# The Reihenschieber – A cold war encryption device

Of course, my model is considerably less secure than the original. The original Reihenschieber was certainly hard to break with the means of the 1950s.

Further reading: How a crypto mystery from the Cold War was solved – or was it?

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## Kommentare (15)

1. #1 Norbert
23. Januar 2017

The plaintext is:

ITSASMALLSTEPFORAMAN

2. #2 Tony
23. Januar 2017

I think it says ‘klausenjoysplayingwithlego’

3. #3 George Lasry
24. Januar 2017

Hi Klaus

Two questions:
1) Is the yellow stencil the same as in your picture?
2) How did you compute the key space?
Assuming the stencil stays the same, this should be
(8!/5!) * (4**5) * (11**5), which is
55,411,851,264 I believe.
George

4. #4 Klaus Schmeh
24. Januar 2017

@Norbert:
Your solution is correct! Congratulations, great job! How did you find it so fast?

5. #5 Klaus Schmeh
24. Januar 2017

@George:
I’m afraid you’re right. I didn’t mind that not only the choice of the bars but also the order of the bars needs to be taken into account. So, it’s even more than 10 billions.

6. #6 Klaus Schmeh
24. Januar 2017

@George:
Yes, I used the same yellow stencil.

7. #7 Klaus Schmeh
24. Januar 2017

@Tony:
>I think it says ‘klausenjoysplayingwithlego’
Good try, but it’s not the correct solution.

8. #8 Norbert
24. Januar 2017

If I didn’t miscount, the key is:
pink 1,2 / red 3,10 / black 2,6 / yellow 3,2 / cyan 4,5

@Klaus: I think the number of possible keys for a given stencil is somewhat higher than 10^10: We have 8 possibilities for choosing the first bar, 7 possibilities for the second bar and so on. Each bar may be rotated (4 possibilities) and shifted into 11 positions:

8 * 7 * 6 * 5 * 4 * (4**5) * (11**5) = 1,108,237,025,280

But, as you pointed out, each bar only affects four contiguous characters, leaving only 8 * 4 * 11 = 352 possibilities for these. Thus, a program may compute the 352 plaintext possibilities for character 1-4, 5-8, and so on, try all combinations of adjacent quadgrams and keep the most promising in memory. My “quick and dirty” program code did not really reveal the solution at once, but when examining the output, I discovered “ITSASMALLSTE” among the best 50 candidates in the list for combined bars 1, 2 and 3. This was clear enough to do the rest by hand.

9. #9 George Lasry
24. Januar 2017

Correction:
This is (8!/3!) * (4**5) * (11**5) =
1,108,237,025,280
Congrats Norbert!, very impressive.
Hope to see you at the HCC and have the opportunity to chat 🙂
George

10. #10 Klaus Schmeh
24. Januar 2017

@Norbert:
I’m absolutely impressed!

>If I didn’t miscount, the key is:
>pink 1,2 / red 3,10 / black 2,6 / yellow 3,2 / cyan 4,5
Yes, this is correct.

11. #11 Fliegenschubser
24. Januar 2017

Interesting device. I see the weakness (as noted) that the first letters only depend on the first bar.
How would it change the game if you read your random sequence not in rows but in columns? Looking at the last picture, it would be (with a space after each column)
YFUR SXA D JHVZ O VJBG TCI
With that you have way more possibilities for each 4-letter-block since its depending on more than one bar. You could also use different stencils to get more possibilities.

12. #12 George Lasry
24. Januar 2017

Hi Klaus

With a real device, it seems that numbers were used and not letters, right?

Also some of the places have a . (a dot) instead of a number. This might imply that such places were to be skipped?

Furthermore – was the stencil fixed for some time or settable?

And finally, were the strips changeable?

Would be interesting to see if the codebreaking algorithm for the Lego version can extended to the real version.
George

13. #13 Klaus Schmeh
24. Januar 2017

@George:
>With a real device, it seems that numbers were
>used and not letters, right?
Yes.

>Also some of the places have a . (a dot) instead of
>a number. This might imply that such places were
>to be skipped?
Probably.

>Was the stencil fixed for some time or settable?
The Reihenschiebers I have seen have only one stencil that is not settabbel.

>And finally, were the strips changeable?
There was a set of 26 bars. Ten had to be chosen for an encryption process.

14. #14 Klaus Schmeh
24. Januar 2017

@Fliegenschubser:
>How would it change the game if you read
>your random sequence not in rows but in columns?
This is the lesson I have learned from this challenge (and from Norbert’s solution). The random letters need to be read top down instead of from left to right. I’m sure the original Reihenschieber was operated this way. I have never seen a Reihenschieber manual, so I didn’t know this.

15. #15 Klaus Schmeh
24. Januar 2017

Here’s an excerpt from an article about the Reihenschieber from Cryptologia 2/96 written by Michael van der Meulen: