# Some Assembly Required 3

## Problem

> <http://mercury.picoctf.net:60154/index.html>

## Solution

1. The new WebAssembly base64 string is as follows: `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`
2. Let's decompile it by first converting it to `wasm` using `write_wasm.py` in `../Some Assembly Required 2` and then using `wasm-decompile`. The decompiled c-like code is in [wasm-decompile-output.c](https://github.com/HHousen/PicoCTF-2021/blob/master/Web%20Exploitation/Some%20Assembly%20Required%203/wasm-decompile-output.c).
3. We can decompile the [compiled wasm](https://github.com/HHousen/PicoCTF-2021/blob/master/Web%20Exploitation/Some%20Assembly%20Required%203/compiled.wasm) to actual c code using `wasm2c`, which is also included in [WebAssembly/wabt](https://github.com/WebAssembly/wabt/). `wasm2c` ([Documentation](https://github.com/WebAssembly/wabt/tree/master/wasm2c)) creates harder to read long c code that will actually run while `wasm-decompile` ([Documentation](https://github.com/WebAssembly/wabt/blob/master/docs/decompiler.md)) creates easier to read c-style pseudocode.
4. We can use `git diff` or [diffchecker.com](https://www.diffchecker.com) to compare the new wasm with the wasm from the previous challenge. We see the addition of a new variable, called `key`, and some changes at the end of the `copy_char`.
5. The encrypted flag is: `0x9d, 0x6e, 0x93, 0xc8, 0xb2, 0xb9, 0x41, 0x8b, 0x9f, 0x90, 0x8c, 0x62, 0xc5, 0xc3, 0x95, 0x88, 0x34, 0xc8, 0x93, 0x92, 0x88, 0x3f, 0xc1, 0x92, 0xc7, 0xdb, 0x3f, 0xc8, 0x9e, 0xc7, 0x89, 0x31, 0xc6, 0xc5, 0xc9, 0x8b, 0x36, 0xc6, 0xc6, 0xc0, 0x90, 0x00, 0x00`. This value is from the [`wasm2c` output file](https://github.com/HHousen/PicoCTF-2021/blob/master/Web%20Exploitation/Some%20Assembly%20Required%203/wasm2c-output.c) because it shows a clear array of hex bytes while the [`wasm-decompile` output file](https://github.com/HHousen/PicoCTF-2021/blob/master/Web%20Exploitation/Some%20Assembly%20Required%203/wasm-decompile-output.c) shows a slight mess.
6. The key is: `0xf1, 0xa7, 0xf0, 0x07, 0xed`.
7. If we analyze the c and c-like code, we will discover that the flag is xored with the reverse of the key. I'm not sure why the key needs to be reversed, but it could have something to do with big endian and little endian since `wasm` is close to machine code.
8. An alternative way to find the key is as follows: We know from step 6 that the key is 5 bytes. We also know that the flag starts with `picoCTF{`. Thus, we can xor the start of the flag (`picoC`) with the encrypted flag `0x9d, 0x6e, 0x93, 0xc8, 0xb2` to find that they key is: `0xed 0x07 0xf0 0xa7 0xf1` ([CyberChef Recipe](https://gchq.github.io/CyberChef/#recipe=XOR\(%7B'option':'Hex','string':'0x9d,%200x6e,%200x93,%200xc8,%200xb2'%7D,'Standard',false\)To_Hex\('Space',0\)\&input=cGljb0M)). This is the reverse of the key found in the code in step 6.
9. Now that we have the key and encrypted flag, we can simply decode the encrypted flag as hex and then xor the encrypted flag and key to decrypt the flag ([CyberChef Recipe](https://gchq.github.io/CyberChef/#recipe=From_Hex\('Auto'\)XOR\(%7B'option':'Hex','string':'ed%2007%20f0%20a7%20f1'%7D,'Standard',false\)\&input=OWQgNmUgOTMgYzggYjIgYjkgNDEgOGIgOWYgOTAgOGMgNjIgYzUgYzMgOTUgODggMzQgYzggOTMgOTIgODggM2YgYzEgOTIgYzcgZGIgM2YgYzggOWUgYzcgODkgMzEgYzYgYzUgYzkgOGIgMzYgYzYgYzYgYzAgOTAgMDAgMDA)).

### Flag

`picoCTF{8aae5dde384ce815668896d66b8f16a1}`