Basic Shift Cipher Decryption Algorithm HELP!
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Hello guys, I'm using matlab to make a function that basically decrypts a shift cipher by taking in the ciphertext string and key integer as parameters and returning the plaintext.
here is the code..
function [ plainText ] = ccdt( c, k )
s = double(c);
for i = 1:numel(s)
s(i) = s(i)-k;
end
plainText = char(s);
return
end
This works fine when the letters don't necessarily need to loop back around to the beginning of the alphabet. For example, if I decrypt the letter 'b' with key = 3, it should give me back 'y', but it's just returning whatever ascii code for 'b' minus 3 which isn't 'y'.
How can I fix this problem? also, how can i modify the code so that lower/ upper case letters don't really matter?
Thanks for your time!
1 commentaire
barjas abu matar
le 3 Nov 2017
i need code basic shift cipher in decryption
Réponse acceptée
Plus de réponses (2)
Aman Gupta
le 27 Juin 2019
function coded = caesar(x,n)
a = double(x);
a = a+n;
for i = 1:length(a)
while (a(i)>126 || a(i)<32)
if a(i)>126
a(i) = 31 + (a(i) - 126);
elseif a(i)<32
a(i) = 127 - (32 - a(i));
end
end
end
coded = char(a);
Rahul Gulia
le 22 Juil 2019
0 votes
function coded = caesar(str,n)
num1 = double(str); %Converting string to double to make the defined shifts
for i = 1 : length(num1)
if num1(i) + n > 126 % If ASCII value goes beyond 126
m = num1(i)-126+n;
p = 31+m;
num1(i) = p;
elseif num1(i)+n < 32 % If ASCII value goes below 32
m = 32 - num1(i) + n;
p = 126 - m;
num1(i) = p;
else m = num1(i) + n; % In a normal condition
num1(i) = m;
end
code(i) = num1(i);
end
coded = char(code);
% I have written this code. Can anyone please expain as what is wrong in here? I know i have made a mistake. But i am not able to figure it out.
% Thanks in advance.
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