Find the greatest product of five consecutive digits in the 1000-digit number.
open System.IO let numbers = File.ReadAllLines(@"C:\TEMP\euler8.txt") |> Seq.concat |> Seq.map (fun c -> int32(c.ToString())) let CalcProduct numbers = numbers |> Seq.fold (fun acc n -> acc * n) 1 let maxProduct = numbers |> Seq.windowed(5) |> Seq.map (fun n -> CalcProduct n) |> Seq.max
To get the 1000 digits into the program, I copied and pasted the digits into a text file and saved it to C:\TEMP\euler8.txt.
Here you see an example of how you can use the File.ReadAllLines static method to read the contents of a text into a string array in F#. However, a string array doesn’t really help us here, so I used the Seq.concat function to merge them into a char array because mapping each char into the integer it represents.
Two things to note here:
1) why does Seq.concat return a char array as opposed to a string?
Because a string can be treated as a char in the same way that a Dictionary<T, V> can be treated as an IEnumerable<KeyValuePair<T, V> > and iterated as such when passed into a method which takes a IEnumerable<KeyValuePair<T, V> > as argument.
Looking at the signature of the Seq.concat function, it takes a seq<‘T list> (a sequence of lists of T, i.e. IEnumerable<IEnumerable<T> >) as argument and therefore the string is interpreted as IEnumerable<IEnumerable<char> > and concatenated into IEnumerable<char>.
The same is applied to all the List/Array/Seq functions when used on a string.
2) why is ToString() needed when casting a char to int32?
Because int32(char) will give you the unicode value of that char! For example,
int32('c') // this is legal and returns 99 int32("c") // this is illegal int32('1') // this returns 49 int32("1") // this returns 1
Moving on, the next line defines a helper function which takes a list of numbers and multiply them together, i.e [x; y; z] => x * y * z. I have done this with the Seq.fold function which applies a function to each element of the list whilst keeping an accumulator throughout, similar to the Enumerable.Aggregate method in Linq. In this particular case, the accumulator starts off at 1, and iteratively multiplied by the elements in the list:
let CalcProduct numbers = numbers |> Seq.fold (fun acc n -> acc * n) 1
It’s worth noting that this function will return 1 if used on an empty array which is not the correct behaviour, but for the purpose of solving the problem at hand it can be safely assumed that this will never happen.
Finally, in the last part of the solution, you will notice yet another new function Seq.windowed, which iterates through the elements in the list yielding a sliding windows of 5 elements:
numbers |> Seq.windowed(5);; val it : seq<int32 > = seq [[|7; 3; 1; 6; 7|]; [|3; 1; 6; 7; 1|]; [|1; 6; 7; 1; 7|]; [|6; 7; 1; 7; 6|]; ...]
I then calculate the product for each of the 5 digit arrays and find the greatest product:
|> Seq.map (fun n -> CalcProduct n) |> Seq.max
7 thoughts on “Project Euler – Problem 8 Solution”
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solution without high level functions:
let find5 list =
let rec loop list’ greatest =
match list’ with
|n1::n2::n3::n4::n5::tl -> loop (n2::n3::n4::n5::tl) (max greatest (n1*n2*n3*n4*n5))
|_ -> greatest
loop list 0
let numbers =
|> Seq.map (fun c -> int32(c.ToString()))
let p8 = find5 numbers
When I went to this problem it asks for the product of the 13 adjacent digits with the largest product. My solution is nearly identical to yours (I used seq.reduce vs seq.fold), and presumably yields the same answer (9x9x8x7x9 = 40824)
I change the argument to “windowed” to 13 and get
([|9; 7; 8; 1; 7; 9; 7; 7; 8; 4; 6; 1; 7|], 2091059712)
But Euler says this is not correct. Weird.
Your solution is probably correct, but with 13 digits it likely overflowed. Try using int64 instead, and a good practice when working on Project Euler problems is to “open Checked” so you don’t get surprised by silent overflows/underflows.
Ah, you’re right, of course. Total rookie mistake I should have thought of on my own.
no worries! I’ve been stung by overflows so many times in the past, coding challenges love to throw that in as a surprise ;-)