rank 和 unrank 组合将 k 个球分配到 n 个不同容量的 bin 中

Posted 2023-03-31

【中文标题】rank 和 unrank 组合将 k 个球分配到 n 个不同容量的 bin 中

【英文标题】：Rank and unrank combinations to distribute k balls into n bins of different capacities

【发布时间】：2020-06-01 18:04:36

【问题描述】：

我想将k 球分配到不同容量的n 箱中。如何对给定 n、k 和 bin 容量的分布进行排名和取消排名？

例子：

n := 3

k := 4

bin capacities := 3,2,1

垃圾箱中的球：

1,2,1, 2,1,1, 2,2,0, 3,0,1, 3,1,0 := 5

有公式吗？

【问题讨论】：

排名和非排名是什么意思？看起来您想计算可以将 k 个球分配到 n 个箱中的方式的数量，每个箱中每个箱都装有不同的有限数量的球？

@doggie_breath 当人们说“rank and unrank”时，他们通常的意思是对球的分布进行排序，然后能够查看分布并说出它是哪个分布，然后查看一个数字m 并说出这是mth 分布。

似乎缺少分布2,1,1。我假设列表是按字典顺序排列的，这应该是列表中的第二个？

@btilly 啊，明白了。感谢您的澄清。

是的，2,1,1 不见了。

【参考方案1】：

我不知道这种技术是否有一个标准名称，但这是一种我已经成功解决了很多次的问题。

我使用动态编程来构建一个可以发生排名/取消排名的数据结构，然后构建逻辑来执行排名/取消排名的事情。

动态编程部分是最难的。

import collections
BallSolutions = collections.namedtuple('BallSolutions', 'bin count balls next_bin_solutions next_balls_solutions');

def find_ball_solutions (balls, bin_capacities):
    # How many balls can fit in the remaining bins?
    capacity_sum = [0 for _ in bin_capacities]
    capacity_sum[-1] = bin_capacities[-1]

    for i in range(len(bin_capacities) - 2, -1, -1):
        capacity_sum[i] = capacity_sum[i+1] + bin_capacities[i]

    cache = 
    def _search (bin_index, remaining_balls):
        if len(bin_capacities) <= bin_index:
            return None
        elif capacity_sum[bin_index] < remaining_balls:
            return None
        elif (bin_index, remaining_balls) not in cache:
            if bin_index + 1 == len(bin_capacities):
                cache[(bin_index, remaining_balls)] = BallSolutions(
                    bin=bin_index, count=1, balls=remaining_balls, next_bin_solutions=None, next_balls_solutions=None)
            else:
                this_solution = None
                for this_balls in range(min([remaining_balls, bin_capacities[bin_index]]), -1, -1):
                    next_bin_solutions = _search(bin_index+1, remaining_balls - this_balls)
                    if next_bin_solutions is None:
                        break # We already found the fewest balls that can go in this bin.
                    else:
                        this_count = next_bin_solutions.count
                        if this_solution is not None:
                            this_count = this_count + this_solution.count
                        next_solution = BallSolutions(
                            bin=bin_index, count=this_count,
                            balls=this_balls, next_bin_solutions=next_bin_solutions,
                            next_balls_solutions=this_solution)
                        this_solution = next_solution
                cache[(bin_index, remaining_balls)] = this_solution
        return cache[(bin_index, remaining_balls)]

    return _search(0, balls)

以下是生成排名解决方案的代码：

def find_ranked_solution (solutions, n):
    if solutions is None:
        return None
    elif n < 0:
        return None
    elif solutions.next_bin_solutions is None:
        if n == 0:
            return [solutions.balls]
        else:
            return None
    elif n < solutions.next_bin_solutions.count:
        return [solutions.balls] + find_ranked_solution(solutions.next_bin_solutions, n)
    else:
        return find_ranked_solution(solutions.next_balls_solutions, n - solutions.next_bin_solutions.count)

这是生成解决方案排名的代码。请注意，如果提供的答案无效，它会爆炸。

def find_solution_rank (solutions, solution):
    n = 0
    while solutions.balls < solution[0]:
        n = n + solutions.next_bin_solutions.count
        solutions = solutions.next_balls_solutions
    if 1 < len(solution):
        n = n + find_solution_rank(solutions.next_bin_solutions, solution[1:])
    return n

这是一些测试代码：

s = find_ball_solutions(4, [3, 2, 1])
for i in range(6):
    r = find_ranked_solution(s, i)
    print((i, r, find_solution_rank(s, r)))

【参考方案2】：

您可以递归地定义此类组合的数量。给定k 球和箱容量q_1, ..., q_n，对于0 和q_1 之间的每个j，将j 球放入q_1 并将剩余的k-j 球分配到其他箱中。

这是一个快速的 Python 实现：

from functools import lru_cache

@lru_cache(None)
def f(n, *qs):
  if not qs:
    return 1 if n == 0 else 0
  q = qs[0]
  return sum(f(n-j, *qs[1:]) for j in range(q+1))

f(4, 3, 2, 1)
# 5

【参考方案3】：

这是一种方法（在伪代码中），虽然它看起来效率不高。在球的数量不适合总剩余容量的地方添加一些短路可能是明智的。如果给定的容量列表将被多次使用，也许一些巧妙的缓存会有所帮助。

所有数字都是非负整数。函数ArrayTail(array a) 是子数组，其元素是第一个之后的输入数组的所有元素。函数ArrayCon(number head, array a)是元素为head后跟a元素的数组。

function Count(array capacities, number balls) -> number
    If balls == 0:
       return 1
    Else if capacities is empty:
       return 0
    Else:
       Let sum: number
       sum <- 0
       For b from 0 to max(balls, capacities[0]):
           sum <- sum + Count(ArrayTail(capacities), b)
       End For
       return sum
    End If/Else
End function

function Rank(array capacities, array counts) -> number
    Precondition: length(capacities) == length(counts)
    Precondition: counts[i] <= capacities[i] for all i < length(counts)
    If counts is empty:
        return 0
    Else:
        Let total: number
        total <- 0
        For c in counts:
            total <- total + c
        End For
        Let r: number
        r <- Rank(ArrayTail(capacities), ArrayTail(counts))
        For b from 0 to (counts[0]-1):
            r <- r + Count(ArrayTail(capacities), total - b)
        End For
        return r
    End If/Else
End function

function Unrank(array capacities, number balls, number rank) -> array
    Precondition: rank < Count(capacities, balls)
    If capacities is empty:
        return empty array
    Else
        Let c0: number
        c0 <- 0
        Loop until "return":
            Let subcount: number
            subcount <- Count(ArrayTail(capacities), balls-c0)
            If subcount <= rank:
                c0 <- c0 + 1
                rank <- rank - subcount
            Else
                return ArrayCon(c0, Unrank(ArrayTail(capacities), balls-c0, rank))
            End If/Else
        End Loop
    End If/Else
End function