Compute tail numbers from ICAO 24bit registration

[n0skill]

Transponders do not, in general, transmit the tail number of the aircraft.

Some aircraft use their tail number as their squawk code but this is not true in general.

Transponders do, however, transmit the ICAO address of the plane.

Some countries use an algorithm that uses the tail number to compute their ICAO adresses. The ICAO address being sent by ADS-B transponders, it should be doable to reverse the ICAO 24 bit to the tail number.

https://groups.yahoo.com/neo/groups/Mode_S/info

[n0skill]

User sno34 on https://discussions.flightaware.com/t/n-number-icao-address/18009/13 shared this bit of code

base9 = '123456789'  # The first digit (after the "N") is always one of these.
base10 = '0123456789' # The possible second and third digits are one of these.
# Note that "I" and "O" are never used as letters, to prevent confusion with "1" and "0"
base34 = 'ABCDEFGHJKLMNPQRSTUVWXYZ0123456789' 
icaooffset = 0xA00001 # The lowest possible number, N1, is this.
b1 = 101711 # basis between N1... and N2...
b2 = 10111 # basis between N10.... and N11....

def suffix(rem):
    """ Produces the alpha(numeric) suffix from a number 0 - 950 """
    if rem == 0:
        suf = ''
    else:
        if rem <= 600: #Class A suffix -- only letters.
            rem = rem - 1
            suf = base34[rem // 25]
            if rem % 25 > 0:
                suf = suf + base34(rem % 25) - 1] # second class A letter, if present.
        else:  #rem > 600 : First digit of suffix is a number.  Second digit may be blank, letter, or number.
            rem = rem - 601
            suf = base10[rem // 35]
            if rem % 35 > 0:
                suf = suf + base34(rem % 35) - 1]
    return suf

def enc_suffix(suf):
    """ Produces a remainder from a 0 - 2 digit suffix. 
    No error checking.  Using illegal strings will have strange results."""
    if len(suf) == 0:
        return 0
    r0 = base34.find(suf[0])
    if len(suf) == 1:
        r1 = 0
    else:
        r1 = base34.find(suf[1]) + 1
    if r0 < 24: # first char is a letter, use base 25
        return r0 * 25 + r1 + 1
    else:  # first is a number -- base 35.
        return r0 * 35 + r1 - 239   

def icao_to_tail(icao):
    if (icao < 0) or (icao > 0xadf7c7):
        return "Undefined"
    icao = icao - icaooffset
    d1 = icao // b1
    nnum = 'N' + base9[d1]
    r1 = icao % b1
    if r1 < 601:
        nnum = nnum + suffix(r1) # of the form N1ZZ
    else:
        d2 = (r1 - 601) // b2  # find second digit.
        nnum = nnum + base10[d2]
        r2 = (r1 - 601) % b2  # and residue after that
        if r2 < 601:  # No third digit. (form N12ZZ)
            nnum = nnum + suffix(r2)
        else:
            d3 = (r2 - 601) // 951 # Three-digits have extended suffix.
            r3 = (r2 - 601) % 951   
            nnum = nnum + base10[d3] + suffix(r3)
    return nnum 

def tail_to_icao(tail):
    if tail[0] != 'N':
        return -1
    icao = icaooffset
    icao = icao + base9.find(tail[1]) * b1
    if len(tail) == 2: # simple 'N3' etc.
        return icao
    d2 = base10.find(tail[2])
    if d2 == -1: # Form N1A
        icao = icao + enc_suffix(tail[2:4])
        return icao
    else: # Form N11... or N111..
        icao = icao + d2 * b2 + 601
        d3 = base10.find(tail[3])
        if d3 > -1: #Form N111 Suffix is base 35.
            icao = icao + d3 * 951 + 601
            icao = icao + enc_suffix(tail[4:6])
            return icao
        else:  #Form N11A
            icao = icao + enc_suffix(tail[3:5])
            return icao

[n0skill]

Partially fixed with afff8c677f8939066908dfa7056bce7e8eb30d81