# Easy Hash (easy)
## Given file and prompt
```Source Code: easy_hash.7zWeb Server: https://crypto01.chal.ctf.westerns.tokyo
For beginners: you can use curl to interact with the web server.
(Example)$  curl https://crypto01.chal.ctf.westerns.tokyo -d 'twctf: hello 2020'```
```pythonimport structimport os
MSG = b'twctf: please give me the flag of 2020'
assert os.environ['FLAG']
def easy_hash(x):    m = 0    for i in range(len(x) - 3):        m += struct.unpack('<I', x[i:i + 4])[0]        m = m & 0xffffffff
    return m
def index(request):    message = request.get_data()    if message[0:7] != b'twctf: ' or message[-4:] != b'2020':        return b'invalid message format: ' + message
    if message == MSG:        return b'dont cheet'
    msg_hash = easy_hash(message)    expected_hash = easy_hash(MSG)    if msg_hash == expected_hash:        return 'Congrats! The flag is ' + "HERE IS THE FLAG"    return 'Failed: easy_hash({}) is {}. but expected value is {}.'.format(message, msg_hash, expected_hash)```
## Approach
I had two approach to this:
- Like MD5, possibly trying to find a collision of the hash (the desired value was `1788732187`)- Finding a character that wouldn't change the hash
### Finding a colision
I tried first to brute force this using the same method as MD5 collision but this took too long and I abandoned.While I do believe that there is a string that will collide and equal the one that we want in value, there was a faster approach.
### A character that wont affect the hash
The function to calculate the hash is the following: 
```pythondef easy_hash(x):    m = 0    for i in range(len(x) - 3):        m += struct.unpack('<I', x[i:i + 4])[0]        m = m & 0xffffffff
    return m```
It takes a window of 4 characters and unpacks their value with `little endian` and `unsigned integer` and sums them all to mod `0xffffffff`. Our goal is to then add one character that would have value of `0` when its unpacked. We can reverse this by using:
```pythonimport structstruct.pack('<I', 0) #==> b'\x00\x00\x00\x00'```
Thus adding the character `\x00` to the string `b'twctf: please give me the flag of 2020'` would make it different from the message while retaining the hash. 
## Solution
Using the python command line:
```pythonimport requests
r = requests.post('https://crypto01.chal.ctf.westerns.tokyo', data=b'twctf: please give me the flag of \x002020')  print(r.text) # 'Congrats! The flag is TWCTF{colorfully_decorated_dream}'```
## Flag
'Congrats! The flag is TWCTF{colorfully_decorated_dream}'