In this blog post, I'll attempt to explain some basic concepts of Functional Programming, using Haskell. This blog post isn't about Haskell per-se, but about one way of approaching this problem, which demonstrates the benefits of functional programming.
You can run most of these examples in ghci, by saving the contents of the example into a .hs file, loading up ghci and running :load file.hs.
Many thanks to Mattox Beckman for coming up with the programming exercise, and Junjie Ying for coming finding a better data structure for this explanation than I came up with.
The Problem
You are Hercules, about to fight the dreaded Hydra. The Hydra has 9 heads. When a head is chopped off, it spawns 8 more heads. When one of these 8 heads is cut off, each one spawns out 7 more heads. Chopping one of these spawns 6 more heads, and so on until the weakest head of the hydra will not spawn out any more heads.
Our job is to figure out how many chops Hercules needs to make in order to kill all heads of the Hydra. And no, it's not n!.
Prelude: Simple Overview Of Haskell Syntax
Before we dive into code, i'll explain a few constructs which are unique to Haskell, so it's easy for non Haskellers.
- List Creation: You can create a list / array using the : operator. This can even be done lazily to get an infinite list.
let firstArray = 0:1:[2, 3]
-- [0, 1, 2, 3]
let infiniteOnes = 1:infiniteOnes
-- [1, 1, 1, 1, 1, ........................]
-- never stops, hit ctrl-C to get your interpreter back
- Defining Function: Looks just like defining a variable, but it takes parameters. One way they are different from other languages is the ability to do pattern matching to simplify your code. Here, I define a method that sums all the elements of a list.
sumOfElements [] = 0
sumOfElements (x:xs) = x + sumOfElements xs
- More List Foo: Adding lists can be done with ++. Checking if a list is empty can be done with null. You can use replicate to create a list with the same elements over and over.
[1] ++ [3] -- [1, 3]
null [] -- True
null [1] -- False
replicate 2 3 -- [3, 3]
Choosing a data structure
Let's choose a simple data structure to represent the hydra. We'll pick an array to represent the heads of the Hydra, using the level of each head as the value. The initial state of the Hydra (with 9 level 9 heads) can be represented as follows: [9, 9, 9, 9, 9, 9, 9, 9, 9].
Chopping off a head
The whole point of functional programming is to build small functions and compose them later. We'll build a few functions, specific to our domain, and a few more general ones to orchestrate.
Let's first build a specific function to chop off the Hydra's head. We know that chopping off one level 9 head should result in 8 level 8 heads (and 8 of the original level 9 heads). This is represented as [8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9]
Let's build the chop function. It takes a single argument, and the current levels of the all live heads. It will return the state of the heads after chopping the first one.
The three lines of code below map to these rules:
- If there are no heads left, just return
[] - If we find that there is a level 1 head at the start of our list, just chop it off and return the rest of the array
- If we find that there is a higher level head at the start of our list, spawn n - 1 heads in it's place
chop [] = []
chop (1:xs) = xs
chop (n:xs) = (replicate (n - 1) (n - 1)) ++ xs
----------------------------------------------------
chop [1]
-- []
chop [4]
-- [3, 3, 3]
chop [3, 3, 3]
-- [2, 2, 3, 3]
chop [9,9,9,9,9,9,9,9,9]
-- [8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9]
Repeatedly chopping heads
Our function chop is a pure function as, given some input, it always returns the same output, without any sort of side effects. Side effects is a general term for modifying inputs / IO Operations / DB Calls, and so on.
Since chop is pure function, we can safely call it over and over. Let's build a list where each element is the result of chopping the previous element.
repeatedlyChop heads = heads:repeatedlyChop (chop heads)
----------------------------------------------------------
repeatedlyChop [3]
-- [[3],[2,2],[1,2],[2],[1], [], [], [] ...]
-- this is an infinite list
This paradigm is so common, that we have a functional construct that does this: iterate. We can replace the above code with the following:
repeatedlyChop heads = iterate chop heads
Truncate that infinite list
Great, we now have built a list of all the states the Hydra is in while Hercules is busy chopping away at it. However, this list goes on forever (we never put in a termination condition in the earlier code), so let's do that now.
We can use a simple empty check (null) to test if the hydra is still alive. Let's keep items as long as the Hydra is alive
takeWhileAlive (x:xs) = if null x then [] else x:(takeWhileAlive xs)
Putting the two together
repeatedlyChopTillDead heads = takeWhileAlive (repeatedlyChopTillDead heads)
----------------------------------------------------------------------------
repeatedlyChopTillDead [4]
-- [[4],[3,3,3],[2,2,3,3],[1,2,3,3],[2,3,3],[1,3,3],[3,3],[2,2,3],[1,2,3],[2,3],[1,3],[3],[2,2],[1,2],[2],[1]]
Again, these patterns are so common, that we can replace the entire thing with a single line. takeWhile keeps things in the list until the first element that doesn't match.
repeatedlyChopTillDead heads = takeWhile (not.null) (iterate chop heads)
Finishing up
Now that we have the sequence of chops needed to kill that Hydra, figuring out the number of chops is just a matter of figuring out how long the sequence is.
countOfChops heads = length (repeatedlyChopTillDead heads)
--------------------------------------------------
countOfChops [1] -- 1
countOfChops [3] -- 5
countOfChops [9,9,9,9,9,9,9,9,9] -- 986409 (this takes a while)
Extending
Now that we've solved the problem, what next? How easy is it to extend this? Let's add a new requirement: Hercules, though a half god, can only fight at most n Hydra heads at a time. If the number of Hydra heads goes above n, then hercules dies. Let's make a function willHerculesDie which takes two parameters, n and the Hydra.
Turns out, this is pretty simple. We just need to count all the heads that are alive. If the count is more than n at any point, then we return true, and Hercules dies.
willHerculesDie n heads = any (> n) (map length (repeatedlyChopTillDead heads))
----------------------------------------------------------------------------
willHerculesDie 37 [9,9,9,9,9,9,9,9,9] -- False
willHerculesDie 36 [9,9,9,9,9,9,9,9,9] -- True
So what next?
We've built a bunch of really composable functions, and we can look at each one independently to tune the system.
You can get Haskell set up with the Haskell Platform and play around with the code here.
Some great books you can check out:
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