jeudi 1 décembre 2011

GCHQ Challenge Part 1 - http://www.canyoucrackit.co.uk/

Hello,

As I have seen that many people already posted their solutions ... I do not see the point of keeping mine :). Here it is.

Today I stumbled upon a challenge that seems to come from GCHQ itself.
GCHQ is basically part of the UK's Secret Service.

An article describing a bit the recruitment campaign:
GCHQ challenges codebreakers via social networks

Anyway, the challenge is located here:
http://www.canyoucrackit.co.uk/

The challenge is divided in 3 parts, the first part of which we see on the front page of the website.

Let's break it!


The challenge

We have a picture with hex digits:




From there, I've got 2 theories of what that is:
- raw data
- machine code

First of all, we need to get those "hex digits" down to binary form and hell yeah ... we're lazy:
$ gocr cyber.png

You get most of the digits but you have to fix it to get this:

eb 04 af c2 bf a3 81 ec  OO 01 OO OO 31 c9 88 Oc
Oc fe c1 75 f9 31 cO ba  ef be ad de 02 04 Oc OO
dO c1 ca 08 8a 1c Oc 8a  3c 04 88 1c 04 88 3c Oc
fe c1 75 e8 e9 5c OO OO  OO 89 e3 81 c3 04 OO OO
OO 5c 58 3d 41 41 41 41  75 43 58 3d 42 42 42 42
75 3b 5a 89 d1 89 e6 89  df 29 cf f3 a4 89 de 89
d1 89 df 29 cf 31 cO 31  db 31 d2 fe cO 02 1c 06
8a 14 06 8a 34 1e 88 34  06 88 14 1e OO f2 30 f6
8a 1c 16 8a 17 30 da 88  17 47 49 75 de 31 db 89
d8 fe cO cd 80 9O 9O e8  9d ff ff ff 41 41 41 41

Okay, let's check the raw data we have:

$ hexdump -C gchq-part1.bin 
00000000  eb 04 af c2 bf a3 81 ec  00 01 00 00 31 c9 88 0c  |............1...|
00000010  0c fe c1 75 f9 31 c0 ba  ef be ad de 02 04 0c 00  |...u.1..........|
00000020  d0 c1 ca 08 8a 1c 0c 8a  3c 04 88 1c 04 88 3c 0c  |........<.....<.|
00000030  fe c1 75 e8 e9 5c 00 00  00 89 e3 81 c3 04 00 00  |..u..\..........|
00000040  00 5c 58 3d 41 41 41 41  75 43 58 3d 42 42 42 42  |.\X=AAAAuCX=BBBB|
00000050  75 3b 5a 89 d1 89 e6 89  df 29 cf f3 a4 89 de 89  |u;Z......)......|
00000060  d1 89 df 29 cf 31 c0 31  db 31 d2 fe c0 02 1c 06  |...).1.1.1......|
00000070  8a 14 06 8a 34 1e 88 34  06 88 14 1e 00 f2 30 f6  |....4..4......0.|
00000080  8a 1c 16 8a 17 30 da 88  17 47 49 75 de 31 db 89  |.....0...GIu.1..|
00000090  d8 fe c0 cd 80 90 90 e8  9d ff ff ff 41 41 41 41  |............AAAA|

I do not recognize any particular magic header value (no MZ or ELF or any executable related.
Let's go on our second hypothesis.

Is it machine code?

The first thing that set up red flags about machine code is the first byte: 0xeb!
0xeb for anyone used to shellcode is the opcode corresponding to an inconditionnal jmp on x86 processors.

I believe there must be some construct like this (as usual) or not too far from it:
jmp savepc

getpc:
     pop pc
     jmp payload

savepc:
     call getpc

payload:

For disassembling the file, I used ndisasm and cleaned a bit with some custom tools and vim.

Let's see the reconstructed code:

bits 32

section .text
    global main

main:
    jmp short begin

key0: dd 0xa3bfc2af

; let's make some space for our buffer!
begin:
    sub esp, 0x100

; we init our buffer with value from 0 to 255
xor ecx, ecx
init_array:
    mov [esp+ecx], cl
    inc cl
    jnz init_array

; shuffle values in the array
xor eax, eax
mov edx, 0xdeadbeef
shuffle:
    add al, [esp+ecx]   ; al += array[ecx]
    add al, dl          ; index = al + dl
    ror edx, 0x8        ; every 4 rotations we get our original value ;)
    ; we swap values
    mov bl, [esp+ecx]   ; bl = array[ecx]
    mov bh, [esp+eax]   ; bh = array[ecx]
    mov [esp+eax], bl   ; array[eax] = bl
    mov [esp+ecx], bh   ; array[ecx] = bh
    inc cl              ; go forward in our array
    jnz shuffle

    jmp dword save_pc

get_encoded:
    mov ebx, esp                ; pointer ebx to ret address on stack
    add ebx, strict dword 0x4   ; address of array (ignoring the ret address on the stack ;))
    pop esp                     ; esp = program counter (point after the call to get_encoded)
    pop eax                     ; first 4 bytes after the call
    cmp eax, 0x41414141
    jnz exit

    pop eax                     ; next 4 bytes after the call
    cmp eax, 0x42424242
    jnz exit

    ; copy message to buffer in stack
    pop edx         ; get length of message 
    mov ecx, edx    ; ecx = len(msg)
    mov esi, esp    ; esi = &msg
    mov edi, ebx    ; edi = address of array
    sub edi, ecx    ; edi = ebx - len(msg) = start of dest area
    rep movsb       ; copying the message to the stack (writing over array)

; init for decoding encoded message
    mov esi, ebx    ; esi = &buffer
    mov ecx, edx    ; ecx = len(msg)
    mov edi, ebx    ; edi = &buffer
    sub edi, ecx    ; edi = ebx - len(msg) = start of dest area
    xor eax, eax
    xor ebx, ebx
    xor edx, edx

; loop for decoding the secret message
decode:
    inc al
    add bl, [esi+eax]   ; get one byte of encoded message
    mov dl, [esi+eax]   ; get one byte of encoded message
    mov dh, [esi+ebx]   ; get one byte of encoded message
    ; swap values back
    mov [esi+eax], dh
    mov [esi+ebx], dl
    add dl, dh          ; get index
    ; decode byte
    xor dh, dh          
    mov bl, [esi+edx]   ; get key
    mov dl, [edi]       ; get encoded byte
    xor dl, bl          ; decode byte
    ; save byte
    mov [edi], dl
    ; loop
    inc edi
    dec ecx
    jnz decode

exit:
    xor ebx, ebx
    mov eax, ebx
    inc al
    int 0x80

save_pc:
    nop
    nop
    call dword get_encoded

junk1: dd 0x41414141

The first part (init_array and shuffle) looks like some RC4 variant.
Then we decode it using our RC4 variant ... but wait ... there is data missing!
Where is the encoded message?

If you look at the beginning of the PNG file you can see an encoded base64 string:

00000000  89 50 4e 47 0d 0a 1a 0a  00 00 00 0d 49 48 44 52  |.PNG........IHDR|
00000010  00 00 02 e4 00 00 01 04  08 02 00 00 00 ef 6a b6  |..............j.|
00000020  2d 00 00 00 01 73 52 47  42 00 ae ce 1c e9 00 00  |-....sRGB.......|
00000030  00 09 70 48 59 73 00 00  0b 13 00 00 0b 13 01 00  |..pHYs..........|
00000040  9a 9c 18 00 00 00 07 74  49 4d 45 07 db 08 05 0e  |.......tIME.....|
00000050  12 33 7e 39 c1 70 00 00  00 5d 69 54 58 74 43 6f  |.3~9.p...]iTXtCo|
00000060  6d 6d 65 6e 74 00 00 00  00 00 51 6b 4a 43 51 6a  |mment.....QkJCQj|
00000070  49 41 41 41 43 52 32 50  46 74 63 43 41 36 71 32  |IAAACR2PFtcCA6q2|
00000080  65 61 43 38 53 52 2b 38  64 6d 44 2f 7a 4e 7a 4c  |eaC8SR+8dmD/zNzL|
00000090  51 43 2b 74 64 33 74 46  51 34 71 78 38 4f 34 34  |QC+td3tFQ4qx8O44|
000000a0  37 54 44 65 75 5a 77 35  50 2b 30 53 73 62 45 63  |7TDeuZw5P+0SsbEc|
000000b0  59 52 0a 37 38 6a 4b 4c  77 3d 3d 32 ca be f1 00  |YR.78jKLw==2....|
000000c0  00 20 00 49 44 41 54 78  da ec bd 79 74 1c d5 b5  |. .IDATx...yt...|

The base64 string "QkJCQjIAAACR2PFtcCA6q2eaC8SR+8dmD/zNzLQC+td3tFQ4qx8O447TDeuZw5P+0SsbEcYR78jKLw=" decodes to the following:

junk2: dd 0x42424242
msg_length: dd 0x00000032
msg_encoded: db `\x91\xd8\xf1\x6d\x70\x20\x3a\xab\x67\x9a\x0b\xc4\x91\xfb\xc7\x66\x0f\xfc\xcd\xcc\xb4\x02\xfa\xd7\x77\xb4\x54\x38\xab\x1f\x0e\xe3\x8e\xd3\x0d\xeb\x99\xc3\x93\xfe\xd1\x2b\x1b\x11\xc6\x11\xef\xc8\xca\x2f`


It is a NASM source which correspond bit to bit to the original gchq.bin ;).:
$ nasm gchq.asm
So the first challenge basically consist in executing the code and dumping the decoded string.

Let's dump the decoded string!

For this part, I did no want to patch the binary code as I would deem it inelegant.
Instead, I chose to write a "launcher" and ptrace our gchq code :).

Here it is:


// @author  m_101
// @year    2011
// @desc    Program for launching GCHQ code
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>

#include <string.h>

// for memalign, mprotect
#include <sys/mman.h>
#include <errno.h>
#include <malloc.h>
#include <signal.h>

// ptrace
#include <sys/ptrace.h>
// for registers
#include <sys/user.h>

#define __NR_exit   1
#define MSG_LENGTH  0x32

// launch a file
int launch_file (char *filename) {
    //
    char *mem;
    void (*func)();
    // file related
    FILE *fp;
    int szFile;

    if (!filename)
        return -1;

    // open file
    fp = fopen(filename, "r");
    if (!fp) {
        fprintf(stderr, "error: Failed opening (r): %s\n", filename);
        exit(1);
    }

    // get file size
    fseek(fp, 0, SEEK_END);
    szFile = ftell(fp);
    fseek(fp, 0, SEEK_SET);
    printf("[+] File size: %d\n", szFile);

    // alloc aligned memory for file
    mem = memalign(PAGE_SIZE, szFile * sizeof(*mem));
    if (!mem) {
        printf("[-] error: %s\n", strerror(errno));
        return 1;
    }
    memset(mem, 0, szFile * sizeof(*mem));

    // fill mem
    fread(mem, sizeof(*mem), szFile, fp);

    // set permissions
    if (mprotect(mem, szFile * sizeof(*mem), PROT_READ | PROT_WRITE | PROT_EXEC)) {
        printf("[-] error: %s\n", strerror(errno));
        return 1;
    }

    // close file
    fclose(fp);

    // execute code
    printf("[+] Executing code at address %p\n", mem);
    func = (void *) mem;
    func();

    return 0;
}

void regs_show (struct user_regs_struct *regs) {
    if (!regs)
        return;

    printf("eax: 0x%08lx\n", regs->orig_eax);
    printf("ebx: 0x%08lx\n", regs->ebx);
    printf("ecx: 0x%08lx\n", regs->ecx);
    printf("edx: 0x%08lx\n", regs->edx);
    printf("esi: 0x%08lx\n", regs->esi);
    printf("edi: 0x%08lx\n", regs->edi);
    printf("ebp: 0x%08lx\n", regs->ebp);
    printf("esp: 0x%08lx\n", regs->esp);
    printf("eip: 0x%08lx\n", regs->eip);
    printf("eflags: 0x%08lx\n", regs->eflags);
    printf("cs: 0x%08lx\n", regs->xcs);
    printf("ds: 0x%08lx\n", regs->xds);
    printf("es: 0x%08lx\n", regs->xes);
    printf("fs: 0x%08lx\n", regs->xfs);
    printf("gs: 0x%08lx\n", regs->xgs);
    printf("ss: 0x%08lx\n", regs->xss);
}

int main (int argc, char *argv[]) {
    int retcode;
    pid_t cpid;
    struct user_regs_struct regs = {0};
    int cstatus;
    // for ptrace
    int syscall;
    // dump memory of child process
    char decoded[MSG_LENGTH * 2] = {0};
    uint32_t word;
    uint32_t addr;
    int offset;


    // check number of arguments
    if (argc < 2) {
        printf("Usage: %s filename\n", argv[0]);
        return -1;
    }

    cpid = fork();
    // child
    if (cpid == 0) {
        ptrace(PTRACE_TRACEME, 0, NULL, NULL);
        raise(SIGTRAP);
        launch_file(argv[1]);
    }
    // parent
    else if (cpid > 0) {
        // wait for child
        waitpid(cpid, &cstatus, 0);

        // trace the process to arrive to the exit syscall
        do {
            // get at syscall entry
            ptrace(PTRACE_SYSCALL, cpid, NULL, NULL);
            waitpid(cpid, &cstatus, 0);

            // get registers of child process
            ptrace(PTRACE_GETREGS, cpid, NULL, ®s);
            syscall = regs.orig_eax;

            // we are at exit syscall
            // then we finish the loop and do more processing
            if (syscall != __NR_exit) {
                // wait for syscall to complete
                // and avoid exiting process
                ptrace(PTRACE_SYSCALL, cpid, NULL, NULL);
                waitpid(cpid, &cstatus, 0);
            }
        } while (syscall != __NR_exit);

        // get registers of child process
        ptrace(PTRACE_GETREGS, cpid, NULL, ®s);
        printf("== Registers\n");
        regs_show(®s);

        // dump decoded string
        addr = regs.esi - MSG_LENGTH;
        for (offset = 0; offset < MSG_LENGTH; offset += 4) {
            word = ptrace(PTRACE_PEEKDATA, cpid, addr + offset, NULL);
            // printf("word: 0x%08x\n", word);
            *((uint32_t *)(decoded+offset)) = word;
        }
        printf("\n[+] Decoded string: '%s'\n", decoded);

        // continue process (so it exits)
        ptrace(PTRACE_CONT, cpid, NULL, NULL);
    }
    // error
    else {
        printf("Failed fork\n");
    }

    return 0;
}

Now compile it using this command line:
$ gcc -m32 -g -o launcher launcher.c

You can now launch it:
$ ./launcher ./gchq 
[+] File size: 218
[+] Executing code at address 0x832f000
[+] Decoded string: 'GET /15b436de1f9107f3778aad525e5d0b20.js HTTP/1.1'

Pawned :).

Conclusion

Here we saw the first challenge of the GCHQ recruitment campaign, not too hard or not to bad for a first level :).

To be continued on the next level ;),

m_101

- My solutions: here

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