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\author{Robin K. S. Hankin\\University of Cambridge}
\title{Urn Sampling Without Replacement: Enumerative Combinatorics In
  \proglang{R}}
%\VignetteIndexEntry{Urn sampling without replacement}

%% for pretty printing and a nice hypersummary also set:
\Plainauthor{Robin K. S. Hankin}
\Plaintitle{Urn sampling without replacement: enumerative combinatorics in
  R}
\Shorttitle{Urn sampling without replacement}

%% an abstract and keywords
\Abstract{This vignette is based on~\cite{hankin2007}.

  This short paper introduces a code snippet in the form of
  two new \proglang{R} functions that enumerate possible draws from an urn
  without replacement; these functions call  \proglang{C}
  code, written by the author.  Some simple combinatorial problems are solved
  using the software.

  For reasons of performance, this vignette uses pre-calculated
  answers.  To calculate everything from scratch, set variable {\tt
    calculate\_from\_scratch} in the first chunk to {\tt TRUE}.
}

\Keywords{Urn problems, drawing without replacement, enumerative
  combinatorics, Scrabble, \proglang{R}}
\Plainkeywords{Urn problems, drawing without replacement, enumerative
  combinatorics, Scrabble, 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\\
  University of Cambridge\\
  19 Silver Street\\
  Cambridge CB3 9EP\\
  United Kingdom\\
  E-mail: \email{hankin.robin@gmail.com}
}
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\begin{document}
<<set_cutdownproblem,echo=FALSE,print=FALSE>>=
<<results=hide>>=
do_from_scratch <- FALSE
@ 

\section{Introduction}

Drawing balls from an urn without replacement is a classical paradigm
in probability~\citep{feller1968}.  It is useful in practice, and many
elementary statistics textbooks use it to introduce the binomial and
hypergeometric distributions.

In this paper, I introduce software, written by the author, that
enumerates\footnote{Enumerate: ``to mention [a number of things] one
by one, as if for the purpose of counting''} all possible draws from
an urn containing specified numbers of balls of each of a finite
number of types.  Order is not important in the sense that drawing AAB
is equivalent to drawing ABA or BAA.  Formally, drawing~$n$ balls
without replacement from an urn containing $f_1,f_2,\ldots,f_S$ balls
of types $1,2,\ldots, S$ is equivalent to choosing a solution
$a_1,\ldots,a_S$ to the Diophantine equation
\begin{equation}
\label{basic_equation}
\sum_{i=1}^Sa_i=n,\qquad 0\leqslant a_i\leqslant f_i,
\end{equation}
with
probability~$\left.\left(\begin{array}{c}f_i\\a_i\end{array}\right)\right/
\left(\begin{array}{c}\sum f_i\\n\end{array}\right)$.

Combinatorial enumeration is often necessary in the context of integer
optimization: the appropriate configurations are enumerated and the
optimal one reported.  Sometimes explicit enumeration is needed to
count solutions satisfying some condition: simply enumerate candidate
solutions, then test them one by one.  

The software discussed here comprises two new \proglang{R}
\citep{rmanual} functions \code{S()} and \code{blockparts()}, which
are currently part of the \pkg{partitions} package~\citep{hankin2006},
version 1.3-3.  These functions call \proglang{C} code, also written
by the author, which is available as part of the package.

All software is available from CRAN, \url{http://CRAN.r-project.org/}.

\section{Examples}

The software associated with this snippet is now used to answer a
variety of combinatorial questions, written in textbook example style,
that require enumerative techniques to solve.

{\bf\em Question\/} A chess player is considering endgames in which
White has a king, no pawns, and exactly three other pieces.  What
combinations of white pieces are possible?  No promotions have
occurred.
 
{\bf\em Answer\/} This is an urn problem with a pool of~7 objects, in
this case non-king chess pieces.  An enumeration of the size-3 draws
is required, which is given by new function {\tt blockparts()}.  This
function enumerates the distinct solutions to
equation~\ref{basic_equation} in columns which appear in
lexicographical order:

<<load_library,echo=FALSE,print=FALSE>>=
library(partitions)
library(polynom)
options(width=85)
<<chess_problem,echo=TRUE,print=TRUE>>=
blockparts(c(Bishops=2,Knights=2,Rooks=2,Queens=1),3)
@ 
The first sample appears as the first column: this is the first
lexicographically, as all draws are from as low an index of \code{f}
as possible.  Subsequent draws are in lexicographical order.  Starting
with a draw \code{d}---a vector of \code{length(y)} elements---the
next draw is obtained by the following algorithm:

\begin{enumerate}
\item Starting at the beginning of the vector, find the first block
that can be moved one square to the right, and move it.  If no such
block exists, all draws have been enumerated: {\bf stop}.
\item Take all blocks to the left of the one that has moved, and place
them sequentially in the frame, starting from the left.
\item Go to item 1.
\end{enumerate}


Figure~\ref{blocks} shows an example of this in action.  The left
diagram shows a draw of a bishop and two knights, corresponding to the
the second column of the matrix returned by \code{blockparts()} above.
The first block that can be moved is one of the knights.  This moves
one place to the right and becomes a rook.  The remaining pieces (that
is, one bishop and one knight) are redistributed starting from the
left; they become two bishops.

There are thus~\Sexpr{S(c(1,2,2,2),3)} combinations: note that the
majority of them have no Queen.  The solutions are in lexicographical
order, which is useful in some contexts. 

\begin{figure}[htbp]
  \begin{center}
\includegraphics[width=10cm]{blocks}
\caption{A pictorial description of the algorithm used in \label{blocks}
  function \code{blockparts()}.  Three blocks (grey squares) are
  arranged in a tableau (B, N, R, Q, representing the chess pieces
  under consideration) in two consecutive configurations, the left
  one first.  The larger, line squares above each piece name show the
  maximum number of chess pieces allowed; thus the two knights in the
  left diagram completely fill the `N' column and this indicates that
  a maximum of two knights may be drawn.  The left diagram thus
  corresponds to one bishop and two knights: this is column two in the
  matrix returned by \code{blockparts()} in the \proglang{R} chunk
  above.  The right diagram shows the next lexicographical
  arrangement, corresponding to column three; the algorithm for the
  change is described in the text} \end{center} \end{figure}

{\bf\em Question\/} ``Scrabble'' is a popular word board game that
involves choosing, at random, a rack of~7 tiles from a pool of~100
with the following frequencies:

<<scrabble_tiles>>=
scrabble <-
c(a=9,b=2,c=2,d=4,e=12,f=2,g=3,h=2,i=9,j=1,k=1,l=4,m=2,n=6,o=8,p=2,q=1,r=6,s=4,t=6,u=4,v=2,w=2,x=1,y=2,z=1," "=2)
<<echo=TRUE>>=
scrabble
@ 

(note the last entry: two of the tiles are blank).
\begin{itemize}
\item[({\bf\em i\/})] how many distinct racks are possible?
\item[({\bf\em ii\/})] what proportion of racks have no blanks?
\item[({\bf\em iii\/})] what is the most probable rack and what is its frequency?
\end{itemize}

{\bf\em Answer}

({\bf\em i\/}).  The number of draws is given by function {\tt S()}.
This function returns the number of solutions to
equation~\ref{basic_equation} by determining the coefficient of $x^n$
in the generating function~$\prod_{i=1}^S\sum_{j=0}^{f_i} x^j$ using
the \pkg{polynom} \citep{venables2006} package:

<<call_S,echo=TRUE>>=
S(scrabble,7)
@ 

({\bf\em ii\/}).  The number of racks with no blank is given by function {\tt
  S()}, applied with a suitably shortened vector argument; the
proportion of racks with no blank is then:
<<call_S_noblanks,print=TRUE,echo=TRUE>>=
S(scrabble[-27],7)/S(scrabble,7)
@ 

Note that this question is distinct from that of determining the {\em
  probability} of drawing no blanks, which is given by elementary
combinatorial arguments as~{\small $\left. 
\left(\begin{array}{c}98 \\7\end{array}\right)\right/
\left(\begin{array}{c}100\\7\end{array}\right)
$}, or about~\Sexpr{100*round(choose(98,7)/choose(100,7),2)}\%.   This
value is larger because racks with one or more blanks have relatively
low probabilities of being drawn.

({\bf\em iii\/}).  To determine the most probable rack, we note that the
probability of a given draw is given by
\[
\frac{
\prod\left(\begin{array}{c}f_i\\a_i\end{array}\right)}
    {\left(\begin{array}{c}100\\7\end{array}\right)}.
\]

The appropriate \proglang{R} idiom would be to enumerate all possible
racks using \code{blockparts()}, and apply a function that calculates
the probability of each rack:

<<probOfRacks,print=FALSE,echo=FALSE,cache=TRUE>>=
f <- function(a){prod(choose(scrabble,a))/choose(sum(scrabble),7)}
if(do_from_scratch){
  racks <- blockparts(scrabble,7)
  probs <- apply(racks,2,f)
  otarine_ans <- rep(names(scrabble), racks[, which.max(probs)])
  max_probs <- max(probs)
  mp <- min(probs)
  a <- floor(log10(mp))
  how_many_racks <- sum(probs==min(probs))
} else {
  load("answers.Rdata")
}
@  

\begin{Schunk}
\begin{Sinput}
> f <- function(a) { prod(choose(scrabble, a))/choose(sum(scrabble), 7) }
> racks <- blockparts(scrabble, 7)
> probs <- apply(racks, 2, f)
\end{Sinput}
\end{Schunk}


The draw of maximal probability is given by the maximal element of
\code{probs}.  The corresponding rack is then:

\begin{Schunk}
\begin{Sinput}
> rep(names(scrabble), racks[, which.max(probs)])
\end{Sinput}
\end{Schunk}

<<print_otarine_letters,echo=FALSE,print=TRUE,cache=FALSE>>=
otarine_ans
@ 

In the context of the full Scrabble problem, there is only one
acceptable anagram of the most probable rack: ``otarine''.  Its
probability is

\begin{Schunk}
\begin{Sinput}
> max(probs)
\end{Sinput}
\end{Schunk} 

<<print_max_probs,echo=FALSE,print=TRUE,cache=FALSE>>=
max_probs
@ 

or just over once per~\Sexpr{ceiling(1/max_probs)} draws.  It is
interesting to note that the {\em least} probable rack is not unique:
there are exactly~\Sexpr{how_many_racks} racks each with
minimal probability (about~$\Sexpr{round(mp/10^a,3)}\times
10^{\Sexpr{a}}$).

\section{Conclusions}

The software discussed in this code snippet enumerates the possible
draws from an urn made without replacement; it is used to answer
several combinatorial questions that require enumeration for their
answer.  Further work might include enumeration of solutions of
arbitrary linear Diophantine equations.

\subsection*{Acknowledgement}
I would like to acknowledge the many stimulating and helpful comments
made by the R-help list while preparing this software.

I would also like to thank an anonymous referee who suggested that the
\code{polynom} package could be used to evaluate the generating
function appearing in \code{S()}.

\bibliography{partitions}
\end{document}
