Question

我有5个草莓，2个柠檬和一个香蕉。对于这些的每种可能的组合（包括选择0），共有对象总数。我最终想要列出这些和出现的频率。

[1 strawberry, 0 lemons, 0 bananas] = 1 objects  
[2 strawberries, 0 lemons, 1 banana] = 3 objects  
[0 strawberries, 1 lemon, 0 bananas] = 1 objects  
[2 strawberries, 1 lemon, 0 bananas] = 3 objects  
[3 strawberries, 0 lemons, 0 bananas] = 3 objects

对于以上5种组合的选择，“ 1”的频率为2，“ 3”的频率为3。

显然，有更多可能的组合，每种组合都会改变频率结果。是否有一种公式化的方法可以解决整个组合的频率问题？

当前，我已经在Python中设置了蛮力功能。

special_cards = {
    'A':7, 'B':1,   'C':1,  'D':1,  'E':1,  'F':1,  'G':1,  'H':1,  'I':1,  'J':1, 'K':1,   'L':1,  
    'M':1,  'N':1,  'O':1,  'P':1,  'Q':1,  'R':1,  'S':1,  'T':1,  'U':1,  'V':1,  'W':1,  'X':1,  
    'Y':1,  'Z':1,  'AA':1, 'AB':1, 'AC':1, 'AD':1, 'AE':1, 'AF':1, 'AG':1, 'AH':1, 'AI':1, 'AJ':1, 
    'AK':1, 'AL':1, 'AM':1, 'AN':1, 'AO':1, 'AP':1, 'AQ':1, 'AR':1, 'AS':1, 'AT':1, 'AU':1, 'AV':1, 
    'AW':1, 'AX':1, 'AY':1
}

def _calc_dis_specials(special_cards):
    """Calculate the total combinations when special cards are factored in"""
    # Create an iterator for special card combinations.
    special_paths = _gen_dis_special_list(special_cards)
    freq = {}
    path_count = 0
    for o_path in special_paths:  # Loop through the iterator
        path_count += 1  # Keep track of how many combinations we've evaluated thus far.
        try:  # I've been told I can use a collections.counter() object instead of try/except.
            path_sum = sum(o_path)  # Sum the path (counting objects)
            new_count = freq[path_sum] + 1  # Try to increment the count for our sum.
            freq.update({path_sum: new_count})
        except KeyError:  
            freq.update({path_sum: 1})
            print(f"{path_count:,}\n{freq}")
    print(f"{path_count:,}\n{freq}")
    # Do things with results yadda yadda

def _gen_dis_special_list(special_cards):
    """Generates an iterator for all combinations for special cards"""
    product_args = []
    for value in special_cards.values():  # A card's "value" is the maximum number that can be in a deck.
        product_args.append(range(value+1))  # Populates product_args with lists of each card's possible count.
    result = itertools.product(*product_args)
    return result

但是，对于大量的对象池（超过50个），阶乘就变得一发不可收拾。数十亿种组合。我需要一种公式化的方法。

看一些输出，我注意到一些事情：

1
{0: 1}
2
{0: 1, 1: 1}
4
{0: 1, 1: 2, 2: 1}
8
{0: 1, 1: 3, 2: 3, 3: 1}
16
{0: 1, 1: 4, 2: 6, 3: 4, 4: 1}
32
{0: 1, 1: 5, 2: 10, 3: 10, 4: 5, 5: 1}
64
{0: 1, 1: 6, 2: 15, 3: 20, 4: 15, 5: 6, 6: 1}
128
{0: 1, 1: 7, 2: 21, 3: 35, 4: 35, 5: 21, 6: 7, 7: 1}
256
{0: 1, 1: 8, 2: 28, 3: 56, 4: 70, 5: 56, 6: 28, 7: 8, 8: 1}
512
{0: 1, 1: 9, 2: 36, 3: 84, 4: 126, 5: 126, 6: 84, 7: 36, 8: 9, 9: 1}
1,024
{0: 1, 1: 10, 2: 45, 3: 120, 4: 210, 5: 252, 6: 210, 7: 120, 8: 45, 9: 10, 10: 1}
2,048
{0: 1, 1: 11, 2: 55, 3: 165, 4: 330, 5: 462, 6: 462, 7: 330, 8: 165, 9: 55, 10: 11, 11: 1}
4,096
{0: 1, 1: 12, 2: 66, 3: 220, 4: 495, 5: 792, 6: 924, 7: 792, 8: 495, 9: 220, 10: 66, 11: 12, 12: 1}
8,192
{0: 1, 1: 13, 2: 78, 3: 286, 4: 715, 5: 1287, 6: 1716, 7: 1716, 8: 1287, 9: 715, 10: 286, 11: 78, 12: 13, 13: 1}
16,384
{0: 1, 1: 14, 2: 91, 3: 364, 4: 1001, 5: 2002, 6: 3003, 7: 3432, 8: 3003, 9: 2002, 10: 1001, 11: 364, 12: 91, 13: 14, 14: 1}
32,768
{0: 1, 1: 15, 2: 105, 3: 455, 4: 1365, 5: 3003, 6: 5005, 7: 6435, 8: 6435, 9: 5005, 10: 3003, 11: 1365, 12: 455, 13: 105, 14: 15, 15: 1}
65,536
{0: 1, 1: 16, 2: 120, 3: 560, 4: 1820, 5: 4368, 6: 8008, 7: 11440, 8: 12870, 9: 11440, 10: 8008, 11: 4368, 12: 1820, 13: 560, 14: 120, 15: 16, 16: 1}
131,072
{0: 1, 1: 17, 2: 136, 3: 680, 4: 2380, 5: 6188, 6: 12376, 7: 19448, 8: 24310, 9: 24310, 10: 19448, 11: 12376, 12: 6188, 13: 2380, 14: 680, 15: 136, 16: 17, 17: 1}
262,144
{0: 1, 1: 18, 2: 153, 3: 816, 4: 3060, 5: 8568, 6: 18564, 7: 31824, 8: 43758, 9: 48620, 10: 43758, 11: 31824, 12: 18564, 13: 8568, 14: 3060, 15: 816, 16: 153, 17: 18, 18: 1}
524,288
{0: 1, 1: 19, 2: 171, 3: 969, 4: 3876, 5: 11628, 6: 27132, 7: 50388, 8: 75582, 9: 92378, 10: 92378, 11: 75582, 12: 50388, 13: 27132, 14: 11628, 15: 3876, 16: 969, 17: 171, 18: 19, 19: 1}
1,048,576
{0: 1, 1: 20, 2: 190, 3: 1140, 4: 4845, 5: 15504, 6: 38760, 7: 77520, 8: 125970, 9: 167960, 10: 184756, 11: 167960, 12: 125970, 13: 77520, 14: 38760, 15: 15504, 16: 4845, 17: 1140, 18: 190, 19: 20, 20: 1}
2,097,152
{0: 1, 1: 21, 2: 210, 3: 1330, 4: 5985, 5: 20349, 6: 54264, 7: 116280, 8: 203490, 9: 293930, 10: 352716, 11: 352716, 12: 293930, 13: 203490, 14: 116280, 15: 54264, 16: 20349, 17: 5985, 18: 1330, 19: 210, 20: 21, 21: 1}
4,194,304
{0: 1, 1: 22, 2: 231, 3: 1540, 4: 7315, 5: 26334, 6: 74613, 7: 170544, 8: 319770, 9: 497420, 10: 646646, 11: 705432, 12: 646646, 13: 497420, 14: 319770, 15: 170544, 16: 74613, 17: 26334, 18: 7315, 19: 1540, 20: 231, 21: 22, 22: 1}
8,388,608
{0: 1, 1: 23, 2: 253, 3: 1771, 4: 8855, 5: 33649, 6: 100947, 7: 245157, 8: 490314, 9: 817190, 10: 1144066, 11: 1352078, 12: 1352078, 13: 1144066, 14: 817190, 15: 490314, 16: 245157, 17: 100947, 18: 33649, 19: 8855, 20: 1771, 21: 253, 22: 23, 23: 1}
16,777,216
{0: 1, 1: 24, 2: 276, 3: 2024, 4: 10626, 5: 42504, 6: 134596, 7: 346104, 8: 735471, 9: 1307504, 10: 1961256, 11: 2496144, 12: 2704156, 13: 2496144, 14: 1961256, 15: 1307504, 16: 735471, 17: 346104, 18: 134596, 19: 42504, 20: 10626, 21: 2024, 22: 276, 23: 24, 24: 1}
33,554,432
{0: 1, 1: 25, 2: 300, 3: 2300, 4: 12650, 5: 53130, 6: 177100, 7: 480700, 8: 1081575, 9: 2042975, 10: 3268760, 11: 4457400, 12: 5200300, 13: 5200300, 14: 4457400, 15: 3268760, 16: 2042975, 17: 1081575, 18: 480700, 19: 177100, 20: 53130, 21: 12650, 22: 2300, 23: 300, 24: 25, 25: 1}
67,108,864
{0: 1, 1: 26, 2: 325, 3: 2600, 4: 14950, 5: 65780, 6: 230230, 7: 657800, 8: 1562275, 9: 3124550, 10: 5311735, 11: 7726160, 12: 9657700, 13: 10400600, 14: 9657700, 15: 7726160, 16: 5311735, 17: 3124550, 18: 1562275, 19: 657800, 20: 230230, 21: 65780, 22: 14950, 23: 2600, 24: 325, 25: 26, 26: 1}
134,217,728
{0: 1, 1: 27, 2: 351, 3: 2925, 4: 17550, 5: 80730, 6: 296010, 7: 888030, 8: 2220075, 9: 4686825, 10: 8436285, 11: 13037895, 12: 17383860, 13: 20058300, 14: 20058300, 15: 17383860, 16: 13037895, 17: 8436285, 18: 4686825, 19: 2220075, 20: 888030, 21: 296010, 22: 80730, 23: 17550, 24: 2925, 25: 351, 26: 27, 27: 1}
268,435,456
{0: 1, 1: 28, 2: 378, 3: 3276, 4: 20475, 5: 98280, 6: 376740, 7: 1184040, 8: 3108105, 9: 6906900, 10: 13123110, 11: 21474180, 12: 30421755, 13: 37442160, 14: 40116600, 15: 37442160, 16: 30421755, 17: 21474180, 18: 13123110, 19: 6906900, 20: 3108105, 21: 1184040, 22: 376740, 23: 98280, 24: 20475, 25: 3276, 26: 378, 27: 28, 28: 1}
536,870,912
{0: 1, 1: 29, 2: 406, 3: 3654, 4: 23751, 5: 118755, 6: 475020, 7: 1560780, 8: 4292145, 9: 10015005, 10: 20030010, 11: 34597290, 12: 51895935, 13: 67863915, 14: 77558760, 15: 77558760, 16: 67863915, 17: 51895935, 18: 34597290, 19: 20030010, 20: 10015005, 21: 4292145, 22: 1560780, 23: 475020, 24: 118755, 25: 23751, 26: 3654, 27: 406, 28: 29, 29: 1}
1,073,741,824
{0: 1, 1: 30, 2: 435, 3: 4060, 4: 27405, 5: 142506, 6: 593775, 7: 2035800, 8: 5852925, 9: 14307150, 10: 30045015, 11: 54627300, 12: 86493225, 13: 119759850, 14: 145422675, 15: 155117520, 16: 145422675, 17: 119759850, 18: 86493225, 19: 54627300, 20: 30045015, 21: 14307150, 22: 5852925, 23: 2035800, 24: 593775, 25: 142506, 26: 27405, 27: 4060, 28: 435, 29: 30, 30: 1}

请注意，我仅在找到新密钥（总和）时才进行打印。我注意到了

仅以2和的幂发现新的和
结果是对称的。

这向我暗示了一种可行的公式化方法。

关于如何进行的任何想法？

Answer 1

好消息；那里有一个公式，如果有任何混淆，我将在其中解释路径。

让我们看一下您的第一个示例：5个草莓（S），2个柠檬（L）和一个香蕉（B）。让我们布置所有水果：

S S S S S L L B

我们现在实际上可以重新表述这个问题，因为例如总数为3的次数就是您可以从此列表中选择3种水果的不同方式的数量。

在统计数据中，选择函数（也称为nCk）仅回答以下问题：从n个项的组中选择k个项的组有多少种方法。计算为n！/（（（n-k）！* k！），其中“！”是阶乘，一个数字乘以所有小于自身的数字。
因此，3s的频率将是（水果数）“选择”（有问题的总数），或者8选择3。这是8！/（5！* 3！）= 56。

是否有一种公式化的方法可以找到组合总和的频率？

1 个答案: