scramble.py

# N.B. The following values work. Technically, you can use any s and p with p a primitive root modulo s.
# It works best if you use an s that is one more than a multiple of three. (e.g. 4, 7, 10, 13)
# It will work for any x up to 2^{s - 1} - 1
# inv calculates is the inverse of p in \mathbb{Z}_{s}.

# s  p
# 1  0
# 2  1
# 3  2
# 4  3
# 5  2,3
# 6  5
# 7  3,5
# 9  2,5
# 10 3,7
# 11 2,6,7,8
# 13 2,6,7,11
# 14 3,5
# 17 3,5,6,7,10,11,12,14
# 18 5,11
# 19 2,3,10,13,14,15
# 22 7,13,17,19
# 23 5,7,10,11,14,15,17,19,20,21
# 25 2,3,8,12,13,17,22,23
# 26 7,11,15,19
# 27 2,5,11,14,20,23
# 29 2,3,8,10,11,14,15,18,19,21,26,27
# 31 3,11,12,13,17,21,22,24
# 34 3,5,7,11,23,27,29,31
# 37 2,5,13,15,17,18,19,20,22,24,32,35
# 38 3,13,15,21,29,33
# 41 6,7,11,12,13,15,17,19,22,24,26,28,29,30,34,35
# 43 3,5,12,18,19,20,26,28,29,30,33,34
# 46 5,7,11,15,17,19,21,33,37,43
# 47 5,10,11,13,15,19,20,22,23,26,29,30,31,33,35,38,39,40,41,43,44,45
# 49 3,5,10,12,17,24,26,33,38,40,45,47
# 50 3,13,17,23,27,33,37,47
# 53 2,3,5,8,12,14,18,19,20,21,22,26,27,31,32,33,34,35,39,41,45,48,50,51
# 54 5,11,23,29,41,47
# 58 3,11,15,19,21,27,31,37,39,43,47,55
# 59 2,6,8,10,11,13,14,18,23,24,30,31,32,33,34,37,38,39,40,42,43,44,47,50,52,54,55,56
# 61 2,6,7,10,17,18,26,30,31,35,43,44,51,54,55,59
# 62 3,11,13,17,21,43,53,55
# 67 2,7,11,12,13,18,20,28,31,32,34,41,44,46,48,50,51,57,61,63
# 71 7,11,13,21,22,28,31,33,35,42,44,47,52,53,55,56,59,61,62,63,65,67,68,69

s = 13
p = 7
x = 473245

# First, take your number and turn it into a bitstring of the appropriate length.
y = bin(int(x))[2:].zfill(s - 1)
print(y)

def encode(y, s, p):
    l = ["*"] * (s - 1)
    
    i = 1
    tmp = y[0]
    for t in range(s):
        l[i - 1] = tmp
        tmp = y[i - 1]
        i = i * p % s
    return ''.join(l)

def inv(s, p):
    for x in range(s):
        if x * p % s == 1:
            return x
    return None

def decode(y, s, p):
    return encode(y, s, inv(s, p))

def arr2int(arr):
    return int(''.join(arr), 2)

# Now, we can encode that bitstring of an appropriate length using encode
print(encode(y, s, p))

# And turn that result into a triplet of numbers (that you can then turn into words the way you had before) up to 2^{(s - 1)/3 - 1}
print(list(map(arr2int, zip(*[iter(encode(y, s, p))]*((s - 1)//3)))))

# Decoding is pretty easy too
print(decode(encode(y, s, p), s, p))