On this page:
6.1 Follow-up from Lab
6.2 Quick Lists

6 6/19: Quick Lists

Due: 6/19, midnight by svn.

Language: Java.

You must complete this assignment with your partner and should not discuss solutions with anyone but your partner and the course staff.

This assignment is due Tuesday at midnight.

6.1 Follow-up from Lab

Complete exercises 1-11 from Lab 10.

6.2 Quick Lists

I have always been underwhelmed by the fact that the cons lists that are so useful only allow a really slow implementation of list-ref, aka get in Java parlance. Why should it take a million rests just to get the millionth element?

To combat this drawback of an otherwise lovely data structure, the list, CS researchers have devised an idea for a new implementation of lists that would let you get the millionth element in about 20 operations. If his idea works, the get operation will take roughly log(i) steps to get the ith element. The other list operations, on the other hand, would remain more or less just as efficient as before; taking the rest of a list, for example, might take a few more steps to compute, but it would be some small constant number of extra steps. In the end, we’d have something that behaves just like a list, but with a much better get operation.

Your task is to take this idea and implement it. Since you’ll be building a new kind of list data structure, let’s first agree on the list interface we want. We’ll follow Hoare’s example and dub these lists “quick lists”:

interface QList<X> {

   // Cons given element on to this list.

   QList<X> cons(X x);


   // Get the first element of this list (only defined on non-empty lists).

   X first();


   // Get the rest of this list (only defined on non-empty lists).

   QList<X> rest();


   // Get the ith element of this list

   // (only defined for lists of i+1 or more elements).

   X get(Integer i);


   // Compute the number of elements in this list.

   Integer size();



// new QEmpty<X>() should construct an empty quick list of Xs.

So now let’s talk about the idea behind this data structure.

Instead of representing a list as a “list of elements” you could do better by representing a list as a “forest of trees of elements”. Moreover, the trees will get bigger and bigger as you go deeper into the forest, and the trees will always be full, meaning if a tree has a left and right subtree, both will be the same size and full. (For the moment, don’t worry about why this makes get fast—think about that after you’ve implemented the idea.)

So here are the key invariants of a quick list:

Now that we have the invariant, let’s talk about the operations and how they both can use and maintain the invariant.

First, first. Since the list must be non-empty, we know the forest has at least one tree, so we can get the first element of the list by getting “the first” element of the tree, which for quick lists, will be the top element.

Now, length. If the forest is empty, the list has length 0. If a forest has a tree, the length of the list is the size of the tree plus the size of the rest of the forest. (It’s useful to store the size of a tree separately from a tree so that you don’t have to compute it every time you need it.)

The get method works as follows: if the index is 0, the list must be non-empty, so take the first element, i.e. the top element of the first tree in the forest. If the index is non-zero, there are two case: if it’s less than the size of the first tree, the element is in that tree, so fetch it from the first tree. If it’s larger, adjust the index, and look in the remaining trees of the forest.

To fetch an element from a tree: if the index is zero, the element is the top element. Otherwise, if the index is less than half the size, it’s on the left side; if the index is greater than half, it’s on the right. (You might do yourself a favor a develop get for full binary trees and get it working and thoroughly tested before attempting get for quick lists.)

These element-producing operations considered so far have used the invariant. Now let’s turn to the list-producing operations which must maintain it.

When an element is consed, if there at least two trees in the forest and the first two are the same size, then make a new tree out of these two and with the given element on top (notice how this tree is definitely full). Otherwise just make a new tree with one element and make it the first tree in the forest.

To take the rest of a list, there must be at least one tree in the forest (since the list is non-empty). We want to split this tree into its left and right and make these the first two trees in the forest. The element that was on top is dropped on the floor and we’re left with a representation of the rest of the list.

And that’s that. When writing your code you want to make sure the invariants are always true. Good code should make this fact obvious; bad code, not so much.

This is a nice little exercise in data structure design and implementation, and although I wish this were really my idea, I actually got it from reading a book by Chris Okasaki, who has designed a bunch of these kinds of data structures. Go forth, and may your get never be slow again.