Tags: rev 

Rating: 4.2

ghidra and IDA didn't like the PLT of this binary, idk why. Rather than static analyzing first, I choose to do this blackbox. The binary read from flag.txt then output some numbers, see syscall trace below.
```
$ strace ./znfl
...
openat(AT_FDCWD, "flag.txt", O_RDONLY) = 3
fstat(3, {st_mode=S_IFREG|0644, st_size=43, ...}) = 0
...
```

Running the binary couple of times with same flag.txt content, you'll notice that the output doesn't stay the same every run, but somewhere if you have run the binary fast enough, (like really fast, 2 times in a second) you'll notice that the output is the same. Here's an example
```
$ ./znfl
1224429843 1224429840 1224429855 1224429840 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 ...
$ ./znfl
1224429843 1224429840 1224429855 1224429840 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 1224429855 1224429855 1224429852 1224429847 ...
```

We could guess It's using something random as a key. To make things easier, I created a `LD_PRELOAD` lib to patch `rand()` nad make it not "random".
```C
// gcc -shared -fPIC -o libunrandom.so unrandom.c
int rand() {
return 0;
}
```
After this the output is not changed anymore after every run.
```
$ LD_PRELOAD=./libunrandom.so ./znfl
4 7 8 7 0 8 8 11 0 8 8 11 0 8 8 ...
$ LD_PRELOAD=./libunrandom.so ./znfl
4 7 8 7 0 8 8 11 0 8 8 11 0 8 8 ...
```

Now, on to how the numbers are generated. I still didn't bother to do static analysis. At this point, I started to try out some different input from flag.txt content.
```
$ echo "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA" > flag.txt && LD_PRELOAD=./libunrandom.so ./znfl
0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11
```
You'll notice that the number is repeated every 4 number. From that Information, we could change our first four character from input to something else
```
$ echo "BBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA" > flag.txt && LD_PRELOAD=./libunrandom.so ./znfl
12 13 1 7 12 13 1 7 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11
```
And to our surprise, It became another repeated number `12 13 1 7 12 13 1 7 ...`. Because of that, we could try to lower our input guess to 2 character patern.
```
$ echo "CCBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA" > flag.txt && LD_PRELOAD=./libunrandom.so ./znfl
7 3 6 15 12 13 1 7 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11
$ echo "CBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA" > flag.txt && LD_PRELOAD=./libunrandom.so ./znfl
6 8 6 7 12 13 1 7 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11
```
From this, we could guess that every 2 character is mapped to 4 number at output. Try changing the input some more if you need to make sure It's the behaviour of the program.

Since every 2 character of input is perfectly mapped to 4 number output, we could use dictionary to recover the flag. There's still a problem though, we still didn't know what's the key used for the output.txt. To get the key, we could start from the flag format, we know that flag format is `FwordCTF{...}` with that we could get the different from our `Fw` output and output.txt.
```
$ echo "FworAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA" > flag.txt && LD_PRELOAD=./libunrandom.so ./znfl
6 7 10 9 4 9 10 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11 0 8 8 11
$ cat output.txt
1155306822 1155306823 1155306826 1155306825 1155306820 1155306825 1155306826 1155306827 1155306821 1155306819 1155306831 1155306823 1155306821 1155306831 1155306816 1155306823 1155306818 1155306820 1155306829 1155306816 1155306823 1155306818 1155306828 1155306822 1155306824 1155306816 1155306826 1155306826 1155306819 1155306821 1155306819 1155306830 1155306819 1155306819 1155306825 1155306826 1155306828 1155306816 1155306826 1155306826 1155306825 1155306821 1155306822 1155306823 1155306821 1155306818 1155306825 1155306819 1155306816 1155306828 1155306822 1155306821 1155306820 1155306824 1155306829 1155306827 1155306816 1155306831 1155306830 1155306825 1155306820 1155306830 1155306818 1155306829 1155306831 1155306829 1155306830 1155306830 1155306827 1155306822 1155306822 1155306828 1155306831 1155306829 1155306816 1155306831 1155306819 1155306817 1155306818 1155306831 1155306821 1155306830 1155306820 1155306825
```
Since `6 7 10 9` has `0 +1 +4 +3` pattern, and the output.txt doesn't have the same addition/subtraction pattern, we could guess that xor is used in this program. Just xor the first number from both output, `1155306822 ^ 6 = 1155306816`, and we got `1155306816` as key to translate output.txt to our patched program output.

From here on, you'll just need to get the dictionary for every 2 character of input, then recover the flag using our dictionary.
```py
from pwn import *

def conn(level="info"):
return process("./znfl", env={"LD_PRELOAD": "./libunrandom.so"}, level=level)

dic = {}

p = log.progress("get dict")
for first in range(0x21, 0x7F):
for second in range(0x21, 0x7F):
with open("flag.txt", "wb") as f:
pload = bytes([first, second]) + b"AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\n"
f.write(pload)
r = conn("warn")
a = int(r.recvuntil(" ", 1))
b = int(r.recvuntil(" ", 1))
c = int(r.recvuntil(" ", 1))
d = int(r.recvuntil(" ", 1))
r.close()
# print(f"{a} {b} {c} {d} = {chr(first)}{chr(second)}")
p.status(f"{first:02x}{second:02x}")
dic[(a,b,c,d)] = (first, second)
p.success("done")

with open("output.txt") as f:
item = f.read().split()

data = []
for c in item:
data.append(int(c))

for pos in range(0, len(data), 4):
key = (data[pos], data[pos+1], data[pos+2], data[pos+3])
if key in dic:
print(f"{chr(dic[key][0])}{chr(dic[key][1])}", end="")
else:
print(key, "not found")
```

Run the script, and we will get the flag at the end.
```
$ python solve.py
[+] get dict: done
FwordCTF{n0t_4_bad_id3a_4ft3r_All_semah!!}
```

glamorous-noobSept. 1, 2020, 9:22 p.m.

Wow that was a great approach!
Thanks for posting and congrats on winning the 1st place ;)