C++ vs C# vs F# vs Haskell

Since I first posted an F# solution to Left-Truncatable Primes, C# and Haskell have entered the frame, and although this is not a good problem for a serious comparison of languages, I think it is still interesting.

So lets start at the beginning, Dominic Penfold initially posed the problem and gave an efficient algorithm implemented in C++ (39 lines):

bool IsPrime(int num)
{
    int sqrt = 1 + sqrtf(num);
    for (int i=2; i<sqrt; i++)
    {
        if (num % i == 0)
            return false;
    }
    return true;
}
 
int NthLeftTruncatablePrime(int target)
{
    std::vector<int> PrimeRoots;
    PrimeRoots.push_back(2);
    PrimeRoots.push_back(3);
    PrimeRoots.push_back(5);
    PrimeRoots.push_back(7);
    int start = 0;
    int tens = 10;    
    while(true)
    {
        int end = PrimeRoots.size();
        for (int i=1; i<10; i++)
        {
            for (int pos = start; pos< end; pos++)
            {
                int newprime = tens * i + PrimeRoots[pos];
                if (IsPrime(newprime))
                {
                    PrimeRoots.push_back(newprime);
                    if (PrimeRoots.size()==target)                                      
                        return newprime;                    
                }
            }
        }
        start = end;
        tens *= 10;
    }
}

Performance was considerably faster when 32-bit integers were switched for 64-bit integers.

The C++ code is very easily converted into C# (35 lines):

static bool IsPrime(int num)
{
    int sqrt = 1 + (int ) System.Math.Sqrt(num);
    blue">for (int i = 2; i < sqrt; i++)
    {
        if (num % i == 0)
            return false;
    }
    return true;
}

static int NthLeftTruncatedPrime(int target)
{
    var primeRoots = new List<int>() { 2, 3, 5, 7};               
    int start = 0;
    int tens = 10;
    while (true)
    {
        int end = primeRoots.Count;
        for (int i = 1; i < 10; i++)
        {
            for (int pos = start; pos < end; pos++)
            {
                int newprime = tens * i + primeRoots[pos];
                if (IsPrime(newprime))
                {
                    primeRoots.Add(newprime);
                    if (primeRoots.Count == target)
                        return newprime;
                }
            }
        }
        start = end;
        tens *= 10;
    }
}

Conversion to F# is only slightly trickier as there is no equivalent return statement, but similar behavior can be accomplished using yield (30 lines):

let IsPrime n =
    if n = 1 then false
    else    
        let max = n |> float |> sqrt |> int
        let rec Test = function
            | x when x > max -> true
            | x -> if (n % x) = 0 then false else Test (x+1)
        Test 2

let NthLeftTruncatablePrime n =
    let oneDigitPrimes = [2;3;5;7]          
    seq {            
        let primes = ref oneDigitPrimes        
        for tens = 1 to 8 do
            let acc = ref []
            for i=1 to 9 do 
                let newPrime = i * pown 10 tens                
                let primes' = ref !primes
                while (!primes').Length > 0 do                            
                    let x = newPrime+(!primes').Head
                    if IsPrime x then 
                        acc := x :: !acc
                        yield x
                    primes' := (!primes').Tail                    
                done                                     
            done         
            primes := !acc |> List.rev    
        done            
    }
    |> Seq.append oneDigitPrimes
    |> Seq.nth (n-1)

The extra overhead of generating a sequence can be avoided using instead recursion (25 lines including IsPrime function):

let NthLeftTruncatablePrime index =
    let rec Find digit factor primes primes' count acc =
        match digit, primes' with
        | 10, _ -> 
            let primes = List.rev acc
            Find 1 (10*factor) primes primes count []
        | _, [] ->
            Find (digit+1) factor primes primes count acc
        | _, prime::tail -> 
            let k = (digit * factor) + prime
            let count, acc =
                if IsPrime k then count+1, k::acc else count, acc 
            if count = index then k
            else Find digit factor primes tail count acc
    let primes = [2;3;5;7]
    if index <= 4 then List.nth primes (index-1)
    else Find 1 10 primes primes 4 []

Later Matt Curran (coincidentally another ex-games developer) got in on the act with a Haskell version (17 lines including a comment – although the lines are a little dense):

isPrime :: Int -> Bool
isPrime 1 = False
isPrime x = null $ take 1 [n | n <- [2..upperBound x], 0 == mod x n]
    where
        upperBound = floor . sqrt . fromIntegral

-- list of all left truncatable primes
ltps :: [Int]
ltps = ps 1 [0]
    where
        ps _ []      = []
        ps m prevMag = thisMag ++ (ps (m*10) thisMag)
            where
                thisMag = primes [x*m | x <- [1..9]] prevMag
                    where
                        primes [] _ = []
                        primes (m:mgs) bases = 
                            (filter isPrime [m+x | x <- bases]) ++ (primes mgs bases)

Results Breakdown by language

Note: the time is to generate the 1000th prime, and the same computer was used for all tests, as was release mode. Haskell time coming soon.

LeftTruncatablePrime.cpp (1.45 kb)

LeftTruncatablePrime.cs (1.58 kb)

LeftTruncatablePrime_seq.fs (1.53 kb)

LeftTruncatablePrime_rec.fs (1.36 kb)