\documentclass{article}  
\usepackage{amssymb}
\usepackage{latexsym}
\title{Counting snail venom}
\author{Geoff Smith, 1998}
\date{}
\begin{document}
\maketitle

\section*{Introduction}

Let $\Sigma= \{B, W\}$ be a two letter alphabet. Let $\Sigma^*$
denote the set of {\em words} on this alphabet -- that is to say
finite sequences each entry of which is drawn from the alphabet. 
The number of terms in a sequence is its {\em length}. Notice that
the number of words of length $n$ is $2^n$ since each of the $n$ terms 
may be chosen independently. The number of words involving exactly
$r$ copies of $B$ and $s$ copies of $W$ is $(r+s)!/r!s!.$ This
quantity is known as the {\em binomial coefficient} $(^{r+s}_{\ s})$
and arises in algebra via
\begin{equation}\label{eq:bin_thm}
(x + y)^n = \sum_{\stackrel{r,s \geq 0}{r+s=n}} (^{r+s}_{\ s}) x^r y^s.
\end{equation}
where $x$ and $y$ are unknowns. These quantities $(^{r+s}_{\ s})$
also form the entries of {\em Pascal's triangle} where 
$r+s$ is the row number and $r$ indexes how far the entry is from the
left (and $s$ from the right). Note that $0! = 1$ in order to make everything
work smoothly. Putting $x=y=1$ in (\ref{eq:bin_thm}) yields that
there are $2^n$ words of length $n,$ a fact which we have already noted.
The numbers from Pascal's triangle crop up frequently when performing
enumeration arguments.

We will need a fact about these binomial coefficients.
Notice that
\[ -1 + (^m_1) - (^m_2) + (^m_3) \ldots (-1)^m (^m_m)= 0
\mbox{ when } m \geq 1\]
because then $-(1-1)^m =0$ and the binomial theorem applies.
This really says that if you add up the entries in any row of
Pascal's triangle (except the top one), alternating the signs 
as you go, the answer is $0.$

\section*{How big is a union of finitely many finite sets?}

Suppose that you have two finite sets $A$ and $B.$
You can find the size of their union using
\[ \left |A \cup B \right| = |A| + |B| - |A \cap B| \]
because when you work out $|A| + |B|$ the elements of 
$A \cap B$ are being `counted twice'. You compensate for this
by subtracting $|A \cap B|.$

Now suppose you have three finite sets. A very careful analysis of counting
will show you that
\begin{equation}\label{eq:toyincexc}
|A \cup B \cup C| = |A| + |B|  + |C| - |A \cap B| - |A \cap C|
- |B \cap C| + |A \cap B \cap C|.
\end{equation}
What is going on here is that you first try $|A| + |B|  + |C|$
but this is wrong because elements in $A \cap B$, $A \cap C$, and
$B \cap C$ have been counted too much. You therefore try to eliminate 
this over-counting by subtracting $|A \cap B| + |A \cap C| + |B \cap C|,$
but then notice that elements of $A \cap B \cap C$ have been `over removed'.
You compensate for this by adding $|A \cap B \cap C|$ and all is well.
We are edging toward the inclusion-exclusion enumeration principle.
Let is look at a concrete example.
Let $A = \{1,2,3,4\},$ $B=\{3,4,5,6\}$ and $C= \{2,3,4,5,6,7 \}.$
Now (\ref{eq:toyincexc}) says
$7 = 4 + 4 + 6 - 2 -3 -4 + 2,$ which happily is true.

We prove validity of the 
Inclusion-Exclusion counting principle.

\noindent{\bf Theorem} Suppose $n \in {\mathbb N}$ and $A_i$ is a finite set
for $1 \leq i \leq n.$ It follows that
\begin{eqnarray*}
\left| \bigcup_{1 \leq i \leq n}  A_i\right|
& = &\sum_{1 \leq {i_1} \leq n} \left| A_{i_1}\right| 
- \sum_{1 \leq {i_1} < {i_2} \leq n} \left| A_{i_1} \cap A_{i_2}\right|\\
& & + \sum_{1 \leq {i_1} < {i_2}  < {i_3} \leq n}
\left| A_{i_1} \cap A_{i_2}\cap A_{i_3}\right| 
- \ldots
+ (-1)^{n+1} \left| \bigcap_{i=1}^n A_i \right|.
\end{eqnarray*}

\vspace{1cm}

\noindent {\bf Proof} 

Induct on 
$| \cup_i A_i|.$ If all $A_i$ are empty, then the theorem holds, and 
the induction starts without difficulty. Now focus on the case where
$| \cup_i A_i|>0.$ Pick any $x \in \cup_i A_i$ and form new sets
by $B_i = A_i \setminus \{x\}$ (i.e. remove $x$ from all the sets
$A_i$ which contain it). Now $| \cup_i B_i| = | \cup_i A_i|-1$
so the theorem holds for the sets $B_i$ by induction. Now pop
$x$ back in to all sets from which you deleted it. 
The re-insertion of $x$ makes a contribution
of $+1$ on the left hand side.
On the right, suppose $x$ occurs in
exactly $m (\geq 1)$ of the $A_i$. The re-insertion of $x$ has the
effect of adding 
exactly $m - (^m_2) + (^m_3) \ldots + (-1)^m (^m_m)$ to the right hand side.
However, we have prepared the ground in the previous section,
so we know that this quantity is 1. The induction step is complete, and thus
the inclusion-exclusion principle is valid.

\subsection*{Example Application 1}

The Euler $\varphi$-function is $\varphi: {\mathbb N} \rightarrow {\mathbb N}$
defined by 
\[ \varphi(n) = \left|\{ k \mid 1 \leq k \leq n,\ \mbox{ g.c.d.}(k,n)
= 1 \}\right|.\] 
Here g.c.d. denotes {\em greatest common divisor}.
We have not made this function up. It is an important function in number 
theory, combinatorics and algebra, and it has sweet properties.
For example, if $a,b \in {\mathbb N}$ and $a,b$ are coprime (i.e. have 
1 as their greatest common divisor), then 
$\varphi(ab) = \varphi(a)\varphi(b).$ In fact this follows immediately from
the next proposition.

\vspace{1cm}

\noindent{\bf Proposition} Let the prime divisors of $n$ be 
$p_1, p_2, \ldots, p_k$ (without repetition). It follows that
\[ \varphi(n) = n \prod_{i=1}^k (1 - p_i^{-1}).
\]

\vspace{1cm}

\noindent {\bf Proof} The theorem is trivially true if $n = 1,$ since
the empty product is 1. Thus we may assume $n > 1,$ and thus $k \geq 1.$
For each $i$ in the range $1 \leq i \leq k$ we put 
\[ A_i = \{ t \mid 1 \leq t \leq n,\  n/p_i \in {\mathbb N} \}.\]
Here we have eschewed the vertical line notation $p_i \mid n$ deliberately
in order to avoid notational collision.

Notice that $|A_i| = n/p_i.$ Moreover, if $i \not = j$ then
$|A_i \cap A_j | = n/p_ip_j$ and so on. The inclusion-exclusion
principle enables us to count the natural numbers between 1 and $n$
(inclusive) which are {\em not} coprime to $n$ in two ways.
\[n - \varphi(n) = n \sum_i 1/p_i - n \sum_{i_1<i_2} n/p_{i_1}p_{i_2}
+ \ldots.\]
Rearrange, and a use a little algebra to obtain the result.

\subsection*{Example Application 2}

Deal two packs of shuffled cards simultaneously. What is the probability
that no pair of identical cards will be exposed simultaneously?

\vspace{1cm}


Fix the first pack, and consider all possible rearrangements of the 
second pack. For each $i$ in the range $1 \leq i \leq 52$
let $A_i$ denote the set of all arrangements of the second pack which
happen to have the property that the card in position $i$ matches
the card in position $i$ of the first pack. Obviously
$|A_i| = 51!$ for every $i.$ Moreover, if $i \not = j,$
then $|A_i \cap A_j| = 50!$ and so on. Let $X = \cup_i A_i,$ so 
the probability
of at least one match is $|X|/52!.$ We calculate this 
using the inclusion-exclusion principle.
\begin{eqnarray*}
|X|/52!& = & (52!)^{-1}\left( (^{52}_{\ 1}) 51! - (^{52}_{\ 2}) 50!
+ (^{52}_{\ 3}) 49! - \ldots  - (^{52}_{52}) 0!\right)\\ 
& = & 1 - 1/2! + 1/3! - \ldots -1/52! \approx 1 - (\sum_{i=0}^\infty (-1)^i/i!)
= 1 - 1/e.
\end{eqnarray*}
Thus the probability of no coincidences is (to an excellent approximation)
$1/e.$

Here we have use the fact that
\[ e^x = 1 + x + x^2/2! + \ldots = \sum_{i=0}^\infty x^i/i! \]
and put $x = -1.$ 


\section*{Enumerations related to snail venom.}

We define a {\em language} $L$ to be an arbitrary subset of $\Sigma^*$.
Associated to a language we have a power series $p_L$ in two variables
$b$ and $w$ where 
\[ p_L = p_L(b,w) = \sum_{n=0}^\infty \sum_{m=0}^\infty c_{mn}b^m w^n\] and 
$c_{mn}$ is the number of words in $L$ which involve exactly $m$ copies 
of the letter $B$ and $n$ copies of the letter $W.$

\begin{enumerate}
\item[(1)] If the language $L_1$ consists of all words which do not involve
$W$ then $c_{mn} = 0$ whenever $n>0,$ and $c_{m0} = 1$ for every $m$
since there is exactly one word in the language of length $m$ involving
no $W.$ Thus 
the power series is \[p_{L_1} = \sum_{m=0}^\infty  b^m = (1-b)^{-1}. \]
\item[(2)] If the language $L_2$ consists of all words in $\Sigma^*$
of length $t$ where $t$ is fixed, then 
\[p_{L_2} = \sum_{\stackrel{m, n \geq 0}{n+m=t}} \frac{t!}{n!m!}b^mc^n.\]
If we further specify that $t=3$ to give the language $\hat L_2$ we have
\[p_{\hat L_2} = b^3w^0 + 3b^2w^1 + 3b^1w^2 + b^0w^3 
= b^3 + 3b^2w + 3bw^2 + w^3.\]
In this case the power series collapse to being a polynomial expression,
since all save a finite number of coefficients vanish. Notice that
the coefficients here are drawn from a row of {\em Pascal's triangle},
and another way of describing $p_{L_2}$ is $(b+w)^3.$ To understand this,
think of $+$ as being ``or'' and multiplication (juxtaposition) as being 
``and then''. In this sense $(b+w)^3 = (b+w)(b+w)(b+w)$ exactly describes 
the $\hat L_2.$ Each word in $\hat L_2$ is of the form:
$B$ or $W,$ then $B$ or $W,$ then $B$ or $W.$
\item[(3)] If the language $L_3$ consists of all words where
there are no juxtaposed $W$s and no juxtaposed $B$s then there
is one word in $L_3$ of length 0, the empty word, and exactly two 
words of every other length. Thus \[p_{L_3} = 1 + \sum_{n=0}^\infty 
b^n w^{n+1} + \sum_{n=0}^\infty b^{n+1}w^n + \sum_{n=1}^\infty 2b^nw^n.\]  
\end{enumerate}

The particular languages of interest in the investigation of snail
venom are the ones satisfying the following five conditions.
\newpage 
\begin{enumerate} 
\item[(i)]The words involve an even number of occurrences of $B.$
\item[(ii)] The words begin and end with a letter $B.$
\item[(iii)] Words must not contain the subword $WWWWWWW.$
\item[(iv)] Words must not contain the subword $BBB$ (i.e. you must not 
have three consecutive occurrences of $B).$
\item[(v)] Suppose a word involves exactly $2k$ occurrences of 
the letter $B$. Label them from the left the $1st, 2nd, 
\ldots 2k\mbox{-th}.$ Between at least one $(2i-1)$-th and $2i$-th occurrence
of $B,$ there must at least two consecutive occurrences of $W.$ 
\end{enumerate}

The language $S$ specified by the first three conditions
are all possible words associated with snail venom.
The language defined by all five these conditions are the
so-called inaccessible words, and are exactly those
which do not admit of synthesis by a specific manufacturing process.

The language $S$ is defined by conditions
$(i),\ (ii)$ and $(iii).$ Let $f = (1 + w + w^2 + w^3 + w^4 + w^5 + w^6).$
The definition of $p_S$ yields that
\[ p_S = f + bfb + bfbfbfb + bfbfbfb + \ldots \]
or more formally
\[ p_S = \sum_{i=0}^\infty b^{2i}f^{2i-1} \] 

If you wish to see you many words there are in 
this language of length 30 and involving 8 occurrences of
$B$ (and so 22 occurrences of $W$) you look at the coefficient
of $b^8w^{22}$ and this is precisely the coefficient of $w^{22}$
in $f^7.$ This can be determined in a trice by almost any computer
algebra system (AXIOM, MAPLE, REDUCE etc).

Now we proceed to the main calculation, where we also impose conditions
$(iv)$ and $(v),$ and are particularly interested in the cases where
there are 4, 6 and 8 occurrences of $B,$ and when the word
lengths does not happen to exceed 30. This final constraint can
be imposed at the very end of the calculations by truncating 
the appropriate polynomial if necessary.

\section*{The Main Calculation}

All these calculations are the same in spirit, but the details become a little
more complicated as we proceed.

We introduce a formal and compact way of describing languages.
The letters $B$ and $W$ stand for themselves. The letter $G$ stands
for {\em either $W$ or no letter}, the letter $F$ stands for
{\em consecutive $W$s, somewhere between 0 and 6 in number}.
We interpret $\cup$ to mean the union of languages. 

\subsection*{Four occurrences of $B$}

Consider the three languages  
$C_1 = BFBFBFB,$ $C_2 = BBBFB \cup BFBBB$ and  $C_3 = BGBFBGB.$
The language (of inaccessible words) we are interested in is 
\[ C = \left(C_2 \cup C_3 \right).\]

The polynomial for $C_1$ is $b^4f^3.$
$C_2$ is described by $b^4(2f -1)$ and 
$C_3$ is described by $b^4g^2f.$
$C_2 \cap C_3$ is described by $b^4(2g -1).$
Thus the inaccessible language $C_2 \cup C_3$ is described by 
\[b^4(2f -1 + g^2f - (2g -1)) = b^4(2f + g^2f -2g).\]

Of course, since every single polynomial in this analysis involves the factor
$b^4,$ we could have suppressed it. We didn't for expository reasons.

\subsection*{Six occurrences of $B$}

In this section we will suppress the ever present factor $b^6.$
Consider the three languages
$D_1 = BFBFBFBFBFB,$ $D_2$ the language consisting of words
containing three or more consecutive $B$s, and $D_3 = BGBFBGBFBGB.$
The polynomial describing $D_1$ is $f^5$ and that describing
$D_3$ is $g^3f^2.$ 

\subsubsection*{The language $D_2$}

$D_2 = R \cup S \cup T \cup U$ where 
\[\begin{array}{cclc}
  & &\mbox{Language}&\mbox{Polynomial}\\
R &=& BBBFBFBFB&f^3\\
S &=& BFBBBFBFB&f^3\\     
T &=& BFBFBBBFB&f^3\\ 
U &=& BFBFBFBBB&f^3
\end{array}\]

The sum of the polynomials of these 4 languages is $4f^3.$
We use the inclusion-exclusion principle to determine the polynomial
describing $D_2$

\[\begin{array}{cclc}
  & &\mbox{Language}&\mbox{Polynomial}\\
R \cap S&=& BBBBFBFB&f^2\\
R \cap T&=& BBBBBFB&f\\
R \cap U&=& BBBFBBB&f\\ 
S \cap T&=& BFBBBBFB&f^2\\
S \cap U&=& BFBBBBB&f\\ 
T \cap U&=& BFBFBBBB&f^2
\end{array}\]

The sum of the polynomials of these 6 languages is $3f+3f^2.$

\[\begin{array}{cclc}
  & &\mbox{Language}&\mbox{Polynomial}\\
R \cap S \cap T&=&BBBBBFB&f\\ 
S \cap T \cap U&=&BFBBBBB&f\\
R \cap T \cap U&=&BBBBBB&1\\
R \cap S \cap U&=&BBBBBB&1
\end{array}\]

The sum of the polynomials of these 4 languages is $2f+2.$

\[\begin{array}{cclc}
  & &\mbox{Language}&\mbox{Polynomial}\\
R \cap S \cap T \cap U&=&BBBBBB&1
\end{array}\]

The polynomial of this language is $1.$

The language $D_2$ of words involving at least 3 
consecutive $B$s is described by the polynomial
$4f^3 - (3f+3f^2)+ (2f+2) -1,$ 
using the inclusion-exclusion principle.

\subsubsection*{The language $D_2 \cup D_3$}

$D_2 \cap D_3$ is the union of the following four languages.
\[\begin{array}{cclc}
  & &\mbox{Language}&\mbox{Polynomial}\\
L &=&BBBGBFBGB& g^2f\\
M &=&BGBBBFBGB& g^2f\\
N &=&BGBFBBBGB& g^2f\\
0 &=&BGBFBGBBB& g^2f
\end{array}.\]

The sum of the polynomials of these 4 languages is $4g^2f.$
We must calculate the polynomials associated with various intersections.

\[\begin{array}{cclc}
  & &\mbox{Language}&\mbox{Polynomial}\\
L \cap M&=& BBBBFBGB& gf\\
L \cap N&=& BBBBBGB&  g\\
L \cap O&=& BBBGBBB&  g\\
M \cap N&=& BGBBBBGB& g^2\\ 
M \cap O&=& BGBBBBB&  g\\
N \cap O&=& BGBFBBBB& gf
\end{array}\]

The sum of the polynomials of these 6 languages is $3g+2gf +g^2.$


\[\begin{array}{cclc}
  & &\mbox{Language}&\mbox{Polynomial}\\
L \cap M \cap N  &=& BBBBBGB& g\\
L \cap M \cap O &=& BBBBBBB& 1\\
L \cap N \cap O &=& BBBBBBB& 1\\
M \cap N \cap O &=& BGBBBBBB& g
\end{array}\]

The sum of the polynomials of these 4 languages is $2g+2.$

\[\begin{array}{cclc}
  & &\mbox{Language}&\mbox{Polynomial}\\
L \cap M \cap N \cap O &=& BBBBBB& 1.
\end{array}\]

The polynomial of this languages is $1.$

Thus $D_2 \cap D_3$ is described by
$4g^2f - (g^2 + 2gf +3g) +(2g+2)-1.$ 

We conclude that the inaccessible language 
$D_2 \cup D_3$ is described by the polynomial

\[4f^3 - (3f+3f^2)+ (2f+2) -1
+
g^3f^2
-
(4g^2f - (g^2 + 2gf +3g) +(2g+2)-1)\]
which simplifies a little to yield
\[ 4f^3 - f - 3f^2 + g^4f^3 - 4g^2f + g^2 +2gf + g.\]

\subsection*{Eight occurrences of $B$}


In this section we will suppress the ever present factor $b^8.$
Consider the three languages
$E_1 = BFBFBFBFBFBFBFB,$ $E_2$ the language consisting of words
containing three or more consecutive $B$s, and $E_3 = BGBFBGBFBGBFBGB.$
The polynomial describing $E_1$ is $f^7$ and that describing
$E_3$ is $g^4f^3.$

\subsubsection*{The language $E_2$}

$E_2 = R \cup S \cup T \cup U \cup V \cup W$ where

\[\begin{array}{cclc}
  & &\mbox{Language}&\mbox{Polynomial}\\
R &=& BBBFBFBFBFBFB&f^5\\ 
S &=& BFBBBFBFBFBFB&f^5\\              
T &=& BFBFBBBFBFBFB&f^5\\ 
U &=& BFBFBFBBBFBFB&f^5\\       
V &=& BFBFBFBFBBBFB&f^5\\
W &=& BFBFBFBFBFBBB&f^5
\end{array}\]
The sum of the polynomials of these 6 languages is $6f^5.$

We must walk the inclusion-exclusion road. From now on, we may sometimes
omit the full language description when it is clear.

\[\begin{array}{cclcc}
  & &\mbox{Language}&\mbox{Polynomial}\\
R \cap S &=&BBBBFBFBFBFB&f^4\\
R \cap T&=& BBBBBFBFBFB& f^3\\   
R \cap U&=& BBBFBBBFBFB& f^3\\ 
R \cap V&=& BBBFBFBBBFB& f^3\\
R \cap W&=& BBBFBFBFBBB& f^3\\
S \cap T&=& BFBBBBFBFBFB&f^4\\ 
S \cap U&=& BFBBBBBFBFB& f^3\\
S \cap V&=& BFBBBFBBBFB& f^3\\
S \cap W&=& BFBBBFBFBBB& f^3\\
T \cap U&=& BFBFBBBBFBFB&f^4\\  
T \cap V&=& BFBFBBBBBFB&f^3\\
T \cap W&=& BFBFBBBFBBB&f^3\\
U \cap V&=& BFBFBFBBBBFB&f^4\\      
U \cap W&=& BFBFBFBBBBB&f^3\\
V \cap W&=& BFBFBFBFBBBB&f^4
\end{array}\]
The sum of the polynomials of these 15 languages is $5f^4 + 10f^3.$

From now on we may group together similar languages under the heading
{\em Siblings:} often languages which are the reversals of one another.

\[\begin{array}{cclcc}
  & &\mbox{Language}&\mbox{Polynomial}&\mbox{Siblings}\\
R \cap S \cap T&=&BBBBBFBFBFB&f^3&4\\ 
R \cap S \cap U&=&BBBBBBFBFB&f^2&6\\
R \cap S \cap V&=&BBBBFBBBFB&f^2&4\\
R \cap T \cap V&=&BBBBBBBFB&f&2\\
R \cap S \cap W&=&BBBBFBFBBB&f^2&2\\
R \cap T \cap W&=&BBBBBFBBB&f&2
\end{array}\]
The sum of the polynomials of these 20 languages is $4f^3 + 12f^2 + 4f.$

\[\begin{array}{cclcc}
  & &\mbox{Language}&\mbox{Polynomial}&\mbox{Siblings}\\
R \cap S \cap T \cap U&=&BBBBBBFBFB&f^2&3 \\
R \cap S \cap T \cap V&=&BBBBBFBBBFB&f^2&4 \\
R \cap S \cap T \cap W&=&BBBBBFBFBBB&f^2&2 \\
R \cap S \cap U \cap V&=&BBBBBBBFB&f&2 \\
R \cap S \cap U \cap W&=&BBBBBBBB&1&2 \\ 
R \cap S \cap V \cap W&=&BBBBFBBBB&f&1 \\
R \cap T \cap U \cap W&=&BBBBBBBB&1&1
\end{array}\]

The sum of the polynomials of these 15 languages is $9f^2+3f+3.$

\[\begin{array}{cclcc}
  & &\mbox{Language}&\mbox{Polynomial}&\mbox{Siblings}\\
R \cap S \cap T \cap U \cap V&=&BBBBBBBFB&f&2\\
\mbox{ 4 others}             &=&BBBBBBBB&1&4
\end{array}\]

The sum of the polynomials of these 6 languages is $2f+4.$

\[\begin{array}{cclcc}
  & &\mbox{Language}&\mbox{Polynomial}&\mbox{Siblings}\\
\mbox{All six}&=&BBBBBBBB&1&1
\end{array}\]

The polynomial of this language is $1.$

The polynomial describing this language is 
\[6f^5-(5f^4+10f^3)+(4f^3+12f^2+4f)
-(9f^2 + 3f + 3)+(2f+4)-1\]

\subsection*{The language $E_2 \cup E_3$}

We use an inclusion-exclusion argument, and express $E_2 \cap E_3$
as the union of the following six languages.
\[\begin{array}{cclc}
  & &\mbox{Language}&\mbox{Polynomial}\\
L&=& BBBGBFBGBFBGB&g^3f^2\\          
M&=& BGBBBFBGBFBGB&g^3f^2\\             
N&=& BGBFBBBGBFBGB&g^3f^2\\            
O&=& BGBFBGBBBFBGB&g^3f^2\\           
P&=& BGBFBGBFBBBGB&g^3f^2\\          
Q&=& BGBFBGBFBGBBB&g^3f^2         
\end{array}\]
The sum of the polynomials of these 6 languages is $6g^3f^2.$

\[\begin{array}{cclcc}
  & &\mbox{Language}&\mbox{Polynomial}\\
L \cap M &=& BBBBFBGBFBGB & g^2f^2&2\\
M \cap N &=& BGBBBBGBFBGB & g^3f &2\\
N \cap 0 &=& BGBFBBBBFBGB & g^2f^2 &1\\
\mbox{ others  }&=& \mbox{various}  & g^2f&10
\end{array}\]
The sum of the polynomials describing these 15 languages
is $10g^2f+3g^2f^2+2g^3f$

\[\begin{array}{cclcc}
  & &\mbox{Language}&\mbox{Polynomial}\\
L \cap M \cap N&BBBBBGBFBGB&g^2f&2\\
M \cap N \cap 0&BGBBBBBFBGB&g^2f&2\\
L \cap M \cap 0&BBBBBBFBGB&gf&2\\
L \cap N \cap 0&BBBBBBFBGB&gf&2\\
M \cap N \cap P&BGBBBBBBGB&g^2&2\\
L \cap M \cap P&BBBBFBBBGB&gf&2\\
L \cap M \cap Q&BBBBFBGBBB&gf&2\\
L \cap O \cap P&BBBGBBBBGB&g^2&2\\
L \cap N \cap P&BBBBBBBGB&g&2\\
L \cap N \cap Q&BBBBBGBBB&g&2
\end{array}\]
The sum of the polynomials describing these 20 languages
is $4g^2f + 8gf + 4g^2 + 4g.$ 

\[\begin{array}{cclcc}
  & &\mbox{Language}&\mbox{Polynomial}\\
L \cap M \cap N \cap 0&=&BBBBBBFBGB&gf&2\\
M \cap N \cap 0 \cap P&=&BGBBBBBBGB&g^2&1\\
L \cap M \cap N \cap P&=&BBBBBBBGB&g&8\\ 
L \cap M \cap P \cap Q&=&BBBBFBBBB&f&1\\
L \cap N \cap O \cap Q&=&BBBBBBBB&1&3 
\end{array}\]
The sum of the polynomials describing these 15 languages
is $2gf+g^2+8g+f+3.$ 

\[\begin{array}{cclcc}
L \cap M \cap N \cap O \cap P&=&BBBBBBBGB&g&2\\
\mbox{ 4 others }            &=&BBBBBBBB&1&4
\end{array}\]

The sum of the polynomials describing these 6 languages
is $2g+4.$

\[\begin{array}{cclcc}
L \cap M \cap N \cap O \cap P \cap Q&=&BBBBBBBB&1&1
\end{array}\]

The polynomial of this language
is $1.$

The polynomial describing the language $E_2 \cap E_3$ 
is therefore 
\[6g^3f^2 -(3g^2f^2+2g^3f + 10g^2f)
+(4g^2f + 4g^2 + 4g + 8gf)\]\[
-(2gf + f + g^2 + 8g + 3)
+(2g+4) -1\]

The polynomial describing the (inaccessible) language $E_2 \cup E_3$
is therefore
\[6f^5-(5f^4+10f^3)+(4f^3+12f^2+4f) -(9f^2 + 3f + 3)+(2f+4)-1 
+ g^4f^3\]\[
-\left[(6g^3f^2 -(3g^2f^2+2g^3f + 10g^2f
+(4g^2f + 4g^2 + 4g + 8gf)\right.\]\[
\left. -(2gf + f + g^2 + 8g + 3)
+(2g+4) -1\right]\]

\section*{The escargot polynomials and associated numbers}

By truncating the polynomials we can focus on the case that the
words in question have length no more than 30. By evaluating
$w$ to be 1, we count the words (we call these words {\em strings}). 
However, in the chemical application,
we can allow for the 19 possible amino acids which may occur when a $W$ is 
present by putting $w=19$ (we call these words {\em sequences}).
We let $e_m$ be the polynomial which enumerates all words satisfying
conditions (i), (ii) and (iii) and involve $m$ copies
of $B$, and are truncated to allow for the 
total length $\leq 30$ condition (the escargot polynomials). 
Let $f_m$ be the polynomials of inaccessibility, which are defined
in the same way, save that conditions (iv) and (v) are also imposed.

\subsection*{Four occurrences of $B$.}

\[ f_4= 1+3\,w+6\,{w}^{2}+6\,{w}^{3}+6\,{w}^{4}+6\,{w}^{5}+6\,{w}^{6}+3\,{w}^{
7}+{w}^{8}\]

\[f_4(1)= 38 \mbox{ inaccessible strings of length at most 30.}\]
\[f_4(19)= 19963135442 \mbox{ inaccessible sequences of length at most 30.}\]

\[e_4(w) = 1+3\,w +6\,{w}^{2} +10\,{w}^{3} +15\,{w}^{4} +21\,{w}^{5} 
+28\,{w }^{6}\]
\[ +33\,{w}^{7} +36\,{w}^{8} +37\,{w}^{9} +36\,{w}^{10} +33\,{w}^{11} +28\,{w}^
{12} +21\,{w}^{13}\]
\[ +15\,{w}^{14} +10\,{w}^{15} +6\,{w}^{16} +3\,{w}^{17} +{w}^{18}\]

\[e_4(1)=343  \mbox{ strings of length at most 30.}\]
\[e_4(19)= 122463904886205958677421 \mbox{ sequences of length at most 30.}\]

\subsection*{Six occurrences of $B$.}

\[f_6 = 1+5\,w +15\,{w}^{2} +35\,{w}^{3} +60\,{w}^{4} +89\,{w}^{5}\]
\[+122\,{w}^{6} +154\,{w}^{7} +175\,{w}^{8} +180\,{w}^{9} +171\,{w}^{10}\] 
\[+154\,{w}^{11} +129\,{w}^{12} +96\,{w}^{13} +65\,{w}^{14} +41\,{w}^{15}\] 
\[+24\,{w}^{16} +12\,{w}^{17} +4\,{w}^{18}\] 

\[f_6(1)= 1532 \mbox{ inaccessible strings of length at most 30.}\]
\[f_6(19)= 489875340706634924622560\] inaccessible sequences of length at 
most 30.

\[e_6 = 
1+5\,w +15\,{w}^{2} +35\,{w}^{3} +70\,{w}^{4} +126\,{w}^{5} +210\,{w}^{6}\]
\[ +325\,{w}^{7} +470\,{w}^{8} +640\,{w}^{9} +826\,{w}^{10} +1015\,{w}^{11}\]
\[ +1190\,{w}^{12} +1330\,{w}^{13} +1420\,{w}^{14} +1451\,{w}^{15} \]
\[+1420\,{w}^{16} +1330\,{w}^{17} +1190\,{w}^{18} +1015\,{w}^{19}\]
\[ +826\,{w}^{20} +640\,{w}^{21} +470\,{w}^{22} +325\,{w}^{23} +210\,{w}^{24}\]

\[e_6(1)= 16555 \mbox{ strings of length at most 30.}\]
\[e_6(19)= 1119403018505084128111935530758381\] sequences of length 
at most 30.

\subsection*{Eight occurrences of $B$}

\[f_8 = 1+7\,w+28\,{w}^{2}+84\,{w}^{3}+210\,{w}^{4}+442\,{w}^{5}+817\,{w}^{6}+
1371\,{w}^{7}\]\[+2125\,{w}^{8}+3064\,{w}^{9}+4129\,{w}^{10}+5234\,{w}^{11
}+6290\,{w}^{12}+7186\,{w}^{13}\]\[+7808\,{w}^{14}+8081\,{w}^{15}+7987\,{w
}^{16}+7550\,{w}^{17}+6818\,{w}^{18}\]\[+5866\,{w}^{19}+4805\,{w}^{20}+
3747\,{w}^{21}+2771\,{w}^{22}\]

\[f_8(1)= 86421 \mbox{ inaccessible strings of length at most 30.}\]
Thus there are 
\[f_8(19) =40471537701497846906771178820797\] inaccessible sequences of 
length at most 30.

\[e_8 = 1+7\,w+28\,{w}^{2}+84\,{w}^{3}+210\,{w}^{4}+462\,{w}^{5}+924\,{w}^{6}+
1709\,{w}^{7}\]\[+2954\,{w}^{8}+4809\,{w}^{9}+7420\,{w}^{10}+10906\,{w}^{
11}+15330\,{w}^{12}\]\[+20664\,{w}^{13}+26769\,{w}^{14}+33390\,{w}^{15}+
40166\,{w}^{16}+46655\,{w}^{17}\]\[+52374\,{w}^{18}+56854\,{w}^{19}+59710
\,{w}^{20}+60691\,{w}^{21}+59710\,{w}^{22}\]

\[e_8(1) = 501827 \mbox{ strings of length at most 30.}\]
\[e_8(19)=
855972319026576304079305969491905\]
sequences of length at most 30.
Thus about 5\% of the sequences are unaccessible.
\end{document}

