Home » HACKER-TECH » Decoding an air conditioner retain a watch on’s checksum with differential cryptanalysis

Decoding an air conditioner retain a watch on’s checksum with differential cryptanalysis

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Abet in 2009 I wrote an Arduino library (IRRemote) to
encode and decode infrared signals for distant controls.
I bought an email no longer too long within the past from somebody searching to retain a watch on an air conditioner.
Evidently air conditioner distant controls are mighty difficult than TV distant controls:
the codes are longer and contain a rather complex checksum.
My reader had composed 35 signals from his air conditioner distant retain a watch on, nonetheless couldn’t work out the checksum algorithm.
I made a decision to make use of differential cryptanalysis to work out the checksum, which used to be overkill nonetheless an absorbing exercise.
In case any individual else needs to decode a identical distant retain a watch on, I’ve written up how I discovered the algorithm.

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My IR distant library could per chance moreover very effectively be weak with the Arduino to ship and web signals. (This is no longer the air conditioner distant.)

The problem is to web a checksum algorithm that once given three bytes of enter (left), computes the kindly one byte checksum (valid).

10100001 10010011 01100011 => 01110111
10100001 10010011 01100100 => 01110001
10100001 10010011 01100101 => 01110000
10100001 10010011 01100110 => 01110010
10100001 10010011 01100111 => 01110011
10100001 10010011 01101000 => 01111001
10100001 10010011 01101001 => 01111000
10100001 10010011 01101010 => 01111010
10100001 10010011 01101011 => 01111011
10100001 10010011 01101100 => 01111110
10100001 10010011 01101101 => 01111111
10100001 10010011 01101110 => 01111100
10100001 10010011 01101111 => 01111101
10100001 10010011 01110001 => 01100100
10100001 10010011 01110010 => 01100110
10100001 10010011 01110011 => 01100111
10100001 10010011 01110100 => 01100001
10100001 10010011 01110101 => 01100000
10100001 10010011 01110111 => 01100011
10100001 10010011 01110111 => 01100011
10100001 10010011 01111000 => 01101001
10100001 10010011 01101000 => 01111001
10100001 00010011 01101000 => 11111001
10100001 00010011 01101100 => 11111110
10100001 10010011 01101100 => 01111110
10100001 10010100 01111110 => 01101011
10100001 10000010 01101100 => 01100000
10100001 10000001 01101100 => 01100011
10100001 10010011 01101100 => 01111110
10100001 10010000 01101100 => 01111100
10100001 10011000 01101100 => 01110100
10100001 10001000 01101100 => 01101100
10100001 10010000 01101100 => 01111100
10100001 10011000 01101100 => 01110100
10100010 00000010 11111111 => 01111110

The foundation leisurely differential cryptanalysis is to glimpse at the outputs as a result of inputs that absorb a runt distinction, to interrogate what patterns emerge (particulars).
Finding air conditioner checksums is more or less a trivial application of differential cryptanalysis, nonetheless the usage of differential cryptanalysis affords a framework for impending the tell.
I wrote a easy program that found enter pairs that differed in one bit and displayed the distinction (i.e. xor) between the corresponding checksums.
The desk under reveals the adaptations.

000000000000000000000001 : 00000001
000000000000000000000010 : 00000010
000000000000000000000010 : 00000011
000000000000000000000100 : 00000100
000000000000000000000100 : 00000110
000000000000000000000100 : 00000111
000000000000000000001000 : 00001100
000000000000000000001000 : 00001110
000000000000000000001000 : 00001111
000000000000000000010000 : 00010000
000000000000100000000000 : 00001000
000000000001000000000000 : 00011000
000000001000000000000000 : 10000000

The first thing to peep is that changing one bit within the enter causes a relatively runt alternate within the output.
If the checksum were something cryptographic, a single bit alternate would solely alternate the output (so that you would peep half the bits flipped on practical).
Thus, we all know we’re facing a easy algorithm.

The second thing to peep is the higher four bits of the checksum alternate merely: changing a runt within the higher four bits of an enter byte changes the corresponding
bit within the higher four bits of the output. This implies that the the three bytes are merely xor’d to generate the higher four bits.
Surely, my reader had already determined that the xor of the enter bytes (alongside with 0x20) yielded the higher four bits of the checksum.

The final thing to peep is there could be an abnormal avalanche sample within the decrease four bits. Altering the lowest enter bit changes the lowest checksum bit.
Altering the second-lowest enter bit changes the second-lowest checksum bit and potentially the last checksum bit.
Likewise changing the fourth-lowest enter bit changes the fourth-lowest checksum bit and potentially the bits to the sing.
And the alternate sample progressively has 1’s potentially adopted by 0’s, no longer a combination. (1100, 1110, 1111)

What easy operation has this create of avalanche make? Comprise in mind adding two binary numbers. In the occasion you alternate a high-account for little bit of an enter, handiest that bit will alternate within the output.
In the occasion you alternate a low-account for enter bit, the low-account for little bit of the output will alternate. However per chance there will seemingly be a raise, and the next bit will alternate. And if there could be a raise from that situation, the 0.33 bit will alternate. Likewise, changing a runt within the guts will alternate that bit and potentially most definitely the most bits to the left (due to carries). So if you alternate the low-account for bit, the alternate within the output could per chance moreover very effectively be 0001 (no raise), or 0011, or 0111, or 1111 (all carries).
This is the an identical sample seen within the air conditioner checksums nonetheless backwards. This raises the possibility that the checksum is the usage of a binary sum, nonetheless we’re the bits backwards.

So I made a program that reversed the bits within the enter and output, and took the sum of the four bits from each and every byte.
The output under reveals the reversed enter, the sum, and the four-bit cost from the kindly checksum.
Display that the sum and the kindly cost in overall add as much as forty six or 30 (two in need of a a pair of of sixteen).
This urged that the checksum is (-sum-2) & 0xf.

110001101100100110000101 32 14
001001101100100110000101 22 Eight
101001101100100110000101 30 0
011001101100100110000101 26 four
111001101100100110000101 34 12
000101101100100110000101 21 9
100101101100100110000101 29 1
010101101100100110000101 25 5
110101101100100110000101 33 13
001101101100100110000101 23 7
101101101100100110000101 31 15
011101101100100110000101 27 Three
111101101100100110000101 35 Eleven
100011101100100110000101 28 2
010011101100100110000101 24 6
110011101100100110000101 32 14
001011101100100110000101 22 Eight
101011101100100110000101 30 0
111011101100100110000101 34 12
111011101100100110000101 34 12
000111101100100110000101 21 9
000101101100100110000101 21 9
000101101100100010000101 21 9
001101101100100010000101 23 7
001101101100100110000101 23 7
011111100010100110000101 17 13
001101100100000110000101 15 0 *
001101101000000110000101 19 12 *
001101101100100110000101 23 7
001101100000100110000101 Eleven Three
001101100001100110000101 12 2
001101100001000110000101 12 Three *
001101100000100110000101 Eleven Three
001101100001100110000101 12 2
111111110100000001000101 23 7

That formula worked with three exceptions (marked with asterisks). Studying the exceptions confirmed that adding in byte 1 bit Three and byte 2 bit 7 yielded the kindly answer in all cases.

Conclusion

Inserting this together yields the following algorithm (beefy code right here):

inbytes = plot(bitreverse, inbytes)
xorpart = (inbytes[0] ^ inbytes[1] ^ inbytes[2] ^ 0x4) & 0xf
sumpart = (inbytes[0] >> four) + (inbytes[1] >> four) + (inbytes[2] >> four) +
  (inbytes[2] & 1) + ((inbytes[1] >> Three) & 1) + 1
sumpart = (-sumpart) & 0xf
result = bitreverse((sumpart << four) | xorpart)

Is this the sing formula? It affords the sing checksum for all of the given inputs.
On the alternative hand, some bits never alternate within the enter recordsdata; in particular all inputs originate with “101000”,
so there is not any methodology of exciting how they’ve an ticket on the algorithm. They’d moreover fair be added to the sum, for instance, changing the fixed +1.
The fixed 0x4 within the xor also makes me a runt suspicious.
It’s somewhat that that you would be capable to moreover absorb of that the checksum formula would must be tweaked if additional enter recordsdata becomes on hand.

I could per chance moreover fair still point out that figuring out the checksum formula is pointless for loads of air-con purposes.
Most users could per chance moreover valid laborious-code the checksums for the handful of instructions they want to ship, in have to figuring out the overall algorithm.

In account for so that you can make use of my IR library, it is on GitHub.
I’m now no longer actively enthusiastic with the library, so please put up complications on the GitHub repository in have to on this weblog put up.
Because of of Rafi Kahn, who has taken over library upkeep and enchancment.

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