On Recursion - Part 1

Master R

By Guangming Lang Comment

Intro

I’ve been thinking about recursion lately. It’s very powerful but it can be hard to reason about sometimes. Through this article and the following ones, hopefully I can make it easier for you to understand it and to write more recursive functions.

Find the first 3 prime numbers

Let’s start with a very simple task: find the first 3 prime numbers. How’d you do it? The intuitive way is to start with 1, and ask if 1 is a prime. The answer is no. So we move to 2 and ask if 2 is a prime, and the answer is yes. We now have our first prime number. We then move onto 3 and ask if 3 is a prime and we repeat the procedure until we find the 3rd prime number. The answer, of course, is the sequence (2, 3, 5). This way of thinking is recursion. Keep reading to find out why.

Find the first N primes that are greater than M

The task is to find the first 3 prime numbers. What’s unsaid but self-evident is that the numbers must be greater than 0 because all prime numbers are greater than 0. So we can rewrite the task: find the first 3 primes that are greater than 0. Further more, let’s be ambitious and consider its general version: find the first N primes that are greater than M.

The basic structure of any recursion involves the base case and the non-base case. In the general task, the base case is when N = 0 and the result will be the empty sequence (). The non-base case is When N > 0, and when N > 0, if M+1 is a prime, the result will be (M+1, the first N-1 primes that are greater than M+1), otherwise, (the first N primes that are greater than M+1)

You probably realized it by now we can view the task as a math function f(N, M) = (the first N primes > M) with input parameters N and M and outputs a sequence of first N primes > M. We can then translate the above recursive description into R code.

f = function(N, M) {
        if (N == 0) { # base case
                return(c())
        } else { # non-base case
                if (is_prime(M+1)) c(M+1, f(N-1, M+1)) 
                else f(N, M+1)
        }
}

is_prime = function(n) {
        if (n == 1L) FALSE
        else n == 2L || all(n %% 2L:max(2, floor(sqrt(n))) != 0)
}

The function is_prime() is taken from this stackoverflow answer, you can read the logics behind there. Alternatively, you can implement it using recursion. Consider it as today’s homework. :)

Let’s use f to find the first 3 prime numbers and first 5 primes greater than 100.

f(3, 0)
## [1] 2 3 5
f(5, 100)
## [1] 101 103 107 109 113

This article is inspired by an example described in this little pamphlet by Panicz.

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