explaination of the code

pattern=['00110011';
    '00110101';
    '00110110';
    '00111001';
    '00111010';
    '01010011';
    '01010101';
    '01010110';
    '01011001';
    '01011010';
    '01011100';
    '01100011';
    '01100101';
    '01100110';
    '01101001';
    '01101010';
    '01101100';
    '10010011';
    '10010101';
    '10010110';
    '10011100';
    '10100011';
    '10100101';
    '10100110';
    '10101001';
    '10101010';
    '10101100';
    '11000011';
    '11000101';
    '11000110';
    '11001001';
    '11001010';
    '11001100']

chars={'a','b','c','d','e','f','g','h','i','j','l','m','o','p','q','r','s','t','u','w','x','y','z','0','2','3','4','5','6','7','8','9'};
pv=input('enter password');
[fname path]=uigetfile('*.jpg','enter an image to be encrypted');
fname=[path,fname];
im=imread(fname);
imshow(im);
title('actual image');
[x y z]=size(im);
for(i=1:length(pv))
  a=pv(1);
  a=find(chars==a);
  a=pattern{a,1};
 num(1)=bin2dec(a);
end

sir ,in the above code, pv is the text password and for each character a specific pattern of binary value is assigned and that binary value should be converted in to decimal value. num is the variable. i am doubtful at the step a=pattern{a,1}; please explain kindly

[Image Analyst]

http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup

Answer 1

[Walter Roberson]

The code is wrong because it uses pv(1) and num(1) instead of pv(i) and num(i) . The "find" is okay.