Question about Kenngruppen

[JMV]

Sorry I don't Understand too well about those :

I your example in the Sticky "How to Encode & Decode Enigma Messages"

The messages used look like this :
2045/28.02.2010/U432/250/1OF1/ZAB ERA
VWFG OUGC YKII HHHJ FYQY OTPT CUXO BSQ

From what I can understand the 3 first letters of the trigram ( ZAB ) are taken randomly. You put ZAB as start position and type ERA which gives you RTZ. You put RTZ as start position and type your message

In the example the Kenngruppen for the 28 are GWU CQP EHG OCK what are they for ? Why are there 4 of them ? Why not have chosen one of them for the message, instead of the random ZAB ? If one was to be chosen for something ( because I suppose there must be a use for them at some stage ), How do I know if I have to choose the first, second, third, or last ?

Could somebody explain to me. ?
Thanks in advance
JMV

[The Enigma]

Those other kenngruppen are for that particular date also a valid choice.

When you start coding a message, you first choose a random combination of three chars (in the example it was ZAB).

Then after looking into the code book, you have the choice of four valid kenngruppen.

In the example, the first of four (GWU) was chosen, but it was also OK if one of the other 3 were chosen (CQP or EHG or OCK)

They have no further meaning in the process.

[JMV]

Sorry I don't get it.
When I type GWU, I get ZOR not ERA.
It doesn't say anywhere in the tutorial, that you make use of the Kenngruppen...

[JMV]

Also, How does the decoder knows which Kenngruppe you're using ? Is it always the 1st one ( though it was never used in the sticky tutorial ) If not, then he can be wrong 3 times, admitting that he can wrongly choose the 3 wrong ones first...?

[The Enigma]

Let's start from the beginning;

Did you setup the enigma machine properly?
Follow the instructions as described here

After done so would you please be so kind to paste the enigma setup in here, just to make sure everything is OK.


You can find the applied settings under the Enigma menu "view key".
The menu will pop up when you hover the cursor over the white tag
(upper right corner).

Chose copy to clipboard and paste it in here.

[JMV]

Thank you for the answer,
I'm very confused, also because you don't seem to be following what I said from my first thread ( No offence there OK ? ) Let's say you're somewhat missing my point, for whatever reason... or I miss something.
I can't answer right now, but will do it all over, and will contact you later, and will tell you how it gets.
But the point also is that the Kenngruppen where not used in this tutorial, and I'm wondering why, as they apparently are in :
http://www.youtube.com/watch?v=BSHfcPbhMCs SubSim Codebook Tutorial

Sorry about that, may be I' m a bit thick, but I don't think we're on the same wave length.

[The Enigma]

Dont worry, I'm not so easily offended.

Well, I tried to answer your question about the other 3 kenngruppen.
Somehow I failed.

The reason why not all kenngruppen are used is that only one of the four kengruppen is needed for coding.

Next you said that the outcome of entering GWU is ZOR and not ERA.
That led me to the conclusion that maybe your Enigma wasn't setup properly.
To check if my conclusion is right I need to know the setup.
Hence my question to paste it in here for checking.

[EDIT]
Just took a quick look into that video.
They explain a different method.
We use one of the 4 kenngruppen.

Just follow the instructions from the Enigma Flotilla.

Thank you for the answer,
I'm very confused, also because you don't seem to be following what I said from my first thread ( No offence there OK ? ) Let's say you're somewhat missing my point, for whatever reason... or I miss something.
I can't answer right now, but will do it all over, and will contact you later, and will tell you how it gets.
But the point also is that the Kenngruppen where not used in this tutorial, and I'm wondering why, as they apparently are in :
http://www.youtube.com/watch?v=BSHfcPbhMCs SubSim Codebook Tutorial

[JMV]

My setup was ok I guess :
UKW: C
Walzen: IV VII VIII
Ringstellung: I-09 S-19 L-12
Stecker: AF BE CQ DN HU IK JY LP OS TW

So I understand they didn't use the Kenngruppen in the tutorial, and I was just wondering why ?

Anyway they are not using GWU at all as should be expected but instead are using the random ZAB from which they get RTZ after entering ERA.

[JMV]

In this Video : http://www.youtube.com/watch?v=BSHfcPbhMCs ( from 5:45 onwards )
They say you can take which ever of the 4 Kenngruppen you must take, but apparently it has to be decided before hand.
What they do then is enter whatever Kenngruppe they have chosen in the start position on the machine, and type the message.
But before they enter the message they type 2 random letters, so that the BDU knows which Kenngruppe you're using they say, then retype the Kenngruppe you have used, and then retype a random letter before sending the text message proper... ( confusing )

I will try to find out exactly.

Anyway it seems it's you to tell the Kaleuns which of the kenngruppen to choose.

[JMV]

Following message encoded with your April 1939 codebuch on date 26.
Message starts with second Kenngruppe NEF and a random letter. So message actualy starts with XXUF.

0115/26.04.2010/UXXX/66/1OF1/GEF VON

ANEF XXUF WIGD ZAYY VUJO VVBB NILF GEWH VCFX QUAX IDOX ZQQB NCFE COZP NTTN BPNF LU

Hope it's some good news for me, and that I can soon be part of the club...
JMV

[toodrunk]

So, I think you're asking the WHY of it, not the HOW of it, if that's the case: I think it's a validation thing.

You encode your messages for the the day using one of the four kengruppen. If you get a kenngruppen that's different, then you might question the validity of the message you've decoded.

As far as I can tell, as long as you set it up right, then it doesn't really matter what your trigram is, as long as whoever's decoding understands the SOP.

If you're asking the HOW for the SOP, here you go:

The first part of the trigram is three random letters. The second part is the three letter RESULT of typing in the three letter group (kengruppen) you've used to encode the main body of your message.
(Anybody feel free to correct me if I'm incorrect)

I've encoded a message with a start point of (kengruppen) of STI. So I pick a random three letter group, WRX, and set my rotors to that. I type in (the kengruppen) STI for the result PNY. My trigram is now WRX PNY.

When you start at WRX, type in PNY, you get the kengruppen STI.

At least, I'm pretty sure that's how it works.

[toodrunk]

Originally Posted by JMV

Following message encoded with your April 1939 codebuch on date 26.
Message starts with second Kenngruppe NEF and a random letter. So message actualy starts with XXUF.

0115/26.04.2010/UXXX/66/1OF1/GEF VON

ANEF XXUF WIGD ZAYY VUJO VVBB NILF GEWH VCFX QUAX IDOX ZQQB NCFE COZP NTTN BPNF LU

IF YOU CAN DECIPHER THIS MESSAGE IT MEANS I PRETTY MUCH GOT THE HANG OF IT NOW

Cool man, one thing though, I'd say don't start your message like that. The first four-letter group needs to be code too.

First time I tried decoding your message, it came out garbled. It wasn't till I re-read your post that I realized what happened. Refer to other post for the kenngruppen explination

[JMV]

OK Toodrunk, thanks for the answer. I think I try to understand that thingy with kenngruppen.

[KL-alfman]

Hello JMV!

and welcome to our Flotilla!
I think you did great in understanding the principles of the Enigma and look forward to your messages.

(I will be the one in charge of sending your encrypted contact-reports to OS).

[The Enigma]

Summarizing the coding steps:
On the left the instructions, on the right the example: