Project Euler Problem 8 in F#

After two easy problems, it’s now time for Problem 8.

At first, this problem also seemed to me more complex than it really is. Let’s have a look at the question:

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

So, basically what we need to do is:

  • consider all chunks of this list of numbers that can be created by taking 5 consecutive digits from the beginning till the end (there will be n-5 chunks if n is the number of digits in the list)
  • for each chunk, multiply all the digits together (that’s n-1 products per chunk)
  • pick the value corresponding to the greatest product.

At first I have considered resolving an easier problem first, like: find the chunk whose sum is bigger and then calculate that product. Assuming that sum is cheaper to calculate than product (and it is), we would have had (n-5) * (n-1) sums and just 1 product. However you can easily find examples where the sum of the digits is the same while the product is different: e.g. 4 + 5 = 7 + 2 = 9 but 4 * 5 = 20 > 7 * 2 = 14. Therefore that was a dead end.

A more mathematically sound approach to find the chunk giving the highest product might have been to first locate the chunk with the highest sum of logarithms of each digit and then calculate the corresponding product. However, a quick test revealed that logarithm is way slower than product.

Finally, a couple of micro-optimizations that occurred to me.

  • You can skip all chunks containing at least one 0 (trivially, the product will be zero)
  • You don’t really have to multiply all the digits for each chunk (that is (n-5) * (n-1) products): once you have computed the product of the first chunk, the product of the next chunk would be the product of the previous chunk times the sixth digit divided by the first digit, and so on for subsequent chunks.

Eventually, I did not try to implement these two things, since brute force, which I am about to show you, is so quick that it’s wasn’t really worth it:

#light

let digits = [| 7; 3; 1; 6; 7; 1; 7; 6; 5; 3; 1; 3; 3; 0; 6; 2; 4; 9;
    1; 9; 2; 2; 5; 1; 1; 9; 6; 7; 4; 4; 2; 6; 5; 7; 4; 7; 4; 2; 3; 5;
    5; 3; 4; 9; 1; 9; 4; 9; 3; 4; 9; 6; 9; 8; 3; 5; 2; 0; 3; 1; 2; 7;
    7; 4; 5; 0; 6; 3; 2; 6; 2; 3; 9; 5; 7; 8; 3; 1; 8; 0; 1; 6; 9; 8;
    4; 8; 0; 1; 8; 6; 9; 4; 7; 8; 8; 5; 1; 8; 4; 3; 8; 5; 8; 6; 1; 5;
    6; 0; 7; 8; 9; 1; 1; 2; 9; 4; 9; 4; 9; 5; 4; 5; 9; 5; 0; 1; 7; 3;
    7; 9; 5; 8; 3; 3; 1; 9; 5; 2; 8; 5; 3; 2; 0; 8; 8; 0; 5; 5; 1; 1;
    1; 2; 5; 4; 0; 6; 9; 8; 7; 4; 7; 1; 5; 8; 5; 2; 3; 8; 6; 3; 0; 5;
    0; 7; 1; 5; 6; 9; 3; 2; 9; 0; 9; 6; 3; 2; 9; 5; 2; 2; 7; 4; 4; 3;
    0; 4; 3; 5; 5; 7; 6; 6; 8; 9; 6; 6; 4; 8; 9; 5; 0; 4; 4; 5; 2; 4;
    4; 5; 2; 3; 1; 6; 1; 7; 3; 1; 8; 5; 6; 4; 0; 3; 0; 9; 8; 7; 1; 1;
    1; 2; 1; 7; 2; 2; 3; 8; 3; 1; 1; 3; 6; 2; 2; 2; 9; 8; 9; 3; 4; 2;
    3; 3; 8; 0; 3; 0; 8; 1; 3; 5; 3; 3; 6; 2; 7; 6; 6; 1; 4; 2; 8; 2;
    8; 0; 6; 4; 4; 4; 4; 8; 6; 6; 4; 5; 2; 3; 8; 7; 4; 9; 3; 0; 3; 5;
    8; 9; 0; 7; 2; 9; 6; 2; 9; 0; 4; 9; 1; 5; 6; 0; 4; 4; 0; 7; 7; 2;
    3; 9; 0; 7; 1; 3; 8; 1; 0; 5; 1; 5; 8; 5; 9; 3; 0; 7; 9; 6; 0; 8;
    6; 6; 7; 0; 1; 7; 2; 4; 2; 7; 1; 2; 1; 8; 8; 3; 9; 9; 8; 7; 9; 7;
    9; 0; 8; 7; 9; 2; 2; 7; 4; 9; 2; 1; 9; 0; 1; 6; 9; 9; 7; 2; 0; 8;
    8; 8; 0; 9; 3; 7; 7; 6; 6; 5; 7; 2; 7; 3; 3; 3; 0; 0; 1; 0; 5; 3;
    3; 6; 7; 8; 8; 1; 2; 2; 0; 2; 3; 5; 4; 2; 1; 8; 0; 9; 7; 5; 1; 2;
    5; 4; 5; 4; 0; 5; 9; 4; 7; 5; 2; 2; 4; 3; 5; 2; 5; 8; 4; 9; 0; 7;
    7; 1; 1; 6; 7; 0; 5; 5; 6; 0; 1; 3; 6; 0; 4; 8; 3; 9; 5; 8; 6; 4;
    4; 6; 7; 0; 6; 3; 2; 4; 4; 1; 5; 7; 2; 2; 1; 5; 5; 3; 9; 7; 5; 3;
    6; 9; 7; 8; 1; 7; 9; 7; 7; 8; 4; 6; 1; 7; 4; 0; 6; 4; 9; 5; 5; 1;
    4; 9; 2; 9; 0; 8; 6; 2; 5; 6; 9; 3; 2; 1; 9; 7; 8; 4; 6; 8; 6; 2;
    2; 4; 8; 2; 8; 3; 9; 7; 2; 2; 4; 1; 3; 7; 5; 6; 5; 7; 0; 5; 6; 0;
    5; 7; 4; 9; 0; 2; 6; 1; 4; 0; 7; 9; 7; 2; 9; 6; 8; 6; 5; 2; 4; 1;
    4; 5; 3; 5; 1; 0; 0; 4; 7; 4; 8; 2; 1; 6; 6; 3; 7; 0; 4; 8; 4; 4;
    0; 3; 1; 9; 9; 8; 9; 0; 0; 0; 8; 8; 9; 5; 2; 4; 3; 4; 5; 0; 6; 5;
    8; 5; 4; 1; 2; 2; 7; 5; 8; 8; 6; 6; 6; 8; 8; 1; 1; 6; 4; 2; 7; 1;
    7; 1; 4; 7; 9; 9; 2; 4; 4; 4; 2; 9; 2; 8; 2; 3; 0; 8; 6; 3; 4; 6;
    5; 6; 7; 4; 8; 1; 3; 9; 1; 9; 1; 2; 3; 1; 6; 2; 8; 2; 4; 5; 8; 6;
    1; 7; 8; 6; 6; 4; 5; 8; 3; 5; 9; 1; 2; 4; 5; 6; 6; 5; 2; 9; 4; 7;
    6; 5; 4; 5; 6; 8; 2; 8; 4; 8; 9; 1; 2; 8; 8; 3; 1; 4; 2; 6; 0; 7;
    6; 9; 0; 0; 4; 2; 2; 4; 2; 1; 9; 0; 2; 2; 6; 7; 1; 0; 5; 5; 6; 2;
    6; 3; 2; 1; 1; 1; 1; 1; 0; 9; 3; 7; 0; 5; 4; 4; 2; 1; 7; 5; 0; 6;
    9; 4; 1; 6; 5; 8; 9; 6; 0; 4; 0; 8; 0; 7; 1; 9; 8; 4; 0; 3; 8; 5;
    0; 9; 6; 2; 4; 5; 5; 4; 4; 4; 3; 6; 2; 9; 8; 1; 2; 3; 0; 9; 8; 7;
    8; 7; 9; 9; 2; 7; 2; 4; 4; 2; 8; 4; 9; 0; 9; 1; 8; 8; 8; 4; 5; 8;
    0; 1; 5; 6; 1; 6; 6; 0; 9; 7; 9; 1; 9; 1; 3; 3; 8; 7; 5; 4; 9; 9;
    2; 0; 0; 5; 2; 4; 0; 6; 3; 6; 8; 9; 9; 1; 2; 5; 6; 0; 7; 1; 7; 6;
    0; 6; 0; 5; 8; 8; 6; 1; 1; 6; 4; 6; 7; 1; 0; 9; 4; 0; 5; 0; 7; 7;
    5; 4; 1; 0; 0; 2; 2; 5; 6; 9; 8; 3; 1; 5; 5; 2; 0; 0; 0; 5; 5; 9;
    3; 5; 7; 2; 9; 7; 2; 5; 7; 1; 6; 3; 6; 2; 6; 9; 5; 6; 1; 8; 8; 2;
    6; 7; 0; 4; 2; 8; 2; 5; 2; 4; 8; 3; 6; 0; 0; 8; 2; 3; 2; 5; 7; 5;
    3; 0; 4; 2; 0; 7; 5; 2; 9; 6; 3; 4; 5; 0|]

let calculate_max_chunk_product chunk_size =
    seq {
        for i in 0 .. (Array.length digits - chunk_size) do
            let chunk = Array.sub digits i chunk_size
            yield Array.fold (fun acc elem -> acc * elem) 1 chunk 
        }
    |> Seq.max

let problem_8 () =
    calculate_max_chunk_product 5

As you can see, we are generating a sequence of products and then pick the maximum value. Each product is created by first taking a chunk of the array (using the Array.sub) and then using Array.fold to compute the product of all elements in the chunk.

Again, I have tried to keep the imperative part to a minimum and I was able to get by without using any mutable variable.

See all Project Euler problems solutions.

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  • bypasser

    Logarithms are really fast, if you’re smart: simply build a lookup table with precomputed logarithms of numbers between 0 and 9.

    Now you have indeed reduced the problem of finding the biggest product of 5 numbers to finding the biggest sum of 5 lookups, which should be much faster.

    (To eventually discover that you have no performance problem at all… Premature optimisation is the root of etc.)

  • bypasser

    Logarithms are really fast, if you’re smart: simply build a lookup table with precomputed logarithms of numbers between 0 and 9.

    Now you have indeed reduced the problem of finding the biggest product of 5 numbers to finding the biggest sum of 5 lookups, which should be much faster.

    (To eventually discover that you have no performance problem at all… Premature optimisation is the root of etc.)

  • stefanoricciardi

    Bypasser,

    I had not thought about pre-computing the logarithms; it’s a clever approach considering that we have a fairly limited set of inputs that we are dealing with.

    But, as you said, given the size of the problem, brute force here is perfectly fine :)

    Thank you for stepping by.

    Stefano

  • stefanoricciardi

    Bypasser,

    I had not thought about pre-computing the logarithms; it’s a clever approach considering that we have a fairly limited set of inputs that we are dealing with.

    But, as you said, given the size of the problem, brute force here is perfectly fine :)

    Thank you for stepping by.

    Stefano

  • Nynn

    //Playing with Seq.windowed

    let nombre = “73167176531330624919…………”
    let produitMax =
    nombre.ToCharArray() //get array of char
    |> Seq.ofArray //get seq of char
    |> Seq.windowed 5 //get seq of arrays of 5 chars
    |> Seq.map (fun a -> (Array.fold (fun p e -> p * (int e – 48)) 1 a)) //get seq of products of 5 digits
    |> Seq.max //get the max product

    printfn “%d” produitMax

    System.Console.ReadLine() |> ignore

  • Nynn

    //Playing with Seq.windowed

    let nombre = “73167176531330624919…………”
    let produitMax =
    nombre.ToCharArray() //get array of char
    |> Seq.ofArray //get seq of char
    |> Seq.windowed 5 //get seq of arrays of 5 chars
    |> Seq.map (fun a -> (Array.fold (fun p e -> p * (int e – 48)) 1 a)) //get seq of products of 5 digits
    |> Seq.max //get the max product

    printfn “%d” produitMax

    System.Console.ReadLine() |> ignore

  • Gryphes

    http://gryphes.tk/
    http://gryphes.tk/ProjectEuler_08FlowChart.jpg

    #include

    using std::cout;
    using std::endl;
    using std::cin;

    int main()
    {
    //decualre the variables
    int x=1,n=0,i=0,temp=0;
    char value[1001] =”73167176531330624919225119674426574742355349194934″
    “96983520312774506326239578318016984801869478851843″
    “85861560789112949495459501737958331952853208805511″
    “12540698747158523863050715693290963295227443043557″
    “66896648950445244523161731856403098711121722383113″
    “62229893423380308135336276614282806444486645238749″
    “30358907296290491560440772390713810515859307960866″
    “70172427121883998797908792274921901699720888093776″
    “65727333001053367881220235421809751254540594752243″
    “52584907711670556013604839586446706324415722155397″
    “53697817977846174064955149290862569321978468622482″
    “83972241375657056057490261407972968652414535100474″
    “82166370484403199890008895243450658541227588666881″
    “16427171479924442928230863465674813919123162824586″
    “17866458359124566529476545682848912883142607690042″
    “24219022671055626321111109370544217506941658960408″
    “07198403850962455444362981230987879927244284909188″
    “84580156166097919133875499200524063689912560717606″
    “05886116467109405077541002256983155200055935729725″
    “71636269561882670428252483600823257530420752963450″;

    while((value[n]))
    {
    n++;
    i++;
    if(i != 6)
    {
    x *= (value[n-1] – ’0′);
    }else
    {
    i = 0;
    n -= 5;
    if(x > temp)
    {
    temp = x;
    }
    x = 1;
    }
    }
    cout << "the greatest product of five consecutive digits is " << temp << endl;
    cin.get();
    return 0;
    }