giuliano-macedo/picottf18-be-quick-or-be-dead-2

PicoCTF 2018 solution for the problem 'Be quick or be dead 2' using the iterative Fibonacci function

PicoCTF 2018 solution for the problem 'Be quick or be dead 2' using the iterative Fibonacci function

disclaimer

i got the binary from here

this solution is very overkill

pre-requisites

• python >=3.6
• pip
• gcc

installation

just pip it after venv

pip install -r requirements.txt

write up

the binary contains a recursive implementation of nth Fibonacci number function, called 'fib', that is calculation for the 1083th Fibonacci number that is used to print the flag, since the recursive function is very slow, and that this number causes 64bit integer overflow the function may run forever.

the correct answer for the problem would be:

9641162182178966878126331027202834784434723577592322830700454745652427494401346945631082965963962317692358822696127040961581675695438118874508418491101822679355067810556808551572644321954159676320600161466564032755133080685122

however, since overflow, the program would accept the following integer from the fib function: -1066907070

the program then waits for the computation for 3 seconds, if the fib function is still running until there it will fail and not print the flag, otherwise it will print the flag.

this solution therefore compiles an iterative version of the nth Fibonacci number computation in another binary extracts it and patch onto the original binary, since this solution is O(1) it will run basically instantly in any machine and print the flag.

Usage

run the compile_fib.py to generate fib_iterative.txt that have the assembly code for the iterative version of the Fibonacci, edit the labels from the jump-based instructions and insert into patch_it and the binary be-quick-or-be-dead-2_patched will be created and will print the flag.