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I have an array $a$ with $n$ elements, all of which have an associated weight. For example:

$a = \{ (A,2), (B,5), (C,9), ..., (Z,1) \}$, such that element $A$ has weight $w_A=2$, element $B$ has weight $w_B=5$ and so on. The array is not necessarily sorted.

Now, I would like to obtain a sample from this array where each element is selected without repetition and the probability is proportional to the weight of the element.

My goal is to sample approximately $k < n$ elements. I am emphasizing approximately because it does not have to be exactly $k$ but something close to $k$ is good enough. More precisely, I don't care if there is a deviation in the size of the final sample of $\pm \delta$ so long as $\delta$ is not too large in comparison to $k$ or $n$.

I know how to do the sampling in time $O(n + k \log n)$ by computing the cumulative weight $W$, then generating a random value in $[0,W]$ followed by a binary search to check which element this corresponds to.

However, my goal here is to understand whether it is possible to make a sampling in $O(n)$ by defining a probability $p_i$ for each element in such a way that I can decide independently for each element whether it should belong to the sample. Meanwhile, in the end I would still have a sample with approximately the size $k$ that I am looking for. If this is possible, then I could make a single pass in array $a$ after defining the probability $p_i$ accordingly which would require the procedure to define $p_i$'s to be $O(n)$ too, of course.

In summary, my question is: is there a way of defining these probabilities $p_i$ to achieve my goal? Or am I looking for an impossible solution here?

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  • $\begingroup$ Can we assume that the elements are distinct? $\endgroup$
    – D.W.
    Jun 18, 2022 at 22:43
  • $\begingroup$ Why do you think that the method you proposed is correct? How does it ensure the "without repetition" part of the problem? $\endgroup$
    – D.W.
    Jun 20, 2022 at 0:06
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    $\begingroup$ Apparently as interpreted by me, if "each element is selected without repetition", then it cannot be true "the probability is proportional to the weight of the element". $\endgroup$
    – John L.
    Jun 23, 2022 at 21:39
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    $\begingroup$ If $k$ does not depend on $n$, $O(n+k\log n)$ is $O(n)$, so why worry ? $\endgroup$
    – user16034
    Jul 19, 2022 at 9:50

2 Answers 2

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A solution should be like:

Assign the probability $p_i$ to the element $i$ equal to the normalized weight, $\dfrac{w_i}{w}$. Then perform $n$ systematic drawings that accept the element $i$ with probability $p_i=kw_i$, in turn. Then the expectation of the number of elements drawn is $k$ and the distribution is honored.

But this fails whenever $w_i>\dfrac1k$, which is no surprise because the elements may not be repeated, which limits their probability of appearance.

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Yes, assuming we can assume that all elements are distinct. Here is one approach. Let $x_1,\dots,x_n$ denote the elements and $w_i$ the weight associated to $x_i$. Consider the following algorithm:

  • For $i:=1,2,\dots,n$:
    • Let $p_i=w_i/(w_i+w_{i+1}+\dots+w_n)$.
    • Draw $m_i$ from a Binomial($k$, $p_i$) distribution.
    • Output $m_i$ copies of $x_i$.
    • Set $k := k - m_i$.

This outputs the right distribution.

What is its running time? Well, you can compute the $p_i$'s in linear time, with a single backward scan. Also, you can sample a value $m$ from a Binomial($k$, $p$) distribution in $O(m)$ time. (Draw a uniform random number $q \in [0,1]$, then compute the tail sums of the Binomial, namely $\Pr[m\ge j]$ for $j:=0,1,2,\dots$, until you find the first tail sum that is $\ge q$. It takes a little bit of care to compute the tails sums, but you can do them in $O(1)$ time per tail sum.) Also, we have $m_1+\dots+m_n=k=O(n)$, so it follows that the total running time of this algorithm is $O(n)$.

I don't know whether this will be faster in practice. The constant factors might well dominate the difference between $O(1)$ vs $O(\log n)$.

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  • $\begingroup$ All elements in the input array are distinct. However, the ~k elements in the sample should also be distinct. The third step says "output m_i copies of x_i" which would violate this characteristic, right? Or did I misunderstood your approach? And oh my, this got a lot more "complex" then I expected. You are right that it may not pay off in the end. $\endgroup$
    – Maltus
    Jun 19, 2022 at 1:32
  • $\begingroup$ @Maltus, sorry, I missed the "without repetition" part. Can you edit the question to clarify that all elements are distinct? $\endgroup$
    – D.W.
    Jun 20, 2022 at 0:06

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