> For the complete documentation index, see [llms.txt](https://soyabeanboi.gitbook.io/bag-of-flags/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://soyabeanboi.gitbook.io/bag-of-flags/practice/cryptohack/general/xor-properties.md).

# XOR Properties

> In the last challenge, you saw how XOR worked at the level of bits. In this one, we're going to cover the properties of the XOR operation and then use them to undo a chain of operations that have encrypted a flag. Gaining an intuition for how this works will help greatly when you come to attacking real cryptosystems later, especially in the block ciphers category.\
> \
> There are four main properties we should consider when we solve challenges using the XOR operator\
> \
> `Commutative: A ⊕ B = B ⊕ A` \
> `Associative: A ⊕ (B ⊕ C) = (A ⊕ B) ⊕ C`\
> `Identity: A ⊕ 0 = A`\
> `Self-Inverse: A ⊕ A = 0`\
> \
> Let's break this down. Commutative means that the order of the XOR operations is not important. Associative means that a chain of operations can be carried out without order (we do not need to worry about brackets). The identity is 0, so XOR with 0 "does nothing", and lastly something XOR'd with itself returns zero.\
> \
> Let's put this into practice! Below is a series of outputs where three random keys have been XOR'd together and with the flag. Use the above properties to undo the encryption in the final line to obtain the flag.\
> \
> `KEY1 = a6c8b6733c9b22de7bc0253266a3867df55acde8635e19c73313`\
> `KEY2 ^ KEY1 = 37dcb292030faa90d07eec17e3b1c6d8daf94c35d4c9191a5e1e`\
> `KEY2 ^ KEY3 = c1545756687e7573db23aa1c3452a098b71a7fbf0fddddde5fc1`\
> `FLAG ^ KEY1 ^ KEY3 ^ KEY2 = 04ee9855208a2cd59091d04767ae47963170d1660df7f56f5faf`\
> \
> \&#xNAN;*Before you XOR these objects, be sure to decode from hex to bytes.*

## Goodbye XOR

The most important property is the *self-inverse* property since it gives us an easy way to undo the keys

The critical thing to note is `(A ⊕ B) ⊕ A = B` so we can remove the effect of `A` XOR on `B`

What we want is `FLAG`, so we need to remove `KEY1`, `KEY2`, and `KEY3` by XORing the respective values

* We have `FLAG ^ KEY1 ^ KEY3 ^ KEY2`
* We are also given `KEY1` and `KEY2 ^ KEY3` and these 2 strings give us one instance of each key
* Thus if we XOR all of these strings together, we can isolate the `FLAG`

-> Due to associativity, the order in which things are XORed does not matter

```python
from pwn import xor

KEY1 = "a6c8b6733c9b22de7bc0253266a3867df55acde8635e19c73313"
KEY2_KEY1 = "37dcb292030faa90d07eec17e3b1c6d8daf94c35d4c9191a5e1e"
KEY2_KEY3 = "c1545756687e7573db23aa1c3452a098b71a7fbf0fddddde5fc1"
FLAG_KEY1_KEY3_KEY2 = "04ee9855208a2cd59091d04767ae47963170d1660df7f56f5faf"

# Convert to bytes
k1 = bytes.fromhex(KEY1)
k2k3 = bytes.fromhex(KEY2_KEY3)
fk1k3k2 = bytes.fromhex(FLAG_KEY1_KEY3_KEY2)

# XOR everything
flag = xor(k1, xor(k2k3, fk1k3k2))

print(flag)
```

## Flag

`crypto{x0r_i5_ass0c1at1v3}`


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