(**updated**; apologies for way too much room left for interpretation in the original post)


Let $\mathcal{A} =A\_{n-1}$ be the $A\_{n-1}$ arrangement in $\mathbb{R}^{n}$, i.e. the set of hyperplanes $H\_{ij}$ specified by equations $x\_i=x\_j$, for $1\leq i\leq j\leq n$. It is well-known that the connected components $c$ of the complement of the intersection of $\mathcal{A}$ with the affine hyperplane $H$={$x\mid \sum\_i x\_i=1$} (let me call them *cells*) are in 1-to-1 correspondence with permutations $\sigma(c)\in S\_{n}$, so that crossing the hyperplane $H\_{ij}$ corresponds to multiplication by the transposition $(i,j)$.
More precisely, $\sigma(c)$ records the order of (Euclidean) distances $d(x,e\_k)$ at which the standard basis vectors $e\_k$ are from every $x\in c$. E.g. for $\sigma(c)=()=12 \dots n$, the identity permutation, one has $d(x,e\_1)$<$d(x,e\_2)$<...<$d(x,e\_n)$, and the arrangement hyperplanes on the boundary of this $c$ are $H\_{k,k+1}$ for $k=1,\dots ,n -1$.


Let $\ell$ be a general position line in $H$. Then $\ell$ intersects $\binom{n}{2}+1$ cells. Two of these intersections are unbounded; let us denote the corresponding cells $c\_0$ and $c\_t$, respectively. As we follow $\ell$ from $c\_0$ to $c\_t$, from one cell to the next, the corresponding permutation $\sigma(c) = i\_1 i\_2...$ $i\_n$ changes from $\sigma(c\_0)$ to $\sigma(c\_t)$; namely, it gets multiplied by $(i\_k,i\_{k+1})$ whenever we cross $H\_{i\_k,i\_{k+1}}$. 
One can also view this as *flipping* the adjacent entries, i.e. applying the transposition $(k,k+1)$ to the sequence $\sigma(c)$.


Assume that $\sigma(c\_0)$ is the identity permutation. Then $\ell$ specifies a sequence of flips $(k,k+1)$ that need to be applied to obtain $\sigma(c\_t)=n, n-1, n-2, \ldots 1$. E.g. for $n=3$ there are two such sequences: 


1. (12), (23), (12)
2. (23), (12), (23)


**Question.** Are all the sequences of $\binom{n}{2}$ *flips* leading from $1,2, \dots, n$ to
$n, n-1, n-2, \ldots 1$ realized by general position lines in $H$? 



lines through A_n reflection arrangement and permutations
I find your description of the combinatorial data you want to use to encode crossing sequences a little unclear. However, I am confident that the answer to your intended question is "no". I will describe combinatorial data which at least as restrictive as what you have described, and show that this is still not restrictive enough.


For simplicity, I will consider paths from the identity region to $n(n-1) \cdots 321$. I will describe such a path by giving (1) the sequence of chambers it passes through: $v\_0 = \mathrm{Id}$, $v\_1$, $v\_2$, ..., $v\_{\binom{n}{2}} = n(n-1) \cdots 321$ and (2) the sequence of walls crossed at every step $(i\_1 \ j\_1)$, $(i\_2 \ j\_2)$, ..., $(i\_{\binom{n}{2}} \ j\_{\binom{n}{2}}$). I require at every step that the wall $(i\_r \ j\_r)$ really does separate regions $v\_{r-1}$ and $v\_{r}$. Using my favorite conventions, this is equivalent to 


$$v\_{r} = (i\_r \ j\_r) v\_{r-1} \ \mbox{and} \ v\_{r-1}^{-1} (i\_r \ j\_r) v\_{r-1} = (k\_r \ (k\_r+1))$$
for some $k\_r$. (Other people may use the inverse convention for labeling regions of the $A\_n$ arrangement by permutations, or for multiplying permutations, but it shouldn't be hard to translate.)


In terms of Fedja's metaphor of the $n$ runners, the second condition says that runners $i\_r$ and $j\_r$ really are next to each other at time $r-1$, in positions $k\_r$ and $k\_r+1$.


Note that you can recover the $v$'s from the $(i \ j)$'s as $v\_r = (i\_r \ j\_r) (i\_{r-1} \ j\_{r-1}) \cdots (i\_2 \ j\_2) (i\_1 \ j\_1)$, so we could just record the sequence of transpositions. However, you would need to recompute the $v$'s to state the condition that $ v\_{r-1}^{-1} (i\_r \ j\_r) v\_{r-1} = (k\_r \ (k\_r+1))$ for some $k\_r$. Even better is just to record the sequence of $k$'s, in which case the $v$'s can be recovered as $v\_r = (k\_1 \ k\_1+1) (k\_2 \ k\_2 +1 ) \cdots (k\_r \ k\_r+1)$. In this language, what I am describing is reduced words for $n(n-1) \cdots 321$.


For example, there are two reduced words for $321$ in $S\_3$:
$$\begin{array}{|l|l|l|}
\hline
v\_r & (i\_r \ j\_r) & k\_r \\
\hline
123,\ 213,\ 231,\ 321 & (1\ 2 ),\ (1\ 3),\ (2\ 3) & 1,\ 2,\ 1 \\
123,\ 132,\ 312,\ 321 & (2\ 3),\ (1\ 3),\ (1\ 2) & 2,\ 1,\ 2 \\
\hline
\end{array}$$


It is not clear to me whether the data you are describing is the same as this, or something more permissive. (For example, $(1\ 2)(2\ 3)(1\ 2)$ also has product $321$, but, would correspond to the sequence of chambers $x \lt y \lt z$, $y \lt x \lt z$, $z \lt x \lt y$, $z \lt y \lt x$, and the middle two chambers are not adjacent. I assume you don't want to permit this, but I don't see any place where you say it is not legitimate.)


However, it doesn't matter, because even reduced words for $n(n-1) \cdots 321$ are not restrictive enough to describe all straight lines through the $A\_{n-1}$ arrangement once $n \geq 9$. 


As fedja suggests, if we have a linear path from the $123\ldots n$ region to the $n (n-1) \cdots 321$ region, we can think of it as describing the positions of $n$ runners who move at constant rates, starting in order $12 \ldots (n-1) n$ and ending in the reverse order.


Similarly, a reduced word for $n(n-1) \cdots 321$ can be realized as a [simple pseduoline arrangement](http://en.wikipedia.org/wiki/Arrangement_of_lines): An arrangement of $n$ paths in $\mathbb{R}^2$ each of which divide the plane in two and only cross in pairs, with each pair only crossing once. And this dictionary is reversable: Every pseduoline arrangment where all $\binom{n}{2}$ pairs of lines cross gives us a reduced word. (You'll often hear pseudoline arrangements called "wiring diagrams" when used in this context.)


Here is the key point:



> 
> For $n \leq 8$, every simple
>  pseudoline arrangement where all pairs
>  of lines cross is realizeable by
>  actual straight lines. However, for
>  $n=9$, this is not true.
> 
> 
> 


Pseudoline arrangements which cannot be realized using straight lines are called non-stretchable. See Section 5.3 of the [Handbook of Discrete and Computational Geometry](http://books.google.com/books?id=QS6vnl8WlnQC&lpg=PA107&ots=4AJBY-sDl8&dq=stretchability%2520of%2520pseudoline%2520arrangements&pg=PA107#v=onepage&q&f=false), or Chapter 7 of Ziegler's Lectures on Polytopes, for examples of nonstretchable arrangements. Peter Shor has [shown](http://math.mit.edu/~shor/papers/Stretchability_NP-hard.pdf) that it is NP-Hard to determine whether a given pseudo-line arrangement is stretchable.


