Project Euler in F# – Problem 22
Unlike pure functional programming languages, F# is a multi-paradigm language, so you can mix object-oriented code with functional one, mainly to exploit the features of the .Net framework.
In Project Euler Problem 22 we are asked to open a file (imperative paradigm) containing a long list of names, sort them and perform some calculation on the characters composing the text.
Here is the description of the exercise:
Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.
For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 × 53 = 49714.
What is the total of all the name scores in the file?
In order to open the given file, we rely on the IO functions provided by the .Net Base Class Library (BCL), that we can access thanks the open statement at line 2 of the following code:
#light
open System.IO
let names = File.ReadAllText("names.txt").Split([|','|]) |> Seq.to_list
let couples = names |> List.map (fun x -> x.Replace("\"", "")) |> List.sort compare |> List.zip [1 .. names.Length]
let score (pos, str) =
let value = str |> Seq.map (fun x -> (1 + Char.code x - Char.code 'A')) |> Seq.fold1 (+)
value * pos
let answer = couples |> Seq.map score |> Seq.fold1 (+)
At line 4 we read the whole content of the names.txt file, split it on the "," (comma) characters and store the result in a list called names.
Since the names written in the file are enclosed by double quotes, we have to remove all of them with the Replace function before we can alphabetically sort the list.
When names is sorted, we combine (zip) it with another list containing the position of each element inside the sequence itself, i.e. the numbers from 1 to the length of the names list.
The result is a sequence of tuples (position, name) such as (938, "COLIN").
At line 7 we define the main function of this exercise, called score. It takes a couple (position, name) and returns the name score, which is computed multiplying the alphabetical value of the name by its position.
The value of a name is obtained by summing the position in the alphabet of each character composing the name itself.
The Char.code function converts the value of a Unicode character to an integer, so we just need to subtract from it the code of the first letter of the alphabet ("A") to get its alphabetical position.
Please notice that using "A" as the first element works because in the input file all names are written with uppercase characters only.
Then we multiply the alphabetical value of the name by its position to return the score of the couple passed as argument.
The exercise asks us to sum the scores of all elements in the input file, so we map the score function and sum the results with fold1 (+), getting the correct answer.
Project Euler in F# – Problem 5 (alternative solution)
There was some buzz concerning my last post on Project Euler Problem 5, since the solution presented implemented a naive brute-force algorithm and its performance was very poor.
In my mind it was only meant to be an introductory article devoted to presenting a first solution to a common problem, the least common multiple (LCM) of a set of numbers, and I planned to write a second post (the one you are reading now) to show how a "smart" algorithm could lead to outstanding performance.
Let's read the text of the exercise another time:
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest number that is evenly divisible by all of the numbers from 1 to 20?
As we already explained, the exercise just asks us to find the least common multiple of all numbers from 1 to 20, and we can compute it by using the greatest common divisor of these numbers.
In fact, we have that:
lcm (a,b) = a * b / gcd (a,b)
In order to compute the LCM of the numbers from 1 to 20, we start computing the LCM of the first two numbers (1 and 2) and then we go on computing the LCM of the result and the third element of the sequence. We go on this way until all elements have been included in the computation.
The application of the same function to all elements of a sequence while keeping an "accumulator" is called folding, and it is a common practice in functional programming.
Let's look at the code:
#light open Microsoft.FSharp.Math.BigInt // Greater Common Divisor let rec gcd a b = match b with | b when b = 0I -> a | b -> gcd b (a % b) // Least Common Multiple let lcm a b = a * b / (gcd a b) let answer = Seq.fold1 lcm [1I .. 20I]
At line 11 we defined the least common multiple as explained above, while at line 5 we have the implementation of the Euclidean algorithm for the Greatest Common Divisor.
Then we simply fold the lcm function on the sequence [1 .. 20] and we get the output.
Do you remember how much time our "optimized" algorithm in the last article takes to perform the same calculation? 25 seconds.
The solution presented here only needs 15 milliseconds!


