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

Question

sadiqa ilyas 2019-7-28

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/473771-how-to-apply-pc1-in-matlab-and-get-answer-in-56-bit-keys

评论： sadiqa ilyas 2019-7-31

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

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Walter Roberson 2019-7-28

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.

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Answer 1

Walter Roberson 2019-7-28

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/473771-how-to-apply-pc1-in-matlab-and-get-answer-in-56-bit-keys#answer_385130

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.