Fast sum of digits in a ternary representation (Python)

You can use itertools.product to generate the digits and then convert to the string representation:

import itertools as it

def new(n):
    s = []
    for digits in it.product((0, 1, 2), repeat=n):
        if sum(digits) == n:
            s.append(''.join(str(x) for x in digits))
    return s

This gives me about 7x speedup:

In [8]: %timeit enumerateSpin(12)
2.39 s ± 7.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [9]: %timeit new(12)
347 ms ± 4.26 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Tested on Python 3.9.0 (IPython 7.20.0) (Linux).

The above procedure, using it.product, also generates numbers from which we know by reasoning that they don't obey the condition (this is the case for half of all numbers since the sum of digits must equal the number of digits). For n digits, we can compute the various counts of digits 2, 1 and 0 that eventually sum up to n. Then we can generate all distinct permutations of these digits and thus only generate relevant numbers:

import itertools as it
from more_itertools import distinct_permutations

def new2(n):
    all_digits = (('2',)*i + ('1',)*(n-2*i) + ('0',)*i for i in range(n//2+1))
    all_digits = it.chain.from_iterable(distinct_permutations(d) for d in all_digits)
    return (''.join(digits) for digits in all_digits)

Especially for large numbers of n this gives an additional, significant speedup:

In [44]: %timeit -r 1 -n 1 new(16)
31.4 s ± 0 ns per loop (mean ± std. dev. of 1 run, 1 loop each)

In [45]: %timeit -r 1 -n 1 list(new2(16))
7.82 s ± 0 ns per loop (mean ± std. dev. of 1 run, 1 loop each)

Note that the above solutions new and new2 have O(1) memory scaling (change new to yield instead of append).

A 10x improvement can be achieved by delegating all calculations to numpy in order to leverage vectorized processing:

def eSpin(n):
    nums    = np.arange(3**n,dtype=np.int)
    base3   = nums // (3**np.arange(n))[:,None] % 3
    matches = np.sum(base3,axis=0) == n
    digits  = np.sum(base3[:,matches] * 10**np.arange(n)[:,None],axis=0)
    return [f"{a:0{n}}" for a in digits]   

How it works (example for eSpin(3)):

nums is an array of all numbers up to 3**n

   [ 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]  

base3 converts it into base 3 digits in an additional dimension:

[[0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2]
 [0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 2 2 2]
 [0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2]]

matches identifies the columns where the sum of base3 digits matches n

 [0 0 0 0 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 0 0 0 0 0]

digits converts the matching columns into a base 10 number formed of the base3 digits

 [ 12  21 102 111 120 201 210]

And finally the matching (base10) numbers are formatted with leading zeros.

performance:

from timeit import timeit
count = 1

print(enumerateSpin(10)==eSpin(10)) # True

t1 = timeit(lambda:eSpin(13),number=count)
print("eSpin",t1) # 0.634 sec

t0 = timeit(lambda:enumerateSpin(13),number=count)
print("enumerateSpin",t0) # 7.362 sec

Tuple version:

def eSpin2(n):
    nums    = np.arange(3**n,dtype=np.int)
    base3   = nums// (3**np.arange(n))[:,None]  % 3
    matches = np.sum(base3,axis=0) == n
    return [*map(tuple,base3[:,matches].T)]

eSpin2(3)
[(2, 1, 0), (1, 2, 0), (2, 0, 1), (1, 1, 1), (0, 2, 1), (1, 0, 2), (0, 1, 2)]

[EDIT] An even faster approach (40x to 80x faster than enumerateSpin)

Using dynamic programming and memoization can provide a lot better performance:

@lru_cache()
def eSpin(n,base=3,target=None):
    if target is None: target = n
    if target == 0: return [(0,)*n]
    if target>base**n-1: return []
    if n==1: return [(target,)]
    result = []
    for d in range(min(base,target+1)):
        result.extend((d,)+suffix for suffix in eSpin(n-1,base,target-d) )
    return result

t4 = timeit(lambda:eSpin(13),number=count)
print("eSpin",t4) # 0.108 sec

eSpin.cache_clear()
t5 = timeit(lambda:eSpin(16),number=count)
print("eSpin",t5) # 2.25 sec

Here's a multiprocessing approach. It'll afford more time savings, the more the problem scales in size

import multiprocessing as mp


def filter(n, qIn, qOut):  # this is the function that will be parallelized
    nums = range(3**n)
    answer = []
    for low,high in iter(qIn.get, None):
        for num in nums[low:high]:
            r = np.base_repr(num, 3)  # ternary representation
            if sum(int(i) for i in r) == num:  # this is your check
                answer.append('0'*(n-len(r)) +r)  # turn it into a fixed length
    qOut.put(answer)
    qOut.put(None)


def enumerateSpin(n):  # this is the primary entry point
    numProcs = mp.cpu_count()-1  # fiddle to taste
    chunkSize = n//numProcs

    qIn, qOut = [mp.Queue() for _ in range(2)]
    procs = [mp.Process(target=filter, args=(n, qIn, qOut)) for _ in range(numProcs)]

    for p in procs: p.start()
    for i in range(0, 3**n, chunkSize):  # chunkify your numbers so that IPC is more efficient
        qIn.put((i, i+chunkSize))
    for p in procs: qIn.put(None)

    answer = []
    done = 0
    while done < len(procs):
        t = qOut.get()
        if t is None:
            done += 1
            continue
        answer.extend(t)

    for p in procs: p.terminate()

    return answer

In general, to get the digits of a number in a specific base we can do:

while num > 0:
    digit = num % base
    num //= base
    print(digit)

When running this with num = 14, base = 3 we get:

2
1
1

Which means that 14 in ternary is 112.
We can extract that into a method digits(num, base) and only use np.base_repr(a,3) when we actualy need to convert the number into a string:

def enumerateSpin(n):
    s = []
    for a in range(0,3**n):
        if sum(digits(a, 3)) == n:
            ternary_rep = np.base_repr(a,3)
            k = len(ternary_rep)
            r = (n-k)*'0'+ternary_rep
            s.append(r)
    return s

Output for enumerateSpin(4):

['0022', '0112', '0121', '0202', '0211', '0220', '1012', '1021', '1102', '1111', '1120', '1201', '1210', '2002', '2011', '2020', '2101', '2110', '2200']