The Number of Prime Parking Functions

Let us suppose that n cars enter a one-way street with n parking spaces marked $1,2,\ldots ,n$, in that order, and each driver intends to park in a particular spot. Say, the ith driver wishes to park in the spot marked $p(i)\in [n]$, where [n] denotes the set $\{1,2,\ldots ,n\}$.

Suppose that if driver i finds his favorite spot free, he parks there, but if he finds it occupied by any of the previous $i-1$ drivers, he looks for the next free space after spot p(i) for parking, and if none exists, he leaves without parking. For which functions ${\textbf{p}}:[n]\rightarrow [n]$ is it possible for all the drivers to park? Let us call such functions parking functions of length n.¹

The answer to this question was given by Konheim and Weiss [3] as follows: Let

$$\begin{aligned} \big (p(1),\ldots ,p(n)\big )^{\uparrow } \end{aligned}$$

denote the nondecreasing rearrangement² of

$$\begin{aligned} \big (p(1),\ldots ,p(n)\big )\,. \end{aligned}$$

If we set

$$\begin{aligned} {\textbf{q}}=(q_1,\ldots ,q_n) =\big (p(1),\ldots ,p(n)\big )^{\uparrow }\,, \end{aligned}$$

then the parking functions are those for which

$$\begin{aligned}&q_i\le i,\text { for every }i\in [n]. \end{aligned}$$

(1)

In this article, we are mainly interested in the subclass of prime parking functions, which are those ${\textbf{p}}$ such that³

$$\begin{aligned}&q_i< i\text { for every }i\in [n]\text { with }i>1\,. \end{aligned}$$

(2)

For an example of a parking function, suppose that $n=15$ and take

$$\begin{aligned} {\textbf{a}}:=(3, 13, 6, 3, 7, 3, 2, 1, 10, 11, 6, 7, 14, 10, 11)\,, \end{aligned}$$

which takes $1\mapsto 3$, $2\mapsto 13$, etc. Then all cars can park in the previously described sense. In fact, the parking goes as follows:

And indeed,

$$\begin{aligned} {\textbf{a}}^\uparrow =(1, 2, 3, 3, 3, 6, 6, 7, 7, 10, 10, 11, 11, 13, 14)\,, \end{aligned}$$

which is componentwise less than or equal to $(1,2,\ldots ,15)$.

Parking functions occur when we consider the set of hyperplanes in ${\mathbb {R}}^n$ defined by the equations $x_i =x_j + k$, where $1\le i<j\le n$ and $k=0,1$, all of which contain a line parallel to the line $\ell$ with equation $x_1=x_2=\cdots =x_n$. Hence, we may represent each hyperplane by its intersection with a given hyperplane orthogonal to $\ell$, for example, the hyperplane with equation $x_1+x_2+\cdots +x_n=0$ (Figure 1).

Figure 1.

Representation of the hyperplane arrangements in dimensions 2 (left) and 3 (right).

Label the region $R_0$ defined by $x_1>x_2>\cdots>x_n>x_1+1$ with $(1,1,\ldots ,1)\in {\mathbb {Z}}^n$. Given regions R and $R'$ separated by the unique hyperplane H such that R and $R_0$ are on the same side of H, give to $R'$ the label of R but add 1 to one coordinate: the ith coordinate if H is defined by an equation of the form $x_i=x_j$, $i<j$, and the jth coordinate if H is defined by an equation of the form $x_i=x_j+1$.

Pak and Stanley showed [4] that the labels thus defined are exactly the parking functions of length n. We proved [1] that the prime parking functions are the Pak–Stanley labels of the relatively bounded regions (shaded in Figure 1, where the cases $n=2$ and $n=3$ are represented, the first one directly, and both by projecting, as explained above).

Our Main Theorem

Let $\text {PF}_{n}$ be the set of parking functions of length n, and $\text {PF}'_{n}\subseteq \text {PF}_{n}$ the set of prime parking functions of length n. We know that

$$\begin{aligned} |\text {PF}_{n}|&=(n+1)^{n-1}\,, \end{aligned}$$

(3)

$$\begin{aligned} |\text {PF}'_{n}|&=(n-1)^{n-1}\,. \end{aligned}$$

(4)

The goal of this paper is to prove (4) in a simple and original manner, as a direct consequence of Theorem 1 below.

Prime parking functions were introduced by Gessel, who proved (4) through generating functions (cf. [5, Exercise 5.49f]). Soon thereafter, Kalikow [2, p. 37] gave a direct proof, which inspired ours.

Theorem 1.

For every ${\textbf{a}}=(a_1,\ldots ,a_n):[n]\rightarrow [n-1]$, there exist a unique $k\in [n-1]$ and a unique prime parking function ${\textbf{b}}=(b_1,\ldots ,b_n)$ such that

$$\begin{aligned} a_i\equiv b_i+k-1\pmod {n-1}\quad \textit { for every }i\in [n]\,. \end{aligned}$$

(5)

Conversely, given $k\in [n-1]$ and the parking function ${\textbf{b}}=(b_1,\ldots ,b_n)$, (5) defines the function ${\textbf{a}}=(a_1,\ldots ,a_n)$ uniquely.

Proof.

Without loss of generality, suppose that ${\textbf{a}}^\uparrow ={\textbf{a}}$, and define, for every $i\in [n-1]$,

$$\begin{aligned} s_i&= (a_{i+1}-a_i)+\cdots +(a_n-a_i)\\&\quad +(a_1-a_i+n-1)+\cdots +(a_{i-1}-a_i+n-1)\\&=\sum _{j=1}^n a_j-n\,a_i+(i-1)(n-1)\,,\\ s_n&=(a_1-a_n+n-1)+\cdots +(a_{n-1}-a_n+n-1)\\&=\sum _{j=1}^n a_j-n\,a_n+(n-1)^2\,, \end{aligned}$$

and let $d\in [n]$ be such that $s_d=\min \{s_i\mid i\in [n]\}$. Let $k=a_d$, and

$$\begin{aligned}&{\textbf{b}}= (1,a_{d+1}-k+1,\ldots ,a_n-k+1,\\&\quad a_1-k+n,\ldots ,a_{d-1}-k+n) \,. \end{aligned}$$

Since for $i\ne d$, we have

$$\begin{aligned}&s_i-s_d=(n-1)(i-d)-n(a_i-a_d)\ge 0\,, \end{aligned}$$

it follows that

$$\begin{aligned}&a_i-k \le (i-d)- \frac{i-d}{n}< {\left\{ \begin{array}{ll}i-d,&{}\text {if }i > d,\\ i-d+1,&{}\text {if }i< d.\end{array}\right. } \end{aligned}$$

Therefore, ${\textbf{b}}$ is a prime parking function, for it is componentwise less than or equal to

$$\begin{aligned} (1,1,2,\ldots ,n-d+1,n-d+2,\ldots ,n-1)\,. \end{aligned}$$

For proving the uniqueness, suppose, contrary to our assumptions, that both ${\textbf{b}}=(b_1,\ldots ,b_n)$ and ${\textbf{c}}=(c_1,\ldots ,c_n)$ were prime parking functions and that they satisfied the hypothesis of our theorem. Then there would exist $\ell \in {\mathbb {N}}$ such that $c_i\equiv b_i+\ell \pmod {n-1}$ for every $i\in [n]$. Without loss of generality, we suppose that ${\textbf{b}}^\uparrow ={\textbf{b}}$. If $b_n+\ell <n$, then $c_1=1+\ell =1$, and $\ell =0$ and ${\textbf{b}}={\textbf{c}}$. Otherwise, let $b_{i-1}<n-\ell \le b_i$. Then

$$\begin{aligned} {\textbf{c}}^\uparrow=\, & {} (\overbrace{b_i+\ell -n+1}^{1},\ldots , \overbrace{b_n+\ell -n+1}^{{}\le n-i}, \\{} & {} \quad \overbrace{1+\ell }^{{}\le n-i+1},\ldots , \overbrace{b_{i-1}+\ell }^{{}\le n-1})\,. \end{aligned}$$

Hence, $b_i=n-\ell <i$ and $\ell +1\le n-i+1$. This contradiction concludes the proof. $\square$

For example, as before, let

$$\begin{aligned} {\textbf{a}}=(3, 13, 6, 3, 7, 3, 2, 1, 10, 11, 6, 7, 14, 10, 11)\in \text {PF}_{15}\,. \end{aligned}$$

Then if $k=10$ and $a_i\equiv b_i+k-1\pmod {n-1}$ for every $i\in [n]$, it follows that

$$\begin{aligned} {\textbf{b}}=(8, 4, 11, 8, 12, 8, 7, 6, 1, 2, 11, 12, 5, 1, 2)\in \text {PF}'_{15}\,. \end{aligned}$$

Note that Theorem 1 shows that $(n-1)^n = (n-1) |\text {PF}'_{n}|$, whence the number of prime parking functions is indeed $(n-1)^{n-1}$.

The Parking Point of View

Note that by (1) and (2), a prime parking function of length n is a parking function of length $n-1$ with an extra 1. More precisely, if ${\textbf{a}}=(a_1,\ldots ,a_n):[n]\rightarrow [n-1]$, $a_i=1$ for some $i\in [n]$, and

$${\textbf{b}}=(a_1,\ldots ,a_{i-1},a_{i+1},\ldots ,a_n):[n-1]\rightarrow [n-1],$$

then ${\textbf{a}}\in \text {PF}'_{n}$ if and only if ${\textbf{b}}\in \text {PF}_{n-1}$. Hence ${\textbf{a}}:[n]\rightarrow [n-1]$ is a prime parking function if all cars can park in a one-way street with n spots labeled 1, 1, 2, . . . , $n-1$. For instance,

$$\begin{aligned} {\textbf{b}}=(8, 4, 11, 8, 12, 8, 7, 6, 1, 2, 11, 12, 5, 1, 2)\in \text {PF}'_{15} \end{aligned}$$

parks as follows:

In fact, if all elements of ${\textbf{a}}$ park as described, the element of ${\textbf{a}}$ that parks in the first spot labeled 1 must be $a_i=1$, so that $a_j>1$ for every $j<i$. Then let ${\textbf{b}}$ be as defined above.

Since ${\textbf{b}}$ is ${\textbf{a}}$ with ith coordinate (equal to 1) deleted, and by hypothesis, ${\textbf{a}}$ parks where the n spots are labeled $1,1,2,\ldots ,n-1$, then ${\textbf{a}}$ parks the ith element in the first spot labeled 1, and afterward, it parks the elements of ${\textbf{b}}$ in the subsequent spots (labeled $1,2,\ldots ,n-1$). So ${\textbf{b}}$ is a parking function.

On the other hand, if again, $a_i=1$ for some $i\in [n]$ such that $a_j>1$ for every $j\in [i-1]$, then the parking procedure parks $a_i$ in the first spot labeled 1 and parks the remaining elements of ${\textbf{a}}$ in the next positions, since they are the elements of ${\textbf{b}}$.

An immediate consequence of Theorem 1 is the following addendum to Konheim and Weiss’s result.

Proposition 2.

Let n cars enter a one-way street with n spots, and suppose driver i wants to park in position $p(i)\le n-1$ . Then there is a unique $k\in [n-1]$ such that all cars can park in the previous sense, provided the spots are labeled consecutively

For example, for

$${\textbf{a}}=(3, 13, 6, 3, 7, 3, 2, 1, 10, 11, 6, 7, 14, 10, 11),$$

we have