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I think list comprehensions are my favorite reason to pull out Haskell.

For example: I was recently asked this:

You have a row of 100 school lockers. For each number from 1 to 100 walk down the line of lockers starting at the beginning, and switch the state of every nth locker. All of the lockers start closed.

For example: we start off at 1 and switch all the lockers to open. (1,2,3,4,5,…) we then go to 2 and switch all of the lockers evenly divisible by 2. (2,4,6,8…) for 3 we flip every 3rd locker (3,6,9,12,…) 4 is every 4th locker. (4,8,12,16,…) etc.

In Haskell this is easily accomplished using a list comprehension.

We can observe through trial (ie: actually flipping every locker) that the lockers are flipped by their factors. (ie: 6 is flipped by 1,2,3,and 6), so we’ll write a quick comprehension to give us the factors of a number:

In this example, we get the factors of 6

haskell

[x | x

links:
[1..6]
modulo
mod
`infix`

Essentially you can read the above code as

haskell

Give me x where x is [1,2,3,4,5,6] and 6 is evenly divisible by x

and breaking it down:

haskell

Give me x[xwhere|xis

We can then realize that if you flip something an even number of times, nothing changes. This means we're looking for the number with an odd number of factors because we want the lockers that are open at the end.

The code to determine an open locker will look like this, where factors is the code we just wrote.

haskell

length factors `mod` 2 == 1

If we stick that in a list comprehension for all numbers [1..100] we have effectively filtered out all of the open lockers.

haskell

[n | n

But wait! There's more! For the low low price of \$19.95 we can realize there's a pattern in the results. The pattern happens to be perfect squares. This is because perfect squares are the only numbers with an odd number of factors!

for example: 9's factors are 1,3 and 9. While 8's factors are 1,2,4 and 8

We can now write a far more efficient list comprehension.

haskell

[x*x | x <- [1..10]]

Which can be read as: give me x*x where x is [1,2,3,4,5,6,7,8,9,10]

We can also expand this into an function using an infinite list, just in case we want to calculate how many lockers are open if we have 50081 lockers.

haskell

let lockers = [x*x | x