CODEWITHSUNDEEP
c++algorithmencryption
Awais Hussain
Awais Hussain

Reputation: 442

Encryption and decryption results are not same in TEA

I have implemented TEA in C++. Following is the code:

#include <stdio.h>
#include <iostream>
#include <iostream>
using namespace std;

int main()
{
    unsigned long y=24,z=32, sum=0, 
        delta=0x9e3779b9, n=16 ;
    int k[4] = { 0x28, 0xc0, 0x20, 0xd0 };
    unsigned long v[2] = {0,0};
    cout<<"original y is "<< y<<". original z is "<<z<<endl;
    while (n-->0) 
    { 
        sum += delta ;
        y += ((z<<4)+k[0]) ^ (z+sum) ^ ((z>>5)+k[1]) ;
        z += ((y<<4)+k[2]) ^ (y+sum) ^ ((y>>5)+k[3]) ;
    } 
    v[0]=y ; v[1]=z ;
    cout<<"encrypted y is "<< v[0]<<". encrypted z is "<<v[1]<<endl;
    n=16;
    y=v[0];
    z=v[1];         
    delta=0x9e3779b9 ;
    sum=delta<<5 ;
    /* start cycle */
    while (n-->0) 
    {
        z-= ((y<<4)+k[2]) ^ (y+sum) ^ ((y>>5)+k[3]) ;
        y-= ((z<<4)+k[0]) ^ (z+sum) ^ ((z>>5)+k[1]) ;
        sum-=delta ;
    }
        v[0]=y ; v[1]=z ;

    cout<<"decrypted y is "<< v[0]<<". decrypted z is "<<v[1]<<endl;    
    getchar();
    return 0;
}

But in the end, the results are not same i.e. the decrypted value is different than the original inputs (y and z). Please point out the mistake which I am unable to see.

Upvotes: 0

Views: 550

Answers (2)

CodesInChaos
CodesInChaos

Reputation: 108820

sum = n * delta

You compute sum as delta << 5 which corresponds to delta * 32. That works for the default value of n = 32 but you reduced n to 16 without fixing the computation of sum. Replace sum=delta<<5; by sum = delta * n.

Also note that reducing the number of rounds weakens the resistance against cryptoanalysis.

Upvotes: 3

melak47
melak47

Reputation: 4850

You set n to 16, while the example implementation on Wikipedia runs the loop 32 times.

If I change n to 32 in your code, it works: See it live on Coliru

Upvotes: 4

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