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\begin{document}
    \titleone{Equal Relabelings}
    \titletwo{for $PQ$-sided dice}
    \doctype{Thesis}
    \doctypeUp{Thesis}
    \degree{Master of Science}
    \field{Mathematics}
    \Author{Alec Lewald}
    \Advisor{Amber Rosin}
    \MemberA{Berit Givens}
    \MemberB{John Rock}
    \Year{2017}
    \quarter{Winter}
    
    
%Type your abstract here.

\Abstract{\addcontentsline{toc}{chapter}{Abstract}
	For $m$ $n$-sided dice, we will call the sums $m, m+1, m+2, \ldots, mn$ the \emph{standard sums}, and we will say that $m$ $n$-sided dice have an \emph{equal relabeling} if the dice can be labeled with positive integers in such a way that the standard sums are equally likely to occur.  We will consider the case when $n=pq$ for distinct primes $p<q$.  The $m$ for which $n=pq$ sided dice have an equal relabeling have previously been characterized and it has been shown that all such equal relabelings require stupid dice (dice with 1's on every face). We find the minimum and maximum number of stupid dice that are needed for an equal relabeling of $pq$-sided dice and show that such a relabeling is possible for any number of stupid dice between the maximum and minimum. 
   }  
   
    % Type your acknowledgments here.
\Acknowledgments{\addcontentsline{toc}{chapter}{Acknowledgements}
        This thesis is dedicated to my family, friends, and professors who have supported me throughout my education.  A special thank you to my parents and Samier who always pushed me even when it was abundantly clear that I had no intention of working.
    }

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% Change the titles of Chapters and Sections to whatever you would like.  
\chapter{Background}

Consider rolling a pair of $6$-sided dice.  The possible sums are $2,3,4,5,6,7,8,9,10,11,$ and $12$ where the likelihood of rolling each of these sums is $\frac{1}{36},\frac{2}{36},\frac{3}{36},\frac{4}{36},\frac{5}{36},\frac{6}{36},\frac{5}{36},\frac{4}{36},\frac{3}{36},\frac{2}{36},$ and $\frac{1}{36}$ repectively.  Can we relabel the dice so that we still get the sums $2,3,\hdots,12$ but the sums have an equal chance of being rolled?  The answer is no due to the fact that there are $6\cdot 6=36$ possible die rolls and $11$ different sums.  Since $11$ does not divide $36$ it is impossible for each sum to have an equally likely chance of being rolled.

Now consider rolling seven $6$-sided dice.  The possible sums are $7,8,9,\hdots,41,42$  where again the sums are not all equally likely.  For example, to roll a $7$ we would need to roll a $1$ on all seven dice, which has a $\frac{1}{6^7}$ chance of occuring.  Whereas rolling an $8$ would require that six of the dice result in a $1$ but the last die results in a $2$, which has a $\frac{1}{6^6}$ chance of occuring.  Can we relabel the dice so the sums $7,8,\hdots ,42$ are equally likely?  Consider the following relabeling:

\begin{tabular}{lllllll}
Die 1:\quad\quad&1&1&2&2&3&3\\
Die 2:\quad\quad&1&1&4&4&7&7\\
Die 3:\quad\quad&1&1&1&10&10&10\\
Die 4:\quad\quad&1&1&1&19&19&19\\
Die 5:\quad\quad&1&1&1&1&1&1\\
Die 6:\quad\quad&1&1&1&1&1&1\\
Die 7:\quad\quad&1&1&1&1&1&1\\
\end{tabular}

\vspace{4mm}

If we roll the first two dice, the sums will be $2, 3, 4, 5, 6, 7, 8, 9,10$ with each combination having $4$ ways of occuring because there are two choices on each die of which face to use, so there are $2\cdot 2=4$ ways of getting each sum.  If we then combine these with die $3$ we get the sums $3,4,\hdots,11$ if we roll a $1$, and we get the sums $12,13,\hdots,20$ if we roll a $10$.  Each of these sums has $3\cdot 2\cdot 2=12$ ways of occuring.  Further combining this with die $4$ we get the sums $4,5,\hdots,21$ and $22,23,\hdots,39$ if we roll a $1$ or $19$ respectively.  Each of these sums has $3\cdot 3\cdot 2\cdot 2=36$ ways of occuring.  Finally, when combined with dice $5,6$, and $7$, each sum increases by exactly $3$.  Therefore, this set of dice yield the sums $7,8,9,\hdots,42$  where each sum has $6\cdot 6\cdot 6\cdot 3\cdot 3\cdot 2\cdot 2=7776$ ways of occuring.  Since there are $6^7$ possible outcomes and each has $6^5=7776$ ways of being rolled, there is a probability of $\frac{6^5}{6^7}=\frac{1}{6^2}=\frac{1}{36}$ for each sum to be rolled.  

This now generates the question of when such a relabeing is possible.  More generally we can look at $m$ dice that have $n$ sides each.  The question of can $m$ $n$-sided dice be relabelled so that we still get the sums $m,m+1,\hdots,mn$ with each sum being equally likely was answered in \cite{ber} when $n=pq$ where $p$ and $q$ are distinct prime numbers with $p<q$.  We have seen that for $n=6$, $m$ can equal $7$ but not $2$.  We will build on the results shown in \cite{ber}.

When rolling $m$ $n$-sided dice we call the sums $m,m+1,m+2,m+3,\dots,mn$ the \emph{standard sums}.  This is due to the fact that if all $m$ die were labeled with the usual labels of $1,2,\hdots,(n-1),n$ the possible sums would be $m, m+1,\dots, mn$.  Note that the number of standard sums is $mn-m+1=m(n-1)+1$.  We say that $m$, $n$-sided dice have an \emph{equal relabeling} if they can be relabeled so that the standard sums are equally likely.  For many equal relabelings, such as the one in the example above, we will need dice that have a label of $1$ on every face; we will refer to these as \emph{stupid dice}.  Before continuing the work done in \cite{ber}, we summarize the main results in \cite{ber}. The principal result from the paper is the following:

\begin{lem}

A set of $m$ $pq$-sided dice have an equal relabeling if and only if 

\begin{equation}\label{lem1.1}
    m=\frac{n^xp^{\gamma y}-1}{n-1}\tag{*}
\end{equation}
or
\begin{equation}\label{lem1.2}
    m=\frac{n^xq^{\gamma y}-1}{n-1}\tag{**}
\end{equation}
where $n=pq$ for distinct primes $p$ and $q$ with $p<q$,  $x$ and $y$ are nonnegative integers such that $(x,y)\neq (0,1)$, and $\gamma$ is the multiplicative order of $p$ modulo $n-1$.

\end{lem}

From Lemma 1, we see that the number of standard sums is \\$m(n-1)+1=n^xp^{\gamma y}=p^{x+\gamma y}q^x$ or $n^xq^{\gamma y}=p^xq^{x+\gamma y}$.  We will use this result later, so we will record this as a lemma.

\begin{lem}

For  $m$ $pq$-sided dice the number of standard sums is $p^{x+\gamma y}q^x$ when $m=\frac{n^xp^{\gamma y}-1}{n-1}$ and $p^xq^{x+\gamma y}$ when $m=\frac{n^xq^{\gamma y}-1}{n-1}$.

\end{lem}

Lastly, it is shown that with the exception of a few possibilities for $x$ and $y$, all equal relabelings require stupid dice.  In particular we have the following:

\begin{lem}

So long as $(x,y)\neq (0,0)$, $(0,1)$ or $(1,0)$, stupid dice are necessary for any equal relabeling of $m$ $n=pq$ sided dice.  

\end{lem}

This provides a great deal of information, but it does not address how many (or how few) stupid dice are required.  We will consider this question.  There is also an unjustified statement in [1], namely that an equal relabeling requires that the number $1$ appears on a given die $1,p,q,$ or $pq$ times, and we will look at this statement.


\newpage

\chapter{A Partial Proof of a Previously Unjustified Statement}

As mentioned before, the proof of Lemma $3$ given in \cite{ber} relies on the statement that:

\textit{if $m$ $pq$-sided dice have an equal relabeling, then the number of $1$'s on any given die must divide $pq$.}

However, no justification for this statement is given.  We attempt to justify it.

Since we need to have our $m$ dice sum to $m$ each die must have at least one face labled as $1$.  We are going to look at two specific dice: one being a regular die that we can choose from our given $m$ dice and the other having faces that equal the sums of the remaining dice.  Die 1 will have $pq$ sides.  We need to prove that each number that appears on die 1 appears $1,p,q,$ or $pq$ times.  For the following proofs we will use the notation that $a_i$ and $b_i$ are the number of faces with value $i$ on dice one and two respectively.  Furthermore, let $s$ and $t$ be the largest values to appear on dice one and two respectively.  Finally, we will denote $k$ being on die 1 or die 2  as $k\in D_1$ and $k\in D_2$ respectively.

\begin{lem}

If the smallest number to appear on die 1 is $1$, each other number that appears on die 1 appears less than or equal to the number of times that $1$ appears on die 1, and $s$ appears the same number of times as $1$ does on die 1.  

\end{lem}

\begin{proof}

Given our $m$ dice we are required to find combinations that add up to the standard sums of $m$, $m+1$, $m+2$, $\ldots$, $mn$.  Assume to the contrary that there exists some die, among the original $m$ dice, without a face of value $1$.  Then consider the sum of the smallest faces on each of the die.  Since $m-1$ of the dice have at least $1$ as their smallest value and one of the dice has at least $2$ as its smallest value, we have that the smallest sum we could get is 
$$1*(m-1)+2*1=m-1+2=m+1.$$
This means that the smallest sum that we can get is $m+1$ which is a contradiction due to the fact that we are required to get a summation of $m$.  Therefore each of our dice must have at least one face of value $1$.  Note that since each die has a $1$, then the smallest value that appears on die 2 is $m-1$ since we have $m-1$ dice remaining after selecting a die for die 1.

To generate equally likely standard sums with our dice we must have that each sum appears the same number of times.  By way of contradiction let $k\in D_1$ with $a_k>a_1$.  Then we know that $k+m-1$ appears at least $a_k* b_{m-1}$ times since $m-1$ appears $b_{m-1}$ times on die $2$.  Next consider that the only values we can use that sum to $m$ are $1\in D_1$ and ${m-1}\in D_2$ since these are the minimum values on each die.  Therefore the number of times that the sum $m$ appears is exactly $a_1*b_{m-1}$.  Since $a_1*b_{m-1}<a_k*b_{m-1}$ the sum $k+m-1$ is more likely to occur than the sum $m$, which is a contradiction.  Therefore we must have that for all $k\in D_1$, $a_k\leq a_1$.  Note that the same reasoning applied to die 2 implies that $b_j\leq b_{m-1}$ for all $j\in D_2$.

Finally, we will consider the sum $s+t$.  If we use a value from die 1 that is less than $s$, say $s-j$, then the value we would need from die 2 is $s+t-(s-j)=t+j$.  Since $t$ is the max value on die 2 it must be that $j\leq 0$ making our value for die 1 at least $s$.  Similarly, if we used $t-i$ for our value from die 2 we would need to use $s+i$ from die 1 making $i\leq 0$.  Since $s$ is our largest value on die 1 we have that $i\leq 0$ making our value for die 2 at least $t$.  Since our value for die 1 is at least $s$ and our value for die 2 is at least $t$ we have that they must be exactly $s$ and $t$.  Therefore the only values that sum to $s+t$ are $s$ on die 1 and $t$ on die 2.  Since all sums appear an equal number of times, we have that $a_1*b_1=a_s*b_t$.  Using our above facts we have that $a_s\leq a_1$ and $b_t\leq b_1$.  If $a_s<a_1$ or $b_t<b_1$ then we would have that $a_s*b_t<a_1*b_1$ which would contradict our previous statement.  Therefore, $a_s=a_1$ and $b_t=b_1$.

It is important to note that since die 1 was arbitrary we have that all of these results hold no matter which die is choosen to be die 1.

\end{proof}

For the rest of the paper we will need terminology to describe the cases $a_i=a_1$ or $b_j=b_{m-1}$.  If this happens, then we refer to $i$ and $j$ as being $full$.  Likewise, if $a_i=0$ or $b_j=0$ then the value $i$ or $j$ is $empty$.  Furthermore, if a sum, say $d$, has the same number of ways of being obtained as $m$ then we will say $d$ is $maxed$.  These will be important due to the fact that if a value is full then the die which it is on can no longer have any more faces with said value.  Similarly, if a combination is maxed then if there are any other two values that make that combination which have not yet been acounted for, one must be empty.  Finally we will refer to values of our dice as $mirroring$ if $a_i=a_{s-(i-1)}$ and $b_j=b_{t-(j-(m-1))}$.  

\begin{lem}

Die 1 is mirrored from $1$ to $1+k_1$ where $1+k_1$ is the smallest non-empty number on die 1 other than $1$; furthermore, $1+k_1\in D_1$ is full.  We will also show that die 2 is mirrored from $m-1$ to $m-1+k_1$ with $m-1$ through $m-2+k_1$ being full while $m-1+k_1$ is empty.

\end{lem}
\begin{proof}

Let $1+k_1$ be the smallest non-empty number on die 1 other than $1$.  Then we know that to get the sums of $m,m+1,\hdots,m-1+k_1$ we must use the value $1\in D_1$ since $(1+k_1)+(m-1)$ is larger than any of these values.  Then for any value we just listed, say $x$,  we know that $x-1\in D_2$ and $x-1$ must be full since there are no other combinations that can result in $x$ other than $1+x-1$.  Therefore the numbers $m-1,m,m+1,\hdots,m-2+k_1$ on die 2 are full.  Since $s+m-1$ is a maxed combination, due to $s$ and $m-1$ being full, we have that any number smaller than $s$ on die 1 that sums to $s+m-1$ which uses the numbers $m,m+1,m+2,\hdots,m-2+k_1$ as summands from die 2 is empty.  This list of empty numbers on die 1 is $s-1,s-2,\hdots,s-k_1+1$.  

Similarly to the above $s-k_1$ is the largest non-empty number on die $1$ other than $s$.  Therefore to get the sums of $s+t-k_1+1,s+t-k_1+2,\hdots,s+t-1$ we must use $s\in D_1$ since $s-k_1+t=s+t-k_1$ is too small.  Then for any value we just listed, say $y$, we know that $y-s\in D_2$ and $y-s$ must be full since there are no other combinations that can result in $y$ other than $s+y-s$.  Therefore $t-k_1+1, t-k_1+2,\hdots, t-1$ on die $2$ are all full.  Since $1+t$ is a maxed combination, due to $1$ and $t$ being full, we have that any number larger than $1$ on die $1$ that sums to $t+1$ which uses the numbers $t-k_1+2, t-k_1+3,\hdots, t-1$ as the summands from die $2$ is empty.  This list of empty numbers on die $1$ is $2,3,\dots,k_1-1,k_1$.

Since $t-k_1+1, t-k_1,\hdots, t-2, t-1,t$ are full, we have the sums of $t-k_1+2,t-k_1+3,\hdots,t-1,t,t+1$ being maxed due to the fact that we can add these values to $1\in D_1$.  Since we know that $1+k_1\in D_1$ is non-empty but $t+1$ is maxed, we know that $t-k_1\in D_2$ is empty or else the sum $t+1$ would occur too many times.

Consider the sum of $s+t-k_1$.  The labels $t-k_1+1,t-k_1+2,\hdots,t-1,t$ are too large to sum with $s$ to attain this, but any value on die $1$ that is less than $s-k_1$ would be too small to combine with $t$.  Therefore we must use some value on die 1 that is greater than or equal to $s-k_1$ but smaller than $s$.  Since all of these values except for $s-k_1$ are empty we must only use $s-k_1$ for this sum.  Thus, $s-k_1$ is full or else $s+t-k_1$ would not be maxed.  Also notice that if we sum $s-k_1\in D_1$ with $m-1+k_1\in D_2$ we get $s+m-1$ which we have already established is maxed, leaving us with the conclusion that $m-1+k_1\in D_2$ is empty.

This brings us to our final point.  Consider the sum $m+k_1$.  This can only be achieved through using $1$ or $1+k_1\in D_1$ since any other number would be too big or is empty.  These reqiure the summands of $m-1+k_1$ or $m-1\in D_2$ respectively.  Since $m-1+k_1\in D_2$ is empty we must have that $1+k_1\in D_1$ is full to make $m+k_1$ maxed.  

In conclusion, die 1 contains at least $1, 1+k_1, s-k_1$, and $s$ which are all full while having $ 2,3,\hdots,k_1$ and $s-k_1+1,s-k_1+2,\hdots, s-1$ be empty.  Similarly we have that die 2 contains at least $m-1,m,m+1,\hdots,m-2+k_1$ and $t-k_1+1,t-k_1+2,\hdots t-1,t$ full while having $m-1+k_1$ and $t-k_1$ be empty.  Using our definition, we have that die 1 mirrors from at least $1$ to $1+k_1$ while die 2 mirrors from at least $m-1$ to $m-1+k_1$.

\end{proof}

\begin{lem}
If there exists a non-empty element of a die that is also not full, then there is at least one die that doesn't mirror.  
\end{lem}

\begin{proof}

Let $e$ and $g$ be the smallest elements on $D_1$ and $D_2$ respecitvely such that they are neither empty nor full.  By way of contradiction assume that both dice mirror.  

We first show that $e-1=g-(m-1)$.  Assume by way of contradiction that $e-1<g-(m-1)$ making $e+(m-1)<g+1$.  Since $e$ is not full we have that $e+(m-1)$ is not maxed from just $e\in D_1$ and $m-1\in D_2$.  Since $m-1$ is the smallest number on $D_2$ we have that there exists some $x\in D_1$ such that $1\leq x<e$ and a $y\in D_2$ such that $m-1< y\leq t$ where $x+y=e+(m-1)$.  Due to the fact that $e+(m-1)<g+1$ we have that $m-1<y<g$.  Since $x<e$ and $y<g$ we have that both are either full or empty.  If one of them is empty then their sum does not contribute to maxing $e+(m-1)$ but if both are full then $e+(m-1)$ would occur too many times.  Therefore $e-1\geq g-(m-1)$.  We can similarly show that $e-1\leq g-(m-1)$ making $e-1=g-(m-1)$.  

Now let $f$ and $h$ be such that $f=s-(e-1)$ and $h=t-(g-(m-1))$.  Since the dice mirror $a_e=a_f$ and $b_g=b_h$.  By the above equations we get $g-(m-1)=e-1=s-f$ which makes $g+f-(m-1)=s$.  Now if we consider the sum of $s\in D_1$ and $(m-1)\in D_2$, we can see that both of the elements are full making $s+(m-1)$ maxed.  Using our above equation we get that 
$$s+(m-1)=g+f-(m-1)+(m-1)=g+f,$$
but since that sum is already maxed we must have that either $g$ or $f$ is empty.  By assumption $g$ and $e$ are nonempty and $a_f=a_e$, therefore $f$ must be nonempty, which gives us a contradiction.  Therefore, if there does exist a smallest such $e\in D_1$ such that $e$ is neither empty nor full, then $D_1$ can't mirror.
\end{proof}

\begin{conj}

Dice 1 and 2 are mirrored with each element being either full or empty. (not finished proving)

\end{conj}

\begin{proof}

(\textit{partial}) We will induct on the number of gaps on $D_1$.  Let lemma $5$ be our base case.  For our inductive case let the elements $1,1+k_1,1+k_1+k_2,\hdots,1+k_1+\hdots+k_n\in D_1$ be full where each $k_i$ represents a gap between each face labling value, $i.e.$ all values between each $1+k_1+k_2+\hdots+k_i$ and $1+k_1+k_2+\hdots+k_i+k_{i+1}$ are empty.  Furthermore let $1+k_1+\hdots+k_n+k_{n+1}\in D_1$ be non-empty and all other values on $D_1$ between $1+k_1+\hdots+k_n$ and $1+k_1+\hdots+k_n+k_{n+1}$ be empty.  Next, let all elements, $m-1\leq x\leq m-1+k_n$ on $D_2$, be full or empty.  Finally, we let all elements of $D_1$ between $1$ and $1+k_1+\hdots,+k_n$ and all elements of $D_2$ between $m-1$ and $m-1+k_n$ mirror.

We then let $m-1+k_n<v\leq m-1+k_{n+1}$ with $v\in D_2$ such that $j$ is the smallest such element where it is neither full nor empty.  Since it is neither full nor empty the sum of $1+v$ is not maxed from just adding $1\in D_1$ and $v\in D_2$.  Therefore there exists some non-empty $x\in D_2$ such that $1+k_1+\hdots +k_j+x=1+v$ for some $1\leq j\leq n$.  The reason that we have this restriction on $j$ is due to the fact that if $j=n+1$ then we would be using $1+k_1+\hdots,+k_n+k_{n+1}\in D_1$ but the smallest sum we could attain from this would be $(1+k_1+\hdots,+k_n+k_{n+1})+(m-1)=m+k_1+\hdots,+k_n+k_{n+1}$ but the max that $1+v$ could be is $m+k_{n+1}$.  

If we use our formula of $1+k_1+\hdots,+k_j+x=1+v$ we get $x=v-k_1-\hdots -k_j<v$.  Since $x<v$ we have that $x$ is either full or empty from or hypothesis.  If $x$ is full then $v$ is empty, otherwise $1+v$ would be over our max.  However, if $x$ is empty then contradict our requirement that $x$ is non-empty.  Since we assumed that $v$ was the smallest non-empty or full element between $m-1+k_n$ and $m-1+k_{n+1}$, we have that all elements on $D_2$ between $m-1$ and $m-1+k_{n+1}$ are either full or empty.

\end{proof}

So far we have shown that $1$ appears on all $m$ dice and appears the most often, each die mirrors from $1$ to at least $1+k_1$ where $1+k_1$ is the smallest non-empty number on the die, and if there exists a non-empty element of a die that is also not full, then the die does not mirror.  The end goal of these Lemmas was to show that each non-empty $a_i$ was equal.  To complete the proof, we would need to show that the dice mirror completely.  We have not yet been able to show this, so for now we state the following conjecture:

\begin{conj}

If $m$ $pq$-sided dice have an equal relabeling, then the number of $1$'s on any given die must divide $pq$.

\end{conj}

We will assume Conjecture $2$ is true in the next chapter.  


\chapter{The Greatest and Least Number of Stupid Dice}

\section{Stupid Dice Facts}

By Lemma $3$ we know that equal relabelings require stupid dice, but how many stupid dice are needed?
Suppose $m$ $pq$-sided dice have an equal relabeling.  Then the number of standard sum is equal to either $p^{x+\gamma y}q^x$ or $p^x q^{x+\gamma y}$ by Lemma $2$.  The number of rolls that sum to each standard sums is calculated by dividing the total number of outcomes by the number of standard sums.  In either case the total number of sums is $n^m=p^mq^m$.  In the first case the number of times each sum appears is $\frac{p^mq^m}{p^{x+\gamma y}q^x}=p^{m-x-\gamma y}q^{m-x}$. In the second case, the number of times each sum appears is $\frac{p^mq^m}{p^xq^{x+\gamma y}}=p^{m-x} q^{m-x-\gamma y}$.  We will consider the case where we have $p^{x+\gamma y}q^{x}$ standard sums, the other case is very similar.   The least standard sum is $m$, and if we only label faces with positive integers, then in order to achieve a sum of $m$ using $m$ dice, every die must have at least one face labeled as $1$.  Let $c_i$ be the number of times $1$ appears on die $i$.  Then each $c_i$ must equal $1,p,q,$ or $pq$ from Conjecture $2$.  Since the sum $m$ can only be achieved by using a $1$ from each die, the number of times the sum $m$ appears is
$$c_1c_2\hdots c_m=p^{m-x-\gamma y}q^{m-x}.$$
For a die to be a stupid die $c_i=n=pq$ .  

To find the maximum and minimum numbers of stupid dice that are possible in an equal relabeling we need to find the least and greatest possible amount of $c_i$'s that equal $pq$.  In all cases to get $c_1,c_2,\hdots ,c_m=p^{m-x-\gamma y}q^{m-x}$, $m-x-\gamma y$ of the $c_i$'s must have a factor of $p$, and $m-x$ $c_i$'s must have a factor $q$.  To find the largest amount of stupid dice, we want as much overlap as possible, so of the $m-x$ $c_i$'s that have a factor of $q$, assume that $m-x-\gamma y$ of them also have $p$ as a factor.  Therefore we have have $m-x-\gamma y$ dice where $c_i=pq$, $m-x-(m-x-\gamma y)=\gamma y$ dice where $c_i=q$, and $m-(m-x-\gamma y)-\gamma y=x$ dice with $c_i=1$, making the making the maximum amount of stupid dice $m-x-\gamma y$.

To find our least number of stupid dice we want as little overlap as possible.  Since $m-x-\gamma y$ $c_i$'s have $p$ as a factor, there are $m-(m-x-\gamma y)=x+\gamma y$ $c_i$'s that don't have $p$ as a factor.  If these $x+\gamma y$ $c_i$'s  have $q$'s as factors then there are still $(m-x)-(x+\gamma y)=m-2x-\gamma y$ factors of $q$ remaining.  Since each $c_i$ can only have at most $1$ factor of $q$, the remaining factors of $q$ must be factors of $c_i$'s which also have $p$ as a factor.  Therefore, there are at least $m-2x-\gamma y$ dice whose $c_i=pq$, making the minimum amount of stupid dice $m-2x-\gamma y$.

Thus, the minimum number of stupid dice is $m-2x-\gamma y$ and the maximum number of stupid dice is $m-x-\gamma y$.

\newpage

\section{An Example}

Before tryinig to find a general relabeling, we consider a specific example.  Let $p=3$, $q=5$, $x=4$, and $y=2$. To find $\gamma$, consider

\begin{tabular}{l}
\quad\quad\quad\quad$3^1\equiv 3 \pmod{14}$\\
\quad\quad\quad\quad$3^2\equiv -5 \pmod{14}$\\
\quad\quad\quad\quad$3^3\equiv -1 \pmod{14}$\\
\quad\quad\quad\quad$3^4\equiv -3 \pmod{14}$\\
\quad\quad\quad\quad$3^5\equiv 5 \pmod{14}$\\
\quad\quad\quad\quad$3^6\equiv 1 \pmod{14}$,\\
\end{tabular}
\vspace{.1in}

thus $\gamma =6$. This makes $n=15$ and
$$m=\frac{15^4 3^{6\cdot 2}-1}{14}=1,921,728,616.$$  
The largest amount of stupid dice we could use is $m-x-\gamma y=1,921,728,600$ and the least amount is $m-2x-\gamma y=1,921,728,596$.  We will show a labeling for $1,921,728,599$ stupid dice.  Consider the following labeling:


The first 14 dice are labeled as 

\begin{tabular}{lll}
Die 1:&\quad\quad1,2,3,4,5&\quad\quad Each value appears 3 times\\
Die 2:&\quad\quad1,6,11&\quad\quad Each value appears 5 times\\
Die 3:&\quad\quad1,16, 31&\quad\quad Each value appears 5 times\\
Die 4:&\quad\quad1, 46, 91&\quad\quad Each value appears 5 times\\
Die 5:&\quad\quad1, 136, 271&\quad\quad Each value appears 5 times\\
Die 6:&\quad\quad1, 406, 811&\quad\quad Each value appears 5 times\\
Die 7:&\quad\quad1, 1216, 2431&\quad\quad Each value appears 5 times\\
Die 8:&\quad\quad1, 3646, 7291&\quad\quad Each value appears 5 times\\
Die 9:&\quad\quad1, 10936, 21871&\quad\quad Each value appears 5 times\\
Die 10:&\quad\quad1, 32806, 65611&\quad\quad Each value appears 5 times\\
Die 11:&\quad\quad1, 98416, 196831&\quad\quad Each value appears 5 times\\
Die 12:&\quad\quad1, 295246, 590491&\quad\quad Each value appears 5 times\\
Die 13:&\quad\quad1, 885736, 1771471&\quad\quad Each value appears 5 times\\
Die 14:&\quad\quad1, 2657206, 5314411&\quad\quad Each value appears 5 times\\
\end{tabular}

\vspace{4mm}

Combining dice $1$ and $2$ we can see that the labelings from die $1$ fill in the gaps of die $2$ creating the sums $2,3,\dots,16$ with each sum appearing 15 times.  Consider the fact that the gaps between the faces on die $3$ are each of size $15$ and dice $1$ and $2$ combine for $15$ consecutive sums.  This means that when we combine die $3$ with our previous combination the gaps will get filled in and the new sums will be $3,4,\dots,47$.  We can continue this process for all $14$ dice and get that the sums $14,15,16, \hdots, 7971628$ occur $3\cdot 5^{13}=3662109375$ times.  

The values on the next 3 dice appear exactly once each.  The 15th die is labeled as

\vspace{-10mm}

$$1, 7971616, 15943231, 23914846, 31886461, 39858076, 47829691, 55801306, 63772921, 71744536,$$
$$79716151, 87687766, 95659381, 103630996, 111602611,$$

the 16th die is labeled as
$$1, 119574226, 239148451, 358722676, 478296901, 597871126, 717445351, 837019576, 956593801,$$
$$1076168026, 1195742251, 1315316476, 1434890701, 1554464926, 1674039151,$$

and the 17th die is labeled as
$$1, 1793613376, 3587226751, 5380840126, 7174453501, 8968066876, 10761680251, 12555293626,$$
$$14348907001, 16142520376, 17936133751, 19729747126, 21523360501, 23316973876, 25110587251.$$

Similar to before, the gaps between the faces on die $15$ are of size $7971616$.  When we combine this with our $7071615$ consecutive sums from the first $14$ dice we get the sums of $15, 16, 17,\hdots, 119574239$ where each sum still appears $3662109375$ times.  

Repeating this with dice $16$ and $17$ makes the sums $17, 18, 19,\hdots, 26904200641$ where each sum appears $3662109375$ times.  The remaining $1,921,728,599$ dice are stupid dice.  When combined with our other $17$ dice we get the sums\\
$1921728616, 1921728617, 1921728618,\hdots, 28825929240$ where each sum appears\\
 $3\cdot 5^{14} \cdot 15^{1921728599}$ times.  


%The first 14 dice are labeled as
%$$1,2,3,4,5\textnormal{(each value appears on 3 faces)}$$
%\vspace{-14mm}
%$$1,6,11\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1,1+15, 1+30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3\cdot 15, 1+3\cdot 30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3^2\cdot 15, 1+3^2\cdot 30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3^3\cdot 15, 1+3^3\cdot 30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3^4\cdot 15, 1+3^4\cdot 30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3^5\cdot 15, 1+3^5\cdot 30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3^6\cdot 15, 1+3^6\cdot 30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3^7\cdot 15, 1+3^7\cdot 30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3^8\cdot 15, 1+3^8\cdot 30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3^9\cdot 15, 1+3^9\cdot 30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3^{10}\cdot 15, 1+3^{10}\cdot 30\textnormal{(each value appears on 5 faces)}$$
%\vspace{-13mm}
%$$1, 1+3^{11}\cdot 15, 1+3^{11}\cdot 30\textnormal{(each value appears on 5 faces)}$$
%The 15th die is labeled as
%$$1,1+3^{12}\cdot 15,1+2\cdot 3^{12}\cdot 15,1+3\cdot 3^{12}\cdot 15,1+4\cdot 3^{12}\cdot 15,1+5\cdot 3^{12}\cdot 15,1+6\cdot 3^{12}\cdot 15,1+7\cdot 3^{12}\cdot 15,$$
%\vspace{-13mm}
%$$1+8\cdot 3^{12}\cdot 15,1+9\cdot 3^{12}\cdot 15,1+10\cdot 3^{12}\cdot 15,1+11\cdot 3^{12}\cdot 15,1+12\cdot 3^{12}\cdot 15,1+13\cdot 3^{12}\cdot 15,1+14\cdot 3^{12}\cdot 15,$$
%the 16th die is labeled as
%$$1,1+3^{12}\cdot 15^2,1+2\cdot 3^{12}\cdot 15^2,1+3\cdot 3^{12}\cdot 15^2,1+4\cdot 3^{12}\cdot 15^2,1+5\cdot 3^{12}\cdot 15^2,1+6\cdot 3^{12}\cdot 15^2,1+7\cdot 3^{12}\cdot 15^2,$$
%\vspace{-13mm}
%$$1+8\cdot 3^{12}\cdot 15^2,1+9\cdot 3^{12}\cdot 15^2,1+10\cdot 3^{12}\cdot 15^2,1+11\cdot 3^{12}\cdot 15^2,1+12\cdot 3^{12}\cdot 15^2,1+13\cdot 3^{12}\cdot 15^2,1+14\cdot 3^{12}\cdot 15^2,$$
%and the 17th die is labaled as
%$$1,1+3^{12}\cdot 15^3,1+2\cdot 3^{12}\cdot 15^3,1+3\cdot 3^{12}\cdot 15^3,1+4\cdot 3^{12}\cdot 15^3,1+5\cdot 3^{12}\cdot 15^3,1+6\cdot 3^{12}\cdot 15^3,1+7\cdot 3^{12}\cdot 15^3,$$
%\vspace{-13mm}
%$$1+8\cdot 3^{12}\cdot 15^3,1+9\cdot 3^{12}\cdot 15^3,1+10\cdot 3^{12}\cdot 15^3,1+11\cdot 3^{12}\cdot 15^3,1+12\cdot 3^{12}\cdot 15^3,1+13\cdot 3^{12}\cdot 15^3,1+14\cdot 3^{12}\cdot 15^3.$$



\section{An Equal Labeling}

We now show that for any $m-2x-\gamma y\leq \lambda\leq m-x-\gamma y$, an equal albeling with $\lambda$ stupid dice is possible.  Let $\lambda=m-2x-\gamma y +t$ where $0\leq t\leq x$.  Label the first $x-t$ dice as follows:

$1,1+q^i,1+2q^i,\hdots ,1+(q-1)q^i$ for $0\leq i\leq x-t-1$

where each label is repeated $p$ times.  To find the sums this group of dice produces, first subtract $1$ from each of the labelings.  The new labeling would be 

$0,q^i,2q^i,\hdots ,(q-1)q^i$ for $0\leq i\leq x-t-1.$\\
These are exactly the numbers we need to write the numbers $0, 1, \hdots, q^{x-t}-1$ in base $q$.  Since each number can be written uniquely in base $q$, each of the sums

$0, 1, 2, \hdots, q^{x-t}-1$\\
occur the same number of times.  Note that this is a set of $q^{x-t}$ consecutive numbers.  Adding the $1$'s back to the labeling adds $x-t$ to each sum yeilding the equally likely sums

$x-t, x-t+1, \hdots, x-t+q^{x-t}-1.$\\
Since there are $p$ choices of which face to use on each die, each of these sums occurs $p^{x-t}$ times.  Label the second $x-t$ dice as follows:

$1,1+p^j q^{x-t},1+2p^j q^{x-t},\hdots ,1+(p-1)p^j q^{x-t}$ for $0\leq j\leq x-t-1$

where each label is repeated $q$ times.  To find the sums this group of dice produces, first subtract $1$ from each of the labels and divide each label by $q^{x-t}$.  This new labeling would be

$0,p^i,2p^i,\hdots ,(p-1)p^i$ for $0\leq i\leq x-t-1.$\\
These are exactly the numbers we need to write $0, 1, \hdots, p^{x-t}-1$ in base $p$.  Since each number can be written uniquely in base $p$, then each of their sums,

$0,1,2,\hdots ,p^{x-t}-1$,

occur the same number of times.  Note that this is a set of $p^{x-t}$ consecutive numbers.  After multiplying each label by $q^{x-t}$, we get the sums $0, q^{x-t}, \hdots, q^{x-t}(p^{x-t}-1).$  Then adding $1$'s to each label, we get the sums

$x-t, x-t+q^{x-t},\hdots, x-t+n^{x-t}-q^{x-t}$\\
which is $p^{x-t}$ numbers each separated by a gap of $q^{x-t}$.  Thus, when we combine the the $q^{x-t}$ consecutive sums from the first $x-t$ dice with the sums from the second $x-t$ dice, the gaps from our second sums will be filled in giving us the $n^{x-t}$ consecutive sums

$2(x-t),2(x-t)+1,\hdots, 2(x-t)+n^{x-t}-1.$\\
Since there are $q$ choices on each die of which face to use, these sums occur $p^{x-t}q^{x-t}=n^{x-t}$ times.  Label the next $\gamma y$ dice as follows:

$1,1+p^k n^{x-t},1+2p^k n^{x-t},\hdots ,1+(p-1)p^k n^{x-t}$ for $0\leq k\leq \gamma y-1$

where each label is repeated $q$ times.  To find the sums this group of dice produces, first subtract $1$ from each of the labels and divide each label by $n^{x-t}$.  This new labeling would be

$0, p^i, 2p^i,\hdots, (p-1)p^i$ for $0\leq i\leq \gamma y-1.$\\
These are exactly the numbers we need to write the numbers $0, 1,\hdots, p^{\gamma y}-1$ in base $p$.  Since each number can be written uniquely in base $p$, then each of their sums,

$0,1,2,\hdots, p^{\gamma y}-1,$ \\
are attained the same number of times.  Note that this is a set of $p^{\gamma y}$ consecutive numbers.  After multiplying each label by $n^{x-t}$, we get the sums \\$0, n^{x-t},2n^{x-t},\hdots, n^{x-t}(p^{\gamma y}-1)$.  Then adding $1$'s to each label, we get the sums

$\gamma y, \gamma y+n^{x-t},\hdots, \gamma y+p^{\gamma y} n^{x-t}-n^{x-t},$\\
which is $p^{\gamma y}$ numbers each separated by a gap of $n^{x-t}$.  Thus, when we combine the $n^{x-t}$ consecutive sums from the first $2(x-t)$ dice with the sums from the next $\gamma y$ dice, the gaps will be filled in giving us the $p^{\gamma y}n^{x-t}$ consecutive sums

$2(x-t)+\gamma y, 2(x-t)+\gamma y+1, \hdots, 2(x-t)+\gamma y+p^{\gamma y}n^{x-t}-1.$

Since there are $q$ choices of which face to use on each die, the sums from this group occur $q^{\gamma y}$ times so the above sums occur $n^{x-t}q^{\gamma y}$ times.  Label the last $t$ dice as follows:

$1,1+n^\delta p^{\gamma y} n^{x-t},1+2n^\delta p^{\gamma y} n^{x-t},\hdots ,1+(n-1)n^\delta p^{\gamma y} n^{x-t}$ for $0\leq \delta \leq t-1$

where each label appears on exactly one face.  To find the sums this group of dice produces, first subtract $1$ from each of the labelings and divide each labeling by $p^{\gamma y}n^{x-t}$.  The new labelings would be

$0,n^\delta,2n^\delta,\hdots ,(n-1)n^\delta$ for $0\leq \delta \leq t-1.$\\
These are exactly the numbers we need to write the numbers $0,1,2,\hdots,n^t-1$ in base $n$.  Since each number can be written uniquely in base $n$, then each of the sums

$0,1,2,\hdots, n^t-1$\\
occur the exactly once.  Note that this is a set of $n^t$ consecutive numbers.  After multiplying each label by $p^{\gamma y}n^{x-t}$ we get the sums \\ $0,p^{\gamma y}n^{x-t},2p^{\gamma y}n^{x-t},\hdots,p^{\gamma y}n^{x-t}(n^t-1).$  Then adding the $1$'s to each label, we get the sums

$t, t+p^{\gamma y} n^{x-t},\hdots, t+p^{\gamma y} n^x-p^{\gamma y} n^{x-t},$\\
which is $n^t$ numbers each separated by gaps of $p^{\gamma y}n^{x-t}$.  Thus, when we combine the $p^{\gamma y}n^{x-t}$ consecutive sums from the first $2(x-t)+\gamma y$ dice with the sums from the next $t$ dice, the gaps will be filled in giving us the $p^{\gamma y}n^x$ consecutive sums

$2x-t+\gamma y,2x-t+\gamma y+1,\hdots,2x-t+\gamma y+p^{\gamma y}n^x-1.$

Each of these sums appears $n^{x-t}q^{\gamma y}$ times.  Finally, label the remaining\\ $m-2x-\gamma y+t$ dice with $1$'s on every face.  This yields the sum $m-2x-\gamma y+t$.  Since there are $n$ choice of which face to use on each of the stupid dice, the sums

$m, m+1, \hdots, m+p^{\gamma y}n^x-1$\\
all appear $n^{x-t}q^{\gamma y}n^{m-2x-\gamma y+t}=n^{m-x-\gamma y}q^{\gamma y}=p^{m-x-\gamma y}q^{m-x}$ times.

We need to show that this labeling results in the standard sums.  It is clear that the lowest sum we can get is $m$ and that each successive sum is increased by $1$.  This means all we need to show is that the largest sum, $m+p^{\gamma y}n^x-1$, is in fact equal to $mn$.  Since we know that $mn-m+1=n^xp^{\gamma y}$ we have that $mn=m+n^xp^{\gamma y}-1$ making the sums created by our dice the standard sums.

We will now show that the standard sums are equally likely by showing that there is only one way to achieve each sum.  Let $\delta$ be one of the standard sums minus $m-2x-\gamma y+t$ (i.e. without the stupid dices' contribution).  From our above sums we know that the first $x-t$ dice always result in a sum of the form $x-t+r$ where $0\leq r\leq q^{x-t}-1$, the next $x-t$ result in a sum of the form $x-t+sq^x$ where $0\leq s\leq p^{x-t}-1$, the next $\gamma y$ dice result in a sum of the form $\gamma y+wn^{x-t}$ where $0\leq w\leq p^{\gamma y}-1$, and the final $t$ result in a sum of the form $t+vp^{\gamma y}n^{x-t}$ where $0\leq v\leq n^t-1$.  So $\delta$ can be written as 
$$\delta =(x-t+r)+(x-t+sq^{x-t})+(\gamma y+wn^{x-t})+(t+vp^{\gamma y}n^{x-t}).$$
To show that this sum is unique assume there exists two sums that yield $\delta$, i.e.
$$(x-t+r_1)+(x-t+s_1q^{x-t})+(\gamma y+w_1n^{x-t})+(t+v_1p^{\gamma y}n^{x-t})$$
$$=(x-t+r_2)+(x-t+s_2q^{x-t})+(\gamma y+w_2n^{x-t})+(t+v_2p^{\gamma y}n^{x-t}).$$

By subtracting constants that appear on both sides, we obtain
$$r_1+s_1q^{x-t}+w_1n^{x-t}+v_1p^{\gamma y}n^{x-t}=r_2+s_2q^{x-t}+w_2n^{x-t}+v_2p^{\gamma y}n^{x-t}.$$
Then we can consider these sums mod $q^{x-t}$.  Since $q^{x-t}|n^{x-t}$ we get that $r_1\equiv r_2 \pmod{q^{x-t}}$.  Then, since $0\leq r_1, r_2\leq q^{x-t}-1$ we have that $r_1=r_2$.  

Next we consider our sums mod $n^{x-t}$.  Since $p\neq 1$ we have that $n^{x-t}\not| q^{x-t}$ making $r_1+s_1q^{x-t}\equiv r_2+s_2q^{x-t} \pmod{n^{x-t}}$.  Using $r_1=r_2$ we get that \\$s_1q^{x-t}\equiv s_2q^{x-t}\pmod{n^{x-t}}$.  Finally, since $0\leq  s_1, s_2\leq p^{x-t}$ we have that \\$0\leq s_1q^{x-t}, s_2q^{x-t}<n^{x-t}-1$ making $s_1q^{x-t}=s_2q^{x-t}$ which shows $s_1=s_2$.  

We will now consider our sums mod $p^{\gamma y}n^{x-t}$.  Since $p\neq 1$ we have that \\$p^{\gamma y}n^{x-t}\not| n^{x-t}$ making $r_1+s_1q^{x-t}+w_1n^{x-t}\equiv r_2+s_2q^{x-t}+w_2n^{x-t} \pmod{p^{\gamma y}n^{x-t}}$.  Using $r_1=r_2$  and $s_1=s_2$ we get that $w_1n^{x-t}\equiv w_2n^{x-t}\pmod{p^{\gamma y}n^{x-t}}$.  Finally, since $0\leq  w_1, w_2\leq p^{\gamma y}-1$ we have that $0\leq w_1n^{x-t}, w_2n^{x-t}< p^{\gamma y}n^{x-t}$ making $2_1n^{x-t}=w_2n^{x-t}$ resulting in $w_1=w_2$.  

Looking back at our original equation 
$$r_1+s_1q^{x-t}+w_1n^{x-t}+v_1p^{\gamma y}n^{x-t}=r_2+s_2q^{x-t}+w_2n^{x-t}+v_2p^{\gamma y}n^{x-t}$$
and using the fact that the $r$'s, $s$'s, and $w$'s are equal we get that $v_1p^{\gamma y}n^{x-t}=v_2p^{\gamma y}n^{x-t}$ making $v_1=v_2$.  Therefore $\delta$ is a unique sum from these groups of dice.  

Thus the minimum number of stupid dice is $m-2x-\gamma y$, the maximum number of stupid dice is $m-x-\gamma y$, and it is possible to find an equal relabeling that uses any number of stupid dice between $m-2x-\gamma y$ and $m-x-\gamma y$.

\chapter{Further Questions}

To continue this work we have left a few open ends that are worth investigating.  Can Conjecture $1$ be proven? How many relabelings are there for a given $m$?  The labeling in $3.3$ can be modified if we let our exponents be $p^{m-x}q^{m-x-\gamma y}$ instead of $p^{m-x-\gamma y}q^{m-x}$, is it a substantial difference?  Are there other substantially different ways to make general relabelings?  What about the case when $n$ is not a product of $2$ distinct primes?

\begin{thebibliography}{99}
\addcontentsline{toc}{chapter}{{Bibiliography}}
\bibitem{ber} F. Bermudez, A. Medina, A. Rosin, E. Scott. {\it Are Stupid Dice Necessary?}, The College Mathematics Journal, Vol. 44, No. 4, September 2013.

\bibitem{gal} J. Gallian, {\it Contemporary Abstract Algebra}, Houghton Mifflin Company, 2008.

\bibitem{gas} W. Gasarch, C. Kruskal, {\it When Can One Load a Set of Dice so That the Sum Is Uniformly Distributed?}, Mathematics Magazine 72, 1999.

\end{thebibliography}

\end{document}