchcard orientations (symmetrical along the leading diagonal), 8 rotor configurations (order and inversion of each rotor), 36x36 rotor positions and 36 reflector positions, or nearly 750,000 possibilities. Too many! BUT as the ciphertext is 1296 characters long, this covers a complete cycle of rotor positions (36x36). That means that starting with ANY rotor position should give plaintext - and only 576 settings to try! Problem is, recognising plaintext.
By copying several (actually 24) Excel Pringle Machines all offset by one step in Rotor 1, we "deciphered" every possible group of 24 successive characters of ciphertext to see if any made sense. (Didn't Bletchley Park do something similar, ganging up lines of Enigmas?) Still too much like hard work for Wombats, but most plaintext has no numbers, whereas 24 characters of random gibberish should, on average, include 6.7 numbers. Get Excel to pick out just the "deciphered" sequences that contain no numbers and statistically there should be 1296/[(26/36)^24] for each configuration, which (pleasingly) is about 0.5. We might still have missed the solution if the sequence chosen included the "notch" that moves Rotor 2, but shifting the starting position of Rotor 1 by 12 and 24 steps ensured one sequence avoiding the notch (but increased the number of settings to test to 1728). There was also a (small) risk of missing the right setting by hitting some plaintext containing numbers - well cryptanalysts need a bit of luck (worked for us!).
This method totally fails if you don't know the rotor wirings. Any incorrect wiring guesses are very likely to throw up numbers and so lead to missing the correct setting. We could have tried all the wiring possibilities, but quickly got to be just too many for us poor Wombats - so we just waited until you gave us the right wirings.
Final hint - try the most obvious punchcard and rotor configurations first.
The Wombats.
Posted by: The Wombats at January 3, 2006 10:33 AM
This is why it's good to have a Pringle machine - lets you see exactly how the rotors turn and so on. Even if you don't draw on the wiring, it's worth making a machine just to be able to work out the kinks in the rotation and reflector bits of your program.
Kati
Posted by: Katriel Cohn-Gordon at January 3, 2006 09:18 AM