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.. avmetadata::
   :title: An Introduction to Recurrent Relations
   :author: Cliff Shaffer
   :institution: Virginia Tech
   :requires: summation
   :satisfies: recurrence
   :topic: Math Background
   :keyword: Recurrence Relations
   :naturallanguage: en
   :programminglanguage: N/A
   :description: Review of basic material on solving recurrence relations needed as background for a Data Structures and Algorithms course.

An Introduction to Recurrence Relations
=======================================

Recurrence Relations
--------------------

The running time for a recursive algorithm is most easily expressed by
a recursive expression because the total time for the recursive
algorithm includes the time to run the recursive
call(s).
A :term:`recurrence relation` defines a function by means of an
expression that includes one or more (smaller) instances of itself.
A classic example is the recursive definition for the
factorial function:

.. math::

   n! = (n-1)! \cdot n\ \mbox{for}\ n>1; \quad 1! = 0! = 1.

Another standard example of a recurrence is the Fibonacci
sequence:

   .. math::

      \mbox{Fib}(n) = \mbox{Fib}(n-1) + \mbox{Fib}(n-2)\ \mbox{for}\ n>2;
      \quad\mbox{Fib}(1) = \mbox{Fib}(2) = 1.

From this definition, the first seven numbers of the
Fibonacci sequence are

.. math::

   1, 1, 2, 3, 5, 8,\ \mbox{and}\ 13.

Notice that this definition contains two parts: the general
definition for :math:`\mbox{Fib}(n)` and the base cases for
:math:`\mbox{Fib}(1)` and :math:`\mbox{Fib}(2)`.
Likewise, the definition for factorial contains a recursive part and
base cases.

Recurrence relations are often used to model the cost of recursive
functions.
For example, the number of multiplications required by a recursive
version of the factorial function for an input of size
:math:`n` will be zero when :math:`n = 0` or :math:`n = 1` (the base
cases), and it will be one plus the cost of calling ``fact`` on a
value of :math:`n-1`.
This can be defined using the following recurrence:

.. math::

   \mathbf{T}(n) = \mathbf{T}(n-1) + 1\ \mbox{for}\ n>1;
   \quad \mathbf{T}(0) = \mathbf{T}(1) = 0.

As with summations, we typically wish to replace the recurrence
relation with a closed-form solution.
One approach is to expand the recurrence by replacing any
occurrences of :math:`\mathbf{T}` on the right-hand side with its
definition.

.. inlineav:: LinearRecurrencesCON ss
   :long_name: Linear Recurrences Slideshow
   :links: AV/Background/LinearRecurrencesCON.css
   :scripts: AV/Background/LinearRecurrencesCON.js
   :output: show
   :keyword: Recurrence Relations


A slightly more complicated recurrence is

.. math::

   \mathbf{T}(n) = \mathbf{T}(n-1) + n; \quad \mathbf{T}(1) = 1.

Again, we will use expansion to help us find a closed form solution.

.. inlineav:: LinearRecurrencesNCON ss
   :long_name: Linear Recurrences Slideshow (n)
   :links: AV/Background/LinearRecurrencesNCON.css
   :scripts: AV/Background/LinearRecurrencesNCON.js
   :output: show
   :keyword: Recurrence Relations