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synacor.py/electric_boogaloo.md

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2017-02-05 20:45:04 +11:00
# Synacor Challenge 2: Electric Boogaloo
So we've finished the official parts of the challenge. What now?
## Self-test
Hex values refer to the instruction lines, not the actual ranges spanned by the data in memory
* `0000` to `013e`: Startup message
* `0140` to `01e3`: Tests `jmp`
* From `0160` to `01c9`, some clever code is used to ‘amplify’ the effect of any error in `jmp` to allow the precise size of the error to be determined.
* If successful, executes `0140`, `015b` to `0160`, then `0166`.
* `01e4` to `01f2`: Tests `jt` and `jf`
* If successful, jumps to `01f4`.
* `01f4` to `0209`: Tests that the registers are initialised to zero
* `020c` to `0215`: Tests `set`
* `0218` to `0233`: Tests `add`
* The test is rudimentary, however, and would not detect many simple errors. In fact it tests only if 1 + 1 ≠ 0.
* `0234` to `024d`: Tests `eq`
* Surprisingly, this is where it is checked that 1 + 1 = 2, but if `add` gives an incorrect non-zero result for 1 + 1, the test will erroneously report that it is `eq` which is not supported!
* It would probably have been a better idea to test `eq` first, before `add`, then use `eq` to test `add`.
* `024e` to `0261`: Tests `push` and `pop`
* Since only `R1` and `R2` are checked, this would not detect errors involving other registers.
* This test, like the last one, reuses the results of previous tests, since that worked out so well for `eq`
* `0264` to `0276`: Tests `gt`
* The tests performed seem quite reasonable, but yet again reuse the results of previous tests…
* `0279` to `02ab`: Tests `and` and `or`
* Confusingly, the error handling is located in different places for each test.
* `02ac` to `02bd`: Tests `not`
* Okay, I admit this one was pretty helpful. What the hell is a ‘15-bit bitwise inverse’? Well the test passes if I do just do mod 32768, so that works I guess…
* `02c0` to `02e8`: Tests `call`
* Although, notably, not `ret`. The tests operates by `jmp`ing back and forth to test the various values of `R1`.
* `02eb` to `0308`: Checks that `add` overflows correctly
* `030b` to `0313`: Checks that 6 × 9 ≠ 42.
* I suspect there is a mistake in this test. Since Adams (1979) demonstrated unequivocally that 6 × 9 is equal to 42, I believe the `jt` should in fact be a `jf`.
* `0316` to `0346`: Continues checking `mult`
* `0349` to `034b`: Two values `4e20` and `2710` are stored in memory here for reference by the following test
* `034d` to `03d1`: Tests `rmem` and `wmem`
* If successful, causes the words from `03a9` to `03ac` to instead read `nop`, `jt 0013 03d2`
* There is more to the portion starting `0375` than meets the eye: see below.
* `0432` to `05b1`: Various error messages
## Decryption
As we know from earlier, most of the strings in the binary are encrypted (or at the very least obfuscated) in some way, and decrypted following the self-test. It is therefore desirable to study this encryption before further study of the binary.
After a wild goose chase examining the code after the self-test, we find that the decryption actually happens *during* the `rmem`/`wmem` test! Very sneaky!
0375 call 06bb
This is the magic line. Digging into the `06bb` subroutine:
06bb push R1
06bd push R2
06bf set R2 17b4
06c2 rmem R1 R2
06c5 push R2
06c7 mult R2 R2 R2
06cb call 084d
06cd set R2 4154
06d0 call 084d
06d2 pop R2
06d4 wmem R2 R1
06d7 add R2 R2 0001
06db eq R1 7562 R2
06df jf R1 06c2
06e2 pop R2
06e4 pop R1
06e6 ret
Inspecting the `084d` subroutine reveals that this is simply an XOR function: `R1 XOR R2`. Crypto rating: 1/10
Rewriting `06bb` function using higher-level syntax reveals that the ‘encryption’ algorithm is really very simple:
```c
06bb() {
R2 = 17b4;
for (R2 = 17b4; R2 != 7562; R2++) {
R1 = [R2];
R1 ^= R2 * R2;
R1 ^= 4154;
[R2] = R1;
}
}
```
*Very* simple.
[By emulating this function in Python](https://github.com/RunasSudo/synacor.py/blob/master/dbg_fastboot.py), we can skip the self-test and computationally-expensive decryption process entirely, and get straight into the good stuff next time we want to play!
## Encrypted strings
So earlier, we produced a tool-assisted speed-run that would complete and dump the codes for any given challenge binary, but where's the fun in that? Why not extract the codes from the binary directly? Of course, this is easier said than done. None of the codes, nor any of the strings relating to them, are visible in the disassembled binary, whether before or after the decryption from the previous section.
Looking through the code following the self-test, we find:
0413 set R1 17c0
0416 call 05ee
Digging deeper, `05ee` calls `05b2` with `R2` set to `05f8`. `05b2` appears to iterate over the characters in a string whose length is stored in address `R1`, and calls `R2` for each character, storing the character in `R1`. `05f8` (the callback provided by `05ee`) simply outputs every character in `R1` it gets.
Immediately after this call to `05ee`, we find:
041e set R1 68e3
0421 set R2 05fb
0424 add R3 XXXX XXXX
0428 call 05b2
In other words, a similar string-handling subroutine is called, but instead of `05f8` (which would simply print the string), `05fb` is called. `05fb` also outputs the character, but only after calling `084d` (XOR) with `R2` set to `R3`.
Now we have everything we need to [extract these encrypted (double-encrypted??) strings](https://github.com/RunasSudo/synacor.py/blob/master/tools/decrypt_strings.py) from the binary!
Only the self-test completion code appears to be stored there, though, so I'm not sure what the point of encrypting those was…