How to convert arrayfun to for loop

I found a source code online for a vigenere cipher and was wondering how to convert it to a for loop.

 function Array = Operator
    count = 27;                         %Primary function of the vignere encryptions by creating a 27x27 array

    Length = 1:count;                    %array of length 27
    %   ie, Create a matrix with 27 shifted substitution alphabets
    %   1 2 3 4 5 ... 26 27
    %   2 3 4 5 6 ... 27  1
    %   3 4 5 6 7 ...  1  2
    % etc.
    Array = arrayfun(@(n) circshift(Length, [0, -n]), 0:count-1, ... 
                 'UniformOutput', false);
    Array = reshape([Array{:}], count, count);
end

 function cipher_text = vigenere_cipher(origionalText,key)
    Array = Operator;

    key = lower(key) - double('a') + 1;     %Converts all text to lowercase
    key(key < 0) = 27;
    origionalText = lower(origionalText) - double('a') + 1;
    origionalText(origionalText < 0) = 27;

    keyLength = rem(0:(numel(origionalText)-1), numel(key))+1;       %Converst the key to the length of the origional text
    k = key(keyLength);

    % Encrypt: C(n) = V(k(n), plaintext(n))
    cipher_text = arrayfun(@(m,n) Array(m,n), k, origionalText) - 1;
    cipher_text(cipher_text == 26) = double(' ') - double('a');
    cipher_text = upper(char(cipher_text + double('a')));
end

Answer 1

[Thorsten]

 count = 27;
 Length = 1:count;
 for n = 0:count-1
   Array(n+1,:) = circshift(Length, [0, -n]);
 end

[Zachary Holmes]

Your output is slightly wrong. Attackatdawn with the key lemon outputs LXEOPVEERNGR, where its meant to be LXFOPVEFRNHR

[Thorsten]

Then there is something wrong with your original code, that also outputs LXEOPVEERNGR.

I also checked that Array computed with your original function and my Array are equal.

[Zachary Holmes]

That output is different slightly, whit your code, when the letters cross from z to a they seem to be going a letter to far. Ie in yours the letter F is meant to be an E

[Zachary Holmes]

All good. i got it working thanks!!

[Thorsten]

So if you're happy, please accept my answer.

Answer 2

[Andrei Bobrov]

Array = hankel(1:27,[27,1:26]);

or

count = 27;
L = (1:count)';
Array1 = zeros(count);
for jj = 1:count
    Array1(jj,:) = rem(L+jj-2,count)+1;
end

or

Array = hankel(L,circshift(L,[0,1]));

[Zachary Holmes]

I don't want to create a new function, i just want to rewrite

Array = arrayfun(@(n) circshift(Length, [0, -n]), 0:count-1, ... 'UniformOutput', false); Array = reshape([Array{:}], count, count);

that bit of code differently

[Andrei Bobrov]

:)

Array = hankel(1:27,[27,1:26]);