Yan Cui

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#### Problem

A Pythagorean triplet is a set of three natural numbers,a<b<c, for which,

a^{2}+b^{2}=c^{2}

For example, 3^{2}+ 4^{2}= 9 + 16 = 25 = 5^{2}.

There exists exactly one Pythagorean triplet for whicha+b+c= 1000.

Find the productabc.

#### Solution

let isPythagoreanTriplet(numbers : int list) = match List.sort(numbers) with | [a; b; c] -> a*a + b*b = c*c | _ –> false let getTriplets = seq { for a = 1 to 1000 do for b = 1 to 1000 do for c = 1 to 1000 do if a + b + c = 1000 then yield [a; b; c] } let pythagoreanTriplet = getTriplets |> Seq.filter isPythagoreanTriplet |> Seq.head let product = pythagoreanTriplet |> Seq.fold (fun acc x -> acc * x) 1

Here I’ve created a function called *isPythagoreanTriplet*, which takes a integer list and checks whether it’s a pythagorean triplet. If this is the first time you’ve seen the **match …. with** syntax, it’s used to match a **pattern** in F#. The patterns are evaluated from top to bottom, the first pattern **decomposes** a sorted version of the supplied *int list* into three values *a*, *b* and *c*, and evaluates whether a, b and c makes a pythagorean triplet:

| [a; b; c] -> a*a + b*b = c*c

If the list does not have exactly three elements, then the next pattern is evaluate. The **_** character is the wildcard character and in this case it simply returns false always. e.g.:

isPythagoreanTriplet [3;4;5;0];; val it : bool = false isPythagoreanTriplet [];; val it : bool = false isPythagoreanTriplet [3;4;5];; val it : bool = true

The *getTriplets* function on the hand, supplies a sequence of all triplets which matches the a + b + c = 1000 requirement. Note that I used the **yield** keyword to produce values that become part of the returned sequence.

The next line of code simply identifies the first (Seq.head) triplet from the above sequence which is a pythagorean triplet.

And finally, the product of the pythagorean triplet is calculated using the Seq.fold function (identical to the function used in the problem 8 solution, which describes the function in more detail).

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Pingback: Project Euler — Problem 21 Solution | theburningmonk.com

Omuyou don’t have to go until 1000 cuz they have to have a sum of 1000 so obviously the first one can’t be more than 1000/3 and second has to be bigger than first but at least -1 smaller than last

this is my solution:

let p8 = [for a in [1..1000/3] do

for b in [(a+1)..(1000-a)/2] do

let c = 1000 – (a+b);

if(a*a+b*b = c*c) then yield (a,b,c,a*b*c)]