 # Google CTF 2017 - Counting (RE)

We’re given two files: an ELF binary counter, and some data code. The goal of the challenge is then to find the output of ./counter 9009131337. Keywords: reverse engineering, virtual machine, hailstone sequence, fibonnaci modulo

### Description

This strange program was found, which apparently specialises in counting. In order to find the flag, you need to find what the output of ./counter 9009131337 is.

File1: counter

File2: code

### Details

Points: 246

Category: reverse

Validations: 33

### Solution

#### Part 1 - Initial analysis

We’re given 2 files: a stripped x64 ELF counter, and some data code:

$file code counter code: data counter: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.24, BuildID[sha1]=ac70b7c58cc7989f829c0f0d50431ea0a92cbefb, stripped  Our goal is to find the output of counter for 9009131337, however testing it for small inputs like 10, 20, 30, 40 already takes quite some time… $ time ./counter 40
CTF{0000000000000280}

real	2m1.388s
user	2m0.744s
sys	0m0.008s


Fortunately, opening counter in IDA reveals a pretty small quantity of code. It’s mainly composed of 3 functions:

• the main function
• a 2nd function reading and parsing the content of code
• a 3rd function iterating over the read content and performing some computations

This scheme very much evokes a virtual machine, where code would be a program, and counter its interpreter. This is confirmed by the reverse engineering, that reveals the following pseudo-algorithm for the main function:

int main(int argc, argv) {
read_and_parse_code();

uint64_t *registers = malloc(26 * sizeof(uint64_t));
memset(registers, 0, 26 * sizeof(uint64_t);
registers = strtol(argv, NULL, 10);

uint32_t entry_point = 0;
execute_code(registers, entry_point)

printf("CTF{%016llx}\n", registers);
return 0;
}


#### Part 2 - Virtual machine

Analyzing the parsing function (which I named read_and_parse_code above) gives us initial information on the VM instructions format. It also contains strings like "Invalid reg" or "Invalid ins", that add semantic for some checked data. In the end we can deduce the following:

• each instruction is of size 0x10
• there are only 3 types of instructions: 0, 1, and 2
• there are 26 registers (from R0 to R25)
• some fields for an instruction are named next, reg, or amo
• instructions have the following fields, depending on their type:
• all instructions: type (+0x0)
• type 0: reg (+0x4) and next (+0x8)
• type 1: reg (+0x4), next1 (+0x8), and next2 (+0xC)
• type 2: amo (+0x4), next1 (+0x8), and next2 (+0xC)
• other offsets are unused

Inspecting the execution function (which I named execute_code above) then allows us to find the implementation of these 3 instructions:

• instruction type 0 is INC_THEN_GOTO:
• inconditionnaly increment the pointed register
• then go to the next instruction
  regs[ instr.reg ] += 1;
pc = instr.next;

• instruction type 1 is IF_DEC_THEN:
• if the pointed register is not 0, decrement it and go to next1
• otherwise, only go to next2
  if( instr.reg ] > 0 ) {
regs[ instr.reg ] -= 1;
pc = instr.next1;
} else {
pc = instr.next2;
}

• instruction type 2 is slightly more complicated:
• copy the registers values in a newly allocated array
• recursively call execute_code with entry point next1
• import amo numbers of registers from the results of the call (meaning a function can have several return values)
• then go to next2
  memcpy(new_regs, regs, 26 * sizeof(uint64_t));
execute_code(new_regs, instr.next1);
memcpy(regs, new_regs, instr.amo * sizeof(uint64_t));
pc = entry.next2;


It’s a sub-function CALL!

Quite unusual. But we now have all the needed information to understand what code implements.

Before starting, we can still deduce a few important things from execute_code:

• a function can have several return values (specified in the amo field of a CALL instruction, and returned in registers {R0..RN})
• execution of a function returns when pc == count_instr where count_instr is the total number of instructions in code (0x77).

#### Part 3 - VM program

The program in code contains 0x77 (119) instructions. Displaying them in hex after removing the unused parts, gives something that is good enough to analyze:

| PC  | TYPE    | REG/AMO | NEXT1  | NEXT2
-------------------------------------------------
(:00, '0100 .... 0000 .... 01000000 02000000')     > IF_DEC_THEN       R0, :01, :02
(:01, '0000 .... 0100 .... 00000000 ........')     > INC_THEN_GOTO     R1, :01
(:02, '0000 .... 0200 .... 03000000 ........')     > INC_THEN_GOTO     R2, :03
(:03, '0000 .... 0200 .... 04000000 ........')     > INC_THEN_GOTO     R2, :04
...
(:0d, '0200 .... 0100 .... 6c000000 0e000000')     > CALL              {R0}, :6c, :0e
...
(:10, '0100 .... 0200 .... 10000000 11000000'),    > IF_DEC_THEN       R2, :10, :11
(:11, '0100 .... 0000 .... 12000000 13000000'),    > IF_DEC_THEN       R0, :12, :13
(:12, '0000 .... 0200 .... 11000000 ........'),    > INC_THEN_GOTO     R2, :11
...
(:31, '0200 .... 0200 .... 5c000000 32000000')     > CALL              {R0..R1}, :5C, :32
...
(:75, '0100 .... 0100 .... 76000000 77000000')
(:76, '0000 .... 0000 .... 75000000 ........')


Static analysis

By manualy analysing the hex-displayed program, we learn several things:

• Functions:
• We can split the program in sub-functions by looking at the next1 field of all the CALL instructions
• We can also learn how many return values these sub-functions have by looking at the amo field of the CALL instructions

For instance, the 2 CALLs instructions above at :0D and :31 indicate that there’s a sub-function at :6C that have only one return value (returned in {R0}), and another one at :5C that have 2 return values (returned in {R0..R1})

We thus identify 9 sub-functions (in addition to the main af :00), at offsets :14, :1D, :2D, :40, :54, :5C, :63, :6C and :71, all having a unique return value, except for :5C that has 2.

• Patterns:

We can identify instruction patterns for commonly used operations:

• A IF_DEC_THEN jumping to itself can be understood as “decrement reg until it’s 0”: it is thus SET reg=0.

  :selfpc     IF_DEC_THEN  <reg>, :selfpc, <next>


For instance, the above instruction at :10 is SET R2=0

• The following IF_DEC_THEN + INC_THEN_GOTO couple can be understood as “increment regY until regX– is 0”. It is thus ADD regY, regX.

  :pc1    IF_DEC_THEN    <regX>, :pc2, <next>
:pc2    INC_THEN_GOTO  <regY>, :pc1


For instance, the above couple at :11 and :12 is ADD R2, R0

• Chaining these 2 first patterns (SET regX, 0 + ADD regX, regY) can be simplified to MOV regX, regY

For instance, the previously inspected instructions at :10, :11 and :12 actually implements MOV R2, R0

We can then start individually reversing the identified functions. Starting with the small ones reveals that :54 is only return R1 as R0, or that :6C is return (R1 < R2), but also that this is quite time consuming, and we better move to a dynamic analysis.

Dynamic analysis

The good part of having a 3-instructions VM is that it’s pretty simple to implement. The script exec.py is a simple interpreter that executes a function from an entry point, and that can print the registers values and the instruction details at each step. It can be used to perform black-box analysis of individual sub-functions, by only observing the outputs for given inputs:

# function :71 below seems to be return (R1 - R2)

In : exec_code(pc=0x71, regs=[0,10,2]) 
Out: 0x8 (8)

In : exec_code(pc=0x71, regs=[0,123,45]) 
Out: 0x4e (78)


Using this interpreter mixed with some static analysis for confirmation, we can fully understand the goal of each function, detailed below. We start from the simple functions that don’t contain any calls, to then move up to the bigger ones, and finally reach main. The file code_disass_full.txt contains the fully commented disassembly of code.

• function :71: R1_SUB_R2 { RETURN R0 = MAX(0, R2-R1) }
• function :6C: R1_LT_R2 { RETURN R0 = (R1 < R2) }
• function :63: R1_MOD_R2 { RETURN R0 = (R1 % R2) }
• function :5C: DIV_R2_2 { RETURN R0 = (R2 / 2), R1 = (R2 % 2) }
• function :54: RET_R1 { RETURN R0 = R1 }
• function :40:
  F40 {
RETURN R0=0 if R1==0
RETURN R0=1 if R1==1
ELSE RETURN R0 = (F40(R1-1, R2) + F40(R1-2, R2)) % R2
}


We recognize here a recursive function summing its 2 preceding values, but using a modulo. It’s the Fibonacci suite, so F40 = FIBONACCI_MODULO.

• function :2D:
  F2D = {
RETURN R0 = (R1 / 2) IF R1 is pair
RETURN R0 = (R1 * 3 + 1) IF R1 is odd
}


This time, we recognize in this function the formula to compute the next element of the Hailstone sequence, so F2D = HAILSTONE_NEXT. The Collatz conjecture is that this sequence eventually reaches 1 for all positive integer, but it has not been proven. However this has been verified for at least all values up to 5×2^60.

• function :1D:
  F1D {
R2 = 0
DO
R1 = HAILSTONE_NEXT(R1)
R2 += 1
WHILE (R1 > 1)
RETURN R0 = R2
}


This function increments a counter for each iteration of the hailstone sequence for a given parameter, until it reaches 1: in other words, it computes the length of the hailstone sequence for this parameter. So F1D = LEN_HAILSTONE

• function :14:
  F14 {
R2 = 0
DO
R2 += LEN_HAILSTONE(R1)
R1 -= 1
WHILE (R1 > 0)
RETURN R0 = R2
}


This function sums the lengths of the hailstone sequences of all values from 1 to a given parameter. So F14 = SUM_LEN_HAILSTONE

• function :00 (main):

  MAIN {
IF PARAM < 11
RETURN R0 = 0
ELSE
MOD = SUM_LEN_HAILSTONE(PARAM)
FLAG = FIBONACCI_MODULO(PARAM, MOD)
RETURN R0 = FLAG
}


Finally, the main function computes a fibonnaci modulo of the given parameter, using the result of SUM_LEN_HAILSTONE for this parameter as modulo. That’s it. That’s the end of the reverse-engineering part: The entire challenge can now be resumed in the following one-liner:

  flag = fibonnaci_modulo( N = 9009131337, MOD = SUM{i=1..N}(length(hailstone_sequence(i))))


#### Part 4 - Solving

Both the hailstone sequence and fibonnaci sequence modulo have already been discussed online: link1, link2, or link3.

The interesting part to know about fibonnaci, is that it is cyclic when computed to a modulo. The period of this cycle for a given modulo is called the Pisano period (see links 2 and 3 above).

The problem can then be decomposed in 3 parts:

1. Finding the modulo by summing the length for all hailstone sequences up to 9009131337
2. Finding the pisano period for this given modulo
3. Computing the fibonnaci number for 9009131337 with the given modulo and pisano period

Fortunately, publicly available implementations from the above links can be adapted to perform these 3 tasks. See files hailstone.c, pisano.c, and fibo_mod.c

$time ./hailstone 9009131337 <MOD:0x1da61603168> ./hailstone 9009131337 1164.38s user 5.17s system 99% cpu 19:38.43 total$ time ./pisano $(( 0x1da61603168 )) <PISANO:0x876915450> ./pisano$(( 0x1da61603168 ))  460.00s user 1.95s system 99% cpu 7:45.60 total

$time ./fibo_mod 9009131337$(( 0x1da61603168 )) $(( 0x876915450 )) <FLAG:0x1bae15b6382> ./fibo_mod 9009131337$(( 0x1da61603168 )) \$(( 0x876915450 ))  111.83s user 0.28s system 99% cpu 1:52.68 total


The final flag is then CTF{000001bae15b6382}.

One final note here, is that the pisano period we found was greater than 9009131337. This means that it didn’t bring any benefits, and computing it was actually useless… But it was still interesting to learn it exists :)

Written on June 18, 2017