How to apply pc1 in matlab and get answer in 56 bit keys

I want to apply pc1 in my matlab code and answer required is in binary form but i am getting decimal number. Can some one help me in this.

 
a=0;
b=17;
n=313;
A=[[36 6]; [117 3] ;[198 310] ;[68 86]];
k=size(A,1);
    temp=A(4,:)
    A11=multell(A(4,:),2,a,b,n);
PC1=[57,49,41,33,25,17,9,1,58,50,42,34,26,18,10,2,59,51,43,35,27,19,11,3,60,52,44,36,63,55,47,39,31,23,15,7,62,54,46,38,30,22,14,6,61,53,45,37,29,21,13,5,28,20,12,4]; %parity 
PC2=[14,17,11,24,1,5,3,28,15,6,21,10,23,19,12,4,26,8,16,7,27,20,13,2,41,52,31,37,47,55,30,40,51,45,33,48,44,49,39,56,34,53,46,42,50,36,29,32]; %key comp    
A(5,:)=addell(A11,A(1,:),a,b,n);
    A(6,:)=addell(A(5,:),A(2,:),a,b,n);
    A(7,:)=addell(A(6,:),A(3,:),a,b,n);
    A(8,:)=addell(A(7,:),A(4,:),a,b,n);
    % k=k+4;
% end
 B=[A(5,:); A(6,:); A(7,:); A(8,:)];
 B1=dec2bin(B)
 key1=reshape(B1,1,64)
 Key1=PC1(key1)
 
 
Key1 =

  Columns 1 through 13

    37    37    29    37    37    29    29    29    37    37    29    29    29

  Columns 14 through 26

    37    37    37    37    37    37    29    37    37    37    29    29    29

  Columns 27 through 39

    37    29    37    29    29    29    29    37    37    29    29    29    29

  Columns 40 through 52

    37    37    37    37    37    29    29    29    29    29    37    29    37

  Columns 53 through 64

    29    37    37    29    29    29    29    29    29    37    37    29

[Walter Roberson]

This appears to relate to https://www.mathworks.com/matlabcentral/answers/472880-can-i-fit-a-loop-in-my-code and addell appears to be addition on elliptic curve.

Answer 1

[Walter Roberson]

The result of dec2bin() is '0' and '1' not 0 and 1. '0' is char(48) and '1' is char(49), so when you use the '0' and '1' as indices in PC1(key1) you are indexing at locations 48 and 49.

I doubt that you should be applying parity check on one bit at a time. You should probably be taking a smaller number of bits in a group, computing a decimal equivalent, and using that to look up an entry in the parity check matrix.

[Walter Roberson]

Note that we cannot assist you with encryption theory for legal reasons.

[sadiqa ilyas]

thank u very much. I was using PC1 in reverse order. It should be key(PC1). your answere clarify my concept.