Ninety-Nine Scala Problems is an excellent problem set to work throught if you want to sharpen your scala programming skills as well as brush up on your understanding of data structures and algorithms.

I’ve been slowly chipping away at the problems, and I found P59, constructing all height balanced binary trees for a given height, to be a nice case study on memoization. In this post I show you my initial implementation and how it does redundant work, and then show you how I used scalaz’s Memo to speed it up.

The Problem

In a height-balanced binary tree, the following property holds for every node: The height of its left subtree and the height of its right subtree are almost equal, which means their difference is not greater than one.

Write a method Tree.hbalTrees to construct height-balanced binary trees for a given height with a supplied value for the nodes. The function should generate all solutions.

scala> Tree.hbalTrees(3, "x") res0: List[Node[String]] = List(T(x T(x T(x . .) T(x . .)) T(x T(x . .) T(x . .))), T(x T(x T(x . .) T(x . .)) T(x T(x . .) .)), ...

Approach

Given a balanced tree of height n, we have the following cases for the height of its subtrees:

The left subtree has height n - 1 and the right subtree has height n - 1 The left subtree has height n - 2 and the right subtree has height n - 1 The left subtree has height n - 1 and the right subtree has height n - 2

For the base cases, we return a single node for height 0 and the three possible balanced trees for height 1:

o o o / \ / \ o , o , o o

In the recursive case we generate the trees of height n-1 and n-2. We take all possible left/right subtree pairs from these two lists, excluding pairs where both subtrees have height n-2 (because these trees are already generated in the recursive call for n-1). Here is the full implementation.

def hBalTrees [ T ]( height : Int , value : T ) : List [ Tree [ T ]] = height match { case 0 => List ( Node ( value , End , End )) case 1 => List ( Node ( value , Node ( value , End , End ), End ), Node ( value , End , Node ( value , End , End )), Node ( value , Node ( value , End , End ), Node ( value , End , End ))) case n => { val nLess1Trees = hBalTrees ( height - 1 , value ) val nLess2Trees = hBalTrees ( height - 2 , value ) val allTrees = nLess2Trees ++ nLess1Trees for { ( t1 , i1 ) <- allTrees . zipWithIndex ( t2 , i2 ) <- allTrees . zipWithIndex if (!( i1 < nLess2Trees . length && i2 < nLess2Trees . length )) } yield { Node ( value , t1 , t2 ) } } }

Duplicate work

Can you see the glaring inefficiency in the implementation? We’re calling the function on n - 1 and n - 2. Suppose we start at n = 5. We recursively solve for n = 4 and n = 3. When n = 4 we recursively solve for n = 3 (DUPLICATE) and n = 2, and so on.

The recurrence relation for this function is T(n) = T(n-1) + T(n-2) + O(n^2), which is at least as slow as exponential runtime (the detailed analysis is not important for our discussion).

To fix this we need to cache the values returned from recursive calls. Scalaz provides a trait that makes this trivial.

Scalaz - Memo

Let’s look at the definition of the Memo trait in the Scalaz library

sealed trait Memo [ @specialized ( Int ) K , @specialized ( Int , Long , Double ) V ] { def apply ( z : K => V ) : K => V }

It consumes a function from K to V and produces another function from K to V. We can use Memo to create a function from a tree height (Int) to a list of balanced trees with that height (List[Tree[T]]).

def hBalTreesMemo [ T ]( value : T ) : Int => List [ Tree [ T ]] = Memo . immutableHashMapMemo [ Int , List [ Tree [ T ]]] { case 0 => List ( Node ( value , End , End )) case 1 => List ( Node ( value , Node ( value , End , End ), End ), Node ( value , End , Node ( value , End , End )), Node ( value , Node ( value , End , End ), Node ( value , End , End ))) case n => { val nLess1Trees = hBalTreesMemo ( value )( n - 1 ) val nLess2Trees = hBalTreesMemo ( value )( n - 2 ) val allTrees = nLess2Trees ++ nLess1Trees for { ( t1 , i1 ) <- allTrees . zipWithIndex ( t2 , i2 ) <- allTrees . zipWithIndex if (!( i1 < nLess2Trees . length && i2 < nLess2Trees . length )) } yield { Node ( value , t1 , t2 ) } } }

The signature of the function was slightly changed to fit the type of Memo but everything else is exactly the same! Let’s see how it performs against the non-memoized implementation.

Benchmarking

Average execution times were gathered for tree heights 1 to 4. Due to the inherent slowness of the algorithm it wasn’t practical to test above 4 with my 2014 MacBook :P.

| Height | Original (ms) | Memoized (ms) | Speedup | |--------|---------------|---------------|---------| | 1 | 2.184389 | 7.043043 | 0.31 | | 2 | 1.546299 | 1.270925 | 1.21 | | 3 | 0.519338 | 0.484961 | 1.07 | | 4 | 27.109193 | 13.389669 | 2.02 |

The memoized version is a low slower for trees of height 1. This is because the additional cost of memoizing is much higher when the function is in its base case, which takes comparatively little time. For trees of size 4 we see a halving in runtime; a pretty solid gain.

Conclusion