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We are given positive integers $n$ and $M$, with $n \ll M$, and a sequence $(a_i)_{i=1}^n$ of distinct integers satisfying $0 \le a_i < M$. Let $k$ be the smallest positive integer such that all the $a_i$ are distinct $\mod k$.

Questions: With $n$ and $M$ fixed, what is the expected value of $k$? And what is its maximum possible value?

Motivation: The $a_i$ are breakpoint addresses in a computer program. The number of breakpoints $n$ will typically be less than 50, and the maximum address $M$ will typically be less than half a million. Given an address, I need to be able to find the breakpoint at that address very quickly. One solution is simply a table of pointers, with $M$ entries, but that is very wasteful of space $-$ most of the entries would be zero. So I would like to use a table of size $k$, if I can be sure of finding a suitable $k$ reasonably quickly.

Once $M$ is large enough, its actual order of magnitude won't affect the expected value very much. In other words, as $M \rightarrow \infty$ with $n$ fixed, the expected value of $k$ tends to a limit. But it will certainly affect the worst case. For instance, given any $m$, we can set $a_i = i \times m!$ to ensure that $k$ is greater than $m$.

I have run some simulations. With $n=50$, and $M=2^{30}$, a run of $100000$ trials gives a mean of about $307$, with a worst case of $k=557$. With $n=100$, the mean is about $972$ and the worst case is $1570$.

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"1570" should be at the end of the last sentence, but on my screen it is displayed three lines above that. I've noticed similar problems before with LaTeX. Does anybody else have this problem? – TonyK Nov 20 '11 at 8:12
TonyK: Yes, I've seen this problem often here; when something in LaTeX is supposed to be at the end of a line, it gets displayed above the beginning of that line or an earlier line. (In this particular case it looks fine for me because on my browser the last line is "is $1570$".) – ShreevatsaR Nov 20 '11 at 8:17
What you really want is a perfect hash function, right? – Henning Makholm Nov 20 '11 at 8:51
@Henning: Yes, but the lookup operation must be very fast, not just $O(1)$ (in my case it's $i$ mod $k$, a single machine instruction). Thanks for the link! – TonyK Nov 20 '11 at 8:57
I think I have a formula for the expectation, as a function of $n$; it's a complicated infinite sum, but converges fast enough to be calculated. It equals about 305 for $n=50$ and about 970 for $n=100$. Do you want to run some simulations for other values of $n$? My formula predicts about 71 for $n=20$, about 598 for $n=75$, and about 1941 for $n=150$. – Greg Martin Nov 22 '11 at 0:40
up vote 1 down vote accepted

We choose $n$ integers independently uniformly at random from $\{1,2,\dots,M\}$ where $M$ is very large. (The original problem specified that they must be distinct integers, but that probably won't affect the final answer, so let's ignore it.) Let ${\mathbf C}(k)$ denote the event that the $n$ integers are not distinct modulo $k$. Define$$ F(k) = \Pr\big( {\mathbf C}(k) \ \big|\ {\mathbf C}(1) \wedge \cdots \wedge {\mathbf C}(k-1) \big) $$ and note that $1-F(k)$ is the probability that $k$ is the least integer modulo which the $n$ integers are distinct, conditional on knowing that they are not distinct modulo any of $1,2,\dots,k-1$. Note also that$$ F(1) F(2) \cdots F(k-1) = \Pr \big( {\mathbf C}(1) \wedge \cdots \wedge {\mathbf C}(k-1) \big), $$ as can be shown by expanding the definition of conditional probability and telescoping the resulting product. Therefore the expectation of the least $k$ for which the $n$ integers are distinct is given by the expression$$\mu= \sum_{k=1}^\infty k \Pr \big( {\mathbf C}(1) \wedge \cdots \wedge {\mathbf C}(k-1) \big) \big( 1 - \Pr\big( {\mathbf C}(k) \ \big|\ {\mathbf C}(1) \wedge \cdots \wedge {\mathbf C}(k-1) \big) \big), $$ or equivalently$$\mu=\sum_{k=1}^\infty \bigg( k (1-F(k)) \prod_{1\le j\le k-1} F(j) \bigg). $$ Therefore it suffices to be able to calculate $F(k)$. All this has been probability so far; now we need to input some number theory.

Define $c(k) = \Pr({\mathbf C}(k))$. It is not hard to show that$$ c(k) = 1 - \frac kk \frac{k-1}k \cdots \frac{k-n+1}k = 1-\frac{k!}{k^k(k-n)!} $$ (where the last expression must be interpreted appropriately when $k<n$ ). (Really we should say that $c(k)$ approaches this expression as $M$ goes to infinity.) Now define a function $f(k)$ as follows: $f(1)=1$; if $k=p^a$ is a power of a prime then $f(p^a) = c(p^a)/c(p^{a-1})$ (this includes $f(p)=c(p)$); otherwise, write $k=p_1^{a_1}\cdots p_r^{a_r}$ with $r\ge2$, in which case $f(k) = c(k)/\big( c(p_1^{a_1}) \cdots c(p_r^{a_r}) \big)$. I claim that $F(k) = f(k)$ (as $M$ goes to infinity).

Proving this statement would be the most technical part of the derivation, I think. [Intuitively, the reason this should be true is: when $k$ and $l$ are relatively prime, then the events ${\mathbf C}(k)$ and ${\mathbf C}(l)$ will turn out to be independent (up to a vanishingly small error as $M$ tends to infinity). So the only events ${\mathbf C}(l)$ that directly influence ${\mathbf C}(k)$ are when $l$ divides $k$. Furthermore, the event ${\mathbf C}(k)$ is a subset of ${\mathbf C}(l)$ whenever $l$ divides $k$. Therefore we only need to consider the maximal prime-power divisors of $k$, and these events are all independent.]

The definition of $f(k)$ can be coded easily, and when substituted into the second equation for $\mu$ above (replacing $F$ by $f$), it leads to the values I mentioned in my comments on the original post.

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Some thoughts:

Your example $a_i = im!$ gives a $k$ whose size is about $(\log(M/n))/\log\log(M/n)$, since that's how large one can choose $m$ before $nm!$ exceeds $M$. One can get a worse $k$ as well, by choosing $a_1=0$, $a_2$ to be the product of all primes up to about $\log M$, $a_3$ to be the product of all primes between $\log M$ and $2\log M$, and so on: then $k$ is at least $(n-1)\log M$ asymptotically. (This is a slight lie, as one needs to put the prime powers in there somewhere too. The asymptotic should still be correct.) Of course this doesn't even take advantage of primes dividing $a_3-a_2$ and so on.

When $M$ is much larger than a prime $p$, the probability that the $a_i$ are distinct modulo $p$ is essentially $(p/p)((p-1)/p)\cdots((p-n+1)/p) = p!/((p-n)!p^n)$; if $p=\lambda n$, then by Stirling's formula this is about $c(\lambda)^n$ where $c(\lambda) = (\lambda/(\lambda-1))^{\lambda-1}/e < 1$. This equals $1/2$ when $\lambda$ is about $n/(2\log 2)$, or $p\approx n^2/(2\log 2)$. But the question of finding even the median (never mind the expectation) of the least $p$ for which they are distinct modulo $p$ seems more daunting. (At least if $M$ is sufficiently large, one should be able to consider these modulo-$p$ events essentially independent.)

[EDITED TO ADD:] I was thinking so much about the asymptotics of the smallest such $k$ that I didn't realize I could at least write down an expression for the expectation of that $k$. Let $D(p) = p!/((p-n)!p^n)$ be the probability that the $a_i$ are distinct modulo $p$; we define $D(p) = 0$ if $p<n$. These events aren't exactly independent: if they're not distinct modulo $p$, then they're slightly less likely to be distinct modulo $p^2$, and so on. If we define $E(p) = D(p)$ and, recursively, $E(p^k) = (D(p^k)-D(p^{k-1}))/(1-D(p^{k-1}))$, then $E(p^k)$ is the probability that they're distinct modulo $p^k$ conditional on the event that they're not distinct modulo $p^{k-1}$. Oi, I just realized that the formula for $E(n)$ for general composite $n$ is going to depend upon $E$ at its prime factors in a complicated way.... Then the expected value of the least $k$ modulo which the $a_i$ are distinct is given by the infinite sum: $$ \displaystyle \sum_{q} \bigg( q E(q) \prod_{r<q} (1-E(r)) \bigg) $$

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Assuming that the $a_i$ are randomly distributed in ${0,...,k-1}$ and that for two different k this distributions are

independent then the problem to found the probability that all $a_i$ are different is the birthday problem. I

implemented in maxima the the probability


that all $n$ numbers randomly chosen from ${0,...,k-1}$ (this needs the load(functs); before executiong this definition)

are different and also an approximation


of $f$.

I estimate the mean by

for k:n thru L do (

and get $305.2774133346098$. Using the approximation $g$ instead of $f$ I get the smaller value $293.0324415803322$. For n=100 I evaluate the for-loop until L:2000 to get $970.2315452158832$ (using f) and $944.4295352864864$ (using p) for

the mean.

For $n=1000$ I only eveluate the approximation until L:86000 and get $55595.16420043662$

To estimate the maximum I calculated

for k:n thru 675 do (

This is the probability that $k$ is between $50$ and $675$. The result is $0.99999999999998$. So the probability that k is larger than $675$ is very small. For $n=100$ and $n=1000$ I calculated values about L:1900 and L:86000 for the upper bound containing most of the $k$.

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