I'd like to use the Damm algorithm to generate check digits for codes with a 32-character alphabet. The algorithm itself is easily applied to any base (except 2 or 6). The difficulty is the necessary look-up table, which must be a totally anti-symmetric quasigroup with a single character (usually 0) down the main diagonal.
The Wikipedia page gives a table for base 10, and the Python implementation gives a table for base-16, but I haven't found a base-32 example. Does anyone have a suitable table for base-32?
Inspired by David Eisenstat's answer (and the original dissertation by Damm that he cites), here's a simple Python routine to calculate / verify a Damm checksum for any base 2n for 2 ≤ n ≤ 32:
# reduction bitmasks for GF(2^n), 2 <= n <= 32
masks = (3, 3, 3, 5, 3, 3, 27, 3, 9, 5, 9, 27, 33, 3, 43, 9,
9, 39, 9, 5, 3, 33, 27, 9, 27, 39, 3, 5, 3, 9, 141)
# calculate Damm checksum for base 2^n
def damm2n (digits, n):
modulus = (1 << n)
mask = modulus | masks[n - 2]
checksum = 0
for digit in digits:
checksum ^= digit
checksum <<= 1
if checksum >= modulus: checksum ^= mask
return checksum
The routine takes a list (or, more generally, an iterable) of digits, which are assumed to be integers in the range 0 to 2n−1 inclusive, and the number n of bits per digit (assumed to be in the range 2 to 32 inclusive).
The totally asymmetric quasigroup used by this implementation of the Damm algorithm is given by the map (a, b) ↦ 2 ⊗ (a ⊕ b), where ⊕ denotes addition in the finite field GF(2n) (which is simply bitwise XOR), ⊗ denotes multiplication in GF(2n), and 2 denotes the element represented by the bitstring 0...0102 in the usual n-bit representation of GF(2n).
This is equivalent to the map (a, b) ↦ (2 ⊗ a) ⊕ b given by Damm in Example 5.2 of his thesis, except that the input digits are b permuted (by multiplying them with 2 in GF(2n)) to ensure that (a, a) ↦ 0 for all a. This is equivalent to permuting the columns of the quasigroup operation table so that the diagonal is all zeros, and allows the checksum to be verified simply by appending it to the original input and checking that the new checksum of the extended input is zero.
The GF(2n) multiplication by 2 is implemented using the usual trick of shifting left by one and, if the n-th bit of the result is set, XORing it with a bitmask corresponding to a monic irreducible polynomial of order n. The specific bitmasks used are taken from the Table of Low-Weight Binary Irreducible Polynomials by Gadiel Seroussi (1998). If you (for some reason) need checksums for bases larger than 232, their table goes up to a whopping 210,000. The Seroussi table lists the exponents of the non-zero coefficients of each reduction polynomial, excluding the constant term; the bitmasks in the code above are obtained by discarding the highest exponent (which is always n), summing together 2k for the other exponents k and adding 1. (Thus, for example, the entry "8,4,3,1" for n = 8 yields the mask 24 + 23 + 21 + 1 = 16 + 8 + 2 + 1 = 27.)
In particular, the code above yields, for n = 4, results matching the base-16 Damm checksum implementation by Johannes Spielmann. This is not guaranteed in general, since there are many possible ways to implement a Damm checksum for a given base, but in this case, the quasigroups used by the two implementations happen to match.
Ps. Here's some Python code to print out the look-up table in a format similar to what the Wikipedia article uses. (Thanks to CJM for the initial version.)
alphabet = '0123456789ABCDEFGHJKLMNPQRTUVWXY' # avoids easy-to-confuse characters
bits = 5
# find out which first single-digit input gives which checksum
r = [-1] * 2**bits
for i in range(2**bits): r[damm2n([i], bits)] = i
# print header
print ' |', ' '.join(alphabet)
print '--+' + '--' * len(alphabet)
# print rest of table
for i in range(2**bits):
row = (alphabet[damm2n([r[i], j], bits)] for j in range(2**bits))
print alphabet[i], '|', ' '.join(row)
To summarize the discussion below: we would like a table that has zeros on the main diagonal. Niklas and I have the impression that, rather than being an essential part of the algorithm, this property is merely to avoid solving the equation x*y = 0 in y for x given, where * is the quasigroup operation. With zeros on the main diagonal, we have x = y, but without, we can compute y via one lookup in a 32-element table.
The construction that Damm describes is problematic because it has the desired property if and only if a = -1, but, in characteristic 2, we have 1 = -1. The constraint solver Z3 was not helpful.
Damm's dissertation (in German) is here. The relevant definition is that a Latin square is totally anti-symmetric iff [for all x and y, the (x,y) element equals the (y,x) element iff x = y]. Damm gives a construction for prime power n other than 2 (including the case n = 32) by taking the (x,y) element to be a*x + y, where a is neither 0 nor 1 and * is multiplication over an n-element Galois field (Beispiel 5.2).
Below is the instantiation of this method for n = 32.
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31],
[2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13, 18, 19, 16, 17, 22, 23, 20, 21, 26, 27, 24, 25, 30, 31, 28, 29],
[4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11, 20, 21, 22, 23, 16, 17, 18, 19, 28, 29, 30, 31, 24, 25, 26, 27],
[6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9, 22, 23, 20, 21, 18, 19, 16, 17, 30, 31, 28, 29, 26, 27, 24, 25],
[8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7, 24, 25, 26, 27, 28, 29, 30, 31, 16, 17, 18, 19, 20, 21, 22, 23],
[10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5, 26, 27, 24, 25, 30, 31, 28, 29, 18, 19, 16, 17, 22, 23, 20, 21],
[12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3, 28, 29, 30, 31, 24, 25, 26, 27, 20, 21, 22, 23, 16, 17, 18, 19],
[14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1, 30, 31, 28, 29, 26, 27, 24, 25, 22, 23, 20, 21, 18, 19, 16, 17],
[16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[18, 19, 16, 17, 22, 23, 20, 21, 26, 27, 24, 25, 30, 31, 28, 29, 2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13],
[20, 21, 22, 23, 16, 17, 18, 19, 28, 29, 30, 31, 24, 25, 26, 27, 4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11],
[22, 23, 20, 21, 18, 19, 16, 17, 30, 31, 28, 29, 26, 27, 24, 25, 6, 7, 4, 5, 2, 3, 0, 1, 14, 15, 12, 13, 10, 11, 8, 9],
[24, 25, 26, 27, 28, 29, 30, 31, 16, 17, 18, 19, 20, 21, 22, 23, 8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7],
[26, 27, 24, 25, 30, 31, 28, 29, 18, 19, 16, 17, 22, 23, 20, 21, 10, 11, 8, 9, 14, 15, 12, 13, 2, 3, 0, 1, 6, 7, 4, 5],
[28, 29, 30, 31, 24, 25, 26, 27, 20, 21, 22, 23, 16, 17, 18, 19, 12, 13, 14, 15, 8, 9, 10, 11, 4, 5, 6, 7, 0, 1, 2, 3],
[30, 31, 28, 29, 26, 27, 24, 25, 22, 23, 20, 21, 18, 19, 16, 17, 14, 15, 12, 13, 10, 11, 8, 9, 6, 7, 4, 5, 2, 3, 0, 1],
[5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10, 21, 20, 23, 22, 17, 16, 19, 18, 29, 28, 31, 30, 25, 24, 27, 26],
[7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8, 23, 22, 21, 20, 19, 18, 17, 16, 31, 30, 29, 28, 27, 26, 25, 24],
[1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14, 17, 16, 19, 18, 21, 20, 23, 22, 25, 24, 27, 26, 29, 28, 31, 30],
[3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12, 19, 18, 17, 16, 23, 22, 21, 20, 27, 26, 25, 24, 31, 30, 29, 28],
[13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2, 29, 28, 31, 30, 25, 24, 27, 26, 21, 20, 23, 22, 17, 16, 19, 18],
[15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16],
[9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6, 25, 24, 27, 26, 29, 28, 31, 30, 17, 16, 19, 18, 21, 20, 23, 22],
[11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4, 27, 26, 25, 24, 31, 30, 29, 28, 19, 18, 17, 16, 23, 22, 21, 20],
[21, 20, 23, 22, 17, 16, 19, 18, 29, 28, 31, 30, 25, 24, 27, 26, 5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10],
[23, 22, 21, 20, 19, 18, 17, 16, 31, 30, 29, 28, 27, 26, 25, 24, 7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8],
[17, 16, 19, 18, 21, 20, 23, 22, 25, 24, 27, 26, 29, 28, 31, 30, 1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14],
[19, 18, 17, 16, 23, 22, 21, 20, 27, 26, 25, 24, 31, 30, 29, 28, 3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12],
[29, 28, 31, 30, 25, 24, 27, 26, 21, 20, 23, 22, 17, 16, 19, 18, 13, 12, 15, 14, 9, 8, 11, 10, 5, 4, 7, 6, 1, 0, 3, 2],
[31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0],
[25, 24, 27, 26, 29, 28, 31, 30, 17, 16, 19, 18, 21, 20, 23, 22, 9, 8, 11, 10, 13, 12, 15, 14, 1, 0, 3, 2, 5, 4, 7, 6],
[27, 26, 25, 24, 31, 30, 29, 28, 19, 18, 17, 16, 23, 22, 21, 20, 11, 10, 9, 8, 15, 14, 13, 12, 3, 2, 1, 0, 7, 6, 5, 4]]
Below is the extremely crappy Haskell code that made this. It might work for other powers of two. The argument 5 in main is for 2^5.
module Main where
import Data.Char
import Data.List
xs +. ys = simplify (xs ++ ys)
xs *. ys
= simplify $
do x <- xs
y <- ys
return (x + y)
simplify xs
= (reverse . map head . filter (odd . length) . group . sort) xs
subseqs [] = [[]]
subseqs (x : xs) = let xss = subseqs xs in xss ++ map (x :) xss
polys n = subseqs [n, n - 1 .. 0]
reduce [] ys = ys
reduce xs [] = []
reduce xs@(x : _) ys@(y : _)
= if x > y then ys else reduce xs (map ((y - x) +) xs +. ys)
irred [] = False
irred ys@(y : _)
= let xss = polys (y `div` 2) \\ [[0]] in
(not . any null . map (flip reduce ys)) xss
irreds n = filter irred (polys n)
ip n = (head . filter irred . map (n :)) (polys (n - 1))
eval xs = (sum . map (2 ^)) xs
timesTable n
= let ms = ip n
zs = polys (n - 1) !! 2
in
do xs <- polys (n - 1)
return $
do ys <- polys (n - 1)
return (reduce ms ((zs *. xs) +. ys))
verify t
= all ((1 ==) . length . filter id) $
zipWith (zipWith (==)) t (transpose t)
main = print $ map (map eval) $ timesTable 5
If you have a TA quasigroup you can simply rearrange the columns in such a way that the 0 are on the main diagonal. Then the quasigroup is (in general) a WTA quasigroup and can be used for the Damm algorithm. I've done this for order 32, see result below, and this is possible for every order except 2 and 6.
I think the quasigroup of order 10, which can be found in Wikipedia, is constructed by Lemma 5.2 of Damm's theses. This is because it should still detect the phonetic errors after rearranging the columns, so the elements need to be renamed and the rows need to be rearranged accordingly.
Finally, here is the WTA quasigroup of order 32 for the Damm algorithm:
00 02 04 06 08 10 12 14 16 18 20 22 24 26 28 30 03 01 07 05 11 09 15 13 19 17 23 21 27 25 31 29
02 00 06 04 10 08 14 12 18 16 22 20 26 24 30 28 01 03 05 07 09 11 13 15 17 19 21 23 25 27 29 31
04 06 00 02 12 14 08 10 20 22 16 18 28 30 24 26 07 05 03 01 15 13 11 09 23 21 19 17 31 29 27 25
06 04 02 00 14 12 10 08 22 20 18 16 30 28 26 24 05 07 01 03 13 15 09 11 21 23 17 19 29 31 25 27
08 10 12 14 00 02 04 06 24 26 28 30 16 18 20 22 11 09 15 13 03 01 07 05 27 25 31 29 19 17 23 21
10 08 14 12 02 00 06 04 26 24 30 28 18 16 22 20 09 11 13 15 01 03 05 07 25 27 29 31 17 19 21 23
12 14 08 10 04 06 00 02 28 30 24 26 20 22 16 18 15 13 11 09 07 05 03 01 31 29 27 25 23 21 19 17
14 12 10 08 06 04 02 00 30 28 26 24 22 20 18 16 13 15 09 11 05 07 01 03 29 31 25 27 21 23 17 19
16 18 20 22 24 26 28 30 00 02 04 06 08 10 12 14 19 17 23 21 27 25 31 29 03 01 07 05 11 09 15 13
18 16 22 20 26 24 30 28 02 00 06 04 10 08 14 12 17 19 21 23 25 27 29 31 01 03 05 07 09 11 13 15
20 22 16 18 28 30 24 26 04 06 00 02 12 14 08 10 23 21 19 17 31 29 27 25 07 05 03 01 15 13 11 09
22 20 18 16 30 28 26 24 06 04 02 00 14 12 10 08 21 23 17 19 29 31 25 27 05 07 01 03 13 15 09 11
24 26 28 30 16 18 20 22 08 10 12 14 00 02 04 06 27 25 31 29 19 17 23 21 11 09 15 13 03 01 07 05
26 24 30 28 18 16 22 20 10 08 14 12 02 00 06 04 25 27 29 31 17 19 21 23 09 11 13 15 01 03 05 07
28 30 24 26 20 22 16 18 12 14 08 10 04 06 00 02 31 29 27 25 23 21 19 17 15 13 11 09 07 05 03 01
30 28 26 24 22 20 18 16 14 12 10 08 06 04 02 00 29 31 25 27 21 23 17 19 13 15 09 11 05 07 01 03
03 01 07 05 11 09 15 13 19 17 23 21 27 25 31 29 00 02 04 06 08 10 12 14 16 18 20 22 24 26 28 30
01 03 05 07 09 11 13 15 17 19 21 23 25 27 29 31 02 00 06 04 10 08 14 12 18 16 22 20 26 24 30 28
07 05 03 01 15 13 11 09 23 21 19 17 31 29 27 25 04 06 00 02 12 14 08 10 20 22 16 18 28 30 24 26
05 07 01 03 13 15 09 11 21 23 17 19 29 31 25 27 06 04 02 00 14 12 10 08 22 20 18 16 30 28 26 24
11 09 15 13 03 01 07 05 27 25 31 29 19 17 23 21 08 10 12 14 00 02 04 06 24 26 28 30 16 18 20 22
09 11 13 15 01 03 05 07 25 27 29 31 17 19 21 23 10 08 14 12 02 00 06 04 26 24 30 28 18 16 22 20
15 13 11 09 07 05 03 01 31 29 27 25 23 21 19 17 12 14 08 10 04 06 00 02 28 30 24 26 20 22 16 18
13 15 09 11 05 07 01 03 29 31 25 27 21 23 17 19 14 12 10 08 06 04 02 00 30 28 26 24 22 20 18 16
19 17 23 21 27 25 31 29 03 01 07 05 11 09 15 13 16 18 20 22 24 26 28 30 00 02 04 06 08 10 12 14
17 19 21 23 25 27 29 31 01 03 05 07 09 11 13 15 18 16 22 20 26 24 30 28 02 00 06 04 10 08 14 12
23 21 19 17 31 29 27 25 07 05 03 01 15 13 11 09 20 22 16 18 28 30 24 26 04 06 00 02 12 14 08 10
21 23 17 19 29 31 25 27 05 07 01 03 13 15 09 11 22 20 18 16 30 28 26 24 06 04 02 00 14 12 10 08
27 25 31 29 19 17 23 21 11 09 15 13 03 01 07 05 24 26 28 30 16 18 20 22 08 10 12 14 00 02 04 06
25 27 29 31 17 19 21 23 09 11 13 15 01 03 05 07 26 24 30 28 18 16 22 20 10 08 14 12 02 00 06 04
31 29 27 25 23 21 19 17 15 13 11 09 07 05 03 01 28 30 24 26 20 22 16 18 12 14 08 10 04 06 00 02
29 31 25 27 21 23 17 19 13 15 09 11 05 07 01 03 30 28 26 24 22 20 18 16 14 12 10 08 06 04 02 00
来源：https://stackoverflow.com/questions/23431621/extending-the-damm-algorithm-to-base-32