I’m back again with another post in my Yet Another Project Euler Series. As a reminder, my approach to these problems is to try to find the most beautiful solution that I can using F#. While performance is an important, I’m looking for the most elegant solutions that I can find.

Project Euler problem ten is another dealing prime numbers.

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

As you can guess, there’s not too much to this problem. In fact, we’ll declare just one simple supporting function.

Using the above function and our existing prime number generator, our solution is short and simple.

|> Seq.take_while (lessThan 2000000L)

|> Seq.sum

I feel a little guilty that this post is so short, but that’s the benefit of spending two whole blog posts working out a prime number generator for problem seven.

In Visual Basic .NET, there are several cases in which the statement completion list will present the user with a list of values rather than the standard completion set. Most often, this occurs when assigning to a variable of one of the common System.Drawing types, Color, Brush or Pen.

At first glance, the screenshot above might seem as if the Visual Basic IDE has hard-coded a set of values into IntelliSense, but that’s not the case. In fact, this is caused by a seldom-used feature of XML Documentation that is supported by Visual Basic .NET, but isn’t currently supported by C#^{1}. By cracking open the the XML Documentation file for System.Drawing.dll (located at C:\Windows\Microsoft.NET\Framework\v2.0.50727\en\System.Drawing.xml on my machine), we’ll see a curious XML tag on the System.Drawing.Color definition.

<summary>Represents an ARGB (alpha, red, green, blue) color.</summary>

<filterpriority>1</filterpriority>

<completionlist cref="T:System.Drawing.Color" />

</member>

The highlighted completionlist tag above is used by Visual Basic to populate the completion list with the public shared^{2} fields and properties from the specified class or module. In this particular case, the XML documentation causes Visual Basic to populate the completion list with the public shared properties of System.Drawing.Color.

Don’t believe me? Just comment out the System.Drawing.Color completionlist tag above, save and restart Visual Studio to see how this influences the statement completion list.

Many of you are probably thinking, “so, is this just a nifty implementation detail, or is something I can actually use?” The answer is, yes, this something you can use today to customize Visual Basic’s statement completion. The code below demonstrates how the completionlist tag can be used.

Public Class Operation

Private ReadOnly _execute As Func(Of Integer, Integer, Integer)

Public Sub New(ByVal execute As Func(Of Integer, Integer, Integer))

_execute = execute

End Sub

Public Function Execute(ByVal arg1 As Integer, ByVal arg2 As Integer) As Integer

Return _execute(arg1, arg2)

End Function

End Class

Public NotInheritable Class CommonOperations

Public Shared ReadOnly Add = New Operation(Function(x, y) x + y)

Public Shared ReadOnly Subtract = New Operation(Function(x, y) x - y)

Public Shared ReadOnly Multiply = New Operation(Function(x, y) x * y)

Public Shared ReadOnly Divide = New Operation(Function(x, y) x / y)

End Class

Visual Basic will automatically pick up the completionlist tag in the code above and use it to populate the completion list like so.

While a bit limited, it’s pretty easy to customize the statement completion list experience for Visual Basic to make certain types of APIs more discoverable. It’s as simple as a single XML tag.

^{1}Kevin Pilch-Bisson (C# IDE Developer Lead and swell guy) has a clever idea for supporting the completionlist tag in the C# statement completion list while staying true to the C# IntelliSense model.^{2}The VB "Shared" keyword = static in C#.

Welcome back F# junkies! In this installment of YAPES^{1}, we’re tackling Project Euler problem nine.

A Pythagorean triplet is a set of three natural numbers,a<b<c, for which,

a^{2}+b^{2}=c^{2}

For example, 3^{2}+ 4^{2}= 9 + 16 = 25 = 5^{2}.

There exists exactly one Pythagorean triplet for whicha+b+c= 1000. Find the productabc.

Like we’ve done before, we’ll break this problem down into steps:

- Generate a stream of Pythagorean triples.
- Find the Pythagorean triple whose sum is equal to 1000.
- Calculate the product of that triple.

Sound good?

The first order of business is to determine how to generate Pythagorean triples. It turns out that there are many ways to do this. However, to keep it simple, we will use Euclid’s classic formula:

Given a pair of positive integersmandnwithm>n

a= 2mn,b=m^{2}–n^{2},c=m^{2}+n^{2}

Truthfully, this formula doesn’t actually produce every possible Pythagorean triple. However, since the solution is within the set that it generates, that really isn’t a problem.

**NeRd Note**

*every*Pythagorean triple, we would have to introduce an additional parameter,

*k*, into the formula.

a= 2kmn,b=k(m^{2}–n^{2}),c=k(m^{2}+n^{2})

First, we’ll declare an F# function that calculates a Pythagorean triple using Euclid’s formula.

let a = 2*m*n

let b = m*m - n*n

let c = m*m + n*n

a,b,c

Using the above function, it’s easy to produce a stream of Pythagorean triples. Our strategy will be to start with the pair, *m* = 2 and *n* = 1. To generate the next pair, we’ll first check if *n* + 1 is less than *m*. If it is, we’ll use *m* and *n* + 1 as the next pair, otherwise we’ll use *m* + 1 and 1. Using Seq.unfold, we can use this algorithm to produce an infinite stream of Pythagorean triples:

Seq.unfold (fun (m,n) ->

let res = getTriple m n

let next = if n+1 < m then m,n+1 else m+1,1

Some(res, next))

(2,1)

With our generator in place, we can easily solve the problem.

|> Seq.find (fun (a,b,c) –> a+b+c = 1000)

|> (fun (a,b,c) –> a*b*c)

While the above solution works well, we can abstract further to produce a more elegant solution. First, we’ll declare three functions to handle the three operations that we’re performing explicitly, *sum*, *product* and *equals*.

let product (a,b,c) = a*b*c

let equals x y = x = y

Now, we can restate our solution using these functions with a bit of function composition to produce a truly beautiful solution.

|> Seq.find (sum >> equals 1000)

|> product

As we’ve seen once before, function composition is beautiful.

^{1}Yet Another Project Euler Series

Welcome coding ninjas!^{1} Continuing with Yet Another Project Euler Series (YAPES™), we come upon problem eight.

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

Being one of the earlier Project Euler problems, this is pretty trivial to solve. We’ll approach the problem by first breaking it into a set of simple steps:

- Transform the 1000-digit number into a list of digits.
- Transform the list of digits into a list of quintuplets—that is, a list containing tuples of each five consecutive elements.
- Transform the list of quintuplets into a list containing the product of each.
- Find the maximum element in the list.

Some or all of the steps above could be combined to produce a more optimal solution. However, the beauty of F#—and functional programming in general—is the ability to easily break a problem like this into a series of data transformations that run quickly enough for most situations. When a problem is broken down into simple operations, it is easier to optimize in other ways (e.g. parallelization).

The first step is to transform the 1000-digit number into a list of digits. To facilitate working with such a large number in code, we’ll represent it as a multi-line string. Like before, we’ll break the problem of converting a string into a list of digits into a set of simple steps:

- Transform the string into a list of chars.
- Because this was a multi-line string, some of the chars will be line breaks and whitespace. So, we’ll filter the list to include only the chars that represent digits.
- Transform the list of digit chars into a list containing the numerical value of each.

Let’s define a few library functions to to help with each step above.

/// Takes a string and produces a list of chars.

let toChars (s : string) =

s.ToCharArray() |> Array.to_list

module Char =

/// Determines whether a char represents a digit.

let isDigit (c : char) =

System.Char.IsDigit(c)

/// Converts a char representing a digit into its numerical value.

let toNumber (c : char) =

int c - int '0'

Armed with these, we can perform the 3 steps above in a declarative fashion.

"73167176531330624919225119674426574742355349194934

96983520312774506326239578318016984801869478851843

85861560789112949495459501737958331952853208805511

12540698747158523863050715693290963295227443043557

66896648950445244523161731856403098711121722383113

62229893423380308135336276614282806444486645238749

30358907296290491560440772390713810515859307960866

70172427121883998797908792274921901699720888093776

65727333001053367881220235421809751254540594752243

52584907711670556013604839586446706324415722155397

53697817977846174064955149290862569321978468622482

83972241375657056057490261407972968652414535100474

82166370484403199890008895243450658541227588666881

16427171479924442928230863465674813919123162824586

17866458359124566529476545682848912883142607690042

24219022671055626321111109370544217506941658960408

07198403850962455444362981230987879927244284909188

84580156166097919133875499200524063689912560717606

05886116467109405077541002256983155200055935729725

71636269561882670428252483600823257530420752963450"

|> String.toChars

|> List.filter Char.isDigit

|> List.map Char.toNumber

The next step is to transform the list of digits into a list of quintuplets. To achieve this, we’ll write a simple recursive function.

match l with

| x1::(x2::x3::x4::x5::_ as t) -> (x1,x2,x3,x4,x5)::(toQuintuplets t)

| _ -> []

Notice that the first pattern match above is slightly more complicated as we bind the list starting with the next element in the list to t, to make the recursion cleaner.

**NeRd Note**

let rec loop l cont =

match l with

| x1::(x2::x3::x4::x5::_ as t) -> loop t (fun l –> cont ((x1,x2,x3,x4,x5)::l))

| _ -> cont []

loop l (fun x -> x)

The last function we’ll define is a small helper to produce the product of all of the values in a quintuplet.

All of the pieces are now in place, and the steps can be combined to solve Project Euler problem eight.

|> toQuintuplets

|> List.map product

|> List.max

It’s that simple!

^{1}Or pirates.