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Codegate CTF 2015 pwn 1000 sokoban

Codegate CTF , Off-by-one

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The challenge is a game that called sokoban. lol
After we passed the first levels, we entered a menu and were able to choose the game mode.


The menu looked like:

  1. Go to random infinite challenge mode
  2. Go to next stage

Next, we opened IDA pro and reversed the binary.
We could easily find the code like:

if (playgame() == 1)
    get_flag();
puts(s);
return;

Then we traced how is the return value assigned......

// just pseudo code
if (win)
  passed++;
return (passed == 228) ? 2 : 0;

228 is the amount of all levels.
It seems impossible to arrive get_flag(). XD
But our goal is very clear, control the EIP and go to get_flag().

We accidentally found the game sometimes generates a blank map in random mode then checks the rule of movement, it restricts the character by the element in the map, not the size of map.
Therefore, once we could get the blank map, we were able to move the character to anyware in bss segment and GOT segment.

There is the defination of elements:

  • \x00: nothing
  • \x01: destination of box
  • \x02: wall
  • other: it's not important.

According to the rule of sokoban, we could push a byte onto \x00 or \x01.
It's very difficult to use......
I tried to move the content of GOT at first, but I found GOT looks like:

time@got.plt: 0xf7ffafa0 0x00007fff 0x00400dd6 0x00000000
wgetch@got.plt: 0xf7bc2f90 0x00007fff 0x00400df6 0x00000000
noecho@got.plt: 0xf7bc0a50 0x00007fff 0x00400e16 0x00000000
wmove@got.plt: 0xf7bc4e40 0x00007fff 0xf799de70 0x00007fff
mvprintw@got.plt: 0xf7bc7db0 0x00007fff 0xf7bc0ad0 0x00007fff

Almost all bytes are adjacent to each other.
Therefore, we couldn't change the GOT area at most situation except ASLR was enable.
For example, it's possible to make a libc address likes 0x7fffff00xx.
So we could modify a byte on GOT to somewhere in libc.

Still seem useless....
But after I checked all possible gadgets, I found a magic gadget at 0x3e260.
That is add rsp, 0x28; (pop XX)*6; ret.
Furthermore, the address of rand() is 0x3d060.
If we modify rand() to that magic gadget, the return address is 0x401a9a after we execute rand() again.
Luckily, there are a hidden function in the game.
If we press v, it will add 0x12 on 0x60c120.
And, 0x3e260 - 0x3d060 = 0x12......

So, hence we had already bypassed the action of assign value to EAX.
If we could control EAX and set EAX = 1, we entered the function get_flag().
Lucklily, if the argument of wgetch is \x00, the return value will be 1.
On x86 architecture, the return value will be stored in EAX.
EAX won't be modified until we call rand().
Finally, the program will print the flag. :)

flag: WH0n in OOme, ZZ as 12e RolanS