Linear Hashing is Awesome

TL;DR

This paper proves that a specific class of linear hash functions, when used with chaining in hashing tables, results in a longest chain length of approximately n^{1/3}, demonstrating the effectiveness of linear hashing.

Contribution

It provides a rigorous analysis of linear hash functions' performance, establishing a bound on the expected maximum chain length in hashing with chaining.

Findings

Expected longest chain length is n^{1/3} for linear hash functions.

The analysis extends to multiply-shift hash functions.

Supports the effectiveness of linear hashing in practical applications.

Abstract

We consider the hash function $h(x) = ((ax+b) \bmod p) \bmod n$ where $a,b$ are chosen uniformly at random from $\{0,1,\ldots,p-1\}$. We prove that when we use $h(x)$ in hashing with chaining to insert $n$ elements into a table of size $n$ the expected length of the longest chain is $\tilde{O}\!\left(n^{1/3}\right)$. The proof also generalises to give the same bound when we use the multiply-shift hash function by Dietzfelbinger et al. [Journal of Algorithms 1997].

Equations

$$M(h,X)=\underset{y\in [m]}{\max }\left|\left\{x\in X\mid h(x)=y\right\}\right|.$$

$$I_m(r,s)=\left\{{\left[r\right]}_m,{\left[r+1\right]}_m,\dots ,{\left[r+s-1\right]}_m\right\}.$$

$$\sum _{b\in B}\sum _{j\in [\iota (b)]}\frac{(\tau (b,j){)}^2}{\delta (b,j)}$$

$$\left(\sum _{j\in [\iota (b)]}\delta (b,j)\right)\left(\sum _{j\in [\iota (b)]}\frac{(\tau (b,j){)}^2}{\delta (b,j)}\right)\ge {\left(\sum _{j\in [\iota (b)]}\tau (b,j)\right)}^2$$

$$M=M(X)=\underset{y\in {\mathbb{Z}}_m}{\max }\left|\left\{x\in X\mid h(x)=y\right\}\right|.$$

$$E\left[M\right]=O\left(\sqrt[3]{n\log n}\right).$$

$$E\left[M\right]=\frac{p-1}{p}E\left[M\mid a\ne 0\right]+\frac{1}{p}E\left[M\mid a=0\right]<E\left[M\mid a\ne 0\right]+1.$$

$$f^{-1}(y)=\left\{{\left[i+km\right]}_p\mid k\in \mathbb{Z},0\le k<\frac{p-i}{m}\right\},$$

$$n(n-1)\cdot \frac{2p}{m\alpha (p-1)}\le \frac{2n}{\alpha }.$$

$$\mathrm{Pr}\left[M\ge 4\alpha \right]=\delta =O\left(\frac{n\log n}{{\alpha }^3}\right).$$

$$E\left[M\right]$$

$$=\sum _{k=1}^{\lfloor \sqrt[3]{n\log n}\rfloor }\mathrm{Pr}\left[M\ge k\right]+\sum _{k=\lfloor \sqrt[3]{n\log n}\rfloor +1}^n\mathrm{Pr}\left[M\ge k\right]$$

$$\le \lfloor \sqrt[3]{n\log n}\rfloor +\sum _{k=\lfloor \sqrt[3]{n\log n}\rfloor +1}^{n}O\left(\frac{n\log n}{k^3}\right)$$

$$=O\left(\sqrt[3]{n\log n}\right)$$

$$E\left[M\right]=O\left(\sqrt[3]{n\log n}\right).$$

$$n(n-1)\cdot \frac{4}{m\alpha }\le \frac{4n}{\alpha }.$$

Full text

1 Introduction

In this paper we study the hash function $h:[p]\to[m]$ (where $[m]=\left\{0,1,\ldots,m-1\right\}$) defined by $h(x)=((ax+b)\bmod p)\bmod m$, where $a,b\in[p]$ are chosen uniformly at random from $[p]$. Here, $p$ is a prime and $p\geq m$. We assume that we have a set $X\subseteq[p]$ of $n$ keys with $n\leq m$ and use $h$ to assign a hash value $h(x)$ to each key $x\in X$. We are interested in the frequency of the most popular hash value, i.e. we study the random variable $M(h,X)$ defined by

[TABLE]

In Theorem 1 we prove that $\operatorname{E}\!\left[{M(h,X)}\right]=O\!\left(\sqrt[3]{n\log n}\right)$. We also consider the hash function $\bar{h}:[q]\to[m]$ defined by $\bar{h}(x)=\left\lfloor{\frac{(ax)\bmod q}{q/m}}\right\rfloor$, where $q,m$ are powers of $2$, $q\geq m\geq n$ and $a$ is chosen uniformly at random among the odd numbers from $[q]$. The function $\bar{h}(x)$ was first introduced by Dietzfelbinger et al. [3]. In Theorem 2 we prove that it also holds that $\operatorname{E}\!\left[{M(\bar{h},X)}\right]=O\!\left(\sqrt[3]{n\log n}\right)$.

We note that when we use $h(x)=((ax+b)\bmod p)\bmod m$ in hashing with chaining, $M$ is the size of the largest chain. When scanning the hash table for an element the expected time used is $O\!\left(1\right)$ and the worst case time is at most $O\!\left(M(h,X)\right)$.

1.1 Related work

It is folklore that the size of the largest chain is $O\!\left(\sqrt{n}\right)$ and this bounds hold for any $2$-independent hash function.

Alon et al. [1] considers the linear hash function $h_{m,k}:\mathcal{F}^{m}\to\mathcal{F}^{k}$, where $\mathcal{F}$ is a finite field and $n=\left|\mathcal{F}\right|^{k}$. The function is defined by $h_{m,k}(x_{1},\ldots,x_{m})=\sum_{i}x_{i}a_{i}$, where $a_{i}\in\mathcal{F}^{k}$ is chosen uniformly at random. For $m=2,k=1$ the hash function is $h_{2,1}(x,y)=ax+by$ where $a,b\in\mathcal{F}$ are chosen uniformly at random. It is shown in [1] that there exists a set $X\subseteq\mathcal{F}^{2}$ such that $\operatorname{E}\!\left[{M(h_{2,1},X)}\right]>\sqrt{n}$ if $n$ is a square and $\operatorname{E}\!\left[{M(h_{2,1},X)}\right]=\Omega\!\left(\sqrt[3]{n}\right)$ if $n$ is a prime power that is not a square. In [1] it is also shown that when $\mathcal{F}$ is the field of two elements the expected length of the longest chain is $O\!\left(\log n\log\log n\right)$ improving the results in [4, 5].

Broder et al. [2] considered $h(x)=(ax+b)\bmod p$ in the context of min-wise hashing.

2 Preliminaries

$\mathbb{Z}$ denotes the integers, and $\mathbb{Z}_{n}=\mathbb{Z}/n\mathbb{Z}$ denotes the integers $\bmod\ n$. $\mathbb{Z}_{n}^{*}$ is the set of elements of $\mathbb{Z}_{n}$ having a multiplicative inverse. $[n]$ is the set of integers from [math] to $n-1$, that is $[n]=\left\{0,1,2,\ldots,n-1\right\}$. For a pair of integers $n,m\in\mathbb{Z}$ such that $(n,m)\neq(0,0)$ we let $\gcd(n,m)$ denote the greatest common divisor of $n$ and $m$. If $\gcd(n,m)=1$ then $n$ and $m$ are said to be coprime.

For integers $x,r\in\mathbb{Z}$ we let $\left[x\right]_{r}\in\mathbb{Z}_{r}$ denote the residue class of $x\bmod r$. We let $\iota_{r}:\mathbb{Z}_{r}\to[r]$ be the unique mapping that satisfies $\left[\iota_{r}\!\left(x\right)\right]_{r}=x$. For $x\in\mathbb{Z}_{r}$ we let $\left|\left|x\right|\right|_{r}=\min\left\{\iota_{r}\!\left(x\right),\iota_{r}\!\left(-x\right)\right\}$.

For a set $S$ and an element $x$ the sets $S+x$ and $xS$ are defined as $\left\{s+x\mid s\in S\right\}$ and $\left\{xs\mid s\in S\right\}$, respectively.

For integers $r,s,m$, let $I_{m}(r,s)\subseteq\mathbb{Z}_{m}$ denote the set

[TABLE]

The set $I_{m}(r,s)$ is called an interval. A non-empty set $X\subseteq\mathbb{Z}_{m}$ is an interval if there exists $r,s$ such that $X=I_{m}(r,s)$.

3 Main Result

In this section we prove the main results of this paper, namely Theorem 1 and Theorem 2. The proofs of the two theorems are very similar and both rely on Lemma 1 below.

Lemma 1**.**

Let $n,M,r$ be integers satisfying $4M\leq n\leq r$ and let $A\subseteq\mathbb{Z}_{r}$ be a set of size $n$. Let $B\subseteq\mathbb{Z}_{p}^{*}$ be a set of size $\leq M$ satisfying the following conditions:

• 1

$\iota(b)\in\left(M,2M\right)$* for all $b\in B$.*

• 2

$\iota(b),\iota(b^{\prime})$, and $r$ are pairwise coprime for every $b,b^{\prime}\in B$ with $b\neq b^{\prime}$.

Assume that for every $b\in B$ there exists an interval $I_{b}$ of size $\left\lceil{\frac{r}{n}}\right\rceil$ such that $I_{b}\cap bA$ contains at least $4M$ elements. Then there exists at least $M\left|B\right|$ ordered pairs of different elements $a,a^{\prime}\in A$ such that $\left|a-a^{\prime}\right|<\frac{r}{nM}$.

Proof.

We note that for every $b$ the set $b^{-1}I_{b}$ is the union of $\iota(b)$ disjoint intervals of size $\leq\left\lceil{\frac{r}{n\iota(b)}}\right\rceil$, and we write it as such a union $b^{-1}I_{b}=\bigcup_{j=0}^{\iota(b)-1}I_{b,j}$. For any $b,b^{\prime}\in B,b\neq b^{\prime}$ the set $b^{-1}I_{b}\cap{b^{\prime}}^{-1}I_{b^{\prime}}$ is either empty or an interval. So for each $b,b^{\prime}\in B,b\neq b^{\prime}$ there is at most one index $j\in[\iota(b)]$ such that the intersection $I_{b,j}\cap{b^{\prime}}^{-1}I_{b^{\prime}}$ is non-empty. For every $b\in B$ and $j\in[\iota(b)]$, let $\delta(b,j)$ denote the number of elements $b^{\prime}\in B$ such that $I_{b,j}\cap{b^{\prime}}^{-1}I_{b^{\prime}}$ is non-empty. Note that $\delta(b,j)\geq 1$ since $b\in B$. Furthermore $\sum_{j=0}^{\iota(b)-1}\delta(b,j)<\left|B\right|+\iota(b)\leq 3M$ since each ${b^{\prime}}^{-1}I_{b^{\prime}}$ has a non-empty intersection with at most one of the sets $I_{b,j},j\in[\iota(b)]$.

The number of ordered pairs of different elements $(a,a^{\prime})\in A\cap I_{b,j}$ such that $\left|a-a^{\prime}\right|<\frac{r}{nM}$ is exactly $\left|A\cap I_{b,j}\right|\cdot\left(\left|A\cap I_{b,j}\right|-1\right)$ since $I_{b,j}$ is an interval of size $\leq\left\lceil{\frac{r}{n\iota(b)}}\right\rceil$ and $\iota(b)>M$. Let $\tau(b,j)=\max\left\{0,\left|A\cap I_{b,j}\right|-1\right\}$, then the number of pairs is at least $(\tau(b,j))^{2}$. We can lower bound the number of such pairs in $A$ by considering the pairs in $A\cap I_{b,j}$ for each $b\in B$ and $j\in[\iota(b)]$ and note that each pair we count is counted at most $\delta(b,j)$ times. This gives that the number of ordered pairs $(a,a^{\prime})\in A$ such that $\left|a-a^{\prime}\right|<\frac{r}{nM}$ is at least:

[TABLE]

For any $b\in B$, by the Cauchy-Schwartz inequality we have that:

[TABLE]

We clearly have that $\sum_{j\in[\iota(b)]}\tau(b,j)\geq 4M-\iota(b)\geq 2M$. Also recall, that we have that $\sum_{j\in[\iota(b)]}\delta(b,j)\leq 3M$. Combining this with (2) and (3) gives that $A$ contains at least $\frac{4M\left|B\right|}{3}\geq M\left|B\right|$ of the desired pairs. ∎

Below is a proof of Theorem 1.

Theorem 1**.**

Let $n,m,p$ be integers with $p$ a prime and $p\geq m\geq n$. Let $X\subseteq\mathbb{Z}_{p}$ be a set of $n$ elements. Let $h:\mathbb{Z}_{p}\to\mathbb{Z}_{m}$ be defined by $h(x)=\left[\iota_{p}\!\left(ax+b\right)\right]_{m}$ where $a,b\in\mathbb{Z}_{p}$ are chosen uniformly at random. Let $M=M(X)$ be the random variable counting the number of elements $x\in X$ that hash to the most popular hash value, that is

[TABLE]

Then

[TABLE]

Proof.

We note that $\operatorname{E}\!\left[{M\mid a=0}\right]=n$ since $h$ is constant when $a=0$. Therefore:

[TABLE]

Therefore it suffices to bound the expected value of $M$ when $a$ is chosen uniformly at random from $\mathbb{Z}_{p}\setminus\left\{0\right\}=\mathbb{Z}_{p}^{*}$ and not from $\mathbb{Z}_{p}$. So from now on, assume that $a$ is chosen uniformly at random from $\mathbb{Z}_{p}^{*}$.

The random variables $a$ and $a^{-1}b$ are independent. Note, that $h(x)$ can be rewritten as $h(x)=\left[\iota_{p}\!\left(a(x+a^{-1}b)\right)\right]_{m}$. It clearly suffices to bound the expected value of $M$ conditioned on all possible values $a^{-1}b$. For any fixed value of $a^{-1}b=c$, the expected value of $M$ conditioned on $a^{-1}b=c$ is the same as the expected value of $M(X+c)$ conditioned on $b=0$. Therefore it suffices to give the proof under the assumption that $b=0$. So we assume that $b=0$.

Let $A=m^{-1}aX$, then there exists an interval $I_{a}$ of size at most $\left\lceil{\frac{p}{m}}\right\rceil$ that contains $M$ elements of $A$ for the following reason: Let $f:\mathbb{Z}_{p}\to\mathbb{Z}_{m}$ be defined by $x\to\left[\iota_{p}\!\left(x\right)\right]_{m}$. By definition, there exists a random variable $y\in\mathbb{Z}_{p}$ such that $\left|f^{-1}(y)\cap aX\right|\geq M$. And there exists a $i\in[m]$ such that

[TABLE]

and hence $I_{a}=m^{-1}f^{-1}(y)$ is an interval of size $\leq\left\lceil{\frac{p}{m}}\right\rceil$ that contains $M$ elements of $A$.

Let $\alpha\in\left[1,\frac{n}{4}\right]$. We are now going to bound the probability that $M\geq 4\alpha$. Let $\delta=\Pr\!\left[{M\geq 4\alpha}\right]$ and let $\mathcal{A}$ be the set of all elements $a_{0}\in\mathbb{Z}_{p}^{*}$ such that $M\geq 4\alpha$ if $a=a_{0}$.

Let $S\subseteq\mathbb{Z}_{p}^{*}$ be the set of all elements $s\in\mathbb{Z}_{p}^{*}$ that satisfies that $\iota_{p}\!\left(s\right)$ is a prime in the interval $(\alpha,2\alpha)$. Let $B\subseteq S$ be the set of all elements $s\in S$ such that $as\in\mathcal{A}$. Note, that $B$ is a random variable. By linearity of expectation, we have that $\operatorname{E}\!\left[{\left|B\right|}\right]=\left|S\right|\delta$. Recall, that $A=m^{-1}aX$. For any $b\in B$ we have that $ab\in\mathcal{A}$ and therefore there exists an interval of size $\left\lceil{\frac{r}{n}}\right\rceil$ that contains at least $4\alpha$ elements of $bA$. By Lemma 1, this implies that there are $\alpha\left|B\right|$ ordered pairs of different elements $x,x^{\prime}\in X$ such that $\left|\left|ax-ax^{\prime}\right|\right|_{p}<\frac{p}{m\alpha}$. So the expected number of elements $x,x^{\prime}\in X$ such that $\left|\left|a(x-x^{\prime})\right|\right|_{p}<\frac{p}{m\alpha}$ is at least $\alpha\operatorname{E}\!\left[{\left|B\right|}\right]=\alpha\delta\left|S\right|$. On the other hand, for each ordered pair of different elements $x,x^{\prime}\in X$ the probability that $\left|\left|a(x-x^{\prime})\right|\right|_{p}<\frac{p}{m\alpha}$ is at most $\frac{2p}{m\alpha(p-1)}$, and by linearity of expectation the expected number of such ordered pairs is at most

[TABLE]

We conclude that $\alpha\delta\left|S\right|\leq\frac{2n}{\alpha}$. By the prime number theorem, $\left|S\right|=\Theta\!\left(\frac{\alpha}{\log\alpha}\right)=\Omega\!\left(\frac{\alpha}{\log n}\right)$. Reordering gives us that:

[TABLE]

The expected value of $M$ can now be bounded in the following manner:

[TABLE]

which was what we wanted. ∎

The proof of Theorem 2 is very similar to the proof of Theorem 1 but we include it for completeness.

Theorem 2**.**

Let $n,\ell,r,q,m$ be integers with $q=2^{r},m=2^{\ell}$ and $q\geq m\geq n$. Let $X\subseteq\mathbb{Z}_{q}$ be a set of $n$ elements. Let $h:\mathbb{Z}_{q}\to[m]$ be defined by $h(x)=\left\lfloor{\iota_{q}\!\left(ax\right)\cdot 2^{\ell-r}}\right\rfloor$ where $a\in\mathbb{Z}_{q}^{*}$ are chosen uniformly at random. Let $M=M(X)=\max_{y\in[m]}\left|\left\{x\in X\mid h(x)=y\right\}\right|$. Then

[TABLE]

Proof.

Let $y$ be a random variable such that $\left|h^{-1}(y)\cap X\right|=M$, and let $A=aX$. The set $ah^{-1}(y)$ is an interval of size $\frac{q}{m}$ that contains exactly $M$ elements of $A$.

Let $\alpha\in\left[1,\frac{n}{4}\right]$. We are now going to bound the probability that $M\geq 4\alpha$. Let $\delta=\Pr\!\left[{M\geq 4\alpha}\right]$, and let $\mathcal{A}$ be the set of all elements $a_{0}\in\mathbb{Z}_{p}^{*}$ such that $M\geq 4\alpha$ if $a=a_{0}$.

Let $S\subseteq\mathbb{Z}_{q}^{*}$ be the set of all elements $s\in\mathbb{Z}_{q}^{*}$ that satisfies that $\iota_{q}\!\left(s\right)$ is a prime in the interval $(\alpha,2\alpha)$. Let $B\subseteq S$ be the set of all elements $s\in S$ such that $as\in\mathcal{A}$. Note, that $B$ is a random variable. By linearity of expectation, we have that $\operatorname{E}\!\left[{\left|B\right|}\right]=\left|S\right|\delta$. Recall, that $A=m^{-1}aX$. For any $b\in B$ we have that $ab\in\mathcal{A}$ and therefore there exists an interval of size $\frac{q}{m}$ that contains at least $4\alpha$ elements of $bA$. By Lemma 1, this implies that there are $\alpha\left|B\right|$ ordered pairs of different elements $x,x^{\prime}\in X$ such that $\left|\left|ax-ax^{\prime}\right|\right|_{q}<\frac{q}{m\alpha}$. So the expected number of elements $x,x^{\prime}\in X$ such that $\left|\left|a(x-x^{\prime})\right|\right|_{q}<\frac{q}{m\alpha}$ is at least $\alpha\operatorname{E}\!\left[{\left|B\right|}\right]=\alpha\delta\left|S\right|$. On the other hand, for each ordered pair of different elements $x,x^{\prime}\in X$ the probability that $\left|\left|a(x-x^{\prime})\right|\right|_{q}<\frac{q}{m\alpha}$ is at most $\frac{4}{m\alpha}$, and by linearity of expectation the expected number of such ordered pairs is at most

[TABLE]

We conclude that $\alpha\delta\left|S\right|\leq\frac{4n}{\alpha}$, and now we can bound the expected value exactly as in Theorem 1. ∎