The next Project Euler problem I want to tackle is number 22:

What is the total of all the name scores in the file of first names?

The file in question contains data that looks like this:

"MARY","PATRICIA","LINDA","BARBARA",...

It's a nice list but I will need to clean it up to get it into a form that is easy to process. To that end I define a couple of helper functions. These allow me to use .NET's Regex class in a more functional manner:

open System.Text.RegularExpressions

let split pattern input =

    (new Regex(pattern)).Split(input)

let replace pattern (replacement:string) input =

    (new Regex(pattern)).Replace(input, replacement)

The value for a name is computed by adding the values for each letter ( a = 1, b = 2, c = 3, etc.). This is pretty easy in F# because a string can be treated as a sequence of characters. Thus the value of a name can be computed like this:

let value (name:string) =

    name |> Seq.map (fun c -> (int)c - (int)'A' + 1) |> Seq.sum

All the names in the file are already in uppercase so we don't have to do any normalisation there.

Once all the names have a value I need to compute a score for each name. This score is the name's value multiplied by its position in the alphabetically sorted list.

So, in words, here's what we do:

1. Read in the names.
2. Sanitize them by getting rid of the extraneous quotes.
3. Split the resulting string into an array of name strings.
4. Sort the array.
5. Compute the score for each name.
6. Sum the resulting scores.
7. Display the result.

Using the previously defined functions this algorithm translates directly into F# code:

open System.IO

File.ReadAllText(@"e:\Project Euler\names.txt")

    |> replace "\"" ""

    |> split ","

    |> Array.sort

    |> Array.mapi (fun i name -> (i + 1)*(value name))

    |> Array.sum

    |> printfn "total = %A"

The twenty-first Project Euler problem is:

Evaluate the sum of all amicable pairs under 10000.

An amicable pair is two numbers such that the sum of the proper divisors of the first number adds up to the second and vice versa. Each member of the pair is considered to be an amicable number. As always Wikipedia has a good article on the subject.

Following the suggestion from the statement of the problem I define a function d(n) that computes the sum of n's divisors:

let d n =

    {1..n/2} |> Seq.filter (fun i -> n%i = 0) |> Seq.sum

Now I need to test whether a given number is amicable:

let amicable n =

    let sum = d n

    (sum <> n) && (d sum = n)

Note that we exclude so called perfect numbers where the sum of n's divisors is n.

Finally, I can compute the sum of all the amicable numbers under 10,000 as follows:

{1..10000} |> Seq.filter (fun i -> amicable i) |> Seq.sum |> printfn "total = %A"

Numerically, the next Project Euler problem is number 20:

Find the sum of digits in 100!

Hmmm... what to do, what to do? Well, as always, lets start with the helper functions. 100! is a very big number. Thus we will have to use F#'s BigInt type. Here's a factorial function:

let factorial n = Seq.fold ( * ) 1I [1I .. n]

Of course, F# already has a factorial function for BigInts but since this is a learning exercise I felt that I had to write my own.

Next I want to turn a big number into a series of digits. This is the function I used for that purpose in my solution to problem number 16:

let rec digits n =

    seq {

        if n > 0I then

            yield n%10I

            yield! digits (n/10I)

    }

Finally we can put these together to create a solution:

factorial 100I |> digits |> Seq.sum |> printfn "sum = %A"

On my computer I get the answer in less time than it takes to threaten the monitor with my fist.

The next Project Euler problem on my list is number 19:

How many Sundays fell on the first of the month during the twentieth century?

For this problem it seemed to me that the easiest attack would be to create a sequence containing all the first days of each month in the given range and then filter it by testing whether the day in question was a Sunday. To make this work I would need a way of calculating the day of the week from the date. Fortunately, once again Wikipedia came to the rescue.

Following the algorithm from this article we define a few helper functions. To calculate the century we say:

let century y = 2 * (3 - (y/100)%4)

To determine whether we have a leap year:

let leap y =

    if y%400 = 0

        then true

        else

            if y%100 = 0

                then false

                else

                    if y%4 = 0

                        then true

                        else false

For the month table I used a pattern match expression which takes the month number and whether it's a leap year or not:

let month l m =

    match m with

    |  0 -> if l then 6 else 0

    |  1 -> if l then 2 else 3

    |  2 -> 3

    |  3 -> 6

    |  4 -> 1

    |  5 -> 4

    |  6 -> 6

    |  7 -> 2

    |  8 -> 5

    |  9 -> 0

    | 10 -> 3

    | 11 -> 5

Note that this match generates a warning to the effect that there are cases which may not be covered:

      match m with
  ----------^
stdin(19,11): warning FS0025: Incomplete pattern matches on this expression. For example, the value '12' may indicate a case not covered by the pattern(s).

In this case I am controlling all the inputs and I know that I would never make a mistake so I decided not to worry about it.

Next I needed to compute the year value:

let year y = y%100 + (y%100)/4

With these helpers, given a date tuple in the form (y, m, d) the day of the week it falls on is given by:

let day (y, m, d) = (century y + year y + month (leap y) m + d)%7

In this scheme Sunday is day 0.

The list of dates I want to examine are given by this sequence:

let dates =

    seq {

        for y in 1901..2000 do

            for m in 0..11 do

                yield (y, m, 1) }

Putting the pieces together to solve the puzzle:

dates |> Seq.filter (fun x -> day x = 0) |> Seq.length |> printfn "count = %A"

Returning to my irregularly scheduled series on Project Euler solutions I'd like to look at problem 18:

Find the maximum sum travelling from the top of the triangle to the base.

Here is a little example:

1

3 2

4 1 6

I have marked the maximum sum with red numbers. For a small triangle of numbers like this you could easily brute force the solution and just check all paths from the top to the bottom. For a large triangle, that would be too labour intensive. A better solution, I think, is to coalesce the rows of the triangle working from the bottom up. So, in this simple example, you would first replace the second row with the maximum value possible by adding a number in that row to the two numbers below it in the triangle. In this case, 3 + 4 > 3 + 1 so you would replace 3 by 7 and 2 + 6 > 2 + 1 so you'd replace 2 by 8. That would reduce the triangle to this:

1

7 8

Performing the same step on the reduced triangle would give you the final answer (since 1 + 8 > 1 + 7).

9

To solve this problem I put the data into an array of arrays:

let data =

    [|

    [|04; 62; 98; 27; 23; 09; 70; 98; 73; 93; 38; 53; 60; 04; 23|];

    [|63; 66; 04; 68; 89; 53; 67; 30; 73; 16; 69; 87; 40; 31|];

    [|91; 71; 52; 38; 17; 14; 91; 43; 58; 50; 27; 29; 48|];

    [|70; 11; 33; 28; 77; 73; 17; 78; 39; 68; 17; 57|];

    [|53; 71; 44; 65; 25; 43; 91; 52; 97; 51; 14|];

    [|41; 48; 72; 33; 47; 32; 37; 16; 94; 29|];

    [|41; 41; 26; 56; 83; 40; 80; 70; 33|];

    [|99; 65; 04; 28; 06; 16; 70; 92|];

    [|88; 02; 77; 73; 07; 63; 67|];

    [|19; 01; 23; 75; 03; 34|];

    [|20; 04; 82; 47; 65|];

    [|18; 35; 87; 10|];

    [|17; 47; 82|];

    [|95; 64|];

    [|75|]

    |]

You will notice that I inverted the triangle so that the widest part is at the bottom. I wanted my algorithm to proceed from the start of the array to end and not vice versa.

Next I defined a helper function:

let max x y = if x < y then y else x

Now I want a function that will coalesce two arrays into one:

let coalesce (a:int[]) (b:int[]) =

    [| for i in 0..b.Length-1 -> max (a.[i]+b.[i]) (a.[i+1]+b.[i]) |]

So, the idea here is that two one-dimensional arrays are passed in. The first one is assumed to be one longer than the second (in production code you'd probably want to enforce the precondition). This function creates a new one-dimensional array with the same length as the second input array. Each entry in the new array is the max of the entry in the second array with either of the numbers above it in the first array.

Finally, I want to apply the coalesce operation to the entire data array:

data |> Array.fold coalesce (Array.zeroCreate (data.[0].Length+1)) |> printfn "%A"

I have created an array with length one greater than the maximum line in the data array and I'm using that as the state accumulator passed to the fold function. At each step in the fold operation it will be replaced by an array one entry smaller until, when it's done, there will just be an array with a single element which will be the final answer.

I should note that one potential problem with this solution arises if you want to know the actual path that led to the answer. 

Many years ago, in the late 1980's when I lived in Vancouver, I took night school courses to learn conversational Cantonese. I stopped when I moved to California in 1990 and all the vocabulary I acquired then has been slowly eroding. However, my previous post on spelling out numbers reminded me of how one says the numbers in Cantonese. Here are the numbers with their Chinese character and Cantonese pronunciation:

one	1	yāt	一
two	2	yih	二
three	3	sàam	三
four	4	sei	四
five	5	nǵh	五
six	6	luhk	六
seven	7	chat	七
eight	8	baat	八
nine	9	gáu	九
ten	10	sahp	十

Subsequent numbers are built logically from these. For example, eleven (11) is 十一, twenty-one (21) is 二十一 and thirty-one (31) is 三十一. In Cantonese you would pronounce 三十一 as "sàam sahp yāt".


A program equivalent to the one that counted the number of letters in the English spelling of numbers would probably count the strokes of the Chinese numbers. It would have a lot fewer special cases than the program I presented earlier.


I use the Yale Romanization for Cantonese since that's what I was taught long back. There is a lot of information about Chinese Numerals online.