Recursion is a term used to describe a function calling itself. It is an important concept in programming and doubly so in Lisp and its dialects. To understand recursion, we turn to **Scheme** - a minimalistic dialect of Lisp. Since this article assumes basic familiarity with Lisp/Scheme syntax, we'll directly jump into looking at our first code.
(define (sum-of-list my-list)
(cond
[(empty? my-list) 0]
[else (+ (first my-list) (sum-of-list (rest my-list)))]))
The above function **sum-of-list** consumes a list of numbers and produces an output which is the sum of all numbers in the list. Thus,
(sum-of-list '(1 34 5)) => 40
Note that in the above call, we use list abbreviations using the quoted syntax instead of using the *cons* syntax.
Dissecting *sum-of-list* we see that it defines a conditional where the output of an empty list, i.e. '() would evaluate to 0. This makes sense because giving it an input of a list with no elements would mean that the sum of the numbers of this list would be 0.
The interesting part comes in the *else* part of the conditional. If the list is non-empty like in our case of **(1 34 5)**, the list is broken up into two - an atomic value consisting of the first element of the list extracted by the **first** operator, and the second part is again a list consisting of everything but the first element of the list in question. This is done using the **rest** operator.
Note that the *first* and *rest* operators are the Scheme equivalents of *car* and *cdr* of Lisp.
The operation that is being carried out here is **addition** since we want the sum of the elements of the list. The first atomic element is to be added to the resultant of (sum-of-list (rest my-list)) which is by our definition a recursive call. Note that the same procedure is being called again without the first element. So, after the first pass, the operation which will be called will be
(sum-of-list '(34 5))
This function call will then in turn go through the same process, each time splitting the list into its first element and the rest of the list till it splits the list into its last element and an empty list, thereby satisfying the first conditional and returning 0. These results are then added up starting from 0 to the first element of the list in reverse order. This process in the end would return us the summation of the list in question, i.e. 40.
We now look at another problem which is commonly solved by recursion - computing the factorial of a number. Here is a recursive Scheme program for the same using the **lambda** procedure.
(define factorial
(lambda (n)
(if (= n 0) 1
(* n (factorial (- n 1))))))
In this example, using recursion we're reducing the problem to a simpler one by recursively calculating the factorial of n - 1. For the case where n equals to 0, we have a base case condition where the problem cannot be simplified further and the recursion stops. This is the same as the case of an empty list in the first example we saw while calculating the sum of a list of numbers.
Since Scheme and other purely functional languages don't give us classical iterative approaches like looping, the most natural way to solve such classes of problems boils down to recursion, where a problem is simplified into smaller pieces and a simple operation like addition or multiplication (the above two examples respectively) and a base case is identified. This base case then has a trivial solution which allows us to build a solution to the problem at hand bottom-up.
For further reference on recursion in Scheme, How to Design Programs by Felleisen et al and Concrete Abstractions: An introduction to computer science using Scheme by Hailperin et al are excellent books.
*Written by Rae* |