Here is an answer to questions of Baruch Oxman, which may be useful
to all:

> I have some questions about Homework 2 :
> 1. What do you mean by saying in question 3 that a complete transposition 
> was made ? I followed the link that appears there and read about the cypher 
> used, but there they are talking about an irregular transpostion, so I
> want  to understang the differences between the two.

Difference between complete and incomplete columnar transposition:

suppose I have a transposition key of length 10,
then in a complete columnar transposition the length
of the transposed text must be divisible by 10 (if the text
is not divisible by 10 then it is padded in order to be divisible by 10).
[see also handout given in the class] 

In irregular (or incomplete columnar transposition) the text is not 
padded, and so some of the transposed columns are shorter than the
others, since you don't know which are which and also since length of 
the message does not leak the possible key length -- this case is much
harder than the one you have in your homework.

> 
> 2. In question 4.A :
> I am not sure I correctly understood the "batons" attack method on the 
> enigma. I understood that me must take the plaintext (the crib we have) and
> the cyphertext and apply R1 on both of them and then we get pairs of letters 
> that are transpositions through Z (the rest 2 rotors + the reflector). If 
> there are no contradictions and if we actually see that Z is an envolution 
> (if Z(A)= C then Z(C)=A) then we found the correct position of the rotor R1.
> Also please tell me if it is correct to move the letters to the left,
> e.c. if we start with the given position than it is:
> A B C D E
> J G D Q O
> 
> then the next rotor position will be
> A B C D E
> G D Q O X

The movement of the rotor is to the left, BUT your reading of the rotor
rotation is not correct.

There is an absolute (fixed) keyboard setting 
Keyboard:         A B C D E F ...

Rotor R1:         A B C D E F ...
                  J G D Q O X

so if you type A -> [A->J] -> J [absolute]
The rotor maps A to J, B to G, etc. When it is moved to the left:

Keyboard:         A B C D E F G...

Rotor R1:       A B C D E F G ...
                J G D Q O X U

so if you type A -> [B->G] -> F [absolute] !! (the alphabet of the rotor
is shifted against the absolute position of the keyboard)

[the way you did  the shift is called a "Ring Setting", which was part
of the real Enigma but is not present in our simple variant of Enigma. It
was possible to rotate the disk with the wires, separately from the disk
with the alphabet, in order to change the stepping function. The
stepping notch was attached to the alphabet ring and not to the
wired-disk. Without this feature Enigma would be going through the
same sequence of 17576 permutations all the time. Real Enigma was 
going through a new sequence of permutations with every new "ring
setting". There were 26^3 possible ring settings (three rotors) and
26^2 of them produced unique permutation sequences (since rotor R3 
is not stepping anybody, its ring setting can be compensated
by a proper indicator setting for R3). Just a historic remark.].

What should be provided in your answer:

1. Explanation of what you did an all the trials, like in the handout.
2. The correct R1 position, for ex. if it is:

Keyboard:         A B C D E F ...

Rotor R1:       A B C D E F ... 
                G D Q O X U 

the answer is B (the letter on the alphabet ring of the rotor
which stands against letter A of the Keyboard, this will be seen in the
indicator window for R1).

Helpful hint: you can simulate the whole process by strips of paper
(as was shown in the class), or on the screen of the computer, in any
simple text Editor (!) --- it is easy to shift strings against each
other just by erasing characters. 

good luck,
Alex Biryukov