Haskell: Creating a sliding window over a collection
Date: 2012-02-28 00:21:59

A couple of years ago when I was playing around with F# I came across the +++<cite>+++http://msdn.microsoft.com/en-us/library/ee340420.aspx[Seq.windowed]+++</cite>+++ function which allows you to http://www.markhneedham.com/blog/2010/01/14/f-refactoring-to-sequencefor-expressions/[create a sliding window of a specific size over a collection].

Taking an example from the F# documentation page:

[source,ocaml]
----

let seqNumbers = [ 1.0; 1.5; 2.0; 1.5; 1.0; 1.5 ] :> seq<float>
let seqWindows = Seq.windowed 3 seqNumbers
----

We end up with this:

[source,text]
----

Initial sequence:
1.0 1.5 2.0 1.5 1.0 1.5

Windows of length 3:
[|1.0; 1.5; 2.0|] [|1.5; 2.0; 1.5|] [|2.0; 1.5; 1.0|] [|1.5; 1.0; 1.5|]
----

Problem 8 of Project Euler is defined like so:

____
Find the greatest product of five consecutive digits in the 1000-digit number. 73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450
____

In order to do that I need to convert that number into arrays of size 5 which first involves converting the number into an array of the individual digits:

[source,haskell]
----

charToArray :: [Char] -> [Int]
charToArray [] = []
charToArray (x:xs) = (read [x]::Int) : (charToArray xs)							
number = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
----

As far as I know there isn't a +++<cite>+++windowed+++</cite>+++ function in Haskell so I wrote the following one to get the arrays of 5 consecutive elements:

[source,haskell]
----

windowed :: Int -> [a] -> [[a]]
windowed size ls =
	(case ls of
		[] -> []
		x:xs ->
			if length ls >= size then
				(take size ls) : windowed size xs
			else windowed size xs)
----

We can then use it like this:

[source,haskell]
----

*Main> maximum (map (\ls -> foldl (*) 1 ls) (windowed 5 (charToArray number)))
40824
----