python - 将递归 python 代码转换为非递归版本

Question

除非我们开始增加 distinct 和 n-symbols 和长度，否则此处提供的代码有效，例如，在我的计算机上 n_symbols=512, length=512, distinct=300 最终出现此错误 RecursionError: maximum比较时超出递归深度，如果我增加 lru_cache 值，则会出现溢出错误。
我想要的是拥有此代码的非递归版本。

from functools import lru_cache
@lru_cache
def get_permutations_count(n_symbols, length, distinct, used=0):
    '''
     - n_symbols: number of symbols in the alphabet
     - length: the number of symbols in each sequence
     - distinct: the number of distinct symbols in each sequence
    '''
    if distinct < 0:
        return 0
    if length == 0:
        return 1 if distinct == 0 else 0
    else:
        return \
          get_permutations_count(n_symbols, length-1, distinct-0, used+0) * used + \
          get_permutations_count(n_symbols, length-1, distinct-1, used+1) * (n_symbols - used)

然后

get_permutations_count(n_symbols=300, length=300, distinct=270)

在大约 0.5 秒内给出答案

2729511887951350984580070745513114266766906881300774347439917775
7093985721949669285469996223829969654724957176705978029888262889
8157939885553971500652353177628564896814078569667364402373549268
5524290993833663948683375995196081654415976659499171897405039547
1546236260377859451955180752885715923847446106509971875543496023
2494854876774756172488117802642800540206851318332940739395445903
6305051887120804168979339693187702655904071331731936748927759927
3688881301614948043182289382736687065840703041231428800720854767
0713406956719647313048146023960093662879015837313428567467555885
3564982943420444850950866922223974844727296000000000000000000000
000000000000000000000000000000000000000000000000

Answer 1

这是我的:

def get_permutations_count_improved(n_symbols, length, distinct):
    if distinct > length or distinct > n_symbols:
        return 0
    ways = [1]
    for _ in range(length):
        ways = [used * (distinct - d) + new
               for d, used, new in zip(range(distinct+1), [*ways, 0], [0, *ways])]
    return ways[distinct] * comb(n_symbols, distinct) * factorial(distinct)

一些参数集的速度比较:

n_symbols length distinct   yours    mine
   300      300    270      0.62 s   0.012 s (~51 times faster)
   512      512    300        -      0.035 s
  1024     1024    600        -      0.22 s
  3000     3000   2700        -      6.0 s

在我的最后一行中，您看到我将总体结果分为三个因素:

• comb(n_symbols, distinct) 用于选择实际使用 n_symbols 符号中的哪些 distinct。这基本上摆脱了 n_symbols 参数，或者将其视为补偿设置 n_symbols = distinct。
• factorial(distinct) 符号首先使用的顺序。这消除了您重复出现的 * (n_symbols - used)。
• ways[distinct] 是构建长度为 length 且具有完全不同的 distinct 不同符号的序列的方法数，其中顺序为他们首先使用的是固定的。

将 ways 表视为二维表可能更容易:ways[length][distinct]。但是为了提高内存效率，我逐行计算它并且只保留最新的一行。

基准测试和一些正确性检查(Try it online!):

from timeit import timeit
from functools import lru_cache
from math import comb, factorial

@lru_cache
def get_permutations_count(n_symbols, length, distinct, used=0):
    '''
     - n_symbols: number of symbols in the alphabet
     - length: the number of symbols in each sequence
     - distinct: the number of distinct symbols in each sequence
    '''
    if distinct < 0:
        return 0
    if length == 0:
        return 1 if distinct == 0 else 0
    else:
        return \
          get_permutations_count(n_symbols, length-1, distinct-0, used+0) * used + \
          get_permutations_count(n_symbols, length-1, distinct-1, used+1) * (n_symbols - used)

def get_permutations_count_improved(n_symbols, length, distinct):
    if distinct > length or distinct > n_symbols:
        return 0
    ways = [1]
    for _ in range(length):
        ways = [used * (distinct - d) + new
               for d, used, new in zip(range(distinct+1), [*ways, 0], [0, *ways])]
    return ways[distinct] * comb(n_symbols, distinct) * factorial(distinct)

funcs = get_permutations_count, get_permutations_count_improved

# Check correctness
stop = 20
for a in range(stop):
    for b in range(stop):
        for c in range(stop):
            expect = get_permutations_count(a, b, c)
            result = get_permutations_count_improved(a, b, c)
            assert result == expect, (a, b, c, expect, result)

# Benchmark
n_symbols, length, distinct = 300, 300, 270
#n_symbols, length, distinct = 512, 512, 300
#n_symbols, length, distinct = 1024, 1024, 600
#n_symbols, length, distinct = 3000, 3000, 2700
for func in funcs[0:] * 3:
    funcs[0].cache_clear()
    t = timeit(lambda: func(n_symbols, length, distinct), number=1)
    print('%.3f seconds ' % t, func.__name__)