Intro to Recurrence Relations in Discrete Mathematics, Slides of Discrete Mathematics

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Discrete Mathematics

CS 2610

November 13, 2008

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Earlier in the semester, we saw how we could define sequences recursively or functionally.

Specifically, we learned how to take functionally- defined sequences and transform them to recursively-defined sequences.

Example: an = 2 n^ becomes

a 0 = 1 an+1 = 2 n+1^ = 2⋅ 2 n = 2an, for n ≥ 1.

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Without initial conditions, a recurrence relation defines a set, or family, of sequences.

Consider an+1 = 2an.

If a 0 = 1, an = 2n^. But if a 0 = 3, an = 3⋅ 2 n^.

These two sequences are clearly similar. This is because an+1 = 2an defines a family of sequences, an = a 0 ⋅ 2 n, for n ≥ 1.

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A recurrence relation along with initial conditions specify a single sequence. Any such sequence is a

solution to the relation.

We can check solutions using substitution.

Consider the recurrence relation an = 2an-1 - an-2.

Is an = 3n a solution for n ≥ 1? Try it out!

an = 2an-1 - an-2 = 2⋅3(n-1) – 3(n-2) = 6n – 6 -3n + 6 = 3n = an

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We can also use this method to describe the famous “Towers of Hanoi” problem…

The recurrence is an = 2an-1 + 1. Can you see why?

Let’s solve the relation for a 0 = 1…

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Finally, let’s see how we can apply recurrence relations and their solutions to a tough counting problem.

How many bitstrings of length n do not contain consecutive 0’s?

The techniques we’ve studied so far can’t solve this without ridiculous amounts of effort!

One solution is 5 -½( (1+5½)/2 )n+2^ - 5 -½( (1-5½)/2 )n+^.

We can find a more elegant and easier solution!!!