CODEWITHSUNDEEP
matlab
Hesham
Hesham

Reputation: 41

RSA variable overflow through multiplication

I am trying to write a code for RSA Algorithm, my code produce (e,d,n) correctly, but the problem occurs when i calculate the cipher. cipher = (plain^e) mod n. because (plain^e) is a very large number. the Calculator gives my correct result, while Matlab Don't. any one have ideas?

clc
e=61;
d=21;
n= 187;
Neuler=160;
Plain='ABCD';
Plain=uint64(Plain);
%%Encrypting Progress
for i=1:length (Plain);
Cypher(i)=mod((Plain(i)^e),n);
end
Numeri_Cypher=uint64(Cypher);
for i=1:length (Numeri_Cypher);
RPlain(i)=mod((Numeri_Cypher(i)^d),n);
end
Result=char(RPlain)

Upvotes: 1

Views: 86

Answers (1)

obchardon
obchardon

Reputation: 10792

The built-in function mod can't deal with big integer. So I created a small implementation of the modular exponentiation, if you use this function you should have no problem.

function result = modpow(base,exp,m) 
   result = 1;
   while (exp > 0) 
      if bitand(exp,1) > 0
         result = mod((result * base),m);
      end
      exp = bitshift(exp,-1);
      base = mod(base^2,m);
   end
end

EXAMPLE:

With the built-in mod function:

mod(3^233249,4)

ans = 0 %incorrect result

With the modpow function

modpow(3,233249,4) 

ans = 1 %correct result

Upvotes: 2

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