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NinetyNine F# Problems  Problems 31  41  Arithmetic]
These are F# solutions of NinetyNine Haskell Problems which are themselves translations of NinetyNine Lisp Problems and NinetyNine Prolog Problems. The solutions are hidden so you can try to solve them yourself.
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/// NinetyNine F# Problems  Problems 31  41
///
/// These are F# solutions of NinetyNine Haskell Problems
/// (http://www.haskell.org/haskellwiki/H99:_NinetyNine_Haskell_Problems),
/// which are themselves translations of NinetyNine Lisp Problems
/// (http://www.ic.unicamp.br/~meidanis/courses/mc336/2006s2/funcional/L99_NinetyNine_Lisp_Problems.html)
/// and NinetyNine Prolog Problems
/// (https://sites.google.com/site/prologsite/prologproblems).
///
/// If you would like to contribute a solution or fix any bugs, send
/// an email to paks at kitiara dot org with the subject "99 F# problems".
/// I'll try to update the problem as soon as possible.
///
/// The problems have different levels of difficulty. Those marked with a single asterisk (*)
/// are easy. If you have successfully solved the preceeding problems you should be able to
/// solve them within a few (say 15) minutes. Problems marked with two asterisks (**) are of
/// intermediate difficulty. If you are a skilled F# programmer it shouldn't take you more than
/// 3090 minutes to solve them. Problems marked with three asterisks (***) are more difficult.
/// You may need more time (i.e. a few hours or more) to find a good solution
///
/// Though the problems number from 1 to 99, there are some gaps and some additions marked with
/// letters. There are actually only 88 problems.

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/// Example:
/// * (isprime 7)
/// T
///
/// Example in F#:
///
/// > isPrime 7;;
/// val it : bool = true
(Solution 1)
(Solution 2)

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/// Example:
/// * (gcd 36 63)
/// 9
///
/// Example in F#:
///
/// > [gcd 36 63; gcd (3) (6); gcd (3) 6];;
/// val it : int list = [9; 3; 3]
(Solution)

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/// Two numbers are coprime if their greatest common divisor equals 1.
///
/// Example:
/// * (coprime 35 64)
/// T
///
/// Example in F#:
///
/// > coprime 35 64;;
/// val it : bool = true
(Solution)

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/// Euler's socalled totient function phi(m) is defined as the number of
/// positive integers r (1 <= r < m) that are coprime to m.
///
/// Example: m = 10: r = 1,3,7,9; thus phi(m) = 4. Note the special case: phi(1) = 1.
///
/// Example:
/// * (totientphi 10)
/// 4
///
/// Example in F#:
///
/// > totient 10;;
/// val it : int = 4
(Solution)

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/// Construct a flat list containing the prime factors in ascending order.
///
/// Example:
/// * (primefactors 315)
/// (3 3 5 7)
///
/// Example in F#:
///
/// > primeFactors 315;;
/// val it : int list = [3; 3; 5; 7]
(Solution)

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///
/// Construct a list containing the prime factors and their multiplicity.
///
/// Example:
/// * (primefactorsmult 315)
/// ((3 2) (5 1) (7 1))
///
/// Example in F#:
///
/// > primeFactorsMult 315;;
/// [(3,2);(5,1);(7,1)]
(Solution)

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/// See problem 34 for the definition of Euler's totient function. If the list of the prime
/// factors of a number m is known in the form of problem 36 then the function phi(m)
/// can be efficiently calculated as follows: Let ((p1 m1) (p2 m2) (p3 m3) ...) be the list of
/// prime factors (and their multiplicities) of a given number m. Then phi(m) can be
/// calculated with the following formula:
/// phi(m) = (p1  1) * p1 ** (m1  1) +
/// (p2  1) * p2 ** (m2  1) +
/// (p3  1) * p3 ** (m3  1) + ...
///
/// Note that a ** b stands for the b'th power of a.
///
/// Note: Actually, the official problems show this as a sum, but it should be a product.
/// > phi 10;;
/// val it : int = 4
(Solution)

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/// Use the solutions of problems 34 and 37 to compare the algorithms. Take the
/// number of reductions as a measure for efficiency. Try to calculate phi(10090) as an
/// example.
///
/// (no solution required)
///

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/// Given a range of integers by its lower and upper limit, construct a list of all prime numbers
/// in that range.
///
/// Example in F#:
///
/// > primesR 10 20;;
/// val it : int list = [11; 13; 17; 19]
(Solution)

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/// Goldbach's conjecture says that every positive even number greater than 2 is the
/// sum of two prime numbers. Example: 28 = 5 + 23. It is one of the most famous facts
/// in number theory that has not been proved to be correct in the general case. It has
/// been numerically confirmed up to very large numbers (much larger than we can go
/// with our Prolog system). Write a predicate to find the two prime numbers that sum up
/// to a given even integer.
///
/// Example:
/// * (goldbach 28)
/// (5 23)
///
/// Example in F#:
///
/// *goldbach 28
/// val it : int * int = (5, 23)
(Solution)

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/// In most cases, if an even number is written as the sum of two prime numbers, one of
/// them is very small. Very rarely, the primes are both bigger than say 50. Try to find
/// out how many such cases there are in the range 2..3000.
///
/// Example:
/// * (goldbachlist 9 20)
/// 10 = 3 + 7
/// 12 = 5 + 7
/// 14 = 3 + 11
/// 16 = 3 + 13
/// 18 = 5 + 13
/// 20 = 3 + 17
/// * (goldbachlist 1 2000 50)
/// 992 = 73 + 919
/// 1382 = 61 + 1321
/// 1856 = 67 + 1789
/// 1928 = 61 + 1867
///
/// Example in F#:
///
/// > goldbachList 9 20;;
/// val it : (int * int) list =
/// [(3, 7); (5, 7); (3, 11); (3, 13); (5, 13); (3, 17)]
/// > goldbachList' 4 2000 50
/// val it : (int * int) list = [(73, 919); (61, 1321); (67, 1789); (61, 1867)]
(Solution)

//naive solution
let isPrime n =
let sqrtn n = int < sqrt (float n)
seq { 2 .. sqrtn n } > Seq.exists(fun i > n % i = 0) > not
// MillerRabin primality test
open System.Numerics
let pow' mul sq x' n' =
let rec f x n y =
if n = 1I then
mul x y
else
let (q,r) = BigInteger.DivRem(n, 2I)
let x2 = sq x
if r = 0I then
f x2 q y
else
f x2 q (mul x y)
f x' n' 1I
let mulMod (a :bigint) b c = (b * c) % a
let squareMod (a :bigint) b = (b * b) % a
let powMod m = pow' (mulMod m) (squareMod m)
let iterate f = Seq.unfold(fun x > let fx = f x in Some(x,fx))
///See: http://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test
let millerRabinPrimality n a =
let find2km n =
let rec f k m =
let (q,r) = BigInteger.DivRem(m, 2I)
if r = 1I then
(k,m)
else
f (k+1I) q
f 0I n
let n' = n  1I
let iter = Seq.tryPick(fun x > if x = 1I then Some(false) elif x = n' then Some(true) else None)
let (k,m) = find2km n'
let b0 = powMod n a m
match (a,n) with
 _ when a <= 1I && a >= n' >
failwith (sprintf "millerRabinPrimality: a out of range (%A for %A)" a n)
 _ when b0 = 1I  b0 = n' > true
 _ > b0
> iterate (squareMod n)
> Seq.take(int k)
> Seq.skip 1
> iter
> Option.exists id
///For MillerRabin the witnesses need to be selected at random from the interval [2, n  2].
///More witnesses => better accuracy of the test.
///Also, remember that if MillerRabin returns true, then the number is _probable_ prime.
///If it returns false the number is composite.
let isPrimeW witnesses = function
 n when n < 2I > false
 n when n = 2I > true
 n when n = 3I > true
 n when n % 2I = 0I > false
 n > witnesses > Seq.forall(millerRabinPrimality n)
// let isPrime' = isPrimeW [2I;3I] // Two witnesses
// let p = pown 2I 4423  1I // 20th Mersenne prime. 1,332 digits
// isPrime' p > printfn "%b";;
// Real: 00:00:03.184, CPU: 00:00:03.104, GC gen0: 12, gen1: 0, gen2: 0
// val it : bool = true
let rec gcd a b =
if b = 0 then
abs a
else
gcd b (a % b)
// using problem 32
let coprime a b = gcd a b = 1
// naive implementation. For a better solution see problem 37
let totient n = seq { 1 .. n  1} > Seq.filter (gcd n >> (=) 1) > Seq.length
let primeFactors n =
let sqrtn n = int < sqrt (float n)
let get n =
let sq = sqrtn n
// this can be made faster by using a prime generator like this one :
// https://github.com/paks/ProjectEuler/tree/master/Euler/Primegen
seq { yield 2; yield! seq {3 .. 2 .. sq} } > Seq.tryFind (fun x > n % x = 0)
let divSeq = n > Seq.unfold(fun x >
if x = 1 then
None
else
match get x with
 None > Some(x, 1) // x it's prime
 Some(divisor) > Some(divisor, x/divisor))
divSeq > List.ofSeq
// using problem 35
let primeFactorsMult n =
let sqrtn n = int < sqrt (float n)
let get n =
let sq = sqrtn n
// this can be made faster by using a prime generator like this one :
// https://github.com/paks/ProjectEuler/tree/master/Euler/Primegen
seq { yield 2; yield! seq {3 .. 2 .. sq} } > Seq.tryFind (fun x > n % x = 0)
let divSeq = n > Seq.unfold(fun x >
if x = 1 then
None
else
match get x with
 None > Some(x, 1) // x it's prime
 Some(divisor) > Some(divisor, x/divisor))
divSeq > Seq.countBy id > List.ofSeq
// using problem 36
let phi = primeFactorsMult >> Seq.fold(fun acc (p,m) > (p  1) * pown p (m  1) * acc) 1
// using problem 31
let primeR a b = seq { a .. b } > Seq.filter isPrime > List.ofSeq
// using problem 31. Very slow on big numbers due to the implementation of primeR. To speed this up use a prime generator.
let goldbach n =
let primes = primeR 2 n > Array.ofList
let rec findPairSum (arr: int array) front back =
let sum = arr.[front] + arr.[back]
match compare sum n with
 1 > findPairSum arr (front + 1) back
 0 > Some(arr.[front] , arr.[back])
 1 > findPairSum arr front (back  1)
 _ > failwith "not possible"
Option.get < findPairSum primes 0 (primes.Length  1)
let goldbachList a b =
let start = if a % 2 <> 0 then a + 1 else a
seq { start .. 2 .. b } > Seq.map goldbach > List.ofSeq
let goldbachList' a b limit = goldbachList a b > List.filter(fst >> (<) limit)
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