Question

另一种措辞是:任意分割同样的物品进入了 k,使一些桶空。

讨论:

an "integer partition of N", to match the usual definition and counting, can be considered as:
- a tuple of positive integers, in decreasing order, which sums to N
a vector of unsigned integers is a "partition" of N, if the sum of its elements (without integer overflow) is N.

我想写出一种功能 f(N,k),即随机和统一地选择了N分治可能长效的病媒,并回收选定的病媒。

如果能找到一个适用于所有k>、=1,但我特别关心k >、N。因此,如果它有助于关注或限制这种条件,即证明。而且,如果我们必须研究近似/湿度,那么把大部分病媒分录为零(至少是千兆字节;2N)就算得太高。

我的初步想法是:

If N is small enough that it is reasonable to calculate (or look up in a table?) the number of integer partitions of N, then maybe we could proceed as:
- Create a vector of k unsigned ints initialized to zero
- Make a random integer partition of N. Let m be the length of this tuple.
- Place those values in the initial m positions of the vector.
- Randomly shuffle the vector.

这种做法将尽可能地对待生产媒介将含有N的条目与含有1的N条目一样。这是正确的。但是,可能有一种简单的权重,可以适用于“任意分割N”的情况,而这种权衡是正确的?

Another approach which feels cleaner, but would likely still need "re-weighting" somewhere:
- Create a vector of k unsigned ints initialized to zero
- do the following N times:
  - randomly choose an element of the vector and increment it

虽然这感觉更清洁,但我认为,这将大大提高“再加权”的难度。虽然第1部分的权重像对我来说是一个困难的算法问题,但我至少可以想象需要计算什么。在这方面,我甚至不敢确定需要重新加权的内容和方式。

The reason I think it likely still needs reweighting is that there is exactly one sequence of random choices that would lead to the vector looking like [N,0,0,...,0], and N! sequences of random choices that would lead to the vector starting with N ones [1,1,...,1,0,0,...,0]. Calculating the ratios of these "incorrect weighting" of the final result sounds doable, but I don t know how I d go about reweighting the individual steps to correct for it.

Or maybe there is another approach entirely, that I have not thought of?

Answer 1

Generate a random int in the range 0-n k-1 times. Treat these as partitions of [1, 1, ..., 1] <- size n. Then the sums between partitions (& endpoints) are your vector.

例如,N=2,k=5:

say we get 0, 1, 1, 2

我们认为:

我们将其解释为[0、1、0、1、0](缩小差距为零)。

如果我们拥有0,2,2,2,那么我们就拥有[<0>,1>,22,00]。

Here s Ruby code for this:

def f(n, k)
  arr = [0]
  ans = []
  (k-1).times do
    arr.append(rand(n+1))
  end
  arr.append(n)
  arr.sort!
  1.upto(k) do |i|
    ans.append(arr[i] - arr[i-1])
  end
  return ans
end

运行时间是O(klog k),因为情况不同。我们可能能够避免这种局面,产生按顺序排列的随机数字。

——————————

这并不统一。这里有100万只(4,2个)

1_000_000.times do
  m[f(2,4)] += 1
end
=> 1000000
> m
=> 
{[1, 0, 1, 0]=>205646,
 [0, 1, 1, 0]=>411624,
 [0, 0, 1, 1]=>205916,
 [0, 1, 0, 1]=>205144,
 [0, 0, 2, 0]=>205961,
 [0, 0, 0, 2]=>68718,
 [1, 0, 0, 1]=>68178,
 [0, 2, 0, 0]=>205736,
 [2, 0, 0, 0]=>68275,
 [1, 1, 0, 0]=>205783}

-——————————-

Stars和驳船工程。这里是《鲁比法典》和《另100万》:

def g(n,k)
  arr = [1]*n + [0]*(k-1) # 1 s represent stars (what we re counting), and 0 s represent bars (separators)
  arr.shuffle!
  ans = []
  sum = 0
  arr.each do |val|
    if val == 0
      ans.append(sum)
      sum = 0
    else
      sum += val
    end
  end
  ans.append(sum)
end

1_000_000.times do
  m[g(2,4)] += 1
end
=> 1000000
> m
=> 
{[1, 1, 0, 0]=>99977,
 [0, 2, 0, 0]=>100150,
 [1, 0, 0, 1]=>100201,
 [0, 1, 1, 0]=>100034,
 [0, 1, 0, 1]=>99422,
 [0, 0, 0, 2]=>99865,
 [2, 0, 0, 0]=>99662,
 [1, 0, 1, 0]=>100359,
 [0, 0, 1, 1]=>100332,
 [0, 0, 2, 0]=>99998}

Answer 2

How about Multinomial distribution 。单行

import numpy as np

z=np.random.multinomial(20, [1/40.]*40)

np.sum(z)

附录

ok,你希望病媒具有同样的可能性。从多方面的最惠国待遇来看,情况显然并非如此。

PMF=(n!/x₁! 即便是所有p_k=1/k因子剂而致人死亡。

因此,可以做的是使用 Dirichlet-Multinomial,该物质有_k=1 PMF,即NOT x/subk,意思是所有病媒都能得到证明。

PMF = Γ(K) Γ(n+1)/Γ(n+K)

The is Dirichlet-multinomial in SciPy ( https://docs.scipy.org/doc/scipy/fer/pubs.dirichlet_multinomial.html,但其中没有固定取样方法(rvs(a)。

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