Lecture 12


MCS 275 Spring 2022
David Dumas

Lecture 12: Recursion

Course bulletins:

  • Homework 4 due Tuesday at Noon CST
  • Anonymous feedback survey coming this week
  • Project 1 grading underway, ETA Thursday


In computer science, recursion refers to a method of solving a problem that depends on solving smaller versions of the same problem.

Usually, recursion involves a function calling itself.

Strategies using recursion

  • Divide and conquer: A problem can be split into pieces; solutions for the pieces can be combined to the full solution.
    • e.g. Mergesort
  • Decrease and conquer: Reduce a problem for a given input (e.g. n) to the answer for a slightly smaller input (e.g. n-1) and a bit of extra work.
    • e.g. Factorial


Recursive solutions are often contrasted with iterative solutions.

  • Iterative: Loops and local variables keep track of all state (work to be done, work completed, next ...)
  • Recursive: Arguments keep track of current state; call stack stores work in progress; return values send back results.

Recursive solutions can always be converted to iterative ones, often at the cost of more complex code.

Stop condition

A function that always calls itself will never finish!

Recursion must include some kind of stop condition—a case in which the function can directly return an answer instead of calling itself.

Today's examples

  • Factorial
  • Fibonacci numbers
  • Paper folding sequence


The classic first example of recursion, computing $$n! = n \times (n-1) \times \cdots \times 2 \times 1.$$

The argument to the function decreases with each subsequent call, so it eventually reaches the stop condition ($n \leq 1$).


The Fibonacci numbers are defined by $$F_0=0,\; F_1=1,\; \text{and }F_n = F_{n-1} + F_{n-2}$$

So the sequence begins $0,1,1,2,3,5,8,13,...$

The definition immediately suggests a recursive implementation.

Paper folding sequence

Paper folding sequence

Paper folding sequence

Let's use $\oplus$ to mean concatenation of binary sequences, so $0110 \oplus 11 = 011011$.

If $A$ is a binary sequence, let $\bar{A}$ denote the sequence with $0$ and $1$ switched, e.g. $\overline{11101} = 00010$

Finally, let $A^r$ denote the sequence in opposite order, e.g. $10010^r = 01001$.

$$PFS(n) = PFS(n-1) \oplus 1 \oplus \overline{PFS(n-1)^r}$$


Revision history

  • 2022-02-07 Initial publication