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\author{Robin K. S. Hankin\\Auckland University of Technology\And
Luke J. West\\National Oceanography Centre, Southampton}
\title{Set Partitions in \proglang{R}}
%\VignetteIndexEntry{Set partitions in R}

%% for pretty printing and a nice hypersummary also set:
\Plainauthor{Robin K. S. Hankin, Luke G. West}
\Plaintitle{Set partitions in R}
\Shorttitle{Set partitions in \proglang{R}}

%% an abstract and keywords
\Abstract{This vignette is based on~\cite{hankin2007a}.

This short paper introduces a code snippet in the form of an
  \proglang{R} function that enumerates all possible partitions of a
  finite set given the sizes of the equivalence classes.  Three
  combinatorial problems are solved using the software: one from
  bioinformatics, one from scheduling theory, and one from forensic
  science.}

\Keywords{Set Partitions, Forensic Science,
Enumerative Combinatorics, \proglang{R}}
\Plainkeywords{Set Partitions, Forensic Science,
Enumerative Combinatorics, R}


%% publication information
%% NOTE: This needs to filled out ONLY IF THE PAPER WAS ACCEPTED.
%% If it was not (yet) accepted, leave them commented.
%% \Volume{13}
%% \Issue{9}
%% \Month{September}
%% \Year{2004}
%% \Submitdate{2004-09-29}
%% \Acceptdate{2004-09-29}

%% The address of (at least) one author should be given
%% in the following format:
\Address{
  Robin K. S. Hankin\\
  Auckland University of Technology\\
  AUT Tower\\
  Wakefield Street\\
  Auckland, New Zealand\\
  E-mail: \email{hankin.robin@gmail.com}\\
  {}\\
  Luke J. West\\
  National Oceanography Centre, Southampton\\
  European Way\\
  Southampton SO14 3ZH\\
  United Kingdom\\
  }
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%% Fax: +43/1/31336-734

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\SweaveOpts{echo=FALSE}
\begin{document}
<<load.library,echo=FALSE,print=FALSE>>=
<<results=hide>>=
require(partitions)
@ 

\section{Introduction}

A {\em partition} of a set~$S=\{1,2,\ldots,n\}$ is a family of
sets~$T_1,T_2,\ldots T_k$ satisfying
\begin{enumerate}
\item $T_i\cup T_j=\emptyset$ if $i\neq j$
\item $\bigcup_{i=1}^k T_k=S$
\item $T_i\neq\emptyset$ for $i=1,\ldots,k$.
\end{enumerate}



There are exactly fifteen ways to partition a set of four elements,
shown in Table~\ref{partitions.4elements} using standard notation.
Here~$(13)(2)(4)$ represents the same partition as~$(2)(4)(31)$: the
induced equivalence relation is identical.  Either one may be
represented as \code{1213}, indicating that elements one and three are
in the first equivalence class (``\code{1}''), element two in the
second, and element four in the third.  In
Table~\ref{partitions.4elements}, two partitions appear in the same
column if their `shape' (that is, the size distribution of the
equivalence classes, formally defined in Section~\ref{algorithms}
below) is the same.  The shapes correspond to integer partitions in
standard form~\citep{hankin2006}, and appear in standard
order~\citep{andrews1998}.  This suggests a natural method for
enumerating the equivalence relations on a set of cardinality~$n$:
enumerate the integer partitions of~$n$, and for each of these,
enumerate the distinct equivalence relations of that shape.

\begin{table}[h]
  \centering
  \begin{tabular}{ccccc}
    (1234) & (123)(4) & (12)(34) & (12)(3)(4) & (1)(2)(3)(4)\\
           & (124)(3) & (13)(24) & (13)(2)(4) &             \\
           & (134)(2) & (14)(23) & (14)(2)(3) &             \\
           & (234)(1) &          & (23)(1)(4) &             \\
           &          &          & (24)(1)(3) &             \\
           &          &          & (34)(1)(2) &             \\
  \end{tabular}
  \caption{The fifteen partitions of a set of
    four \label{partitions.4elements} elements}
\end{table}


This paper introduces software that enumerates\footnote{The transitive
  verb ``to enumerate'' can be used in two senses: firstly, ``to
  calculate the number of''; and secondly, ``to list each occurrence
  of, as if for the purpose of counting''.  This paper adopts the
  second usage.} all set partitions of a set of a specified (finite)
size.  The new functionality presented here is given by
\code{setparts()}, an \proglang{R} \citep{rcore2007} wrapper for
\proglang{C++} code, currently part of the \pkg{partitions}
package~\citep{hankin2006,hankin2007}, version 1.7-3, available under
the GPL from CRAN, \url{http://www.r-project.org/}.

Function \code{setparts()} takes a single argument \code{a}, coerced
to integer mode.  The default case is that of \code{a} being a vector
of length greater than one, in which case it returns all partitions of
a set of \code{sum(a)} elements into equivalence classes of the sizes
of the elements of \code{a}; in the case of \code{a} being of length
one, it returns all partitions of a set of \code{a} elements.  If
\code{a} is a matrix, it returns all equivalence classes with sizes of
the columns of \code{a}.

\section{Algorithms}
\label{algorithms}

In this paper, we present an algorithm that lists all partitions of a
given ``shape''.  Formally, the {\bf shape}
$\lambda=(\lambda_1,\ldots\lambda_k)$ of a partition is given by the
sizes of the~$T_i$; these are conventionally given in standard (i.e. 
non-increasing) order.  Thus~$\lambda$ is an integer partition of~$n$
\citep{hankin2006}, written~$\lambda\vdash n$.

The distinct equivalence relations may be enumerated, for a given
integer partition, using a recursive algorithm that operates on
partially filled partitions.

 Consider an integer partition~$\lambda=
 \left(\lambda_1,\ldots\lambda_k\right)$, also
 written~$\left(1^{f_1},\ldots,n^{f_n}\right)$; the latter notation
 indicates that~$f_i$ elements of~$\lambda$ are equal to~$i$.  The
 algorithm used here fills the equivalence classes sequentially from
 left to right using a technique originally due
 to~\citet{killworth2007}.

The partitions are ordered from largest to smallest; non-empty ties
are broken lexicographically.  This prevents the reporting of
identical equivalence classes (to see this, consider the two identical
equivalence relations~$(123)(45)(67)(8)$ and~$(123)(67)(45)(8)$: the
second descriptor does not occur in the algorithm because the third
equivalence class precedes the second).

\begin{figure}[htbp]
  \begin{center}
\includegraphics{algorithm}
\caption{The\label{algorithm} algorithm used to enumerate set
  partitions of a given shape.  Unfilled positions are indicated by a
  dot.  Of the partitions at a given level, the children of one
  are shown.
  The initial empty configuration is filled from left to right from a
  `pool' of unused elements; equivalence classes are ordered by size
  (largest first) and ties are broken lexicographically at runtime to
  suppress reporting of identical equivalence relations in which two
  classes are transposed.
  Within a class, the elements of partially- and completely- filled
  equivalence classes are ordered in increasing order.  
  A cross indicates an illegal configuration in which the partially
  filled class cannot be completed as insufficient elements remain in
  the pool; the algorithm prunes such configurations as soon as they
  are detected.
  Successful termination of the algorithm, viz. a filled partition, is
  indicated by a tick (``check mark'') at which point the filled
  equivalence class is reported, and the recursion bottoms out}
  \end{center}
\end{figure}

Thus, starting with an empty partition of a given size, one may apply
the algorithm detailed in Figure~\ref{algorithm} recursively until
each full partition is recorded.

\subsection{Performance and complexity}

The total number of set partitions corresponding to a particular
integer partition~$\lambda$ is given by elementary combinatorial
arguments as

\begin{equation}\label{combinatoric_setpartition}
\frac{n!}{\prod_{i=1}^{k}\lambda_i!\cdot\prod_{j=1}^n{f_j}!}.
\end{equation}

However, the number of integer partitions increases rapidly with~$n$;
the exact value is given by the partition function~\code{P(n)} of the
package~\citep{hankin2006}, but the asymptotic form given
by~\citet{hardy1918}, viz.
\begin{equation}\label{asymptotic_setpartition}
P(n)=\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\cdot\left(1+O(n^{-1})\right)
\end{equation} shows that the rate
of growth is rapid. 
The number of {\em set} partitions of a set of size~$n$ is given by
the Bell numbers~$B(n)$~\citep{rota1964}, which grow much more rapidly
than the partition function; an asymptotic form was given
by~\cite{bender1974}:
\begin{equation}\label{asymptoticbell}
B(n) = t^{n-t}e^{t-1}\left(\log t\right)^{-1/2}\cdot\left(1+o(n)\right)
  \end{equation}
where~$t=e^{W_0(n-1/2)}$.  Here~$W_0(\cdot)$ denotes the principal
branch of the Lambert W function~\citep{corless1996}, provided with
the \pkg{gsl} package~\citep{hankin2006b}.
Equation~\ref{asymptoticbell} is easily evaluated and can be used to
give a feel for the magnitude of the Bell numbers.  Note that
convergence is quite slow; for example, $B(10)=115975$ (compare
about~$1.48\times 10^3$ from Equation~\ref{asymptoticbell}),
and~$B(100)\simeq 4.76\times 10^{115}$ (compare~$5.43\times
10^{115}$).


\section{Applications}

Many problems in science and industry involve the optimization of some
desideratum over a finite number of options.  Such optimization is
frequently soluble within the discipline of computational
combinatorics: simply enumerate the possible solutions and choose the
best.  One advantage of this method over heuristic optimization
techniques such as tabu search \citep{glover1997} is that by
enumerating all possible solutions, one is \emph{certain} to find the
global optimum.  In this section we present three examples of
\code{setparts()} in use: one from bioinformatics, one from
multiprocessor scheduling, and one from forensic science.

\subsection{Bioinformatics}

In studies of allelic segregation under disomic inheritance, one
situation encountered is that of enumerating the possible arrangements
of alleles on genetic loci\footnote{A \emph{chromosome} is a double
  helix of DNA, typically occurring in matched pairs in cell nuclei.  A
  \emph{locus} is an identifiable position on a chromosome pair; a
  \emph{gene} is a distinct sequence of DNA base pairs occurring at a
  locus; an \emph{allele} is one of a number of distinguishable forms
  for a particular gene; the \emph{genome} is the complete set of
  genes present in the chromosomes; a \emph{monosomic locus} contains
  two copies of one, and a \emph{disomic locus} two different,
  alleles.  Two alleles at the same locus are said to be
  \emph{colocated}.  A monosomic (disomic) allele is one that occurs
  at a monosomic (disomic) locus in a given genome.  A \emph{gamete}
  is a sequence of alleles chosen from the genome: monosomic and
  disomic alleles are chosen with probability~1 and~0.5
  respectively.}.  \cite{rodzen2002} consider the white sturgeon;
there is much uncertainty surrounding the organization of this
species's genome.

Many alleles of interest may be identified in the genome of the white
sturgeon.  Sometimes these are known to be either monosomic or
disomic, but it is not known which alleles are colocated, nor which
alleles are monosomic and which are disomic.  If researchers have
identified, say, five alleles (conventionally labelled 1-5), and
further that two pairs of these are colocated at two disomic loci and
one is monosomic, then there are a number of possible arrangements for
these alleles on the chromosome.  In the language of the package, we
wish to enumerate the partitions of a set of five elements (the
alleles) into equivalence classes of size~2,2,1 (the loci).  The five
alleles are arranged into two from one disomic locus, two from another
disomic locus, and one from a monosomic locus.

Enumerating the possible arrangements is accomplished using the
\code{setparts()} function:

<<alleles,echo=TRUE>>=
setparts(c(2,2,1))
@ 

Each column comprises two~\code{1}s, two~\code{2}s, and one~\code{3}.
Thus the first column, \code{x=12321}, say, indicates that alleles~1
and~5 [\code{which(x==1)}] occur at the first locus (which is
disomic); alleles~2 and~4 [\code{which(x==2)}] at the second (also
disomic); and allele~3 [\code{which(x==3)}] at the third locus, which
is monosomic.  This could be written~$(15)(24)(3)$; the \proglang{R}
idiom would be

<<use_achims_idea,echo=TRUE>>=
a <- c(1,2,3,2,1)
split(seq_along(a),a)
@ 

[use \code{listParts(c(2,2,1))} to see all possible equivalence classes].
Two examples of electrophoresis gels, corresponding to the partitions
of the first and last columns respectively, are shown in
figure~\ref{electrophoresis_gels}, which is analogous to Figure~1
of~\citet{rodzen2002}.

\begin{figure}[htbp]
  \begin{center}
    \includegraphics{strand}
    \caption{Two electrophoresis gels \label{electrophoresis_gels}
      showing the possible bands that can occur with two different
      arrangements of alleles.  In both gels, two loci are disomic,
      and one is monosomic.  Left, gel corresponding to $(15)(24)(3)$,
      or \code{12321}; this would mean that alleles~1 and~5 are at one
      disomic locus, alleles~2 and~4 at another, and allele~3 at a
      monosomic locus.  Thus band combinations~1-5 and~2-4 are absent.
      The gel on the right corresponds to~$(1)(24)(35)$, or
      \code{31212}}
  \end{center}
\end{figure}

\subsection{Scheduling theory}

The ``augmented multiprocessing model'' is an abstract description of
a certain class of problems that occurs frequently in the field of
scheduling theory~\citep{garey1975}.  We consider a simple variant of
the augmented multiprocessing model (specifically, the ``three
partition problem'', which is known to be NP-complete) and solve it
using the software presented here.

Consider a set of tasks~${\mathcal T}=\{T_1,\ldots,T_m\}$; task~$i$
requires time~$\tau_i$ for its completion in the sense that if it
begins executing at time~$t$ it will complete at time~$t+\tau_i$.
Each of these tasks is to be executed by one of~$n$ processors.  It is
desired to allocate the tasks among the processors so that all tasks
are completed in the minimum possible time.

Thus if there are~$m=4$ tasks with~$T_i$ requiring~$\tau_i=i$ time
units (``seconds'') for completion, and~$n=2$ processors, it is clear
that allocating tasks~1 and~4 to one processor, and tasks~2 and~3 to
the second is optimal.  This would correspond to the set
partition~$(14)(23)$; the time required is~$1+4=2+3=5$ seconds.  Other
allocations such as~$(24)(13)$ would take longer ($2+4=6$ seconds).

More complex examples are not so straightforward.  Consider the case
where~$m=9$ and again~$\tau_i=i$; now there are~$n=3$ processors.  The
possible task schedules may be enumerated using the
\code{restrictedparts()} function\footnote{Recall that
  \code{restrictedparts(a,b)} enumerates the possible partitions of
  \code{a} into at most \code{b} parts.}  of the \pkg{partitions}
package:

<<echo=FALSE,print=FALSE>>=
options(width=63)
m <- 9
n <- 3
@ 

<<setpartsrestrictedparts,echo=TRUE>>=
jj <- setparts(restrictedparts(m,n,include.zero=FALSE))
summary(jj)
@ 

Each column corresponds to a task allocation schedule.  The first
column, for example, indicates that tasks 1,2,3,4,7,8,9 are allocated
to processor~1, task~5 to processor~2, and task~6 to processor~3 and
the set partition might be written~$(1234789)(5)(6)$.  In the matrix,
the maximum entry of each column is~3, showing that all three
processors are always used (setting \code{include.zero} to \code{TRUE}
would relax this requirement and allow some columns to correspond to
allocations in which only one or two processors are allocated a
nonzero number of tasks).

The objective function to be minimized is just the time for the
processor which finished last:

<<definepenaltyfunctionF,echo=TRUE>>=
tau <- 1:9
slowest <-  function(x) max(tapply(tau, x, sum))
@ 

The minimal objective function is clearly 45/3=15.  Identifying
allocations that attain this bound is straightforward:

<<show_the_best,echo=TRUE>>=
time.taken <- apply(jj,2,slowest)
minimal.time <- sum(tau)/n
jj[,time.taken == minimal.time]
@ 

Thus~\Sexpr{sum(time.taken==minimal.time)} allocations attain the maximal
possible efficiency: a conclusion difficult to achieve without
enumeration.  These optima may be differentiated using some secondary
characteristic, such as even distribution of the {\em number} of tasks
among the processors.  This would prescribe either the eighth or ninth
columns, which both allocate three tasks to each processor.

\subsection{Forensic science}

In the field of forensic science, one situation sometimes encountered
is the following.  Crimes~$C_j$, where~$i=1,\ldots,n$ have been
committed; it is desired to infer how many perpetrators are responsible
and which crimes each perpetrator is responsible for.  One prominent
example would be multiple unsolved homicides~\citep{roberts2003}.

Evidence at each of the~$n$ crime scenes is available in the form
of~$m$ Bernoulli random variables that indicate various forensic
characteristics at the crime scene (examples might include presence or
absence of a weapon, forced entry, etc).  The evidence may be
organized in a~$m$-by-$n$ matrix~$E$ of zeroes and ones indicating
whether forensic characteristic~$i$ was true at crime scene~$j$,
where~$1\leqslant i\leqslant m$ and~$1\leqslant j\leqslant n$.

A given row of~$E$ thus corresponds to a particular type of evidence;
if perpetrator~$k$ committed crime~$j$, we would have~$E_{ij}=1$ with
probability~$p_{ik}$, specific to perpetrator~$k$.  For each
evidence~$i$, it is assumed that each perpetrator has a~$p_{ik}$
chosen at random from a beta distribution with
parameters~$\alpha_i,\beta_i$ for~$1\leqslant i\leqslant m$, and that
all perpetrators have the same~$\alpha_i,\beta_i$; these will be known
(or at least estimated) from previous studies of similar crimes.  An
example is shown below.

<<define.E.matrix, echo=FALSE,print=FALSE>>=
E <- matrix(c(0,0,0,1,1,0,0,0,1,1,1,1,1,0,0), 5,3)
dimnames(E) <-  list(
                     evidence=paste("E",1:5,sep=""),
                     crime=paste("C",1:3,sep="")
                     )
@ 


<<print.E,echo=FALSE, print=TRUE>>=
E
@ 

Thus in this example there are~$n=3$ crimes and~$m=5$ forensic
characteristics.  Observe that~\code{C1} and~\code{C2} have identical
evidence, and that~\code{C3} has contrary evidence.  This would
suggest that there are exactly two perpetrators: one responsible for
\code{C1} and \code{C2}, and one for \code{C3}, corresponding to
partition~$(12)(3)$; the hypothesis that this is the case is
written~$H_{[1,1,2]}$.

Considering a single type of evidence, and a single perpetrator, the
probability of observing~$a$ successes and~$b$ failures is

\[
\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}
\int_{p=0}^1
\left[
\frac{(a+b)!}{a!b!}
p^{a}(1-p)^{b}
\right]
p^{\alpha-1}(1-p)^{\beta-1}\,dp=
\frac{(a+b)!}{a!b!}\cdot
\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\cdot
\frac{\Gamma(a+\alpha)\Gamma(b+\beta)}{\Gamma(a+b+\alpha+\beta)}.
\]

Here the probability~$p$ is integrated over its range, weighted by its
prior (the posterior would also be a beta distribution, with
parameters~$\alpha+a$ and~$\beta+b$).

Now consider a set partition~$\wp=T_1,\ldots,T_{r}$ of crimes;
identify a perpetrator with each equivalence class of crimes and
observe that a single perpetrator has~$m$ specific values of~$p_i$,
$1\leqslant i\leqslant m$ and acts independently of other
perpetrators.  The likelihood of~$\wp$ is then
\[
{\mathcal L}(\wp)=
C\left(\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\right)^r
\cdot\prod_{k=1}^{r}\prod_{i=1}^{m}
\frac{\Gamma(a_{ik}+\alpha_i)\Gamma(b_{ik}+\beta_i)}{\Gamma(a_{ik}+b_{ik}+\alpha_i+\beta_i)}
\]
where~$a_{ik}=\sum_{j\in T_k}E_{ij}$ and~$b_{ik}=\sum_{j\in
  T_k}1-E_{ij}$ are the total successes and failures for evidence~$i$
due to perpetrator~$k$ (under~$H_\wp$).  Here~$C$ denotes an
arbitrary multiplicative constant.

It is interesting to note that the case $\alpha=\beta=1$ is not
uninformative in this context, even though this induces a uniform
prior distribution on~$p$.  To see this, consider the first line of
the five by three matrix above [\code{0,0,1}].

The likelihood function for this row is on five partitions:
\begin{eqnarray*}
{\mathcal L}(H_{[1,1,1]}) &=&
\frac{\Gamma(\alpha+1)\Gamma(\beta+2)}{\Gamma(\alpha+\beta+3)}\cdot
\left(\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\right)^1\\
{\mathcal L}(H_{[1,2,1]}) &=&
\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)}\cdot
\frac{\Gamma(\alpha)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+1)}\cdot
\left(\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\right)^2\\
{\mathcal L}(H_{[1,1,2]}) &=&
\frac{\Gamma(\alpha)\Gamma(\beta+2)}{\Gamma(\alpha+\beta+2)}\cdot
\frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma(\alpha+\beta+1)}\cdot
\left(\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\right)^2\\
{\mathcal L}(H_{[2,1,1]}) &=&
\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)}\cdot
\frac{\Gamma(\alpha)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+1)}\cdot
\left(\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\right)^2\\
{\mathcal L}(H_{[1,2,3]}) &=&
\frac{\Gamma(\alpha)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+1)}\cdot
\frac{\Gamma(\alpha)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+1)}\cdot
\frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma(\alpha+\beta+1)}\cdot
\left(\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\right)^3.
\end{eqnarray*}

Thus if~$\alpha=\beta=1$ the likelihood function is
\begin{eqnarray*}
{\mathcal L}(H_{[1,1,1]}) &=&
C/12\\
{\mathcal L}(H_{[1,2,1]}) &=&
C/12\\
{\mathcal L}(H_{[1,1,2]}) &=&
C/6\\
{\mathcal L}(H_{[2,1,1]}) &=&
C/12\\
{\mathcal L}(H_{[1,2,3]}) &=&
C/8
\end{eqnarray*}

so the highest likelihood is indeed that of~$H_{[1,1,2]}$.  It is also
instructive to assume~$\alpha=\beta$ and consider the limit
as~$\alpha\longrightarrow 0$.  This would correspond to perpetrators
having either a very small~$p$, or~$p$ close to~1, intermediate values
being rare.  To first order in~$\alpha$:
\begin{eqnarray*}
{\mathcal L}(H_{[1,1,1]}) &=&
C\cdot\alpha/4\\
{\mathcal L}(H_{[1,2,1]}) &=&
C\cdot\alpha/4\\
{\mathcal L}(H_{[1,1,2]}) &=&
C\cdot(1-\alpha)/4\\
{\mathcal L}(H_{[2,1,1]}) &=&
C\cdot\alpha/4\\
{\mathcal L}(H_{[1,2,3]}) &=&
C\cdot 1/8
\end{eqnarray*}

so the maximum likelihood partition would again be~$H_{[1,1,2]}$; but
one would not be able to reject the hypothesis that there are three
distinct perpetrators (that is, $H_{[1,2,3]}$) at the two units of
support level.

As a final example, consider the case where~$\alpha=\beta$
and~$\alpha\longrightarrow\infty$.  This corresponds to each
perpetrator leaving positive evidence with probability~$\frac{1}{2}$.
The five likelihoods all tend to the same limit.

\subsubsection{Generalization to arbitrary factors}

These ideas have a natural generalization to arbitrary factors.  The
beta distribution becomes a Dirichlet distribution and the posterior
probability becomes

\[
\frac{\left(\sum a_l\right)!}{\prod a_l!}
\frac{\Gamma\left(\sum\alpha_l\right)}{\prod\Gamma\left(\alpha_l\right)}
\frac{\prod\Gamma\left(a_l+\alpha_l\right)}{\Gamma\left(\sum a_l+\sum\alpha_l\right)}
\]
where outcome~$l$ is observed with frequency~$a_l$ and the
corresponding prior distributions are Dirichlet with
parameters~$\alpha_i=\left(\alpha_{i1},\ldots,
  \alpha_{iu}\right)$.  The likelihood is then
\begin{equation}\label{like.mult}
{\mathcal L}(\wp)=
C\prod_{i=1}^{n}
\left[
\left(
\frac{\Gamma\left(\sum_l\alpha_{il}\right)}{\prod_l\Gamma\left(\alpha_{il}\right)}
\right)^r
\prod_{k=1}^{r}
%\cdot\frac{\left(\sum_{l}a_{ikl}\right)!}{\prod_la_{ikl}!}
\frac{\prod_l\Gamma\left(a_{ikl} + \alpha_{il}\right)}{\Gamma\left(
\sum_l\left(a_{ikl}+\alpha_{il}\right)\right)}
\right]
\end{equation}
where~$a_{ikl}$ is the frequency of outcome~$l$ for evidence~$i$ and
perpetrator~$k$, under~$H_\wp$.

\paragraph{Example}

<<set.evidence.matrix,echo=FALSE,print=FALSE>>=
a <- 
structure(
          c(2, 2, 4, 4, 3, 1, 2, 1, 1, 2, 2, 1, 4, 2, 1, 2, 1, 
            4, 2, 4, 2, 1, 4, 4, 1, 2, 1, 4, 2, 1, 1, 2, 1, 1, 2),
          .Dim = c(5L, 7L),
          .Dimnames = list(
            evidence=paste("E",1:5,sep=""),
            crime=paste("C",1:7,sep="")
            )
)
@ 

Consider a case where seven crimes have been committed; there are five
forensic characteristics.  The evidence matrix is as follows:

<<show.a,echo=FALSE,print=TRUE>>=
a
@ 

Each column corresponds to a crime and each row corresponds to a
forensic characteristic.  Rows~1 and~2 are Bernoulli: each is either~1
or~2.  Rows~3-5 are Dirichlet with four possible outcomes.  In this
case, it is assumed that the prior is uniform,
viz~$\alpha_1=\alpha_2=1$ for rows~1 and~2
and~$\alpha_1=\alpha_2=\alpha_3=\alpha_4=1$ for rows~3-5.

<<calc.likelihoods,cache=TRUE>>=
genbeta <- function(x){gamma(sum(x))/prod(gamma(x))}

lp <- function(evidence, alpha, partition){
  r <- length(unique(partition))
  evidence <- as.factor(evidence)
  levels(evidence) <- seq_along(alpha)
  out <- 1
  for(k in unique(partition)){  #k is a perp
    thisperp <- partition==k
    evidence.thisperp <- evidence[thisperp]
    no.of.crimes.thisperp <- sum(thisperp)
    out <- out/genbeta(alpha+table(evidence.thisperp))
  }
  return(out*genbeta(alpha)^r)
}

sp <- setparts(7)
l1 <- apply(sp,2, function(x){lp(evidence=a[1,],alpha=rep(1,2),partition=x)})
l2 <- apply(sp,2, function(x){lp(evidence=a[2,],alpha=rep(1,2),partition=x)})
l3 <- apply(sp,2, function(x){lp(evidence=a[3,],alpha=rep(1,4),partition=x)})
l4 <- apply(sp,2, function(x){lp(evidence=a[4,],alpha=rep(1,4),partition=x)})
l5 <- apply(sp,2, function(x){lp(evidence=a[5,],alpha=rep(1,4),partition=x)})

likelihood <- l1*l2*l3*l4*l5
likelihood <- likelihood/max(likelihood)
support <- log(likelihood)
@ 

\begin{figure}[htbp]
  \begin{center}
<<plotlikelihoods,fig=TRUE>>=
par(mfcol=c(1,2))
plot(likelihood,ylab="likelihood",xlab="hypothesis")
abline(h = exp(-2))
plot(support,ylab="support",xlab="hypothesis")
abline(h = -2)
@
\caption{The\label{support.function} likelihood (left) and support
  (right) for each of the~877 partitions on a set of seven elements.
  The horizontal line shows two units of support less than the
  maximum}
  \end{center}
\end{figure}

Figure~\ref{support.function} shows the likelihood and support for
each of the distinct set partitions on a set of six elements: variable
\code{support} is from Equation~\ref{like.mult}.  The maximum
likelihood estimator is given by

<<show.maximum.likelihood,echo=TRUE,print=FALSE>>=
sp <- setparts(7)
<<echo=TRUE,print=TRUE>>=
sp[,which.max(support)]
@ 

which could be written~$(3456)(27)(1)$.  This suggests that crimes~3-6
were committed by the same perpetrator, crimes~2 and~7 by another
perpetrator, and crime~1 by a third perpetrator.  Note that the
maximum likelihood hypothesis is unique.

The likelihood function may be used to identify those hypotheses that
cannot be rejected on the grounds of insufficient support.  There
are~9 partitions that have fewer than two units of support less than
the maximum.  These are:

<<likelihood.gt.2>>=
index <- order(support,decreasing=TRUE)
sp <- sp[,index]
support <- support[index]
dimnames(sp) <- list(
                     crime = paste("C",1:7, sep=""),
                     partition =  paste("H",1:ncol(sp),sep="")
                     )
<<echo=TRUE,print=TRUE>>=
 sp[, support > -2]
@ 

and the supports for these hypotheses are

<<show.likelihoods,echo=TRUE,print=TRUE>>=
support[support > -2]
@ 

respectively.

Interpretation of this result is not straightforward.  One may assert
that crimes~\code{C3} and~\code{C6} were committed by the same
perpetrator, and that certain pairs of crimes (specifically
\code{C2}-\code{C3}, \code{C2}-\code{C4}, \code{C2}-\code{C5},
\code{C2}-\code{C6}, \code{C3}-\code{C7}, \code{C4}-\code{C7},
\code{C5}-\code{C7}, and~\code{C6}-\code{C7}) were committed by
different perpetrators: for any hypothesis that does not satisfy these
criteria, one could gain at least two units of support by adopting
instead the best supported hypothesis~$H_{[3,2,1,1,1,1,2]}$,
corresponding to~$(3456)(27)(1)$.

\section{Note: a gotcha}

Note carefully that \code{setparts(c(2,1,1))} does \emph{not}
enumerate the ways of placing four numbered balls in three labelled
boxes of capacities 2,1,1.  This is because there are two boxes of
capacity 1, and swapping the balls between these boxes gives a
different box partition but the same set partition (because sets are
unordered).

<<echo=TRUE>>=
setparts(c(2,1,1))
@ 

Note the absence of a column reading \code{1 1 3 2}.  This would
correspond to placing balls 1 and 2 in box 1 (of size 2), ball 3 in
box 3 (of size 1) and ball 4 in box 2 (also of size 1).  The missing
column is because the two boxes of size 1 are indistinguishable, so
\code{1 1 3 2} is the same \emph{set} partition as \code{1 1 2 3}, as
in each case balls 3 and 4 are placed in ``a box of size 1''.

If you want to enumerate the ways of choosing two workers, a secretary
and a chair from four people (who are conveniently numbered 1,2,3,4),
use \code{multinomial()}:

<<echo=TRUE>>=
multinomial(c(worker=2,secretary=1,chair=1))
@ 

Above, see how the rows are named for their equivalence class.  The
first two columns differ only in the identity of the secretary and the
chair (and would be identical set partitions as given by
\code{setparts(c(2,1,1))}).  The reason I mention this is to avoid the
following gotcha (which looks plausible but is incorrect):


<<echo=TRUE>>=
v <- c(worker=2,secretary=1,chair=1)
a <- apply(setparts(v),2,order)
rownames(a) <- rep(names(v),v)
as.partition(a)
@ 

The above is an incomplete enumeration.  We see 1,4,2,3 [which would
  correspond to set partition 1,2,3,1] but not 1,4,3,2 [which would
  correspond to set partition 1,3,2,1].  These two differ in that the
secretary and chair have swapped roles.

\section{Conclusions}

Enumeration of set partitions is required in several branches of
computational combinatorics.

The software discussed in this code snippet enumerates set partitions
for finite sets, under a variety of restrictions on the sizes of the
induced equivalence classes and was used to solve three problems drawn
from diverse applications in computational statistics.

\subsection*{Acknowledgement}
We would like to acknowledge the many stimulating and insightful
comments made on the \proglang{R}-help list while preparing this
software.

\bibliography{partitions}
\end{document}


