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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: 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, or Chapter 7 of Ziegler's Lectures on Polytopes, for examples of nonstretchable arrangements. Peter Shor has shown that it is NP-Hard to determine whether a given pseudo-line arrangement is stretchable.

1

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 $x However, it doesn't matter, because even reduced words for$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: 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, or Chapter 7 of Ziegler's Lectures on Polytopes, for examples of nonstretchable arrangements. Peter Shor has shown that it is NP-Hard to determine whether a given pseudo-line arrangement is stretchable.