# Running the Greene-Nijenhuis Algorithm Backwards

This question is crossposted from math.stackexchange.com, where it remains unanswered.

Let $Y$ be a Young tableau of shape $\lambda:=(\lambda_1,\ldots,\lambda_n)$, where $\lambda_1\geq\lambda_2\geq\ldots\geq\lambda_n$ and $N:=|\lambda|:=\sum_i\lambda_i$. I'll be using the English notation for Young tableaux. Then the hook of a box $(i,j)$ is every box straight below and straight to the right of $(i,j)$, including $(i,j)$ itself.

The Greene-Nijenhuis algorithm generates a uniformly random Young tableau of the above shape as follows:

1) Set $k=N$.

2) Uniformly pick a random unoccupied cell in $Y$.

3) Do a hook walk algorithm till you reach the edge, and place $k$ there.

4) $k\leftarrow k-1$. Repeat steps 2-4 until $k=0$.

I'm wondering if you can generate a uniformly random Young tableau by running the Greene-Nijenhuis algorithm backwards? So set $k=1$, uniformly pick an unnocupied cell, and now perform a hook walk algorithm, except now the hooks are oriented left and up: the hook of $(i,j)$ is now $(i,j)$ and every box straight above and straight to the left. Obviously 1 ends up in the top-left corner. Now do the same for $k=2$, etc.

The product of the reverse hooks does not equal the product of the usual hooks, so it seems that a modification is required, some sort of re-weighting perhaps.

With no modification this procedure indeed does not give a uniform distribution, the minimal example is $\lambda=(3,2)$, as can be easily verified.
One difficulty is that the distribution of cell $k$ given the first $k-1$ cells depends on the number of skew tableaux of certain shapes, and these numbers are much less well behaved then for Young tableaux. For example in the case of a diagonal strip of width 2, there are formulae (see e.g. arXiv:0709.0498) but the probabilities these give do not suggest that there is any simple modification that will work.