如何使用遞回回圈來查找在擲任意數量的骰子和任意數量的邊之間出現值的次數？

我正在嘗試撰寫一個函式，該函式列印出一個字典，該字典顯示同面骰子的組合可能滾動的特定數量的值。

三個6 面骰子的示例輸出

input: 6,3
Desired output: {3: 1, 4: 3, 5: 6, 6: 10, 7: 15, 8: 21, 9: 25, 10: 27, 11: 27, 12: 25, 13: 21, 14: 15, 15: 10, 16: 6, 17: 3, 18: 1}

四個8 面骰子的示例輸出

input: 8,4
desired output: {4: 1, 5: 4, 6: 10, 7: 20, 8: 35, 9: 56, 10: 84, 11: 120, 12: 161, 13: 204, 14: 246, 15: 284, 16: 315, 17: 336, 18: 344, 19: 336, 20: 315, 21: 284, 22: 246, 23: 204, 24: 161, 25: 120, 26: 84, 27: 56, 28: 35, 29: 20, 30: 10, 31: 4, 32: 1}

這是我的函式，它為每個組合創建一個空字典：

def dice_range(dice_type,dice_count):
    dice_dict = {i 1:0 for i in range(dice_type*dice_count) if i 1 >= dice_count}
    return dice_dict

input: dice_range(8,3)
actual output: {3: 0, 4: 0, 5: 0, 6: 0, 7: 0, 8: 0, 9: 0, 10: 0, 11: 0, 12: 0, 13: 0, 14: 0, 15: 0, 16: 0, 17: 0, 18: 0, 19: 0, 20: 0, 21: 0, 22: 0, 23: 0, 24: 0}

有了這個，我就能弄清楚如何解決任何 dice_type(# of side) 的特定數量的 dice_count 以下是前幾個示例的解決方案：

dice_type = 6
dice_count = 3
temp = dice_range(dice_type,dice_count)
# for 3 dice of any number of sides
for x in range(1,dice_type 1):
    for i in range(1,dice_type 1):
        for j in range(1,dice_type 1):
            for k,v in temp.items():
                if x i j == k:
                    temp[k]  =1
               
        
dice_type = 8
dice_count = 4
temp = dice_range(dice_type,dice_count)
# for 4 dice of any number of sides
for y in range(1,dice_type 1):
    for x in range(1,dice_type 1):
        for i in range(1,dice_type 1):
            for j in range(1,dice_type 1):
                for k,v in temp.items():
                    if y x i j == k:
                        temp[k]  =1

我希望這是這個問題的足夠背景。

這是我嘗試過的，但由于某種原因，遞回沒有按我的預期作業。

def dice_range(dice_type,dice_count):
    dice_dict = {i 1:0 for i in range(dice_type*dice_count) if i 1 >= dice_count}
    return dice_dict

def dice_value_calculator(dice_type,dice_count):
    
    PREDICTION = dice_range(dice_type,dice_count)
    
    # Tallies the number by 1 everytime it shows up in the loop.
    def fill_dict(total):
        for k in PREDICTION.keys():
            if total == k:
                PREDICTION[k]  = 1
        
    def loop(temp_dice_count,loop_total = 0):
        
        for loop_i in range(1,(dice_type 1)):

            # Base Case | if the number of dice(dice_count) reaches 0 that means it reached the end of the recursion
            if temp_dice_count == 0:
                fill_dict(loop_total)
                print(PREDICTION)
                loop_total = 0
                
            # General Case / Recursive
            else: 
                loop_total  = loop_i
                temp_dice_count -= 1
                loop(temp_dice_count, loop_total)
                
    loop(dice_count) # calls the loop function 
    return PREDICTION # returns the temp dictionary

print(dice_value_calculator(6,3))

This is the output I'm getting for three 6-sided dice

{3: 2, 4: 4, 5: 0, 6: 2, 7: 0, 8: 0, 9: 0, 10: 0, 11: 0, 12: 0, 13: 0, 14: 0, 15: 0, 16: 0, 17: 0, 18: 0}

Both the fill_dict() function and the dice_range() function work well on their own so it must be how I'm calling loop()

Any tips would be greatly appreciated. Please let me know if I need to clarify anything, this is my first time asking a question so I may have missed something.

Thanks in advance

uj5u.com熱心網友回復：

這個問題至少有四種可能的方法。

“如何獲得動態控制for回圈次數的效果？”

為此，

Here, grouping is done based on thinking that those indistinguishable items are ones(as we do in partitions)

標籤：python loops recursion probability dice

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