Faster Algorithm for sampling a point from the array

Question

Let $M$ be a collection of elements given in the form of the array such that membership of any element can be done in $O(1)$ time. Which means elements of array $M$ are $\{1,2,\cdots,n\}$ such that $1$ is stored at $M[1]$ and $2$ at $M[2]$ and so on. The goal is to sample an element from the array $M$.

Note that we have to design an algorithm multiple times (say $k$ times) in such a way that in each repetition we get a different element as a result.

Sampling means the probability of picking any element from $M$ is $\frac{1}{|M|}$.

My idea is as follows: create another array $F:=[\frac{1}{|M|},\frac{1}{|M|},\cdots,\frac{1}{|M|}]$, which means assigning equal weights to each element.

There are several existing algorithms now which can sample an element from $M$ in expected time $O(1)$ (Say Algorithm 1 Algorithm).

If I run the same algorithm 1 again then the same element may get picked so I have to update something in the array $F$ this update should be in $O(1)$ time.

I am looking for an algorithm that gives $O(k)$ time for $k$ repetitions other than constructing array $F$.

How to modify the above algorithm such that next time I get a different answer?

EDIT: For any $a\in \{1,2,...,n\}, $Pr[Output_1=a] = $\frac{1}{|M|}$ and Pr[Output_i=a|Output_i-1 =a_i-1,...,Output_1=a_1]. Output_i denotes the output of the algorithm in the $i$th iteration and Pr denotes the probability.

I fail to see how establishing an array of weights explicitly changes (or defines) sampling a subset of size $k$. — greybeard, Sep 15 '22 at 05:34
So your question is how to sample k elements from n elements without replacement? https://stackoverflow.com/questions/311703/algorithm-for-sampling-without-replacement — qwr, Sep 15 '22 at 07:20
@greybeard I did this as I will apply the algorithm given in the pdf. — Rma, Sep 15 '22 at 07:20

score 1 · Accepted Answer · answered Sep 15 '22 at 11:33

1

return M[(i - 1) % n + 1]

for $i=1,2, \cdots k$ does the trick.

answered Sep 15 '22 at 11:33

Thanks but how to prove its correctness – Rma Sep 19 '22 at 08:13
@Rma: how could it be wrong ? – Sep 19 '22 at 08:23
for i=1 it is fine but how to prove probability that for i=2,3,..,k is also equally like as written in the edit part of the question. – Rma Sep 19 '22 at 10:31
Well, systematic sampling does not ensure that the elements have the same chance of being drawn. But if you draw $k$ elements from a single set of $n$, does it matter ? – Sep 19 '22 at 10:52
I will pick the k elements one by one. Suppose that it matters that the elements have the same chance of being drawn then? – Rma Sep 19 '22 at 11:38
Draw randomly and swap the chosen elements to the end of the array. – Sep 19 '22 at 12:14

Faster Algorithm for sampling a point from the array

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