Basic Algorithms

================ Start Lecture #8 ================


The department has asked me to make the following announcements.

  1. The prerequisites for this course are V22.0102 and V63.0120.
  2. The final exam is Monday, December 16 10:00-11:50am.

Complexity of Primitive Operations

Our complexity analysis will proceed in a somewhat unusual order. Instead of starting with the bottom (the tree methods in 2.3.1, e.g., is Internal(v)) or the top (the traversals), we will begin by analyzing some middle level procedures assuming the complexities of the low level are as we assert them to be. Then we will analyze the traversals using the middle level routines and finally we will give data structures for trees that achieve our assumed complexity for the low level.

Let's begin!

Complexity Assumptions for the Tree ADT

These will be verified later.

Middle level routines: depth and height

Definitions of depth and height.

Remark: Even our definitions are recursive!

From the recursive definition of depth, the recursive algorithm for its computation essentially writes itself.

Algorithm depth(T,v)
   if T.isRoot(v) then
      return 0
      return 1 + depth(T,T.parent(v))

The complexity is Θ(the answer), i.e. Θ(dv), where dv is the depth of v in the tree T.

Problem Set #1, Problem 3:
Rewrite depth(T,v) without using recursion.
This is quite easy. I include it in the problem set to ensure that you get practice understanding recursive definitions.

The following algorithm computes the height of a position in a tree.

Algorithm height(T,v):
   if T.isLeaf(v) then
      return 0
      for each w in T.children(v) do
      return h

Remarks on the above algorithm

  1. The loop could (perhaps should) be written as an iterator. Indeed, it is an iterator.
  2. This algorithm is not so easy to convert to non-recursive form
    It is not tail-recursive, i.e. the recursive invocation is not just at the end.
  3. To get the height of the tree, execute height(T,T.root())

Theorem: Let T be a tree with n nodes and let cv be the number of children of node v. The sum of cv over all nodes of the tree is n-1.

Proof: This is trivial! ... once you figure out what it is saying. The sum gives the total number of children in a tree. But this almost all nodes. Indeed, there is just one exception.
What is the exception?
The root.

Corollary: Computing the height of an n-node tree has time complexity Θ(n).

Proof: Look at the code.
The while loop has cv iterations, so by the theorem the total number of iterations executed is n-1.
Everything else is Θ(1) per iteration.

Do a few on the board. As mentioned above, becoming facile with recursion is vital for tree analyses.

Definition: A traversal is a systematic way of "visiting" every node in a tree.

Preorder Traversal

Visit the root and then recursively traverse each child. More formally we first give the procedure for a preorder traversal starting at any node and then define a preorder traversal of the entire tree as a preorder traversal of the root.

Algorithm preorder(T,v):
   visit node v
   for each child c of v

Algorithm preorder(T):


  1. In a preorder traversal, parents come before children (which is as it should be :-)).
  2. If you describe a book as an ordered tree, with nodes for each chapter, section, etc., then the pre-order traversal visits the nodes in the order they would appear in a table of contents.

Do a few on the board. As mentioned above, becoming facile with recursion is vital for tree analyses.

Theorem: Preorder traversal of a tree with n nodes has complexity Θ(n).

Proof: Just like height.
The nonrecursive part of each invocation takes O(1+cv)
There are n invocations and the sum of the c's is n-1.

Homework: R-2.3

Postorder Traversal

First recursively traverse each child then visit the root. More formerly

Algorithm postorder(T,v):
   for each child c of v
   visit node v

Algorithm postorder(T):

Theorem: Preorder traversal of a tree with n nodes has complexity Θ(n).

Proof: The same as for preorder.