CSAW CTF 2014 - "saturn" Exploitation 400 Write-up

Hi,

The description for this task was :

    You have stolen the checking program for the CSAW Challenge-Response-Authentication-Protocol system. Unfortunately you forgot to grab the challenge-response keygen algorithm (libchallengeresponse.so). Can you still manage to bypass the secure system and read the flag?

    nc 54.85.89.65 8888

I grabbed the binary , threw it in IDA and then started looking at the main routine. The first function that was called in main was _fillChallengeResponse and it takes two arguments . I named them : fill_arg0 and fill_arg4.
A quick check reveals that this function is imported from an external library (the one we 'forgot' to grab). Also by checking the arguments passed to the function they appear to be pointers , each pointer points to a 32 bytes array in the bss section.We can also see that the first array is directly followed by the next one.

As fillChallengeResponse is given 2 pointers , we can safely guess that its mission is to fill them with the right data.

Let's carry on :

Next, we will enter this loop. Its was previously initialized to 0 and we'll quit the loop only if the iterator is strictly above 0. In this loop, we are first prompted to supply an input in which only the first byte is read , the byte is saved at [esp+1Bh] and the switch statement only uses the highest order nibble of the read byte.
If the switch statement was supplied 0xA0 , it will lead to retrieving the original read byte (0xA2 for example) and then call a function that will access the Array1 and print the dword at the index described by the lowest order nibble of the read byte multiplied by 4 ((0xA2 & 0xF)*4 = 8 for example).
If the switch statement was supplied 0xB0 , the executed block of code will retrieve the original read byte and then call a function that will wait for user input and then compare that input to the dword indexed by the lowest orded nibble of the original byte multiplied by 4 in Array2. If the 2 values are equal another 8 sized array of bytes will be accessed and 1 is written into the same index indicated by the lowest order nibble.
If the switch statement was supplied 0x80 , it will call a function that walk through the array of bytes checking if all the elements are equal to 1. If it's the case , the function will print the contents of "flag.txt".

The trick here is to take advantage of the read_array1 function , to make it print the Array2 and then pass each dword read from Array2 to the check_array2 function. As we already know Array1 and Array2 are sticked to each other and each ones size is 16 bytes this means that supplying 0xA8 will make us read the first dword of the Array2 . So all we need to do is supply 0xA8 as an input , save the printed value from read_array1 function , supply 0xE0 as an input (switch) then supply the saved printed value as a input (in check_array2) , this will result in setting the first byte of the 8 bytes sized array to 1.
We have to  basically repeat the same 8 times , 0xA8 -> 0xAF and 0xE0 -> 0xE8. When done , we'll supply 0x80 as an input and the "target" function will print the flag for us.
Here's an automated python script which prints the flag :

Binary download : Here

Follow me on twitter : Here
Bye.
- Souhail