Brute force subghz fixed codes using flipper zero, initially inspired by CAMEbruteforcer
This repo aims to collect as many brute force files as possible, so if you can or want to contribute you are more than welcome to do so!
This repo contains a python script to generate bruteforce .sub
files for subghz protocols that use fixed OOK codes. Inside the script it is also possible to specify your own protocol in case it's not present.
To generate all the files simply run:
python3 flipperzero-bruteforce.py
It will generate bruteforce files for all the specified protocols organized in many folders with the following structure:
sub_files/
└── PROTOCOL_NAME
├── SPLIT_FACTOR
│ ├── 000.sub
│ ├── ...
│ └── NNN.sub
└── debruijn.sub
For each protocol there are 6 sub folders, containing 1, 2, 4, 8, 16 and 32 files, SPLIT_FACTOR
indicates the number of keys per .sub
file. This is useful when trying to get a close guess to the key.
Right now the protocols supported are:
- CAME
- NICE
- PT-2240
- PT-2262
More info about them can be found here
Adding a protocol is very straight forward, inside the script protocols are defined at the bottom, inside the protocol list:
protocols = [
Protocol("CAME", 12, {"0": "-320 640 ", "1": "-640 320 "}, "-11520 320 "),
Protocol("NICE", 12, {"0": "-700 1400 ", "1": "-1400 700 "}, "-25200 700 "),
Protocol("8bit", 8, {"0": "200 -400 ", "1": "400 -200 "}), # generic 8 bit protocol
...
]
A protocol is defined by a few parameters passed to the constructor in the following order:
- name: the name of the protocol
- n_bits: the number of bits for a single key
- transposition_table: how 0s and 1s are translated into flipper subghz
.sub
language - pilot_period: aka preamble, a recurring pattern at the beginning of each key, defaults to
None
- frequency: working frequency, defaults to 433.92
To compute the time it takes to perform a bruteforce attack, we need to sum the time it takes to send each code:
(pilot_period + n_bits * bit_period) * repetition * 2^n_bits
For example, computing this for CAME turns out to be:
[(11520 + 320) + 12 * (320 + 640)] * 3 * 2^12 = 287.047.680 microseconds ~ 287 seconds