MacMahon Partition Analysis: A discrete approach to broken stick problems

Abstract

We propose a discrete approach to solve problems on forming polygons from broken sticks, which is akin to counting polygons with sides of integer length subject to certain Diophantine inequalities. Namely, we use MacMahon's Partition Analysis to obtain a generating function for the size of the set of segments of a broken stick subject to these inequalities. In particular, we use this approach to show that for $n\ge k\ge 3$, the probability that a k-gon cannot be formed from a stick broken into n parts is given by n! over a product of linear combinations of partial sums of generalized Fibonacci numbers, a problem which proved to be very hard to generalize in the past.

Introduction

Partition Analysis is a computational method to solve problems that involve linear homogeneous Diophantine inequalities and equations. Introduced by MacMahon in his famous book Combinatory Analysis [8], and later used by Stanley in his proof of the Anand–Dumir–Gupta conjecture [12], the method died down for decades until Andrews et al. devoted a series of papers to applications of Partition Analysis. They started with a new proof of the Lecture Hall Partition Theorem of Bousquet-Mélou and Eriksson [4] and even provided a computer algebra implementation of the method (see [1] for a great exposition to the history and ideas of MacMahon's Partition Analysis).

On the other hand, in broken stick problems, one breaks a stick at random at $n-1$ points to obtain n pieces and then asks diverse geometric probability questions involving this broken stick. Henceforth, we will focus on forming polygons from these pieces. The basis of these questions is the famous problem from the Mathematical Tripos on finding the probability that a triangle can be formed from a stick broken at random in three segments, which was quickly generalized to forming an n-gon from the n pieces of a broken stick. The probability in the latter case is simply $1-n/2^{n-1}$, and the most popular solution to this problem comes from the note The Broken Spaghetti Noodle of D'Andrea and Gómez [5]. For more background and motivation on these kinds of problems, an interested reader may wish to read the introduction of [13].

It is natural to generalize these problems by using $3\le k\le n$ pieces to form a k-gon out of the n pieces, but they turn out to be much harder than expected and only a few have found an answer. In project reports by the Illinois Geometry Lab [7], [10], two extreme cases were considered, namely the probability that given a stick broken up into n pieces, there exist three segments that can form a triangle (we call it the “there exist” problem), as well as the probability that all choices of three segments can form a triangle (the “for all” problem). It was stated without proof that the answers are respectively

$$1-n!\prod _{j=2}^n{(F_{j+2}-1)}^{-1}\text{ and }{\left(\begin{array}{c}2n-2\\ n\end{array}\right)}^{-1},$$

where $F_j$ is the jth Fibonacci number. Crowdmath's 2017 project¹ answered many broken stick type questions, but one of the only problems they could not solve had to do with forming polygons from a broken stick. However, the general case of the “for all” problem was solved recently by using results on random divisions of an interval and order statistics.

Theorem 1

[13] Let $n\ge k\ge 3$ be positive integers. The probability that every choice of k segments from a stick broken up into n parts will form a k-gon is given by

$$\frac{n(n-1)\cdots (n-k+3)}{(n-k+2)}\sum _{j=1}^{n-k+2}\frac{{(-1)}^{j+1}}{j^{k-3}}\left(\begin{array}{c}n-k+2\\ j\end{array}\right){(\frac{n-k+2}{j}+1)}_{k-2}^{-1},$$

where

$${(\frac{n-k+2}{j}+1)}_{k-2}=\prod _{i=0}^{k-3}(\frac{n-k+2}{j}+1+i)$$

stands for the rising factorial function or Pochhammer symbol.

It was pointed out that the method could be adapted to answer similar questions such as the “there exist” problem, but in practice it was much harder to apply.

In this paper, we show that MacMahon's Partition Analysis is well suited to solve broken stick problems by considering a discrete setting and exhibiting a connection with systems of linear Diophantine inequalities. This discrete setting is more generally related to counting polygons with sides of integer length, as studied in [2] for example. In particular, we apply it to solve the “there exist” problem in full generality. As a purely aesthetic choice, we will consider the complimentary probability that no choices of k segments from a stick broken up into n parts will form a k-gon.

We define the generalized Fibonacci numbers $F_n^{(k)}$ as it is typically done:

$$F_n^{(k)}:=\{\begin{array}{cc}0\hfill & 0\le n\le k-2,\hfill \\ 1\hfill & n=k-1,\hfill \\ {\sum }_{i=1}^k{F}_{n-i}^{(k)}\hfill & n\ge k.\hfill \end{array}$$

Next, let

$$f_k(i):=\{\begin{array}{cc}0\hfill & 0\le i\le k-2,\hfill \\ {\sum }_{n=k-1}^i{F}_n^{(k)}\hfill & i\ge k-1\hfill \end{array}$$

be restricted partial sums of the $F_n^{(k)}$, and let

$$g_k(j):=1+\sum _{\ell =2}^j{f}_k(n-\ell )(2\le j\le n-k+1)$$

and

$$h_k(\ell ):=f_k(n)+\sum _{j=2}^{\ell }{g}_k(k+1-j)(2\le \ell \le k-2).$$

The preceding definitions allow us to state our main result in terms of linear combinations of generalized Fibonacci numbers.

Theorem 2

Let $n\ge k\ge 3$ be positive integers. The probability that no choices of k segments from a stick broken up into n parts will form a k-gon is given by

$$\frac{n!}{f_{k-1}(k-2)\cdots f_{k-1}\left(n\right)h_{k-1}(2)\cdots h_{k-1}(k-2)},$$

where the $h_{k-1}(\ell )$ are to be removed for $k=3$.

Unfortunately, there is not a (nice enough) closed form for partial sums of generalized Fibonacci numbers to express the previous expression in a simpler form, but one is well known for typical Fibonacci numbers $F_n^{(2)}$:

$$f_2(i)=\sum _{n=1}^i{F}_n^{(2)}=F_{i+2}^{(2)}-1$$

for $i\ge 2$, and so the expression obtained by the Illinois Geometry lab (1.1) follows. Also note that while the probability might seem hard to calculate, it is simple from a computational point of view, since generalized Fibonacci numbers are easily calculated and the sums $f_k(i)$ are documented on the OEIS.

Example

If we want to make a 4-gon, we have

$${(f_3(i))}_{i\ge 2}=(1,2,4,8,15,28,52,96,177,\dots )$$

(OEIS A008937 [9]), and so, if $P(4,n)$ stands for the probability that no choices of 4 segments from a stick broken up into n parts will form a 4-gon, then

$$P(4,n)=\frac{n!}{1\cdot 2\cdot 4\cdots f_3(n)(1+f_3(n-2)+f_3(n))}.$$

Thus, we can easily calculate

$$P(4,4)=\frac{4!}{1\cdot 2\cdot 4\cdot (1+1+4)}=\frac{1}{2},P(4,5)=\frac{15}{88},P(4,6)=\frac{3}{80},$$

and so on.

Our main tool to prove the previous theorem will be the following lemma on partitions subject to a system of Diophantine inequalities, which we will establish using Partition Analysis.

Lemma 1

The number of partitions of N into n positive parts $a_i$ and under the additional linear constraints

$$\begin{gathered} a_1\ge a_2+a_3+\cdots +a_k,a_2\ge a_3+a_4+\cdots +a_{k+1},⋮a_{n-k+1}\ge a_{n-k+2}+a_{n-k+3}+\cdots +a_n \\ {a}_{n-k+2}\ge a_{n-k+3} \\ {a}_{n-k+3}\ge a_{n-k+4}⋮ \\ {a}_{n-1}\ge a_n \end{gathered}$$

equals the number of partitions of N into parts taken from the set of integers

$$\{f_{k-1}(k-2),f_{k-1}(k-1),\dots ,f_{k-1}\left(n\right),h_{k-1}(2),h_{k-1}(3),\dots ,h_{k-1}(k-2)\}.$$

We end this section with a brief overview of the mathematics behind MacMahon's operator that we will use in Partition Analysis. The linear operator ${\mathrm{\Omega }}_{\ge }$ is defined as acting on multiple Laurent series as follows:

$$\underset{\ge }{\mathrm{\Omega }}\sum _{n_1,\dots ,n_r=-\infty }^{\infty }{A}_{n_1,\dots ,n_r}{\lambda }_1^{n_1}\cdots {\lambda }_r^{n_r}=\sum _{n_1,\dots ,n_r=0}^{\infty }{A}_{n_1,\dots ,n_r}.$$

The $A_{n_1,\dots ,n_r}$ can be thought of as rational functions of several variables independent of the ${\lambda }_i$ and the series are to be treated analytically (because the method relies on unique Laurent series expansion of rational functions). Also note that we could specify a domain in each case to guarantee absolute convergence of the sums in an open neighborhood of the circles $|{\lambda }_i|=1$, so that the ${\mathrm{\Omega }}_{\ge }$ operator is well-defined.

While most of the time ${\mathrm{\Omega }}_{\ge }$ acts on ${\lambda }_i$, we will also use ${\mu }_i$ to indicate that these variables encode a different kind of linear constraint than the ${\lambda }_i$.