3 ms·
Well let's do the math 125 kHz 2 bits/cycle = 250 bits /sec A 2^40 bit string has 1099511627776 combinations, and would require 1099511627776 40 = 43980465111
by TheSoftwareGuy 9y ago
Well let's do the math
125 kHz 2 bits/cycle = 250 bits /sec
A 2^40 bit string has 1099511627776 combinations, and would require 1099511627776 40 = 43980465111040 bits to besent to try all of them. That would take 43980465111040/250 seconds or about 5575 years
And that's under ideal "spherical object in a vacuum" conditions.
So yeah your right, unless they get lucky.
- thatsso1999 9y agoI believe your math is off by a factor of 1000. 125kHz = 125,000 cycles/sec * 2 bits/cycle = 250,000 bits/sec 250,000 bits/sec / 40 bits/code = 6250 codes/sec 2^40 = 1,099,511,627,776 possible codes / 6250 codes/sec ~= 175,921,860 sec ~= 5.578 years While this is still an extremely long time for the worst case, by the looks of other comments, as well as the author's video, it appears extremely doubtful that most RFID readers have anywhere near 40 bits of security - and it takes about 10 seconds (65536/6250) for the fuzzer to brute force all codes 16 bits or less.
- TheSoftwareGuy 9y agoShoot you're right, it's kHz not Hz haha I'll update my comment.
- TheSoftwareGuy 9y agoDarn I don't think HN is letting me update my comment.
- mrb 9y agoBruteforcing is totally doable because in practice the unique identifiers broadcasted by these tags are allocated sequentially. I built an AVRFID a few years ago to bruteforce the HID ProxCard II RFIDs at work (45 bits of data: 20-bit mfg ID, 8-bit facility ID, 16-bit tag ID, 1 odd parity bit). Starting from my ID, and incrementing it one by one, it only took seconds to find many other valid IDs that were unlocking doors of various restricted areas. Imagine a disgruntled employee who has his legit RFID and is looking to access other restricted areas to which he normally doesn't have access... bruteforcing works perfectly fine for him.