chinese_signature.sf

#!/usr/bin/ruby

# Daniel "Trizen" Șuteu
# Date: 02 August 2020
# https://github.com/trizen

# Compute the least Chinese signature of n with respect to the positive integers,
# such that when the Chinese Remainder Theorem (CRT) is aplied, n is constructed.

# Example:
#   The Chinese signature of 53 is [0, 1, 2, 1, 3], which represents:
#       53 == 0 (mod 1)
#       53 == 1 (mod 2)
#       53 == 2 (mod 3)
#       53 == 1 (mod 4)
#       53 == 3 (mod 5)

# By applying the CRT, we have:
#   chinese(Mod(0,1), Mod(1,2), Mod(2,3), Mod(1,4), Mod(3,5)) == 53

func chinese_signature(n) {

    var crt = Mod(0, 1)
    var sgn = [crt.lift]

    for k in (2..Inf) {

        if (crt.lift == n) {
            break
        }

        var r = Mod(n, k)
        crt = chinese(crt, r)
        sgn << r.lift
    }

    return sgn
}

for n in (0..50) {
    say "#{n} = #{chinese_signature(n)}"
}

__END__
0 = [0]
1 = [0, 1]
2 = [0, 0, 2]
3 = [0, 1, 0]
4 = [0, 0, 1]
5 = [0, 1, 2]
6 = [0, 0, 0, 2]
7 = [0, 1, 1, 3]
8 = [0, 0, 2, 0]
9 = [0, 1, 0, 1]
10 = [0, 0, 1, 2]
11 = [0, 1, 2, 3]
12 = [0, 0, 0, 0, 2]
13 = [0, 1, 1, 1, 3]
14 = [0, 0, 2, 2, 4]
15 = [0, 1, 0, 3, 0]
16 = [0, 0, 1, 0, 1]
17 = [0, 1, 2, 1, 2]
18 = [0, 0, 0, 2, 3]
19 = [0, 1, 1, 3, 4]
20 = [0, 0, 2, 0, 0]
21 = [0, 1, 0, 1, 1]
22 = [0, 0, 1, 2, 2]
23 = [0, 1, 2, 3, 3]
24 = [0, 0, 0, 0, 4]
25 = [0, 1, 1, 1, 0]
26 = [0, 0, 2, 2, 1]
27 = [0, 1, 0, 3, 2]
28 = [0, 0, 1, 0, 3]
29 = [0, 1, 2, 1, 4]
30 = [0, 0, 0, 2, 0]
31 = [0, 1, 1, 3, 1]
32 = [0, 0, 2, 0, 2]
33 = [0, 1, 0, 1, 3]
34 = [0, 0, 1, 2, 4]
35 = [0, 1, 2, 3, 0]
36 = [0, 0, 0, 0, 1]
37 = [0, 1, 1, 1, 2]
38 = [0, 0, 2, 2, 3]
39 = [0, 1, 0, 3, 4]
40 = [0, 0, 1, 0, 0]
41 = [0, 1, 2, 1, 1]
42 = [0, 0, 0, 2, 2]
43 = [0, 1, 1, 3, 3]
44 = [0, 0, 2, 0, 4]
45 = [0, 1, 0, 1, 0]
46 = [0, 0, 1, 2, 1]
47 = [0, 1, 2, 3, 2]
48 = [0, 0, 0, 0, 3]
49 = [0, 1, 1, 1, 4]
50 = [0, 0, 2, 2, 0]