German Codes Cracked!

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German Codes Cracked!
Topic Started: Mar 3 2006, 03:57 AM (357 Views)
PulsarSL	Mar 3 2006, 03:57 AM Post #1
Super member Posts: 157 Group: Members Member #7 Joined: September 1, 2005	You guys seen this? http://news.bbc.co.uk/2/hi/technology/4763854.stm Pulsar



insecure	Mar 3 2006, 07:08 AM Post #2
Elite member Posts: 381 Group: Trusted Member Member #10 Joined: September 9, 2005	No, I hadn't. "One down" is excellent news. "Two to go" is even better news, for those who have perhaps just learned about this and would like their own chance at fame and - er, fame. (Nope. Too busy. Sorry.)



Boulemans	Mar 4 2006, 11:21 AM Post #3
Just registered Posts: 18 Group: New Members Member #32 Joined: January 20, 2006	I don't understand completely. Is this normal Vigenere or is this a One Time Pad?



rot13	Mar 4 2006, 02:58 PM Post #4
Elite member Posts: 306 Group: Trusted Member Member #14 Joined: September 29, 2005	If it were a one time pad, then it would not have been broken. Or it could have been "broken" many times by coming up with various possible messages with the same number of characters, but without the original pad, you'd never know which message was actually correct. There are lots of other ciphers other than one time pad and Vigenere. Had it been Vigenere, it would have been cracked probably within minute of it being transmitted back in WWII. The Enigma is a rotor-based cipher. In a way it is like Vigenere, but it is more like triple- or quadruple- vigenere with varying key lengths. Wikipedia gives a good overview of how the machine worked. Having written some programs to crack Enigma message, I can tell you that the thing that really throws it is the plugboard. The more connections there are on the plugboard, the harder it is to crack. I don't know why that should be the case, because it seems like the plugboard is just a simple substitution. During the war, the Germans typically used 10 connections on the plugboard, which really makes it hard to crack short messages. Plus, this particular set of messages was done on a 4-rotor enigma.



insecure	Mar 4 2006, 08:27 PM Post #5
Elite member Posts: 381 Group: Trusted Member Member #10 Joined: September 9, 2005	rot13 Mar 4 2006, 02:58 PM Having written some programs to crack Enigma message, I can tell you that the thing that really throws it is the plugboard. The more connections there are on the plugboard, the harder it is to crack. I don't know why that should be the case, because it seems like the plugboard is just a simple substitution. Substitution ciphers are only easy to crack because of frequency analysis. The effect of the rotors and the reflector completely stuffs up frequency analysis. Substitution is a vital component of typical modern block ciphers. (S-boxes and all that.)



rot13	Mar 4 2006, 11:36 PM Post #6
Elite member Posts: 306 Group: Trusted Member Member #14 Joined: September 29, 2005	insecure Mar 4 2006, 08:27 PM rot13 Mar 4 2006, 02:58 PM Having written some programs to crack Enigma message, I can tell you that the thing that really throws it is the plugboard. The more connections there are on the plugboard, the harder it is to crack. I don't know why that should be the case, because it seems like the plugboard is just a simple substitution. Substitution ciphers are only easy to crack because of frequency analysis. The effect of the rotors and the reflector completely stuffs up frequency analysis. Substitution is a vital component of typical modern block ciphers. (S-boxes and all that.) Well, the funny thing was, I was usng Jim Gillogly's method of cracking via Index of Coincidence. IC remains constant under a simple substitution. I just haven't taken the time to see in what way the plugboard is NOT a simple substitution.



insecure	Mar 6 2006, 01:44 PM Post #7
Elite member Posts: 381 Group: Trusted Member Member #10 Joined: September 9, 2005	rot13 Mar 4 2006, 11:36 PM Substitution ciphers are only easy to crack because of frequency analysis. The effect of the rotors and the reflector completely stuffs up frequency analysis. Substitution is a vital component of typical modern block ciphers. (S-boxes and all that.) [/QUOTE] Well, the funny thing was, I was usng Jim Gillogly's method of cracking via Index of Coincidence. IC remains constant under a simple substitution. I just haven't taken the time to see in what way the plugboard is NOT a simple substitution. [/QUOTE] Well, IOC is more useful in Vigenere-style encryptions, as a means of establishing the key length. The plugboard is more of a monoalphabetic substitution cipher. On its own, the plugboard would be a complete waste of plug and board - but working it into an encryption system with lots else going on has a definite strengthening effect on the cipher.



rot13	Mar 6 2006, 02:41 PM Post #8
Elite member Posts: 306 Group: Trusted Member Member #14 Joined: September 29, 2005	insecure,Mar 6 2006 01:44 PM rot13,Mar 4 2006 11:36 PM Substitution ciphers are only easy to crack because of frequency analysis. The effect of the rotors and the reflector completely stuffs up frequency analysis. Substitution is a vital component of typical modern block ciphers. (S-boxes and all that.) Well, the funny thing was, I was usng Jim Gillogly's method of cracking via Index of Coincidence. IC remains constant under a simple substitution. I just haven't taken the time to see in what way the plugboard is NOT a simple substitution. [/QUOTE] Well, IOC is more useful in Vigenere-style encryptions, as a means of establishing the key length. The plugboard is more of a monoalphabetic substitution cipher. On its own, the plugboard would be a complete waste of plug and board - but working it into an encryption system with lots else going on has a definite strengthening effect on the cipher.[/QUOTE] IOC can be useful for cracking Enigma as well. You may have already read it before, but if not, Jim Gillogly's paper on cracking Enigma without cribs using IOC is quite interesting. The funny thing is, as you add more plugs to the plugboard, it becomes harder to crack. That is the part that is difficult to understand. If you apply a mono substitution to a text, the IOC stays the same, but there is something more going on with the Enigma. For example, suppose I decrypt an English text with no plugs on the plugboard. When I get the rotor, ring and reflector settings figured out, I am done. The IOC of the resulting text should be around the value for normal English text. Now, assume that the rotor, ring and reflector settings are identical, but this time there is one plugboard connection. If I decrypt the text using the right R/R/R settings but don't have anything in the plugboard, the IOC is not close to the norm for English. I just went to put together some example texts to demonstrate this and in doing so, I realized why the plugboard has such a drastic effect on the text. The trick is that the no substitution occurs at 2 places. For example, suppose for a certain position that I have a plug that exchanges A and B. If the only effect of the plugboard was to perform a monoalphabetic substitution, then any A's in the output would change to B's, and vice-versa. Basically, you have a substitution everywhere in the plaintext where A and B occur, which is a typical mono sub. But also, you have a substitution for everywhere that A and B would occur in the ciphertext as well. That means that things spread out more. For example, here is the title of Goethe's Faust: FAUSTDERTRAGOEDIEERSTERTEILJOHANNWOLFGANGVONGOETHE When I encrypt it with A-B in the plugboard and a certain R/R/R setting, I get: EJFNXVKTHXIPPAPKTZSTUPKWCAMMJZXHHLABUXWBSRXHLCNRNV The bold text marks all the places affected by the plugboard. Now if I decrypt the text with no plugboard, you can see that more than just the A's in the original text are scrambled: FBUSTDERTRBGOTDIEERSTERTEXLJOHBNNWFOFGBQGVONGOETHE This may not be helpful for anyone else, but at least I can grasp now why the IC attack doesn't work so well when you get up to 10 plugs.



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