Random Sampling in Transition Based Parsing

When sampling binary trees by sampling Shift-Reduce transitions, using a uniform distribution over transitions will bias certain binary tree structures over others. Rather than uniformly sampling transitions, it is necessary to sample transitions in proportion to how many remaining valid sequences have the transition of interest. This calculation is not computationally feasible for non-trivial tree sizes. Instead, use a novel dynamic programming solution that we present and call the Catalan pyramid distribution.


Shift-Reduce parsing is a useful way to linearize binary trees. The word linearize here means to represent a binary tree as a sequence of elements. In this case, each element is either a Shift or Reduce transition.

This is how Shift-Reduce parsing works. We take the leaves of our binary tree and put them on queue. We also initialize an empty stack. The transitions are defined as follows:

  • Shift - Pop an element from the queue and push it on to the stack.
  • Reduce - Pop two elements from the stack. Create a parent node (the two elements are its children). Push the parent on to the stack.

Simple enough! This is a convenient representation for binary trees for multiple reasons. For one, imagine you were tasked with predicting the most likely binary tree structure given only the values of the leaf elements. We can reduce this problem to predicting the next transition given all the previous ones.

Here is another task: randomly sample a binary tree with $N$ leaves. One way to do this would be to list all possible binary trees of $N$ leaves, and randomly choose one. The Catalan Numbers are a sequence of integers, which among other things, represent how many binary trees can be made with N leaves. Here are the first 15:

$$1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440$$

As is obvious, for any large number $N$, it's not feasible to iterate all possible binary trees and choose one. By 15 leaves we are choosing among more than 2 million options. The Catalan numbers continue to grow exponentially and by 25 leaves, there are more than 1 trillion options. There must be a better way.

There are probably many alternatives, but perhaps we can make use of the Shift-Reduce transitions that we have discussed a moment ago.

Uniformly Sampling Transitions

We can try randomly choosing a Shift or Reduce $2N-1$ times, but will find this doesn't quite give the desired outcome. For this to work at all, it's necessary to only choose a Shift or Reduce if they are valid. The word valid here means that using the transition would retain a tree that has $N$ leaves. Some transitions are obviously invalid (like Reducing in one of the first two steps, or Shifting on the final step). Otherwise, we perform the following checks:

  • Is the stack empty or only has size 1? Then we must Shift.
  • Is the queue empty? Then we must Reduce.

Since we not all transitions are not always valid at each time step (depending on the transitions we've chosen so far), there is bias in the types of trees that are created this way. Specifically, certain trees will be more likely than others. In effect, uniformly sampling Shift-Reduce transitions does not uniformly sample binary trees.

Uniformly Sampling Binary Trees

We can still sample binary trees using Shift-Reduce transitions, we need only be careful about which distribution to use when sampling the Shift-Reduce transitions at each step.

The correct distribution to use at each step can be found with this algorithm:

  1. Generate all possible trees with the current transition prefix. The transition prefix is the sequence of transitions that have been used so far.
  2. Calculate the percentage of possible trees that have a Shift at the next step. This is the probability of Shifting (and the complement is the probability of Reducing).

Do you see the problem? This algorithm is correct but requires that we generate all possible trees, which we already said was infeasible for large values of $N$. What we need is a closed form solution that returns the same value.

Catalan Pyramid Distribution

It is possible to build a lookup table that can return the correct value in constant time. The table is accessed using the current time step and the number of Reduces performed so far. It is generated like this:

$$ \begin{aligned} \text{row}_{i,0,0} &= 1 \\ \text{row}_{i,0,1} &= i + 2 \\ \text{row}_{i,n_i-1,1} &= Catalan(i+1) \\ \text{row}_{i,n_i-1,0} &= \text{row}_{i,n_i-1,1} + \text{row}_{i-1,n_i-2,1} \\ \text{row}_{i,j,0} &= \text{row}_{i-1,j,1} \\ \text{row}_{i,j,1} &= \text{row}_{i,j,0} + \text{row}_{i-1,j,1} \end{aligned} $$