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# 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`
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* Since only `R0` and `R1` are checked, this would not detect errors involving other registers.
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* 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`
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* Although, notably, not `ret`. The tests operates by `jmp`ing back and forth to test the various values of `R0`.
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* `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:
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06bb push R0
06bd push R1
06bf set R1 17b4
06c2 rmem R0 R1
06c5 push R1
06c7 mult R1 R1 R1
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06cb call 084d
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06cd set R1 4154
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06d0 call 084d
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06d2 pop R1
06d4 wmem R1 R0
06d7 add R1 R1 0001
06db eq R0 7562 R1
06df jf R0 06c2
06e2 pop R1
06e4 pop R0
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06e6 ret
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Inspecting the `084d` subroutine reveals that this is simply an XOR function: `R0 XOR R1`. Crypto rating: 1/10
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Rewriting `06bb` function using higher-level syntax reveals that the ‘encryption’ algorithm is really very simple:
```c
06bb() {
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R1 = 17b4;
for (R1 = 17b4; R1 != 7562; R1++) {
R0 = [R1];
R0 ^= R1 * R1;
R0 ^= 4154;
[R1] = R0;
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}
}
```
*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:
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0413 set R0 17c0
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0416 call 05ee
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Digging deeper, `05ee` calls `05b2` with `R1` set to `05f8`. `05b2` appears to iterate over the characters in a string whose length is stored in address `R0`, and calls `R1` for each character, storing the character in `R0`. `05f8` (the callback provided by `05ee`) simply outputs every character in `R0` it gets.
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Immediately after this call to `05ee`, we find:
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041e set R0 68e3
0421 set R1 05fb
0424 add R2 XXXX XXXX
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0428 call 05b2
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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 `R1` set to `R2`.
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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…
## The codes
We may not have the codes themselves, but we can now easily locate where they are printed. The tablet code, for example, is conveniently sandwiched between strings `6ed1` and `6ef1`. Thus the code we are looking for is:
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1290 set R0 1092
1293 set R1 650a
1296 set R2 7fff
1299 set R3 6eed
129c call 0731
Looking into `0731`:
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0731 push R3
0733 push R4
0735 push R5
0737 push R6
0739 set R6 0001
073c add R4 R3 R6
0740 rmem R4 R4
0743 add R5 17ed R6
0747 wmem R5 R4
074a add R6 R6 0001
074e rmem R5 17ed
0751 gt R4 R6 R5
0755 jf R4 073c
0758 set R3 0000
075b set R4 0000
075e rmem R5 17ed
0761 mod R5 R4 R5
0765 add R5 R5 17ed
0769 add R5 R5 0001
076d rmem R6 R5
0770 mult R6 R6 1481
0774 add R6 R6 3039
0778 wmem R5 R6
077b push R0
077d push R1
077f set R1 R6
0782 call 084d
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0784 set R6 R0
0787 pop R1
0789 pop R0
078b rmem R5 R1
078e mod R6 R6 R5
0792 add R6 R6 0001
0796 gt R5 R6 R2
079a jt R5 07a0
079d set R3 0001
07a0 add R6 R6 R1
07a4 rmem R6 R6
07a7 add R4 R4 0001
07ab add R5 R4 17f1
07af wmem R5 R6
07b2 rmem R5 17f1
07b5 eq R5 R4 R5
07b9 jf R5 075e
07bc jf R3 0758
07bf push R0
07c1 set R0 17f1
07c4 call 05ee
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07c6 pop R0
07c8 pop R6
07ca pop R5
07cc pop R4
07ce pop R3
07d0 ret
Umm… Sorry, could you repeat that?
Rewriting this again in more friendly terms:
```c
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// R0: A seed of sorts - the same for all users
// R1: The length and alphabet to use, usually 650a, but 653f for the mirror
// R2: Usually 7fff, but 0004 for the mirror
// R3: An initialisation vector of sorts - contents different for every user - points to the length, but this is always 3
0731(R0, R1, R2, R3) {
// copy the string at R3 to 17ed
R6 = 0001;
do {
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R4 = R3 + R6;
R4 = [R4];
R5 = 17ed + R6;
[R5] = R4;
R6 = R6 + 0001;
R5 = [17ed]; // 3, the length - this never seems to change
} while (R6 <= R5);
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// the string at 17ed is now what was at R3
do {
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R3 = 0000; // done flag
R4 = 0000; // index
do {
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R5 = [17ed]; // 3, the length
R5 = R4 % R5;
R5 = R5 + 17ed;
R5 = R5 + 0001; // will cycle through three addresses of 17ed/R3 string: 17ee, 17ef, 17f0
R6 = [R5];
R6 = R6 * 1481;
R6 = R6 + 3039;
[R5] = R6; // mutate that value of the 17ed string
R6 = R0 ^ R6; // combine R0 into R6
R5 = [R1]; // length of the alphabet
R6 = R6 % R5;
R6 = R6 + 0001; // calculate index in alphabet
if (R6 <= R2) {
R3 = 0001; // we are done with the entire code - this returns immediately for all except the mirror
}
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R6 = R6 + R1; // calculate address of letter to use
R6 = [R6]; // the letter to use
R4 = R4 + 0001; // increment the index
R5 = R4 + 17f1; // index of new letter in code
[R5] = R6; // set the letter
R5 = [17f1]; // length of the code: twelve letters
} while (R4 != R5); // loop until we've generated all the letters
} while (!R3);
print(17f1);
}
```
Re-implementing this in Python, we can now extract the code for the tablet directly from the binary!
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Unfortunately, each of the other codes uses an `R0` based on the solution to the puzzle. In the case of the maze code:
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0f03 rmem R0 0e8e
The value at `0e8e` is derived from which rooms are visited in the maze, as mentioned in the main note file. Armed with our [trusty map](https://github.com/RunasSudo/synacor.py/blob/master/tools/graph.py), and cross-referencing the callbacks with the files, we identify:
* Twisty passages entrance, `0949`: Calls `0e9e`, resets `0e8e` to `0000`.
* West to `095d`: Calls `0ec0`: `OR`s `0e8e` with `0008`.
* South to `0926`: Calls `0eca`: `OR`s `0e8e` with `0010`.
* North to `096c`: Calls `0ede`: `OR`s `0e8e` with `0040`.
Putting it all together, the final value at `0e8e` is `0008 | 0010 | 0040` = `0058`.
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Similarly, the `R0` for the teleporter code is the value of `R7` from the Ackermann function, which we determined earlier to be `6486`:
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1592 set R0 R7
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The remaining two codes, for the coins and the vault, are more complicated still, but follow the same pattern of determining `R0` based on the player's history.
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For the coins, the call to `0731` derives an `R0` from the values at memory locations `69de` to `69e2`. One would presume that this relates to the order in which coins are inserted into the puzzle, and following the trail of callbacks for the coins confirms this. The important lines are:
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13c0 rmem R2 099e # nr of coins inserted
13c3 add R2 R2 69dd
13c7 add R2 R2 0001
13cb wmem R2 R1 # R1 is the number of dots on the coin
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Thus the values should be 9, 2, 5, 7 and 3: the missing numbers in the solved equation.
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It is now trivial to generate the correct value of `R0`. Based on the code beginning, `15fd`:
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```python
data_69de = [9, 2, 5, 7, 3]
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R0 = 0
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for i in range(len(data_69de)):
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R1 = data_69de[i]
R0 = R0 + R1
R0 = (R0 * 0x7bac) % 0x8000
R0 = R0 ^ R1
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```
The result is `0b3b`.
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On to the final code, now: the vault lock. At `1679`, `R0` is computed as `([0f73] ^ [0f74]) ^ [0f75]`. Searching the code for references to these, we find:
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1236 wmem 0f70 0016 // weight
1239 wmem 0f71 0000 // counter
123c wmem 0f72 0000
123f wmem 0f73 0000
1242 wmem 0f74 0000
1245 wmem 0f75 0000
It would be nice if these were just the intermediate weights of the orb, but alas the algorithm appears more complex than that.
Nevertheless, unravelling the various functions which reference those memory addresses:
```python
counter = 0
data_0f73 = 0
data_0f74 = 0
data_0f75 = 0
# 08c8
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def rotatey_thing(R0, R1):
while R1 != 0:
R1 = (R1 + 0x7fff) % 0x8000
R2 = R0 & 0x4000
R0 = (R0 * 0x0002) % 0x8000
if R2 == 0:
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continue
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R0 = R0 | 0x0001
return R0
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# 11a3
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def mutate(R0val, R1, R2):
return rotatey_thing(R0val, R1) ^ R2
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# 1135
def update(value, room):
global counter, data_0f73, data_0f74, data_0f75
if counter <= 0x752f:
counter += 1
data_0f73 = mutate(data_0f73, counter + 2, room)
data_0f74 = mutate(data_0f74, counter * counter, room * room)
data_0f75 = mutate(data_0f75, 0x000d, (value * 9) * (value * 9))
update(0x0000, 0x0008) # north 1000
update(0x0004, 0x0009) # east 1011
update(0x0001, 0x000a) # east 1022
update(0x000b, 0x0006) # north 0fde
update(0x0002, 0x0005) # west 0fcd
update(0x0004, 0x0009) # south 1011
update(0x0001, 0x000a) # east 1022
update(0x0012, 0x000b) # east 1033
update(0x0001, 0x000a) # west 1022
update(0x000b, 0x0006) # north 0fde
update(0x0001, 0x0002) # north 0f98
update(0x0001, 0x0003) # east 0fa9
print(data_0f73)
print(data_0f74)
print(data_0f75)
result = (data_0f73 ^ data_0f74) ^ data_0f75
print('0x{:04x}'.format(result))
```
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And with that, we can now [programmatically extract *every single code* given any challenge .tgz!](https://github.com/RunasSudo/synacor.py/blob/master/tools/generate_codes.py)