This page can be used for writing up results in more or less finished form. See the talk page for discussion and more sketchy/incomplete ideas.

1. Oriented swap process (OSP)
2. Last passage percolation (LPP)
3. Point-to-line and line-to-line LPP
4. Particle and block finishing times
5. The absorbing time

add definitions here

\begin{align*} T(a,b;c,d) &= \max_{\pi \in \Pi_{(a,b)\to(c,d)}} \sum_{(i,j)\in \pi} X_{i,j} \quad \textrm{(last passage percolation time from }(a,b)\textrm{ to }(c,d) \textrm{)} \\ \mathbf{V}^n &= (V^n_1,\ldots,V^n_{n-1}) \quad \qquad \qquad \qquad \qquad \qquad \qquad \textrm{(block finishing times)} \\ \mathbf{U}^n &= (T(1,1; n,1), T(1,1; n-1,2), \ldots, T(1,1; 1,n)) \qquad \textrm{(point to line last passage percolation)} \\ \mathbf{W}^n &= (T(1,1; n,1), T(2,1; n,2), \ldots, T(n,1; n,n)) \qquad \textrm{(line to line last passage percolation)} \end{align*}

Theorem. $ \mathbf{U}^n \stackrel{D}{=} \mathbf{W}^n $

Proof. add this

Conjecture. $ \mathbf{V}^n \stackrel{D}{=} \mathbf{W}^n $

Proposition. Let $f_n(x_1,\ldots,x_n)$ denote the joint density function of the random vector $\mathbf{U}^n$. Then we have the recurrence relation $$ f_{n+1}(x_1,\ldots,x_{n+1}) = \exp\left( - \sum_{k=1}^{n+1} x_k \right) \int_0^{x_1 \wedge x_2} \int_0^{x_2 \wedge x_3} \ldots \int_0^{x_n \wedge x_{n+1}} f_n(y_1,\ldots,y_n) \exp\left( \sum_{k=1}^{n+1} y_{k-1} \vee y_k \right) dy_n \, dy_{n-1} \ldots dy_2 \, dy_1 $$

add this

add an explanation of the algorithm for computing the joint density of the finishing times

Weak version of the main conjecture. For $n\ge2$ and $1\le k< n$, we have the equality in distribution $$ (V^n_k, V^n_{k+1}) \stackrel{D}{=} (U^n_k, U^n_{k+1}). $$

Consider as before an array $(X_{i,j)})_{i,j\ge 1}$ of iid $\operatorname{Exp}(1)$ random variables. Denote $$ X_{i,j}' = \begin{cases} 0 & \textrm{if }(i,j)=(1,1) \\ X_{i,j} & \textrm{otherwise.} \end{cases} $$ Let $T_{i,j}=T(1,1;i,j)$ denote the last passage percolation times (from the corner $(1,1)$) associated with the array $(X_{i,j})_{i,j}$. We have $$ (U^n_k,U^n_{k+1}) = (T_{k, n+1-k}, T_{k+1, n-k}). $$ Similarly, let $T'_{i,j}=T'(1,1;i,j)$ denote the last passage percolation times associated with the modified array $(X_{i,j}')_{i,j}$. We are actually only interested in the rectangular $(k+1)\times (n-k)$ subarray $$ \boldsymbol{\tau}^{(n,k)} = (T'_{i,j})_{1\le i\le k+1, 1\le j\le n-k} $$ Let $$ \mathbf{t}^{(n,k)} = (T''_{i,j})_{1\le i\le k+1, 1\le j\le n-k, (i,j)\neq (k+1,n-k)} $$ be a new array obtained from the array $\boldsymbol{\tau}^{(n,k)}$ by performing a jeu de taquin sliding operation (see figure).

Proposition. We have the equality in distribution $$ (V^n_k, V^n_{k+1}) \stackrel{D}{=} (T_{k, n+1-k}'', T_{k+1, n-k}''). $$

Proof. add this

In view of the proposition, we see that the weakened conjecture can be reformulated as follows:

Weak version of the conjecture, reformulated version. We have the equality in distribution $$ (T_{k, n+1-k}'', T_{k+1, n-k}'') \stackrel{D}{=} (T_{k, n+1-k}, T_{k+1, n-k}). $$

It can be checked that it is not the case that $T_{i,j}'' \stackrel{D}{=} T_{i,j}$ for all $i,j$.