The next problem from Hacker Rank that I’m tackling in F# is a problem called Mansasa and the stones (catchy name). The basics of the problem are that you are given 2 numbers and a length and you have to give all of the distinct sums of those numbers of the lists that can be made up of length l. An example is if stone 1 has the value 2 and stone 2 has the value 4 and the length is 3 we get:

0, 2, 2 = sum 4 0, 2, 4 = sum 6 0, 4, 2 = sum 6 0, 4, 4 = sum 8

Note the problem states that we should start with 0 for each entry. This means that the distinct sums in this example would be 4 6 8 (in order).

My idea to solve this problem was to write a function that takes a list and a length and then gives you all permutations for that list of that length. In the example above if you passed this function 2 and 4 and a length of 2 it would give the 4 possible ways of arranging 2 and 4 shown above. Once I have this function I can simply sum the numbers, do a distinct and then format the result. Simples, or so I thought more on that later.

Now that I’ve got a great way to write tests for my code I started with the unit tests for the function:

module combinations =

module combinations = [] let ``combinations for [2;3] of length 2 should be [[2;2];[2;3];[3;2];[3;3]]``() = combinations [2;3] 2 |> should equal [[2;2];[2;3];[3;2];[3;3]] [] let ``combinations for [2;3] of length 3 should be [[2;2;2];[2;2;3];[2;3;2];[2;3;3];[3;2;2];[3;2;3];[3;3;2];[3;3;3]]``() = combinations [2;3] 3 |> should equal [[2;2;2];[2;2;3];[2;3;2];[2;3;3];[3;2;2];[3;2;3];[3;3;2];[3;3;3]] [] let ``combinations for [1;4] of length 1 should be [[1];[4]]``() = combinations [1;4] 1 |> should equal [[1];[4]]

I think it is so neat that the tests describe themselves. I was stuck on writing this function in a recursive way for a little while. I’m still finding it hard to think functionally. I put this problem to my friend Sam Owens who came up with a great solution:

let combination list length = let rec solve letters len = seq { if len = 1 then for l in letters do yield [l] else let theRest = solve letters (len-1) for l in letters do for r in theRest do yield l::r } solve list length

I thought this solution was ingenious. Before I talk through how it works let me show you the more functional version. I created this by translating Sam’s function above to use pattern matching:

let combinations list length = let rec solve lst len = match len with | 1 -> lst |> Seq.map(fun x -> [x]) | i -> lst |> Seq.map(fun l -> solve lst (len-1) |> Seq.map(fun r -> l::r)) |> Seq.collect(fun x -> x) solve list length

Both of these functions are exactly the same (kudos to Sam for coming up with this way of doing it). I think the 2nd function is much more elegantly written and more readable. It’s much terser and really shows the power of F#’s pattern matching.

This function works by the fact that it knows how to print out every combination of a list of length 1. Which is shown on the first line of the match statement. To do this it simply maps out all of the elements as a list of one item. If the function is called with a length greater than one then it calls itself with length – 1 and appends the results from that on to each element in the list passed in.

With this function written composing the function to solve the whole problem was pretty straight forward:

let findMansaNumbers steps (stone1:int) (stone2:int) = combinations [stone1; stone2] (steps-1) |> Seq.map(fun l -> Seq.sum(l)) |> Seq.distinct |> Seq.sort |> fun x -> String.Join(" ", x)

All we have to do is run the combinations function passing it a list of the two stone numbers and a length. We then sum the numbers from each combination, get a distinct list of the sums, sort them in to order and then print them out in a string where each number is separted by a space. The way that F# lends itself well to composing small building blocks together makes it a fun language to program in.

Here comes the kicker. Although this solution works, it passes the first three test cases on the hacker rank problem page it then times out. The reason for this is that the number of combinations the combinations function will return is given by the formula 2^length. This is exponential growth so for big lengths the time to compute all combinations is too long. As a short aside the fact that hackerrank run the code on their server and then time out if your solution takes too long is a good thing. It stops you from being able to brute force the problems which is one of the things that plagues project Euler in my opinion.

So it’s back to the drawing board with this problem. The code is on the repository under slow solution in the MansasaAndStones.fs file in the Hacker Rank in F# repository on github.

To solve this problem it is actually very straight forward. I was just taking completely the wrong approach. If you think of the case where you have 2 stones of values 2 and 4 and are asked to find 3 steps then you can think of this as:

0, 4, 4 = sum 8 = (0*2) + (2*4) 0, 2, 4 = sum 6 = (1*2) + (1*4) 0, 2, 2 = sum 4 = (2*2) + (0*4)

So the distinct sums are 4, 6 and 8. This is calculated by having two loops, one starting and 0 and incrementing, one starting at the number of steps – 1 (as we the first step is always 0 in the problem) and then multiplying one of the loops by the first stone and one by the second.

let findNumbers steps (stone1:int) (stone2:int) = let list1 = [0..steps-1] let list2 = [0..steps-1] |> List.rev List.zip list1 list2 |> Seq.map(fun (x, y) -> (x*stone1)+(y*stone2)) |> Seq.distinct |> Seq.sort |> fun x -> String.Join(" ", x)

The code above is the full solution to the problem. Instead of looping twice we use the zip function to produce tuples of the loop values we can then feed this into the map function which makes the code neater. After that we just have to make sure the list is distinct (as the problem states no duplicates) and then sort it (as the problem states that the values should be in order) then we have to produce a space separated string for each answer.

I really went around the houses with this problem but I did learn some cool things in F#. Solving the combinations problem was fun and helped me practice my skills of turning C# into functional F#. Once I realised I only needed two loops I was quite happy with how nice the solution is. One thing I’m learning more and more about functional programming is that you can write really expressive beautiful code. Erik Meijer would be proud!

To look at the final solution check out the ManasaAndStones module and the solution function from the github repository. As always I would be keen to hear your feedback.