Rating:
# Rolling on my own
We start opening the binary in ghidra, and get a nice dissassembled main:
```c
void main(void)
{
int iVar1;
undefined8 local_58;
undefined8 local_50;
undefined8 local_48;
undefined8 local_40;
char local_38 [36];
int local_14;
int local_10;
int local_c;
puts("I can\'t just let anyone in, what\'s the password?");
__isoc99_scanf(&DAT_00102039);
local_58 = 0x455d35402f28272b;
local_50 = 0x43277c222d58366a;
local_48 = 0x5650324666795152;
local_40 = 0x5f472c67575b6353;
local_c = 0;
while( true ) {
if (0x1f < local_c) {
printf("https://www.youtube.com/watch?v=TeSRoEjevHI");
return;
}
local_10 = (int)local_38[local_c];
iVar1 = roll(local_c);
iVar1 = iVar1 + local_10;
local_14 = iVar1 + ((iVar1 / 6 + (iVar1 >> 0x1f) >> 4) - (iVar1 >> 0x1f)) * -0x60 + 0x22;
if (local_14 != *(char *)((long)&local_58 + (long)local_c)) break;
local_c = local_c + 1;
}
printf("Imposter!!");
/* WARNING: Subroutine does not return */
exit(0);
}
```
We also have a `roll` function that is generating Fibonacci numbers:
```c
int roll(int param_1)
{
int iVar1;
if (1 < param_1) {
iVar1 = roll(param_1 + -1);
param_1 = roll(param_1 + -2);
param_1 = param_1 + iVar1;
}
return param_1;
}
```
the `local_58`, `local_50`, `local_48` and `local_40` could be our flag, we can try to confirm this trying to encrypt some known plaintext with the function:
```python
def roll(i): # actually fibo
if (i > 1):
tmp = roll(i-1)
i = roll(i-2)
i = i + tmp
return i
# let's try to encrypt know plaintext see what happen
s = b"idek{"
out = b''
for i in range(0, len(s)):
mod = roll(i)
c = mod + s[i]
enc_char = c + ((c // 6 + (c >> 0x1f) >> 4) - (c >> 0x1f)) * -0x60 + 0x22
out += bytes([enc_char])
print(out.hex()) # 2b27282f40, looks like local_58 content in big endian, we're on good track
```
From here we can just bruteforce every printable characters until we find a match in the `local_xx`:
```python
from pwn import *
import string
# extracted from binary, most probably the flag
flag1 = 0x455d35402f28272b
flag2 = 0x43277c222d58366a
flag3 = 0x5650324666795152
flag4 = 0x5f472c67575b6353
# flag = p64(flag1) + p64(flag2) + p64(flag3) + p64(flag4)
flag = p64(flag1) + p64(flag2) + p64(flag3) + p64(flag4)
out = b''
for i in range(0, len(flag)):
mod = roll(i)
expected = flag[i]
for char in string.printable: # iterate on all printable chars
c = mod + ord(char)
enc_char = c + ((c // 6 + (c >> 0x1f) >> 4) - (c >> 0x1f)) * -0x60 + 0x22
if enc_char == expected: # if it matches the encrypted flag byte, we know that `char` is the clear byte.
out += bytes([ord(char)])
print(out)
```
