aj
New Member
Posts: 6
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Post by aj on Jan 14, 2019 21:07:20 GMT
Greetings to you I was trying to decode the below ciphertext and tried a lot but did not work It is the 27th October 1944 and the Message reads ciphertext: WZAXU NDCIP YDQGU YMXYU FLGMR Applying the WW2 process, solve the enigma cipher  How to decode it ? to cleartext? |
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aj
New Member
Posts: 6
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Post by aj on Jan 15, 2019 6:50:38 GMT
anybody here?
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Post by Arnaud on Jan 15, 2019 8:25:44 GMT
Tried what i could but no results.
The text is bit short, not sure if Kennengruppen is in use or not.
Any idea which language the clear text is?
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aj
New Member
Posts: 6
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Post by aj on Jan 15, 2019 9:43:47 GMT
Tried what i could but no results. The text is bit short, not sure if Kennengruppen is in use or not. Any idea which language the clear text is? Language may be English or Germany |
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aj
New Member
Posts: 6
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Post by aj on Jan 15, 2019 10:37:05 GMT
Is this code difficult to solve?
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Post by mauser on Jan 17, 2019 2:40:01 GMT
Give me a little while I might be able to work it. Kenngruppe is definitely not in use here.
Mauser sends
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Post by mauser on Jan 17, 2019 3:47:49 GMT
OK, not sure what was meant by the "WWII process" since all we have is the ciphertext ad no header. I ran this code using the 27th October cipher key in my magic Enigma decoder program to absolutely no avail. I suspect there is somethig missing here. Since no reflector is referenced, I assumed the B reflector.
Mauser sends.
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aj
New Member
Posts: 6
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Post by aj on Jan 17, 2019 10:29:29 GMT
Dear Mauser
Wheel order:V, I, IV
Ring settings: 20, 06, 18
Plugboard connections: KX GJ EP AC TB HL MW QS DV OZ
Indicator groups: bvo sur ccc lqe
ciphertext:
WZAXU NDCIP YDQGU YMXYU FLGMR
I think the ciphertext is double encoded
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Post by Arnaud on Jan 18, 2019 10:39:13 GMT
OK, not sure what was meant by the "WWII process" since all we have is the ciphertext ad no header. I ran this code using the 27th October cipher key in my magic Enigma decoder program to absolutely no avail. I suspect there is somethig missing here. Since no reflector is referenced, I assumed the B reflector. Mauser sends. I tried with reflector C as well but it gave out nothing. I do not have (despite your help) working english ngramm files. Might be a reason why it does not work. (actually the clear text language is not known) If as aj says, the ciphertext is double encoded then I am totally lost |
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Post by brossmann on Jan 18, 2019 10:49:27 GMT
But if it is double encoded we need another sheet to run the text through the enigma again with other settings, or am I just totaly lost?
Where did you get the text and code sheet from aj?
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aj
New Member
Posts: 6
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Post by aj on Jan 18, 2019 17:23:21 GMT
But if it is double encoded we need another sheet to run the text through the enigma again with other settings, or am I just totaly lost? Where did you get the text and code sheet from aj? I think its double encoded, this question from cyber academe |
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Post by mauser on Jan 19, 2019 7:25:51 GMT
Dear Mauser Wheel order:V, I, IV Ring settings: 20, 06, 18 Plugboard connections: KX GJ EP AC TB HL MW QS DV OZ Indicator groups: bvo sur ccc lqe ciphertext: WZAXU NDCIP YDQGU YMXYU FLGMR I think the ciphertext is double encoded There is something missing. I have a piece of proprietary software that if I put in wheel order, ring settings, plugboard, it takes less than 2 seconds to decipher the message. I get garbage. If it was double encoded, what was the other cipher? The 26th, the 28? Something else? Right now, I cant help without the rest of the story as they say. Mauser sends. |
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Post by cleomenes on Feb 14, 2019 0:54:55 GMT
Greetings to you I was trying to decode the below ciphertext and tried a lot but did not work It is the 27th October 1944 and the Message reads ciphertext: WZAXU NDCIP YDQGU YMXYU FLGMR Applying the WW2 process, solve the enigma cipher How to decode it ? to cleartext? In WWII they introduced a keying system, the first three characters will be used to decipher the next 3, these 3 decrypted letters would then be used to decipher the rest. So set your wheels to wza and decrypt xun , Now use these decrypted letters to decrypted 'DCIP YDQGU YMXYU FLGMR' |
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Post by Arnaud on Feb 15, 2019 16:20:09 GMT
In WWII they introduced a keying system, the first three characters will be used to decipher the next 3, these 3 decrypted letters would then be used to decipher the rest. So set your wheels to wza and decrypt xun , Now use these decrypted letters to decrypted 'DCIP YDQGU YMXYU FLGMR' tried it....didn't work |
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Post by mauser on Feb 18, 2019 5:46:17 GMT
Those 6 characters are part of the message header. Part of the standard Army and AirForce System. Unfortunately, if this message had a header, that might have worked. FWIW, I tried using the Kenngruppenbuch to derive the message key and that didn’t work either. As I said before, there is a critical piece missing. I could set the message up on my software tool for a brute force attack even knowing the stecker and the key, it could take days to finally be pulled out. I say that since under normal conditions, if I have the entire key, the results are generally only a few seconds after program start.. no such result. If however the message was somehow double encoded, we have no way of knowing what key was used, as double encoding with the exact same key would end up with the original plain text.
Mauser sends.
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