Could this post be any less related to mIRC? Well; lets see. I could talk about zebras.
Now; intrepid C coders; I am doing a course. That course features some rather nasty lecturers. Occasionally we get something fun. I thought this was both fun and hard; I sitll haven't pulled it all to bits in my head (I don't know all of the keys; which was required; but I got both messages.)
Note: Every C/C++ room I tried to get help in more or less told me "your program is a pile of dog crap". They couldn't or wouldn't grasp the concept of a program that's not meant to work.
On with the show.
---
You have just intercepted an encoded message. The message is a sequence of bits which reads as follows in hexadecimal:
6363636363636363724646636F6D6F72
466D203A65693A7243646E206F54540A
5920453A54756F0A6F6F470A21643A6F
594E2020206F776F797275744563200A
6F786F686E6963736C206765796C656B
2C3365737420346E20216F74726F5966
7565636F202061206C61676374206C6F
20206F74747865656561727632727463
6E617920680A64746F69766120646E69
21687467630020656C6C786178742078
6578206F727478787863617800783174
You have no idea how to decode it, but you know that your grade depends on it, so you are willing to do anything to extract the message. Fortunately, one of your many agents on the field has stolen the source code for the decoder. This agent (007) has put the code and the message in the file secret.cpp.
Q has noticed that the decoder takes four integers as arguments. Executing the decoder with various arguments seems to either crash the program or produce unintelligible output. It seems that the correct four integers have to be chosen in order for the program to produce the decoded message. These four integers are the "secret keys."
007 has been unable to find the keys, but from the desk of the encrypting personnel he was able to cunningly retrieve the first five characters of the unencoded message. These characters are:
From:
Assignment
Your assignment is to decode the message, and find the keys.
This exercise is not extremely difficult. However, the strategy of trying things until something works will be ineffective. Try to understand the material in the course, particularly the following:
Strategy
The designers of this decoder weren't very good. They made it possible for us to attack the keys in two independent parts. Try to break the first two keys first, and do not try to break the third and fourth keys until you have succeeded with the first two.
---
#include <stdio.h>
#include <stdlib.h>
int prologue [] = {
0x5920453A, 0x54756F0A, 0x6F6F470A, 0x21643A6F,
0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
0x20206F74, 0x74786565, 0x65617276, 0x32727463,
0x594E2020, 0x206F776F, 0x79727574, 0x4563200A
};
int data [] = {
0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
0x466D203A, 0x65693A72, 0x43646E20, 0x6F54540A,
0x5920453A, 0x54756F0A, 0x6F6F470A, 0x21643A6F,
0x594E2020, 0x206F776F, 0x79727574, 0x4563200A,
0x6F786F68, 0x6E696373, 0x6C206765, 0x796C656B,
0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
0x20206F74, 0x74786565, 0x65617276, 0x32727463,
0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
0x21687467, 0x63002065, 0x6C6C7861, 0x78742078,
0x6578206F, 0x72747878, 0x78636178, 0x00783174
};
int epilogue [] = {
0x594E2020, 0x206F776F, 0x79727574, 0x4563200A,
0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
0x20206F74, 0x74786565, 0x65617276, 0x32727463
};
char message[100];
void usage_and_exit(char * program_name) {
fprintf(stderr, "USAGE: %s key1 key2 key3 key4\n", program_name);
exit(1);
}
void process_keys12 (int * key1, int * key2) {
*((int *) (key1 + *key1)) = *key2;
}
void process_keys34 (int * key3, int * key4) {
*(((int *)&key3) + *key3) += *key4;
}
char * extract_message1(int start, int stride) {
int i, j, k;
int done = 0;
for (i = 0, j = start + 1; ! done; j++) {
for (k = 1; k < stride; k++, j++, i++) {
if (*(((char *) data) + j) == '\0') {
done = 1;
break;
}
message[i] = *(((char *) data) + j);
}
}
message[i] = '\0';
return message;
}
char * extract_message2(int start, int stride) {
int i, j;
for (i = 0, j = start;
*(((char *) data) + j) != '\0';
i++, j += stride)
{
message[i] = *(((char *) data) + j);
}
message[i] = '\0';
return message;
}
int main (int argc, char *argv[])
{
int dummy = 1;
int start, stride;
int key1, key2, key3, key4;
char * msg1, * msg2;
key3 = key4 = 0;
if (argc < 3) {
usage_and_exit(argv[0]);
}
key1 = strtol(argv[1], NULL, 0);
key2 = strtol(argv[2], NULL, 0);
if (argc > 3) key3 = strtol(argv[3], NULL, 0);
if (argc > 4) key4 = strtol(argv[4], NULL, 0);
process_keys12(&key1, &key2);
start = (int)(*(((char *) &dummy)));
stride = (int)(*(((char *) &dummy) + 1));
if (key3 != 0 && key4 != 0) {
process_keys34(&key3, &key4);
}
msg1 = extract_message1(start, stride);
if (*msg1 == '\0') {
process_keys34(&key3, &key4);
msg2 = extract_message2(start, stride);
printf("%s\n", msg2);
}
else {
printf("%s\n", msg1);
}
return 0;
}
---
I was heaps proud to be able to do most of the work using nothing more than Borland's free command line tools. HURRAH!
If you manage to work out all the keys; I'll prolly pester you for an explanation of key2 as no one has of yet been able to explain it too well.
Last edited by ParaBrat; 11/08/03 04:14 PM.
