I really enjoyed playing this CTF with Spiderz team and we ended at position 23.
This reversing challenge was for 250 points , and here's a brief write-up about it :
The binary expects a string as a command line argument and it starts in the beginning by decrypting a string stored statically at .data section. If the position of the character is an even number the character is xored by 0xD4, if it's an odd number the character is xored with 0xD5.
After decrypting , we get the following large equation :
http://pastebin.com/1sU2B1fz
As you can see, each element a[x] is actually the character of position 'x' in the string that we'll supply. Similarly to the Satellite task we're deal with the boolean satisfiability problem .
If you take a second look at the long decrypted string we've got , you'll see that each character (with the exception of a[220] which not referenced anyway) of the string is referenced exactly 3 times and it is always tested against another character which is static in the 3 cases. So to solve this we'll just rely on studying each 2 characters couples alone.
For example to make this statement true :
(a[12] | a[20]) & ( ! a[12] | !a[20]) & (!a[12] | a[20])
a[12] must equal : 0
a [20] must equal : 1
Another example :
(!a[22] | a[150]) & (a[22] | a[150]) & (a[22] | !a[150])
In this case both a[22] and a[150] must be equal to 1 to make this statement true.
In the string shared in pastebin above you'll see that in some cases there's a double NOT (! !) , we'll just remove it as it doesn't change anything.
So to script this we don't basically need to study the SAT algorithm any longer, we can just separate this long string into 2 arrays. Each element of the 2 arrays is a logical equation (ex : "( ! a[22] | a[50] )" ).
The first array will only have the elements that have a NOT ('!') on the both chars or that doesn't have any NOTs in them (ex : ( a[55] | a[60] ) and this one ( ! a[70] | ! a[95] ) )
The other array will have all the equations that have a NOT preceding either one of the chars. (ex : ( ! a[22] | a[50] ).
The reason why I did this because in the case of the first example I gave (here it is :
(a[12] | a[20]) & ( ! a[12] | !a[20]) & (!a[12] | a[20]) ) there will be 2 occurrences of a[12] in the first array which makes it hard to decide whether it's equal to a 0 or 1 , here comes the 2nd array that I called "decide" that will decide by this equation : (!a[12] | a[20]) whether a[12] is 0 or 1 , which is 0 in this case.
So If only one instance of a given a[x] is found in the first array we can decide it's value directly , but if we have 2 instances we'll need to rely on the decide array.
Oh , I almost forgot , there's a character which isn't referenced in this string and which is a[220] . As the flag is generated based on our input, I tested the flag with this character set to 0 and 1. And it basically worked for 0.
Here's the script I wrote and described in this write-up (got us some bonus points though :D ) :
