隨機桶內放球的場景分析

隨機桶內放球的概率計算及使用指導
2019/7/3

均勻隨機桶內放球，有幾個不同的場景，不同的場景會得到迥異的結果。

場景一：將10k個球均勻隨機放到10個桶裏，各個桶裏面球的個數是否均勻？

仿真代碼如下

import random
import math
import matplotlib.pyplot as plt

bucket_num = 10
bucket = [0 for i in range(bucket_num)]

for i in range(10 * 1024):
    rand = random.random()
    index = math.floor(rand * bucket_num)
    bucket[index] += 1
    
plt.plot(bucket)
plt.grid(True)

結果圖像如下，可以看到基本上能做到均勻隨機分配

場景二：將10k個球均勻隨機放到10k個桶裏，能否做到基本無衝突？

仿真代碼如下

import random
import matplotlib.pyplot as plt

length = 10 * 1024
bucket = [0 for i in range(length)]
count  = [0 for i in range(length + 1)]
cnsum  = [0 for i in range(length + 1)]

for i in range(length):
    value = random.randint(0, 10**9)
    pos = value % length
    bucket[pos] += 1
    
for i in range(length + 1):
    count[i] = bucket.count(i)
    if i > 0:
        cnsum[i] = cnsum[i - 1] + count[i] / length
    else:
        cnsum[i] = count[i] / length
    
fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(count, '-rs', label = 'count')
ax.grid(True)
plt.xlim([-1, 10])

ax2 = ax.twinx()
ax2.plot(cnsum, '-bx', label = 'PDF')

統計結果如下

其中，38%的bucket爲空，38%的bucket有一個球，衝突概率高達25%，(100個桶，有25個桶超過一個球)，衝突概率還是挺高的

場景三：n個球，隨機放到m個桶，在所有的放置方式中，桶內球數最大爲k的概率是怎樣一種分佈？

def FACT(n):
    m = 1
    for i in range(n):
        m = m * (i + 1)
    return m

def COMB(n, m):
    a = FACT(n)
    b = FACT(m)
    c = FACT(n - m)
    d = a // b
    s = d // c
    return s

def ARR(n, m):
    a = FACT(n)
    b = FACT(n - m)
    c = a // b
    return c

def func(n, m, k):
    if n == 0:
        return 1
    if n == 1:
        return m
    if m == 1:
        if n <= k:
            return 1
        else:
            return 0
    if k == 1:
        if n <= m:
            s = ARR(m, n)
            return s
        else:
            print(n,m,k,0)
            return 0
    if n < k :
        s = func(n, m, n)
        return s
    else:
        if n == k:
            s = m + func(n, m, k - 1)
            return s
        else:
            s = func(n, m, k - 1)
            num = 1
            while (n >= num * k):
                t = 1
                for i in range(num):
                    t *= COMB(n - i * k, k) * (m - i)
                t = t // FACT(num)
                s += t * func(n - num * k, m - num, k - 1)
                num += 1
            return s

n = 20
m = 60
B = pow(m, n)

count = [0 for i in range(n + 1)]
proba = [0 for i in range(n + 1)]

sum_count = 0

for k in range(1, n + 1):
    sum_count = func(n, m, k)
    count[k] = sum_count - sum(count)
    proba[k] = count[k]/B
    print('k =', k, ', count =', count[k])

結果如下，可見，絕大多數場景下，桶內球個數不超過4

算法除了遞歸這種形式，簡單轉化一下就可以改爲動態規劃實現，計算個10k內的應該問題不大。