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\textbf{Self Avoiding Walks} \\
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University of Utah \\
Chris Staskewicz
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\textbf{The Scenario:}

Suppose you're standing on the plane and you take integer steps in any
of the four N,S,E,W directions for a total of $n$ steps with the
following restriction: you cannot visit the same grid point more than
once.  A walk of this form is called a \textit{Self Avoiding Walk (SAW)}.
For example, you're walk might look something like this:

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\textbf{The Problem:}

\textit{Suppose every walk begins at the origin.  Let $c_n$ be the
  total number of self avoiding walks of size $n$ in the space
  \textbf{Z}$^{d}, d \geq 2$.
  What is the numerical value for $c_n$?}

\textbf{Some Formalism:}

Think about the $n$ step SAW as a sequence of points $(w_1,
\ldots, w_n)$ where $w_i \in \mbox{Z}^{d}, |w_{i+1} - w_i| = 1,
\ \mbox{and} \ w_i \neq w_j \ \mbox{for all} \ i \neq j$.
\vspace{1cm}

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\textbf{A Quick Bound:}

To get an upper bound that will help us later, consider
walks that do not go through immediate reversals, $w_{i+2} \neq w_i$.
There are $2d$ possible moves for the first step and $(2d - 1)$
choices for the remaining $n-1$ steps.  Note, for large $n$, there
will be plenty of non-SAWs.  TOTAL $= 2d (2d - 1)^{n-1}$.

The lower bound will be taken to be the SAWs that take a step in the
positive direction each time.  There are $d$ positive directions for
dimension $d$, and since we have $n$ total steps, TOTAL $= d^n$.

Thus,
\[ d^n \leq c_n \leq 2d (2d - 1)^{n-1} \]

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\textbf{Monotonicity of SAWs:}

George L. O'Brien (1989) proves that $c_n$ is an increasing function
in $n$.

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\textbf{Random Walks and SAWs:}

Let $X$ be a random varialbe which describes the final location of a
SAW and give $X$ a uniform distribution; $P( X = \omega_n ) =
1/c_n$.  Then \textit{mean squared displacement} from the origin is
given by,

\[ E(X^2) = \frac{1}{c_n} \sum_{\omega_n} |\omega_n|^2 \]

Again, calculations become terribly frustrating, however, it is worth
while to compare $E(X^2)$ with that of an simple \textit{random}
walk.  Here we can let $X_i$ go in any of the $1/2d$ directions
independently with
probability $1/2d$, and let $S_n = X_1 + \cdots + X_n$ denote the
final location of the random walk.  We then have,

\[ E(S_n^2) = \sum_{i,j=1}^{n} E(X_i X_j) = N \]

Since when $i \neq j$ we have $E(X_i X_j) = E(X_i) E(X_j) = 0$ using
independence and $E(X_i^2) = 1$ with respect to the Euclidean norm.

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\textbf{Two Conjectures:}

It is believed that the number and the mean squared
displacement of SAWs behave as,

\begin{eqnarray}
  c_n    & \sim & A \mu^n n^{\gamma - 1} \label{eq:asym1} \\
  \nonumber \\
  E(X^2) & \sim & D n ^{2 \nu} \label{eq:asym2}
\end{eqnarray}

where $A, D, \mu, \gamma, \nu$ are all positive constants which depend
upon which dimension we are working in.

Both equations (\ref{eq:asym1}) and (\ref{eq:asym2}) have been proven
to hold only when $d \geq 5$.

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\textbf{Connective Constant, $\mu$:}

If we let,

\[ \mu = \lim_{n \rightarrow \infty} \sqrt[n]{c_n} \]

this limit exists and is called the \textit{connectivity constant}.
This provides some justification for equation (\ref{eq:asym1})
above.

From this we have a nice bound on the connectivity constant in terms
of the dimension of our space,

\[ d \leq \mu \leq 2d - 1 \]

To see this, note for $a > 0, \sqrt[n]{a} \rightarrow 1$ as $n
\rightarrow \infty$, and let $a = 2d/(2d - 1)$.

A proof is based on the following observation,

\begin{eqnarray*}
  c_{n+m}       & \leq & c_n c_m \\
  && \\
  \log(c_{n+m}) & \leq & \log(c_n) + \log(c_m)
\end{eqnarray*}

The sequence $\{ \log(c_n) \}$ is called \textit{subadditive} due to
the last relation and using the following lemma, we have,

\textbf{Lemma.}  \textit{Let $\{a_n\}$ be a subadditive sequence of
  real numbers.  Then,}

\[ \lim_{n \rightarrow \infty} \frac{a_n}{n} =
   \inf_{n \geq 1} \frac{a_n}{n} \]

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\textbf{Critical Exponent $\gamma$:}

Since $\mu^n \leq c_n$ we have $\gamma \geq 1$.  It has been
proven that for equation (\ref{eq:asym1}), $\gamma = 1$ when
the dimension $d \geq 5$.

Note, an \textbf{open problem} is to prove that $\gamma$ is finite or
find its numerical value when $d = 2, 3, 4$ provided equation
(\ref{eq:asym1}) holds.

The probability (uniform measure) that two SAWs of length $n$
beginning at the same
point \textbf{do not} intersect is given by,

\[ p = \frac{c_{2n}}{c_n c_n} \sim
       \frac{2^{\gamma - 1}}{A}n^{1 - \gamma} \]

To see this, note that if two SAWs do not intersect and begin from the
same point, we must choose from the total number of SAWs of length
$2n$; also use equation (\ref{eq:asym1}) above.

Concerning the cases when $d = 2, 3, 4$ we have,

\begin{enumerate}
  \item if $\gamma > 1, p \rightarrow 0$; their paths cross at
    infinity.  Otherwise, $\gamma = 1, p \sim 1/A > 0,$ and there is
    positive probability that their paths do not cross.

  \item for two simple random walks in an analogous situation, it has
    been proven that $p > 0$ when $d \geq 5$ and $p \rightarrow 0$
    otherwise.
\end{enumerate}

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\textbf{Critical Exponent, $\nu$:}

If we assume the equation (\ref{eq:asym2}), mean squared displacement,
we have

\[ |w_n|^2 \leq n^2 \Rightarrow E(X^2) \leq n^2
                    \Rightarrow \nu \leq 1 \]

Note, an \textbf{open problem} is to prove that

\begin{eqnarray}
  E(X^2) & \geq & K n
\end{eqnarray}

for some constant $K$.  This states that the mean squared distance
grows as fast as or faster than \textit{diffusion} which is associated
to the simple random walk.

This also implies that $\nu \geq 1/2$.  It has been proven that $\nu =
1/2$ when $d \geq 5$.

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\textbf{Some Considerations:}

\begin{enumerate}

\item If $W$ denotes the collection of \textit{all} walks of size $n$
  and $W_i$ denotes the subset of walks that intersect
  themselves exactly $i$ times, then we can partition $W$ as

  \[ W = W_0 \bigcup W_1 \bigcup \cdots \bigcup W_{n-1} \]

  with $W_0$ being the collection of SAWs.  From this we can conclude

  \begin{enumerate}

    \item If we let $c(n)_i = c_i$ be the cardinality of $W_i$ ($c_i$
      is a
      function of the length of the walks under consideration), we have

    \[ c_0 = (2d)^n - \sum_{k=1}^{n-1} c_k \]

    Also note, $c_{n-1} = 2d$ since
    if the walk of size $n$ must intersect itself $n-1$ times, it must
    be of length one and there are only $2d$ of these.  Thus we have,

    \[ c_0 = (2d)^n - \sum_{k=1}^{n-2} c_k - 2d \]

    And I believe that the inner sum terms can be calculated in terms
    of $c(j)_0$ for $j = 1, \ldots, n-2$; that is, we can compute
    $c(n)_0$ recursively.  However, the combinatorics seem to become
    difficult to calculate even for $n \geq 4$.  So this leads me to
    believe that no only is it difficult to find $c_0$, but also $c_1,
    c_2,$ and so on.

  \end{enumerate}

\item Recall that O'Brien proved that $c(n)_0$ is monotone in $n$.
  Can we prove that $c(n)_i$ is monotone in $n$ for each $i$?
  And if so, what will that tell us?

\end{enumerate}

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\textbf{References:}

\begin{enumerate}
  \item Neal Madras and Gordon Slade (1993).  The Self Avoiding
      Walk.  Birkhauser.

  \item George L. O'Brien (1990).  Monotonicity of the Number of
      Self-Avoiding Walks.  \textit{J. Stat. Phys.},
      \textbf{11}:969-979.
\end{enumerate}

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