For
a = b ^ c
Since the XOR operation is trivial to reverse when any 2 values in a, b, c are known, this will only require you to just reverse the order in which operations were done. I had to do this in C since I don't use Java.
int *encode(int *text, int *key) {
int j = 0x0;
int i = 0x33; // value 51 in decimal
for (i = 0x33;;) {
text[4] ^= ((key[j++]) ^ (text[0])) + i;
text[5] ^= ((key[j++]) ^ (text[1])) + i;
text[6] ^= ((key[j++]) ^ (text[2])) + i;
text[7] ^= ((key[j++]) ^ (text[3])) + i;
text[1] ^= ((key[j++]) ^ (text[4])) + i;
if (j >= 0x3c)
break;
text[2] ^= ((text[4]) ^ (text[5])) + i;
text[3] ^= ((key[j++]) ^ (text[6])) + i;
text[0] ^= ((key[j++]) ^ (text[7])) + i;
}
return text;
}
Here's the full code that works.
#include <stdio.h>
int *decode(int *text, int *key) {
int j = 0x3c; // value 60 in decimal
int i = 0x33; // value 51 in decimal
for (i = 0x33;;) {
text[1] ^= ((key[j--]) ^ (text[4])) + i;
text[7] ^= ((key[j--]) ^ (text[3])) + i;
text[6] ^= ((key[j--]) ^ (text[2])) + i;
text[5] ^= ((key[j--]) ^ (text[1])) + i;
text[4] ^= ((key[j--]) ^ (text[0])) + i;
if (j <= 0)
break;
text[0] ^= ((key[j--]) ^ (text[7])) + i;
text[3] ^= ((key[j--]) ^ (text[6])) + i;
text[2] ^= ((text[4]) ^ (text[5])) + i;
}
return text;
}
int *encode(int *text, int *key) {
int j = 0x0;
int i = 0x33; // value 51 in decimal
for (i = 0x33;;) {
text[4] ^= ((key[j++]) ^ (text[0])) + i;
text[5] ^= ((key[j++]) ^ (text[1])) + i;
text[6] ^= ((key[j++]) ^ (text[2])) + i;
text[7] ^= ((key[j++]) ^ (text[3])) + i;
text[1] ^= ((key[j++]) ^ (text[4])) + i;
if (j >= 0x3c)
break;
text[2] ^= ((text[4]) ^ (text[5])) + i;
text[3] ^= ((key[j++]) ^ (text[6])) + i;
text[0] ^= ((key[j++]) ^ (text[7])) + i;
}
return text;
}
int key[] = {0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41,
0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41,
0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41,
0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41,
0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41,
0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41, 0x41};
int text[] = {0x61, 0x62, 0x63, 0x64, 0x65, 0x66, 0x67, 0x68};
int main(int argc, char **argv) {
int i, *d, *e;
for (i = 0; i < 8; i++) {
printf("%x::", text[i]);
}
putchar(10);
d = decode(text, key);
for (i = 0; i < 8; i++) {
printf("%x::", d[i]);
}
putchar(10);
e = encode(text, key);
for (i = 0; i < 8; i++) {
printf("%x::", e[i]);
}
putchar(10);
}
This produces the following output
61::62::63::64::65::66::67::68::
65a::3e9::64a::bd6::5c2::11::1a9::85e::
61::62::63::64::65::66::67::68::
I was able to get back the original array after a decode
and encode
.