# Pythagorean triples

Pythagorean triples
You are encouraged to solve this task according to the task description, using any language you may know.
A Pythagorean triple is defined as three positive integers (a,b,c) where a < b < c, and a2 + b2 = c2. They are called primitive triples if a,b,c are coprime, that is, if their pairwise greatest common divisors gcd(a,b) = gcd(a,c) = gcd(b,c) = 1. Because of their relationship through the Pythagorean theorem, a, b, and c are coprime if a and b are coprime (gcd(a,b) = 1). Each triple forms the length of the sides of a right triangle, whose perimeter is P = a + b + c.

The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.

Extra credit: Deal with large values. Can your program handle a max perimeter of 1,000,000? What about 10,000,000? 100,000,000?

Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.

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## Contents

Translation of efficient method from C, see the WP article. Compiles on gnat/gcc.

with Ada.Text_IO; procedure Pythagorean_Triples is    type Large_Natural is range 0 .. 2**63-1;     -- this is the maximum for gnat    procedure New_Triangle(A, B, C: Large_Natural;                          Max_Perimeter: Large_Natural;                          Total_Cnt, Primitive_Cnt: in out Large_Natural) is      Perimeter: constant Large_Natural := A + B + C;   begin      if Perimeter <= Max_Perimeter then         Primitive_Cnt := Primitive_Cnt + 1;         Total_Cnt     := Total_Cnt + Max_Perimeter / Perimeter;         New_Triangle(A-2*B+2*C,     2*A-B+2*C,    2*A-2*B+3*C,   Max_Perimeter, Total_Cnt, Primitive_Cnt);         New_Triangle(A+2*B+2*C,     2*A+B+2*C,    2*A+2*B+3*C,   Max_Perimeter, Total_Cnt, Primitive_Cnt);         New_Triangle(2*B+2*C-A,     B+2*C-2*A,    2*B+3*C-2*A,   Max_Perimeter, Total_Cnt, Primitive_Cnt);      end if;   end New_Triangle;    T_Cnt, P_Cnt: Large_Natural; begin   for I in 1 .. 9 loop      T_Cnt := 0;      P_Cnt := 0;      New_Triangle(3,4,5, 10**I, Total_Cnt => T_Cnt, Primitive_Cnt => P_Cnt);      Ada.Text_IO.Put_Line("Up to 10 **" & Integer'Image(I) & " :" &                             Large_Natural'Image(T_Cnt) & " Triples," &                             Large_Natural'Image(P_Cnt) & " Primitives");   end loop;end Pythagorean_Triples;

Output:

Up to 10 ** 1 : 0 Triples, 0 Primitives
Up to 10 ** 2 : 17 Triples, 7 Primitives
Up to 10 ** 3 : 325 Triples, 70 Primitives
Up to 10 ** 4 : 4858 Triples, 703 Primitives
Up to 10 ** 5 : 64741 Triples, 7026 Primitives
Up to 10 ** 6 : 808950 Triples, 70229 Primitives
Up to 10 ** 7 : 9706567 Triples, 702309 Primitives
Up to 10 ** 8 : 113236940 Triples, 7023027 Primitives
Up to 10 ** 9 : 1294080089 Triples, 70230484 Primitives

## AutoHotkey

#NoEnvSetBatchLines, -1#SingleInstance, Force ; Greatest common divisor, from http://rosettacode.org/wiki/Greatest_common_divisor#AutoHotkeygcd(a,b) {	Return b=0 ? Abs(a) : Gcd(b,mod(a,b))} count_triples(max) {	primitives := 0, triples := 0, m := 2	while m <= (max / 2)**0.5	{		n := mod(m, 2) + 1		,p := 2*m*(m + n)		, delta := 4*m		while n < m and p <= max			gcd(m, n) = 1				? (primitives++				, triples += max // p)				: ""			, n += 2			, p += delta		m++	}	Return primitives " primitives out of " triples " triples"} Loop, 8	Msgbox % 10**A_Index ": " count_triples(10**A_Index)
Output:
10: 0 primitives out of 0 triples
100: 7 primitives out of 17 triples
1000: 70 primitives out of 325 triples
10000: 703 primitives out of 4858 triples
100000: 7026 primitives out of 64741 triples
1000000: 70229 primitives out of 808950 triples
10000000: 702309 primitives out of 9706567 triples
100000000: 7023027 primitives out of 113236940 triples

## BBC BASIC

The built-in array arithmetic is very well suited to this task!

      DIM U0%(2,2), U1%(2,2), U2%(2,2), seed%(2)      U0%() =  1, -2, 2,  2, -1, 2,  2, -2, 3      U1%() =  1,  2, 2,  2,  1, 2,  2,  2, 3      U2%() = -1,  2, 2, -2,  1, 2, -2,  2, 3       seed%() = 3, 4, 5      FOR power% = 1 TO 7        all% = 0 : prim% = 0        PROCtri(seed%(), 10^power%, all%, prim%)        PRINT "Up to 10^"; power%, ": " all% " triples" prim% " primitives"      NEXT      END       DEF PROCtri(i%(), mp%, RETURN all%, RETURN prim%)      LOCAL t%() : DIM t%(2)       IF SUM(i%()) > mp% ENDPROC      prim% += 1      all% += mp% DIV SUM(i%())       t%() = U0%() . i%()      PROCtri(t%(), mp%, all%, prim%)      t%() = U1%() . i%()      PROCtri(t%(), mp%, all%, prim%)      t%() = U2%() . i%()      PROCtri(t%(), mp%, all%, prim%)      ENDPROC

Output:

Up to 10^1:          0 triples         0 primitives
Up to 10^2:         17 triples         7 primitives
Up to 10^3:        325 triples        70 primitives
Up to 10^4:       4858 triples       703 primitives
Up to 10^5:      64741 triples      7026 primitives
Up to 10^6:     808950 triples     70229 primitives
Up to 10^7:    9706567 triples    702309 primitives
Up to 10^8:  113236940 triples   7023027 primitives


## Bracmat

Translation of: C
(pythagoreanTriples=  total prim max-peri U.       (.(1,-2,2) (2,-1,2) (2,-2,3))        (.(1,2,2) (2,1,2) (2,2,3))        (.(-1,2,2) (-2,1,2) (-2,2,3))    : ?U  & ( new-tri    =     i t p Urows Urow Ucols        , a b c loop A B C      .     !arg:(,?a,?b,?c)          & !a+!b+!c:~>!max-peri:?p          & 1+!prim:?prim          & div$(!max-peri.!p)+!total:?total & !U:?Urows & ( loop = !Urows:(.?Urow) ?Urows & !Urow:?Ucols & :?t & whl ' ( !Ucols:(?A,?B,?C) ?Ucols & (!t,!a*!A+!b*!B+!c*!C):?t ) & new-tri$!t              & !loop            )          & !loop        |    )  & ( Main    =   seed      .   (,3,4,5):?seed        & 10:?max-peri        &   whl          ' ( 0:?total:?prim            & new-tri$!seed & out$ ( str                $( "Up to " !max-peri ": " !total " triples, " !prim " primitives." ) ) & !max-peri*10:~>10000000:?max-peri ) ) & Main$); pythagoreanTriples$;  Output (under Linux): Up to 10: 0 triples, 0 primitives. Up to 100: 17 triples, 7 primitives. Up to 1000: 325 triples, 70 primitives. Up to 10000: 4858 triples, 703 primitives. Up to 100000: 64741 triples, 7026 primitives. Up to 1000000: 808950 triples, 70229 primitives. Up to 10000000: 9706567 triples, 702309 primitives. Up to 100000000: 113236940 triples, 7023027 primitives. Under Windows XP Command prompt the last result is unattainable due to stack overflow. With very few changes we can get rid of the stack exhausting recursion. Instead of calling new-tri recursively, be push the triples to test onto a stack and return to the Main function. In the innermost loop we pop a triple from the stack and call new-tri. The memory overhead is only a few megabytes for a max perimeter of 100,000,000. On my Windows XP box the whole computation takes at least 15 minutes! Given enough time (and memory), the program can compute results for larger perimeters. (pythagoreanTriples= total prim max-peri U stack. (.(1,-2,2) (2,-1,2) (2,-2,3)) (.(1,2,2) (2,1,2) (2,2,3)) (.(-1,2,2) (-2,1,2) (-2,2,3)) : ?U & ( new-tri = i t p Urows Urow Ucols Ucol , a b c loop A B C . !arg:(,?a,?b,?c) & !a+!b+!c:~>!max-peri:?p & 1+!prim:?prim & div$(!max-peri.!p)+!total:?total          & !U:?Urows          & ( loop            =   !Urows:(.?Urow) ?Urows              & !Urow:?Ucols              & :?t              &   whl                ' ( !Ucols:(?A,?B,?C) ?Ucols                  & (!t,!a*!A+!b*!B+!c*!C):?t                  )              & !t !stack:?stack              & !loop            )          & !loop        |    )  & ( Main    =   seed      .   10:?max-peri        &   whl          ' ( 0:?total:?prim            & (,3,4,5):?stack            &   whl              ' (!stack:%?seed ?stack&new-tri$!seed) & out$ ( str                $( "Up to " !max-peri ": " !total " triples, " !prim " primitives." ) ) & !max-peri*10:~>100000000:?max-peri ) ) & Main$); pythagoreanTriples$; ## C Sample implemention; naive method, patentedly won't scale to larger numbers, despite the attempt to optimize it. Calculating up to 10000 is already a test of patience. #include <stdio.h>#include <stdlib.h> typedef unsigned long long xint;typedef unsigned long ulong; inline ulong gcd(ulong m, ulong n){ ulong t; while (n) { t = n; n = m % n; m = t; } return m;} int main(){ ulong a, b, c, pytha = 0, prim = 0, max_p = 100; xint aa, bb, cc; for (a = 1; a <= max_p / 3; a++) { aa = (xint)a * a; printf("a = %lu\r", a); /* show that we are working */ fflush(stdout); /* max_p/2: valid limit, because one side of triangle * must be less than the sum of the other two */ for (b = a + 1; b < max_p/2; b++) { bb = (xint)b * b; for (c = b + 1; c < max_p/2; c++) { cc = (xint)c * c; if (aa + bb < cc) break; if (a + b + c > max_p) break; if (aa + bb == cc) { pytha++; if (gcd(a, b) == 1) prim++; } } } } printf("Up to %lu, there are %lu triples, of which %lu are primitive\n", max_p, pytha, prim); return 0;} output: Up to 100, there are 17 triples, of which 7 are primitive Efficient method, generating primitive triples only as described in the same WP article: #include <stdio.h>#include <stdlib.h>#include <stdint.h> /* should be 64-bit integers if going over 1 billion */typedef unsigned long xint;#define FMT "%lu" xint total, prim, max_peri;xint U[][9] = {{ 1, -2, 2, 2, -1, 2, 2, -2, 3}, { 1, 2, 2, 2, 1, 2, 2, 2, 3}, {-1, 2, 2, -2, 1, 2, -2, 2, 3}}; void new_tri(xint in[]){ int i; xint t[3], p = in[0] + in[1] + in[2]; if (p > max_peri) return; prim ++; /* for every primitive triangle, its multiples would be right-angled too; * count them up to the max perimeter */ total += max_peri / p; /* recursively produce next tier by multiplying the matrices */ for (i = 0; i < 3; i++) { t[0] = U[i][0] * in[0] + U[i][1] * in[1] + U[i][2] * in[2]; t[1] = U[i][3] * in[0] + U[i][4] * in[1] + U[i][5] * in[2]; t[2] = U[i][6] * in[0] + U[i][7] * in[1] + U[i][8] * in[2]; new_tri(t); }} int main(){ xint seed[3] = {3, 4, 5}; for (max_peri = 10; max_peri <= 100000000; max_peri *= 10) { total = prim = 0; new_tri(seed); printf( "Up to "FMT": "FMT" triples, "FMT" primitives.\n", max_peri, total, prim); } return 0;} Output Up to 10: 0 triples, 0 primitives.Up to 100: 17 triples, 7 primitives.Up to 1000: 325 triples, 70 primitives.Up to 10000: 4858 triples, 703 primitives.Up to 100000: 64741 triples, 7026 primitives.Up to 1000000: 808950 triples, 70229 primitives.Up to 10000000: 9706567 triples, 702309 primitives.Up to 100000000: 113236940 triples, 7023027 primitives. Same as above, but with loop unwound and third recursion eliminated: #include <stdio.h>#include <stdlib.h>#include <stdint.h> /* should be 64-bit integers if going over 1 billion */typedef unsigned long xint;#define FMT "%lu" xint total, prim, max_peri; void new_tri(xint in[]){ int i; xint t[3], p; xint x = in[0], y = in[1], z = in[2]; recur: p = x + y + z; if (p > max_peri) return; prim ++; total += max_peri / p; t[0] = x - 2 * y + 2 * z; t[1] = 2 * x - y + 2 * z; t[2] = t[1] - y + z; new_tri(t); t[0] += 4 * y; t[1] += 2 * y; t[2] += 4 * y; new_tri(t); z = t[2] - 4 * x; y = t[1] - 4 * x; x = t[0] - 2 * x; goto recur;} int main(){ xint seed[3] = {3, 4, 5}; for (max_peri = 10; max_peri <= 100000000; max_peri *= 10) { total = prim = 0; new_tri(seed); printf( "Up to "FMT": "FMT" triples, "FMT" primitives.\n", max_peri, total, prim); } return 0;} ## C# Based on Ada example, which is a translation of efficient method from C, see the WP article. using System; namespace RosettaCode.CSharp{ class Program { static void Count_New_Triangle(ulong A, ulong B, ulong C, ulong Max_Perimeter, ref ulong Total_Cnt, ref ulong Primitive_Cnt) { ulong Perimeter = A + B + C; if (Perimeter <= Max_Perimeter) { Primitive_Cnt = Primitive_Cnt + 1; Total_Cnt = Total_Cnt + Max_Perimeter / Perimeter; Count_New_Triangle(A + 2 * C - 2 * B, 2 * A + 2 * C - B, 2 * A + 3 * C - 2 * B, Max_Perimeter, ref Total_Cnt, ref Primitive_Cnt); Count_New_Triangle(A + 2 * B + 2 * C, 2 * A + B + 2 * C, 2 * A + 2 * B + 3 * C, Max_Perimeter, ref Total_Cnt, ref Primitive_Cnt); Count_New_Triangle(2 * B + 2 * C - A, B + 2 * C - 2 * A, 2 * B + 3 * C - 2 * A, Max_Perimeter, ref Total_Cnt, ref Primitive_Cnt); } } static void Count_Pythagorean_Triples() { ulong T_Cnt, P_Cnt; for (int I = 1; I <= 8; I++) { T_Cnt = 0; P_Cnt = 0; ulong ExponentNumberValue = (ulong)Math.Pow(10, I); Count_New_Triangle(3, 4, 5, ExponentNumberValue, ref T_Cnt, ref P_Cnt); Console.WriteLine("Perimeter up to 10E" + I + " : " + T_Cnt + " Triples, " + P_Cnt + " Primitives"); } } static void Main(string[] args) { Count_Pythagorean_Triples(); } }} Output: Perimeter up to 10E1 : 0 Triples, 0 Primitives Perimeter up to 10E2 : 17 Triples, 7 Primitives Perimeter up to 10E3 : 325 Triples, 70 Primitives Perimeter up to 10E4 : 4858 Triples, 703 Primitives Perimeter up to 10E5 : 64741 Triples, 7026 Primitives Perimeter up to 10E6 : 808950 Triples, 70229 Primitives Perimeter up to 10E7 : 9706567 Triples, 702309 Primitives Perimeter up to 10E8 : 113236940 Triples, 7023027 Primitives ## Clojure This version is based on Euclid's formula: for each pair (m,n) such that m>n>0, m and n coprime and of opposite polarity (even/odd), there is a primitive Pythagorean triple. It can be proven that the converse is true as well. (defn gcd [a b] (if (zero? b) a (recur b (mod a b)))) (defn pyth [peri] (for [m (range 2 (Math/sqrt (/ peri 2))) n (range (inc (mod m 2)) m 2) ; n<m, opposite polarity :let [p (* 2 m (+ m n))] ; = a+b+c for this (m,n) :while (<= p peri) :when (= 1 (gcd m n)) :let [m2 (* m m), n2 (* n n), [a b] (sort [(- m2 n2) (* 2 m n)]), c (+ m2 n2)] k (range 1 (inc (quot peri p)))] [(= k 1) (* k a) (* k b) (* k c)])) (defn rcount [ts] ; (->> peri pyth rcount) produces [total, primitive] counts (reduce (fn [[total prims] t] [(inc total), (if (first t) (inc prims) prims)]) [0 0] ts)) To handle really large perimeters, we can dispense with actually generating the triples and just calculate the counts: (defn pyth-count [peri] (reduce (fn [[total prims] k] [(+ total k), (inc prims)]) [0 0] (for [m (range 2 (Math/sqrt (/ peri 2))) n (range (inc (mod m 2)) m 2) ; n<m, opposite polarity :let [p (* 2 m (+ m n))] ; = a+b+c for this (m,n) :while (<= p peri) :when (= 1 (gcd m n))] (quot peri p)))) ## CoffeeScript This algorithm scales linearly with the max perimeter. It uses two loops that are capped by the square root of the half-perimeter to examine/count provisional values of m and n, where m and n generate a, b, c, and p using simple number theory.  gcd = (x, y) -> return x if y == 0 gcd(y, x % y) # m,n generate primitive Pythag triples## preconditions:# m, n are integers of different parity# m > n# gcd(m,n) == 1 (coprime)## m, n generate: [m*m - n*n, 2*m*n, m*m + n*n]# perimeter is 2*m*m + 2*m*n = 2 * m * (m+n)count_triples = (max_perim) -> num_primitives = 0 num_triples = 0 m = 2 upper_limit = Math.sqrt max_perim / 2 while m <= upper_limit n = m % 2 + 1 p = 2*m*m + 2*m*n delta = 4*m while n < m and p <= max_perim if gcd(m, n) == 1 num_primitives += 1 num_triples += Math.floor max_perim / p n += 2 p += delta m += 1 console.log num_primitives, num_triples max_perim = Math.pow 10, 9 # takes under a minutecount_triples(max_perim)  output time coffee pythag_triples.coffee 70230484 1294080089 real 0m45.989s  ## Common Lisp (defun mmul (a b) (loop for x in a collect (loop for y in x for z in b sum (* y z)))) (defun count-tri (lim) (let ((prim 0) (cnt 0)) (labels ((count1 (tr) (let ((peri (reduce #'+ tr))) (when (<= peri lim) (incf prim) (incf cnt (truncate lim peri)) (count1 (mmul '(( 1 -2 2) ( 2 -1 2) ( 2 -2 3)) tr)) (count1 (mmul '(( 1 2 2) ( 2 1 2) ( 2 2 3)) tr)) (count1 (mmul '((-1 2 2) (-2 1 2) (-2 2 3)) tr)))))) (count1 '(3 4 5)) (format t "~a: ~a prim, ~a all~%" lim prim cnt)))) (loop for p from 2 do (count-tri (expt 10 p))) output 100: 7 prim, 17 all1000: 70 prim, 325 all10000: 703 prim, 4858 all100000: 7026 prim, 64741 all1000000: 70229 prim, 808950 all10000000: 702309 prim, 9706567 all... ## D ### Lazy Functional Version With hints from the Haskell solution. void main() { import std.stdio, std.range, std.algorithm, std.typecons, std.numeric; enum triples = (in int n) pure nothrow => iota(1, n + 1) .map!((in z) => iota(1, z + 1) .map!(x => iota(x, z + 1) .map!(y => tuple(x, y, z)))) .joiner.joiner .filter!(t => t[0] ^^ 2 + t[1] ^^ 2 == t[2] ^^ 2 && [t[]].sum <= n) .map!(t => tuple(t[0 .. 2].gcd == 1, t[])); auto xs = triples(100); writeln("Up to 100 there are ", xs.walkLength, " triples, ", xs.filter!q{ a[0] }.walkLength, " are primitive.");} Output: Up to 100 there are 17 triples, 7 are primitive. ### Shorter Version import std.stdio; ulong[2] tri(ulong lim, ulong a=3, ulong b=4, ulong c=5) { immutable l = a + b + c; if (l > lim) return [0, 0]; typeof(return) r = [1, lim / l]; r[] += tri(lim, a - 2*b + 2*c, 2*a - b + 2*c, 2*a - 2*b + 3*c)[]; r[] += tri(lim, a + 2*b + 2*c, 2*a + b + 2*c, 2*a + 2*b + 3*c)[]; r[] += tri(lim, -a + 2*b + 2*c, -2*a + b + 2*c, -2*a + 2*b + 3*c)[]; return r;} void main() { foreach (immutable p; 1 .. 8) writeln(10 ^^ p, " ", tri(10 ^^ p));} Output: 10 [0, 0] 100 [7, 17] 1000 [70, 325] 10000 [703, 4858] 100000 [7026, 64741] 1000000 [70229, 808950] 10000000 [702309, 9706567] ### Short SIMD Version With LDC compiler this is a little faster than the precedent version (remove @nogc to compile it with the current version of LDC compiler). import std.stdio, core.simd; ulong2 tri(in ulong lim, in ulong a=3, in ulong b=4, in ulong c=5)pure nothrow @nogc { immutable l = a + b + c; if (l > lim) return [0, 0]; typeof(return) r = [1, lim / l]; r += tri(lim, a - 2*b + 2*c, 2*a - b + 2*c, 2*a - 2*b + 3*c); r += tri(lim, a + 2*b + 2*c, 2*a + b + 2*c, 2*a + 2*b + 3*c); r += tri(lim, -a + 2*b + 2*c, -2*a + b + 2*c, -2*a + 2*b + 3*c); return r;} void main() { foreach (immutable p; 1 .. 8) writeln(10 ^^ p, " ", tri(10 ^^ p).array);} The output is the same. ### Faster Version Translation of: C import std.stdio; alias Xuint = uint; // ulong if going over 1 billion. __gshared Xuint nTriples, nPrimitives, limit; void countTriples(Xuint x, Xuint y, Xuint z) nothrow @nogc { while (true) { immutable p = x + y + z; if (p > limit) return; nPrimitives++; nTriples += limit / p; auto t0 = x - 2 * y + 2 * z; auto t1 = 2 * x - y + 2 * z; auto t2 = t1 - y + z; countTriples(t0, t1, t2); t0 += 4 * y; t1 += 2 * y; t2 += 4 * y; countTriples(t0, t1, t2); z = t2 - 4 * x; y = t1 - 4 * x; x = t0 - 2 * x; }} void main() { foreach (immutable p; 1 .. 9) { limit = Xuint(10) ^^ p; nTriples = nPrimitives = 0; countTriples(3, 4, 5); writefln("Up to %11d: %11d triples, %9d primitives.", limit, nTriples, nPrimitives); }} Output: Up to 10: 0 triples, 0 primitives. Up to 100: 17 triples, 7 primitives. Up to 1000: 325 triples, 70 primitives. Up to 10000: 4858 triples, 703 primitives. Up to 100000: 64741 triples, 7026 primitives. Up to 1000000: 808950 triples, 70229 primitives. Up to 10000000: 9706567 triples, 702309 primitives. Up to 100000000: 113236940 triples, 7023027 primitives. Using the power p up to 11, using ulong for xuint, and compiling with the dmd -L/STACK:10000000 switch to increase the stack size to about 10MB: Up to 10: 0 triples, 0 primitives. Up to 100: 17 triples, 7 primitives. Up to 1000: 325 triples, 70 primitives. Up to 10000: 4858 triples, 703 primitives. Up to 100000: 64741 triples, 7026 primitives. Up to 1000000: 808950 triples, 70229 primitives. Up to 10000000: 9706567 triples, 702309 primitives. Up to 100000000: 113236940 triples, 7023027 primitives. Up to 1000000000: 1294080089 triples, 70230484 primitives. Up to 10000000000: 14557915466 triples, 702304875 primitives. Total run-time up to 10_000_000_000: about 63 seconds. Waiting less than half an hour: Up to 10: 0 triples, 0 primitives. Up to 100: 17 triples, 7 primitives. Up to 1000: 325 triples, 70 primitives. Up to 10000: 4858 triples, 703 primitives. Up to 100000: 64741 triples, 7026 primitives. Up to 1000000: 808950 triples, 70229 primitives. Up to 10000000: 9706567 triples, 702309 primitives. Up to 100000000: 113236940 triples, 7023027 primitives. Up to 1000000000: 1294080089 triples, 70230484 primitives. Up to 10000000000: 14557915466 triples, 702304875 primitives. Up to 100000000000: 161750315680 triples, 7023049293 primitives. ## Erlang %%%% Pythagorian triples in Erlang, J.W. Luiten%%-module(triples).-export([main/1]). %% Transformations t1, t2 and t3 to generate new triples t1(A, B, C) -> {A-2*B+2*C, 2*A-B+2*C, 2*A-2*B+3*C}.t2(A, B, C) -> {A+2*B+2*C, 2*A+B+2*C, 2*A+2*B+3*C}.t3(A, B, C) -> {2*B+2*C-A, B+2*C-2*A, 2*B+3*C-2*A}. %% Generation of triplescount_triples(A, B, C, Tot_acc, Cnt_acc, Max_perimeter) when (A+B+C) =< Max_perimeter -> Tot1 = Tot_acc + Max_perimeter div (A+B+C), {A1, B1, C1} = t1(A, B, C), {Tot2, Cnt2} = count_triples(A1, B1, C1, Tot1, Cnt_acc+1, Max_perimeter), {A2, B2, C2} = t2(A, B, C), {Tot3, Cnt3} = count_triples(A2, B2, C2, Tot2, Cnt2, Max_perimeter), {A3, B3, C3} = t3(A, B, C), {Tot4, Cnt4} = count_triples(A3, B3, C3, Tot3, Cnt3, Max_perimeter), {Tot4, Cnt4};count_triples(_A, _B, _C, Tot_acc, Cnt_acc, _Max_perimeter) -> {Tot_acc, Cnt_acc}. count_triples(A, B, C, Pow) -> Max = trunc(math:pow(10, Pow)), {Tot, Prim} = count_triples(A, B, C, 0, 0, Max), {Pow, Tot, Prim}. count_triples(Pow) -> count_triples(3, 4, 5, Pow). %% Display a single result.display_result({Pow, Tot, Prim}) -> io:format("Up to 10 ** ~w : ~w triples, ~w primitives~n", [Pow, Tot, Prim]). main(Max) -> L = lists:seq(1, Max), Answer = lists:map(fun(X) -> count_triples(X) end, L), lists:foreach(fun(Result) -> display_result(Result) end, Answer). Output: Up to 10 ** 1 : 0 triples, 0 primitives Up to 10 ** 2 : 17 triples, 7 primitives Up to 10 ** 3 : 325 triples, 70 primitives Up to 10 ** 4 : 4858 triples, 703 primitives Up to 10 ** 5 : 64741 triples, 7026 primitives Up to 10 ** 6 : 808950 triples, 70229 primitives Up to 10 ** 7 : 9706567 triples, 702309 primitives Up to 10 ** 8 : 113236940 triples, 7023027 primitives Up to 10 ** 9 : 1294080089 triples, 70230484 primitives Up to 10 ** 10 : 14557915466 triples, 702304875 primitives Up to 10 ** 11 : 161750315680 triples, 7023049293 primitives  ## Euphoria Translation of: D function tri(atom lim, sequence in) sequence r atom p p = in[1] + in[2] + in[3] if p > lim then return {0, 0} end if r = {1, floor(lim / p)} r += tri(lim, { in[1]-2*in[2]+2*in[3], 2*in[1]-in[2]+2*in[3], 2*in[1]-2*in[2]+3*in[3]}) r += tri(lim, { in[1]+2*in[2]+2*in[3], 2*in[1]+in[2]+2*in[3], 2*in[1]+2*in[2]+3*in[3]}) r += tri(lim, {-in[1]+2*in[2]+2*in[3], -2*in[1]+in[2]+2*in[3], -2*in[1]+2*in[2]+3*in[3]}) return rend function atom max_perimax_peri = 10while max_peri <= 100000000 do printf(1,"%d: ", max_peri) ? tri(max_peri, {3, 4, 5}) max_peri *= 10end while Output: 10: {0,0} 100: {7,17} 1000: {70,325} 10000: {703,4858} 100000: {7026,64741} 1000000: {70229,808950} 10000000: {702309,9706567} 100000000: {7023027,113236940}  ## F# Translation of: OCaml let isqrt n = let rec iter t = let d = n - t*t if (0 <= d) && (d < t+t+1) // t*t <= n < (t+1)*(t+1) then t else iter ((t+(n/t))/2) iter 1 let rec gcd a b = let t = a % b if t = 0 then b else gcd b t let coprime a b = gcd a b = 1 let num_to ms = let mutable ctr = 0 let mutable prim_ctr = 0 let max_m = isqrt (ms/2) for m = 2 to max_m do for j = 0 to (m/2) - 1 do let n = m-(2*j+1) if coprime m n then let s = 2*m*(m+n) if s <= ms then ctr <- ctr + (ms/s) prim_ctr <- prim_ctr + 1 (ctr, prim_ctr) let show i = let s, p = num_to i in printfn "For perimeters up to %d there are %d total and %d primitive" i s p;; List.iter show [ 100; 1000; 10000; 100000; 1000000; 10000000; 100000000 ] Output: For perimeters up to 100 there are 17 total and 7 primitive For perimeters up to 1000 there are 325 total and 70 primitive For perimeters up to 10000 there are 4858 total and 703 primitive For perimeters up to 100000 there are 64741 total and 7026 primitive For perimeters up to 1000000 there are 808950 total and 70229 primitive For perimeters up to 10000000 there are 9706567 total and 702309 primitive For perimeters up to 100000000 there are 113236940 total and 7023027 primitive ## Factor Pretty slow (100 times slower than C)... USING: accessors arrays formatting kernel literals mathmath.functions math.matrices math.ranges sequences ;IN: rosettacode.pyth CONSTANT: T1 { { 1 2 2 } { -2 -1 -2 } { 2 2 3 }}CONSTANT: T2 { { 1 2 2 } { 2 1 2 } { 2 2 3 }}CONSTANT: T3 { { -1 -2 -2 } { 2 1 2 } { 2 2 3 }} CONSTANT: base { 3 4 5 } TUPLE: triplets-count primitives total ;: <0-triplets-count> ( -- a ) 0 0 \ triplets-count boa ;: next-triplet ( triplet T -- triplet' ) [ 1array ] [ m. ] bi* first ;: candidates-triplets ( seed -- candidates )${ T1 T2 T3 } [ next-triplet ] with map ;: add-triplets ( current-triples limit triplet -- stop )  sum 2dup > [   /i [ + ] curry change-total   [ 1 + ] change-primitives drop t   ] [ 3drop f ] if ;: all-triplets ( current-triples limit seed -- triplets )  3dup add-triplets [     candidates-triplets [ all-triplets ] with swapd reduce  ] [ 2drop ] if ;: count-triplets ( limit -- count )  <0-triplets-count> swap base all-triplets ;: pprint-triplet-count ( limit count -- )  [ total>> ] [ primitives>> ] bi   "Up to %d: %d triples, %d primitives.\n" printf ;: pyth ( -- )  8 [1,b] [ 10^ dup count-triplets pprint-triplet-count ] each ;
Up to 10: 0 triples, 0 primitives.
Up to 100: 17 triples, 7 primitives.
Up to 1000: 325 triples, 70 primitives.
Up to 10000: 4858 triples, 703 primitives.
Up to 100000: 64741 triples, 7026 primitives.
Up to 1000000: 808950 triples, 70229 primitives.
Up to 10000000: 9706567 triples, 702309 primitives.
Up to 100000000: 113236940 triples, 7023027 primitives.
Running time: 57.968821207 seconds

## Forth

   \ Two methods to create Pythagorean Triples\ this code has been tested using Win32Forth and gforth : pythag_fibo ( f1 f0 -- )     \ Create Pythagorean Triples from 4 element Fibonacci series     \ this is called with the first two members of a 4 element Fibonacci series     \ Price and Burkhart have two good articles about this method     \ "Pythagorean Tree: A New Species" and     \ "Heron's Formula, Descartes Circles, and Pythagorean Triangles"     \ Horadam found out how to compute Pythagorean Triples from Fibonacci series      \ compute the two other members of the Fibonacci series and put them in     \ local variables.  I was unable to do this with out using locals     2DUP + 2DUP + 2OVER 2DUP + 2DUP +     LOCALS| f3 f2 f1 f0 |      wk_level @  9 .r f0 8 .r  f1 8 .r  f2 8 .r  f3 8 .r      \ this block calculates the sides of the Pythagorean Triangle using single precision     \ f0 f3 * 14 .r                 \ side a  (always odd)     \ 2 f1 * f2 * 10 .r             \ side b  (a multiple of 4)     \ f0 f2 * f1 f3 * + 10 .r       \ side c, the hyponenuse, (always odd)      \ this block calculates double precision values     f0 f3 um* 15 d.r                    \ side a  (always odd)     2 f1 * f2 um* 15 d.r                \ side b  (a multiple of 4)     f0 f2 um* f1 f3 um* d+ 17 d.r  cr   \ side c, the hypotenuse, (always odd)      MAX_LEVEL @ wk_LEVEL @ U> IF   \ TRUE if MAX_LEVEL > WK_LEVEL     wk_level @ 1+ wk_level !      \ this creates a teranary tree of Pythagorean triples     \ use a two of the members of the Fibonacci series as seeds for the     \ next level     \ It's the same tree created by Barning or Hall using matrix multiplication     f3 f1 recurse     f3 f2 recurse     f0 f2 recurse      wk_level @ 1- wk_level !      else     then      drop drop drop drop ; \ implements the Fibonacci series -- Pythagorean triple\ the stack contents sets how many iteration levels there will be: pf_test     \ the stack contents set up the maximum level     max_level !     0 wk_level !     cr      \ call the function with the first two elements of the base Fibonacci series     1 1 pythag_fibo  ; : gcd ( a b -- gcd )  begin ?dup while tuck mod repeat ; \ this is the classical algorithm, known to Euclid, it is explained in many\ books on Number Theory\ this generates all primitive Pythagorean triples \ i -- inner loop index or current loop index\ j -- outer loop index\ stack contents is the upper limit for j\ i and j can not both be odd\ the gcd( i, j ) must be 1\ j is greater than i\ the stack contains the upper limit of the j variable: pythag_ancn  ( limit -- )     cr     1 + 2 do        i 1 and if 2 else 1 then        \ this sets the start value of the inner loop so that        \ if the outer loop index is odd only even inner loop indices happen        \ if the outer loop index is even only odd inner loop indices happen        i swap do             i j gcd 1 - 0> if else  \ do this if gcd( i, j ) is 1             j 5 .r i 5 .r              \ j j * i i * - 12 .r   \ a side of Pythagorean triangle (always odd)             \ i j * 2 * 9 .r        \ b side of Pythagorean triangle (multiple of 4)             \ i i * j j * + 9 .r    \ hypotenuse of Pythagorean triangle (always odd)              \ this block calculates double precision Pythagorean triple values             j j um* i i um* d- 15 d.r    \ a side of Pythagorean triangle (always odd)             i j um* d2* 15 d.r           \ b side of Pythagorean triangle (multiple of 4)             i i um* j j um* d+ 17 d.r    \ hypotenuse of Pythagorean triangle (always odd)              cr then 2 +loop       \ keep i being all odd or all even     loop ;   Current directory: C:\Forth okFLOAD 'C:\Forth\ancien_fibo_pythag.F'  ok  ok   ok  ok3 pf_test         0       1       1       2       3              3              4                5        1       3       1       4       5             15              8               17        2       5       1       6       7             35             12               37        3       7       1       8       9             63             16               65        3       7       6      13      19            133            156              205        3       5       6      11      17             85            132              157        2       5       4       9      13             65             72               97        3      13       4      17      21            273            136              305        3      13       9      22      31            403            396              565        3       5       9      14      23            115            252              277        2       3       4       7      11             33             56               65        3      11       4      15      19            209            120              241        3      11       7      18      25            275            252              373        3       3       7      10      17             51            140              149        1       3       2       5       7             21             20               29        2       7       2       9      11             77             36               85        3      11       2      13      15            165             52              173        3      11       9      20      29            319            360              481        3       7       9      16      25            175            288              337        2       7       5      12      17            119            120              169        3      17       5      22      27            459            220              509        3      17      12      29      41            697            696              985        3       7      12      19      31            217            456              505        2       3       5       8      13             39             80               89        3      13       5      18      23            299            180              349        3      13       8      21      29            377            336              505        3       3       8      11      19             57            176              185        1       1       2       3       5              5             12               13        2       5       2       7       9             45             28               53        3       9       2      11      13            117             44              125        3       9       7      16      23            207            224              305        3       5       7      12      19             95            168              193        2       5       3       8      11             55             48               73        3      11       3      14      17            187             84              205        3      11       8      19      27            297            304              425        3       5       8      13      21            105            208              233        2       1       3       4       7              7             24               25        3       7       3      10      13             91             60              109        3       7       4      11      15            105             88              137        3       1       4       5       9              9             40               41 ok  ok10 pythag_ancn     2    1              3              4                5    3    2              5             12               13    4    1             15              8               17    4    3              7             24               25    5    2             21             20               29    5    4              9             40               41    6    1             35             12               37    6    5             11             60               61    7    2             45             28               53    7    4             33             56               65    7    6             13             84               85    8    1             63             16               65    8    3             55             48               73    8    5             39             80               89    8    7             15            112              113    9    2             77             36               85    9    4             65             72               97    9    8             17            144              145   10    1             99             20              101   10    3             91             60              109   10    7             51            140              149   10    9             19            180              181 ok

## Fortran

Works with: Fortran version 90 and later
Translation of: C efficient method
module triples  implicit none   integer :: max_peri, prim, total  integer :: u(9,3) = reshape((/ 1, -2, 2,  2, -1, 2,  2, -2, 3, &                                 1,  2, 2,  2,  1, 2,  2,  2, 3, &                                -1,  2, 2, -2,  1, 2, -2,  2, 3 /), &                                (/ 9, 3 /)) contains recursive subroutine new_tri(in)  integer, intent(in) :: in(:)  integer :: i  integer :: t(3), p   p = sum(in)  if (p > max_peri) return   prim = prim + 1  total = total + max_peri / p  do i = 1, 3    t(1) = sum(u(1:3, i) * in)    t(2) = sum(u(4:6, i) * in)    t(3) = sum(u(7:9, i) * in)    call new_tri(t);  end doend subroutine new_triend module triples program Pythagorean  use triples  implicit none   integer :: seed(3) = (/ 3, 4, 5 /)   max_peri = 10  do    total = 0    prim = 0    call new_tri(seed)    write(*, "(a, i10, 2(i10, a))") "Up to", max_peri, total, " triples",  prim, " primitives"    if(max_peri == 100000000) exit    max_peri = max_peri * 10  end doend program Pythagorean
Output:
Up to         10          0 triples         0 primitives
Up to        100         17 triples         7 primitives
Up to       1000        325 triples        70 primitives
Up to      10000       4858 triples       703 primitives
Up to     100000      64741 triples      7026 primitives
Up to    1000000     808950 triples     70229 primitives
Up to   10000000    9706567 triples    702309 primitives
Up to  100000000  113236940 triples   7023027 primitives

## Go

package main import "fmt" var total, prim, maxPeri int64 func newTri(s0, s1, s2 int64) {    if p := s0 + s1 + s2; p <= maxPeri {        prim++        total += maxPeri / p        newTri(+1*s0-2*s1+2*s2, +2*s0-1*s1+2*s2, +2*s0-2*s1+3*s2)        newTri(+1*s0+2*s1+2*s2, +2*s0+1*s1+2*s2, +2*s0+2*s1+3*s2)        newTri(-1*s0+2*s1+2*s2, -2*s0+1*s1+2*s2, -2*s0+2*s1+3*s2)    }} func main() {    for maxPeri = 100; maxPeri <= 1e11; maxPeri *= 10 {        prim = 0        total = 0        newTri(3, 4, 5)        fmt.Printf("Up to %d:  %d triples, %d primitives\n",            maxPeri, total, prim)    }}

Output:

Up to 100:  17 triples, 7 primitives
Up to 1000:  325 triples, 70 primitives
Up to 10000:  4858 triples, 703 primitives
Up to 100000:  64741 triples, 7026 primitives
Up to 1000000:  808950 triples, 70229 primitives
Up to 10000000:  9706567 triples, 702309 primitives
Up to 100000000:  113236940 triples, 7023027 primitives
Up to 1000000000:  1294080089 triples, 70230484 primitives
Up to 10000000000:  14557915466 triples, 702304875 primitives
Up to 100000000000:  161750315680 triples, 7023049293 primitives


## Groovy

### Parent/Child Algorithm

Solution:

class Triple {    BigInteger a, b, c    def getPerimeter() { this.with { a + b + c } }    boolean isValid() { this.with { a*a + b*b == c*c } }} def initCounts (def n = 10) {    (n..1).collect { 10g**it }.inject ([:]) { Map map, BigInteger perimeterLimit ->        map << [(perimeterLimit): [primative: 0g, total: 0g]]    }} def findPythagTriples, findChildTriples findPythagTriples = {Triple t = new Triple(a:3, b:4, c:5), Map counts = initCounts() ->    def p = t.perimeter    def currentCounts = counts.findAll { pLimit, tripleCounts -> p <= pLimit }    if (! currentCounts || ! t.valid) { return }    currentCounts.each { pLimit, tripleCounts ->        tripleCounts.with { primative ++; total += pLimit.intdiv(p) }    }    findChildTriples(t, currentCounts)    counts} findChildTriples = { Triple t, Map counts ->    t.with {        [            [ a - 2*b + 2*c,  2*a - b + 2*c,  2*a - 2*b + 3*c],            [ a + 2*b + 2*c,  2*a + b + 2*c,  2*a + 2*b + 3*c],            [-a + 2*b + 2*c, -2*a + b + 2*c, -2*a + 2*b + 3*c]        ]*.sort().each { aa, bb, cc ->            findPythagTriples(new Triple(a:aa, b:bb, c:cc), counts)        }    }}

Test:

printf ('    LIMIT       PRIMATIVE          ALL\n')findPythagTriples().sort().each { perimeterLimit, result ->    def exponent = perimeterLimit.toString().size() - 1    printf ('a+b+c <= 10E%2d  %9d %12d\n', exponent, result.primative, result.total)}

Output:

    LIMIT       PRIMATIVE          ALL
a+b+c <= 10E 1          0            0
a+b+c <= 10E 2          7           17
a+b+c <= 10E 3         70          325
a+b+c <= 10E 4        703         4858
a+b+c <= 10E 5       7026        64741
a+b+c <= 10E 6      70229       808950
a+b+c <= 10E 7     702309      9706567
a+b+c <= 10E 8    7023027    113236940
a+b+c <= 10E 9   70230484   1294080089
a+b+c <= 10E10  702304875  14557915466

## Icon and Unicon

This uses the elegant formula (#IV) from Formulas for generating Pythagorean triples

 link numberslink printf procedure main(A)  # P-triples    plimit := (0 < integer(\A[1])) | 100 # get perimiter limit    nonprimitiveS := set()  # record unique non-primitives triples   primitiveS := set()     # record unique primitive triples    u :=  0   while (g := (u +:= 1)^2) + 3 * u + 2 < plimit / 2 do {      every v := seq(1) do {         a := g + (i := 2*u*v)         b := (h := 2*v^2) + i         c := g + h + i         if (p := a + b + c) > plimit then break           insert( (gcd(u,v)=1 & u%2=1, primitiveS) | nonprimitiveS, memo(a,b,c))          every k := seq(2) do {      # k is for larger non-primitives                      if k*p > plimit then break                  insert(nonprimitiveS,memo(a*k,b*k,c*k) )            }         }      } printf("Under perimiter=%d: Pythagorean Triples=%d including primitives=%d\n",       plimit,*nonprimitiveS+*primitiveS,*primitiveS)  every put(gcol := []  , &collections)printf("Time=%d, Collections: total=%d string=%d block=%d",&time,gcol[1],gcol[3],gcol[4])end  procedure memo(x[]) #: return a csv string of arguments in sorted orderevery (s := "") ||:= !sort(x) do s ||:= ","return s[1:-1]end

The output from some sample runs with BLKSIZE=500000000 and STRSIZE=50000000 is below. It starts getting very slow after 10M at about 9 minutes (times are shown in ms. I suspect there may be faster gcd algorithms that would speed this up.

Output:
Under perimiter=10: Pythagorean Triples=0 including primitives=0
Time=3, Collections: total=0 string=0 block=0

Under perimiter=100: Pythagorean Triples=17 including primitives=7
Time=3, Collections: total=0 string=0 block=0

Under perimiter=1000: Pythagorean Triples=325 including primitives=70
Time=6, Collections: total=0 string=0 block=0

Under perimiter=10000: Pythagorean Triples=4858 including primitives=703
Time=57, Collections: total=0 string=0 block=0

Under perimiter=100000: Pythagorean Triples=64741 including primitives=7026
Time=738, Collections: total=0 string=0 block=0

Under perimiter=1000000: Pythagorean Triples=808950 including primitives=70229
Time=12454, Collections: total=0 string=0 block=0

Under perimiter=10000000: Pythagorean Triples=9706567 including primitives=702309
Time=560625, Collections: total=16 string=8 block=8

pytr n = filter (\(_, a, b, c) -> a+b+c <= n) [(prim a b c, a, b, c) | a <- [1..n], b <- [1..n], c <- [1..n], a < b && b < c, a^2 + b^2 == c^2]    where prim a b _ = gcd a b == 1 main = putStrLn $"Up to 100 there are " ++ (show$ length xs) ++ " triples, of which " ++ (show $length$ filter (\(x,_,_,_) -> x == True) xs) ++ " are primitive."    where xs = pytr 100
Up to 100 there are 17 triples, of which 7 are primitive.

Recursive primitive generation:

triangles :: Int -> [[Int]]triangles max_peri | max_peri < 12 = []            | otherwise = concat tiers where    tiers = takeWhile (not.null) $iterate tier [[3,4,5]] tier = concatMap (filter ((<=max_peri).sum).tmul) tmul t = map (map (sum . zipWith (*) t)) [[[ 1,-2,2],[ 2,-1,2],[ 2,-2,3]], [[ 1, 2,2],[ 2, 1,2],[ 2, 2,3]], [[-1, 2,2],[-2, 1,2],[-2, 2,3]]] triangle_count max_p = (length t, sum$ map ((max_p div).sum) t)    where t = triangles max_p main = mapM_ putStrLn $map (\n -> show n ++ " " ++ show (triangle_count n))$ map (10^) [1..]
Output:
10 (0,0)
100 (7,17)
1000 (70,325)
10000 (703,4858)
100000 (7026,64741)
1000000 (70229,808950)
10000000 (702309,9706567)
...


## J

Brute force approach:

pytr=: 3 :0  r=. i. 0 3  for_a. 1 + i. <.(y-1)%3 do.    b=. 1 + a + i. <.(y%2)-3*a%2    c=. a +&.*: b    keep=. (c = <.c) *. y >: a+b+c    if. 1 e. keep do.      r=. r, a,.b ,.&(keep&#) c     end.  end.  (,.~ prim"1)r) prim=: 1 = 2 +./@{. |:

Example use:

First column indicates whether the triple is primitive, and the remaining three columns are a, b and c.

   pytr 1001  3  4  51  5 12 130  6  8 101  7 24 251  8 15 170  9 12 151  9 40 410 10 24 260 12 16 201 12 35 370 15 20 250 15 36 390 16 30 340 18 24 301 20 21 290 21 28 350 24 32 40   (# , [: {. +/) pytr 100 0   (# , [: {. +/) pytr 10017 7   (# , [: {. +/) pytr 1000325 70   (# , [: {. +/) pytr 100004858 703

pytr 10000 takes 4 seconds on this laptop, and time to complete grows with square of perimeter, so pytr 1e6 should take something like 11 hours using this algorithm on this machine.

A slightly smarter approach:

trips=:3 :0  'm n'=. |:(#~ 1 = 2 | +/"1)(#~ >/"1) ,/ ,"0/~ }. i. <. %: y  prim=. (#~ 1 = 2 +./@{. |:) (#~ y >: +/"1)m (-&*: ,. +:@* ,. +&*:) n  /:~ ; <@(,.~ # {. 1:)@(*/~ 1 + y i.@<.@% +/)"1 prim)

usage for trips is the same as for pytr. Thus:

   (# , 1 {. +/) trips 100 0   (# , 1 {. +/) trips 10017 7   (# , 1 {. +/) trips 1000325 70   (# , 1 {. +/) trips 100004858 703   (# , 1 {. +/) trips 10000064741 7026   (# , 1 {. +/) trips 1000000808950 70229   (# , 1 {. +/) trips 100000009706567 702309

The last line took about 16 seconds.

That said, we do not actually have to generate all the triples, we just need to count them. Thus:

trc=:3 :0  'm n'=. |:(#~ 1 = 2 | +/"1)(#~ >/"1) ,/ ,"0/~ }. i. <. %: y  <.y%+/"1 (#~ 1 = 2 +./@{. |:) (#~ y >: +/"1)m (-&*: ,. +:@* ,. +&*:) n)

The result is a list of positive integers, one number for each primitive triple which fits within the limit, giving the number of triples which are multiples of that primitive triple whose perimeter is no greater than the limiting perimeter.

   (#,+/)trc 1e87023027 113236940

But note that J's memory footprint reached 6.7GB during the computation, so to compute larger values the computation would have to be broken up into reasonable sized blocks.

## Java

### Brute force

Theoretically, this can go "forever", but it takes a while, so only the minimum is shown. Luckily, BigInteger has a GCD method built in.
 import java.math.BigInteger;import static java.math.BigInteger.ONE; public class PythTrip{     public static void main(String[] args){        long tripCount = 0, primCount = 0;         //change this to whatever perimeter limit you want;the RAM's the limit        BigInteger periLimit = BigInteger.valueOf(100),                peri2 = periLimit.divide(BigInteger.valueOf(2)),                peri3 = periLimit.divide(BigInteger.valueOf(3));         for(BigInteger a = ONE; a.compareTo(peri3) < 0; a = a.add(ONE)){            BigInteger aa = a.multiply(a);             for(BigInteger b = a.add(ONE);                    b.compareTo(peri2) < 0; b = b.add(ONE)){                BigInteger bb = b.multiply(b);                BigInteger ab = a.add(b);                BigInteger aabb = aa.add(bb);                 for(BigInteger c = b.add(ONE);                        c.compareTo(peri2) < 0; c = c.add(ONE)){                     int compare = aabb.compareTo(c.multiply(c));                    //if a+b+c > periLimit                    if(ab.add(c).compareTo(periLimit) > 0){                        break;                    }                    //if a^2 + b^2 != c^2                    if(compare < 0){                        break;                    }else if (compare == 0){                        tripCount++;                        System.out.print(a + ", " + b + ", " + c);                         //does binary GCD under the hood                        if(a.gcd(b).equals(ONE)){                            System.out.print(" primitive");                            primCount++;                        }                        System.out.println();                    }                }            }        }        System.out.println("Up to a perimeter of " + periLimit + ", there are "                + tripCount + " triples, of which " + primCount + " are primitive.");    }}

Output:

3, 4, 5 primitive
5, 12, 13 primitive
6, 8, 10
7, 24, 25 primitive
8, 15, 17 primitive
9, 12, 15
9, 40, 41 primitive
10, 24, 26
12, 16, 20
12, 35, 37 primitive
15, 20, 25
15, 36, 39
16, 30, 34
18, 24, 30
20, 21, 29 primitive
21, 28, 35
24, 32, 40
Up to a perimeter of 100, there are 17 triples, of which 7 are primitive.

### Parent/child

Translation of: Perl 6
(with limited modification for saving a few BigInteger operations)
Works with: Java version 1.5+

This can also go "forever" theoretically. Letting it go to another order of magnitude overflowed the stack on the computer this was tested on. This version also does not show the triples as it goes, it only counts them.

import java.math.BigInteger; public class Triples{    public static BigInteger LIMIT;    public static final BigInteger TWO = BigInteger.valueOf(2);    public static final BigInteger THREE = BigInteger.valueOf(3);    public static final BigInteger FOUR = BigInteger.valueOf(4);    public static final BigInteger FIVE = BigInteger.valueOf(5);    public static long primCount = 0;    public static long tripCount = 0;     //I don't know Japanese :p    public static void parChild(BigInteger a, BigInteger b, BigInteger c){        BigInteger perim = a.add(b).add(c);        if(perim.compareTo(LIMIT) > 0) return;        primCount++; tripCount += LIMIT.divide(perim).longValue();        BigInteger a2 = TWO.multiply(a), b2 = TWO.multiply(b), c2 = TWO.multiply(c),                   c3 = THREE.multiply(c);        parChild(a.subtract(b2).add(c2),                 a2.subtract(b).add(c2),                 a2.subtract(b2).add(c3));        parChild(a.add(b2).add(c2),                 a2.add(b).add(c2),                 a2.add(b2).add(c3));        parChild(a.negate().add(b2).add(c2),                 a2.negate().add(b).add(c2),                 a2.negate().add(b2).add(c3));    }     public static void main(String[] args){        for(long i = 100; i <= 10000000; i*=10){            LIMIT = BigInteger.valueOf(i);            primCount = tripCount = 0;            parChild(THREE, FOUR, FIVE);            System.out.println(LIMIT + ": " + tripCount + " triples, " + primCount + " primitive.");        }    }}

Output:

100: 17 triples, 7 primitive.
1000: 325 triples, 70 primitive.
10000: 4858 triples, 703 primitive.
100000: 64741 triples, 7026 primitive.
1000000: 808950 triples, 70229 primitive.
10000000: 9706567 triples, 702309 primitive.

## jq

Works with: jq version 1.4

The jq program presented here is based on Euclid's formula, and uses the same algorithm and notation as in the AutoHotKey section.

The implementation illustrates how an inner function with arity 0 can attain a high level of efficiency with both jq 1.4 and later. A simpler implementation is possible with versions of jq greater than 1.4.

def gcd(a; b):  def _gcd:    if .[1] == 0 then .[0]    else [.[1], .[0] % .[1]] | _gcd    end;  [a,b] | _gcd ; # Return: [total, primitives] for pythagorean triangles having# perimeter no larger than peri.# The following uses Euclid's formula with the convention: m > n.def count(peri):   # The inner function can be used to count for a given value of m:  def _count:    # state [n,m,p, [total, primitives]]    .[0] as $n | .[1] as$m | .[2] as $p | if$n < $m and$p <= peri then        if (gcd($m;$n) == 1)        then .[3] | [ (.[0] + ((peri/$p)|floor) ), (.[1] + 1)] else .[3] end | [$n+2, $m,$p+4*$m, .] | _count else . end; # m^2 < m*(m+1) <= m*(m+n) = perimeter/2 reduce range(2; (peri/2) | sqrt + 1) as$m    ( [1, 2, 12, [0,0]];      (($m % 2) + 1) as$n      | (2 * $m * ($m + $n) ) as$p   # a+b+c for this (m,n)      | [$n,$m, $p, .[3]] | _count ) | .[3] ; # '''Example''':<lang jq>def pow(i): . as$in | reduce range(0; i) as $j (1; . *$in); range(1; 9) | . as $i | 10|pow($i) as $i | "\($i):  \(count($i) )"  Output: $ jq -M -c -r -n -f Pythagorean_triples.jq10:  [0,0]100:  [17,7]1000:  [325,70]10000:  [4858,703]100000:  [64741,7026]1000000:  [808950,70229]10000000:  [9706567,702309]100000000:  [113236940,7023027]

## Lasso

// Brute Force: Too slow for large numbersdefine num_pythagorean_triples(max_perimeter::integer) => {   local(max_b) = (#max_perimeter / 3)*2    return (      with a in 1 to (#max_b - 1)      sum integer(         with b in (#a + 1) to #max_b         let c = math_sqrt(#a*#a + #b*#b)         where #c == integer(#c)         where #c > #b         where (#a+#b+#c) <= #max_perimeter         sum 1      )   )}stdout(Number of Pythagorean Triples in a Perimeter of 100: )stdoutnl(num_pythagorean_triples(100))

Output:

Number of Pythagorean Triples in a Perimeter of 100: 17


## Liberty BASIC

 print time$() for power =1 to 6 perimeterLimit =10^power upperBound =int( 1 +perimeterLimit^0.5) primitives =0 triples =0 extras =0 ' will count the in-range multiples of any primitive for m =2 to upperBound for n =1 +( m mod 2 =1) to m -1 step 2 term1 =2 *m *n term2 =m *m -n *n term3 =m *m +n *n perimeter =term1 +term2 +term3 if perimeter <=perimeterLimit then triples =triples +1 a =term1 b =term2 do r = a mod b a =b b =r loop until r <=0 if ( a =1) and ( perimeter <=perimeterLimit) then 'we've found a primitive triple if a =1, since hcf =1. And it is inside perimeter range. Save it in an array primitives =primitives +1 if term1 >term2 then temp =term1: term1 =term2: term2 =temp 'swap so in increasing order of side length nEx =int( perimeterLimit /perimeter) 'We have the primitive & removed any multiples. Now calculate ALL the multiples in range. extras =extras +nEx end if scan next n next m print " Number of primitives having perimeter below "; 10^power, " was "; primitives, " & "; extras, " non-primitive triples." print time$()next power print "End"end
17:59:34
Number of primitives having perimeter below 10 was 0 & 0 non-primitive triples.
17:59:34
Number of primitives having perimeter below 100 was 7 & 17 non-primitive triples.
17:59:34
Number of primitives having perimeter below 1000 was 70 & 325 non-primitive triples.
17:59:34
Number of primitives having perimeter below 10000 was 703 & 4858 non-primitive triples.
17:59:35
Number of primitives having perimeter below 100000 was 7026 & 64741 non-primitive triples.
17:59:42
Number of primitives having perimeter below 1000000 was 70229 & 808950 non-primitive triples.
18:01:30
End


## Mathematica

Short code but not a very scalable approach...

pythag[n_] := Block[{soln = Solve[{a^2 + b^2 == c^2, a + b + c <= n, 0 < a < b < c}, {a, b, c}, Integers]},        {Length[soln], Count[GCD[a, b] == GCD[b, c] == GCD[c, a] == 1 /. soln, True]}      ]
Output:
pythag[10]
{0,0}

pythag[100]
{17, 7}

pythag[1000]
{325, 70}

## MATLAB / Octave

N=  100;  a = 1:N;  b = a(ones(N,1),:).^2;  b = b+b';  b = sqrt(b);  [y,x]=find(b==fix(b)); % test   % here some alternative tests  % b = b.^(1/k); [y,x]=find(b==fix(b)); % test 2  % [y,x]=find(b==(fix(b.^(1/k)).^k));  % test 3  % b=b.^(1/k); [y,x]=find(abs(b - round(b)) <= 4*eps*b);   z  = sqrt(x.^2+y.^2);  ix = (z+x+y<100) & (x < y) & (y < z);   p  = find(gcd(x(ix),y(ix))==1);   % find primitive triples   printf('There are %i Pythagorean Triples and %i primitive triples with a perimeter smaller than %i.\n',...         sum(ix), length(p), N);

Output:

 There are 17 Pythagorean Triples and 7 primitive triples with a perimeter smaller than 100.

## Modula-3

Note that this code only works on 64bit machines (where INTEGER is 64 bits). Modula-3 provides a LONGINT type, which is 64 bits on 32 bit systems, but there is a bug in the implementation apparently.

MODULE PyTriple64 EXPORTS Main; IMPORT IO, Fmt; VAR tcnt, pcnt, max, i: INTEGER; PROCEDURE NewTriangle(a, b, c: INTEGER; VAR tcount, pcount: INTEGER) =  VAR perim := a + b + c;        BEGIN    IF perim <= max THEN      pcount := pcount + 1;      tcount := tcount + max DIV perim;      NewTriangle(a-2*b+2*c, 2*a-b+2*c, 2*a-2*b+3*c, tcount, pcount);      NewTriangle(a+2*b+2*c, 2*a+b+2*c, 2*a+2*b+3*c, tcount, pcount);      NewTriangle(2*b+2*c-a, b+2*c-2*a, 2*b+3*c-2*a, tcount, pcount);    END;  END NewTriangle; BEGIN  i := 100;   REPEAT    max := i;    tcnt := 0;    pcnt := 0;    NewTriangle(3, 4, 5, tcnt, pcnt);    IO.Put(Fmt.Int(i) & ": " & Fmt.Int(tcnt) & " Triples, " &      Fmt.Int(pcnt) & " Primitives\n");    i := i * 10;  UNTIL i = 10000000;END PyTriple64.

Output:

100: 17 Triples, 7 Primitives
1000: 325 Triples, 70 Primitives
10000: 4858 Triples, 703 Primitives
100000: 64741 Triples, 7026 Primitives
1000000: 808950 Triples, 70229 Primitives


## Nimrod

Translation of: C
const u = [[ 1, -2,  2,  2, -1,  2,  2, -2,  3],           [ 1,  2,  2,  2,  1,  2,  2,  2,  3],           [-1,  2,  2, -2,  1,  2, -2,  2,  3]] var  total, prim = 0  maxPeri = 10 proc newTri(ins) =  var p = ins[0] + ins[1] + ins[2]  if p > maxPeri: return  inc(prim)  total += maxPeri div p   for i in 0..2:    newTri([u[i][0] * ins[0] + u[i][1] * ins[1] + u[i][2] * ins[2],            u[i][3] * ins[0] + u[i][4] * ins[1] + u[i][5] * ins[2],            u[i][6] * ins[0] + u[i][7] * ins[1] + u[i][8] * ins[2]]) while maxPeri <= 100_000_000:  total = 0  prim = 0  newTri([3, 4, 5])  echo "Up to ", maxPeri, ": ", total, " triples, ", prim, " primitives"  maxPeri *= 10

Output:

Up to 10: 0 triples, 0 primitives
Up to 100: 17 triples, 7 primitives
Up to 1000: 325 triples, 70 primitives
Up to 10000: 4858 triples, 703 primitives
Up to 100000: 64741 triples, 7026 primitives
Up to 1000000: 808950 triples, 70229 primitives
Up to 10000000: 9706567 triples, 702309 primitives
Up to 100000000: 113236940 triples, 7023027 primitives

## OCaml

let isqrt n =   let rec iter t =      let d = n - t*t in      if (0 <= d) && (d < t+t+1) (* t*t <= n < (t+1)*(t+1) *)      then t else iter ((t+(n/t))/2)   in iter 1 let rec gcd a b =   let t = a mod b in   if t = 0 then b else gcd b t let coprime a b = gcd a b = 1 let num_to ms =   let ctr = ref 0 in   let prim_ctr = ref 0 in   let max_m = isqrt (ms/2) in   for m = 2 to max_m do      for j = 0 to (m/2) - 1 do         let n = m-(2*j+1) in         if coprime m n then            let s = 2*m*(m+n) in            if s <= ms then               (ctr := !ctr + (ms/s); incr prim_ctr)      done   done;  (!ctr, !prim_ctr) let show i =  let s, p = num_to i in  Printf.printf "For perimeters up to %d there are %d total and %d primitive\n%!" i s p;; List.iter show [ 100; 1000; 10000; 100000; 1000000; 10000000; 100000000 ]

Output:

For perimeters up to 100 there are 17 total and 7 primitive
For perimeters up to 1000 there are 325 total and 70 primitive
For perimeters up to 10000 there are 4858 total and 703 primitive
For perimeters up to 100000 there are 64741 total and 7026 primitive
For perimeters up to 1000000 there are 808950 total and 70229 primitive
For perimeters up to 10000000 there are 9706567 total and 702309 primitive
For perimeters up to 100000000 there are 113236940 total and 7023027 primitive

## PARI/GP

This version is reasonably efficient and can handle inputs like a million quickly.

do(lim)={  my(prim,total,P);  lim\=1;  for(m=2,sqrtint(lim\2),    forstep(n=1+m%2,min(sqrtint(lim-m^2),m-1),2,      P=2*m*(m+n);      if(gcd(m,n)==1 && P<=lim,        prim++;        total+=lim\P      )    )  );  [prim,total]};do(100)

## Pascal

Program PythagoreanTriples (output); var  total, prim, maxPeri: int64; procedure newTri(s0, s1, s2: int64);  var    p: int64;  begin    p := s0 + s1 + s2;    if p <= maxPeri then    begin      inc(prim);      total := total + maxPeri div p;      newTri( s0 + 2*(-s1+s2),  2*( s0+s2) - s1,  2*( s0-s1+s2) + s2);      newTri( s0 + 2*( s1+s2),  2*( s0+s2) + s1,  2*( s0+s1+s2) + s2);      newTri(-s0 + 2*( s1+s2),  2*(-s0+s2) + s1,  2*(-s0+s1+s2) + s2);    end;  end; begin  maxPeri := 100;  while maxPeri <= 1e10 do  begin    prim := 0;    total := 0;    newTri(3, 4, 5);    writeln('Up to ', maxPeri, ': ', total, ' triples, ', prim, ' primitives.');    maxPeri := maxPeri * 10;  end;end.

Output (on Core2Duo 2GHz laptop):

time ./PythagoreanTriples
Up to 100: 17 triples, 7 primitives.
Up to 1000: 325 triples, 70 primitives.
Up to 10000: 4858 triples, 703 primitives.
Up to 100000: 64741 triples, 7026 primitives.
Up to 1000000: 808950 triples, 70229 primitives.
Up to 10000000: 9706567 triples, 702309 primitives.
Up to 100000000: 113236940 triples, 7023027 primitives.
Up to 1000000000: 1294080089 triples, 70230484 primitives.
Up to 10000000000: 14557915466 triples, 702304875 primitives.
109.694u 0.094s 1:50.43 99.4%   0+0k 0+0io 0pf+0w

## Perl

sub gcd {    my ($n,$m) = @_;    while($n){ my$t = $n;$n = $m %$n;        $m =$t;    }    return $m;} sub tripel { my$pmax  = shift;    my $prim = 0; my$count = 0;    my $nmax = sqrt($pmax)/2;    for( my $n=1;$n<=$nmax;$n++ ) {        for( my $m=$n+1; (my $p = 2*$m*($m+$n)) <= $pmax;$m+=2 ) {            next unless 1==gcd($m,$n);            $prim++;$count += int $pmax/$p;        }    }    printf "Max. perimeter: %d, Total: %d, Primitive: %d\n", $pmax,$count, $prim;} tripel 10**$_ for 1..8;
Output:
Max. perimeter: 10, Total: 0, Primitive: 0
Max. perimeter: 100, Total: 17, Primitive: 7
Max. perimeter: 1000, Total: 325, Primitive: 70
Max. perimeter: 10000, Total: 4858, Primitive: 703
Max. perimeter: 100000, Total: 64741, Primitive: 7026
Max. perimeter: 1000000, Total: 808950, Primitive: 70229
Max. perimeter: 10000000, Total: 9706567, Primitive: 702309
Max. perimeter: 100000000, Total: 113236940, Primitive: 7023027


## Perl 6

Here is a straight-forward, naive brute force implementation:

constant limit = 20; for [X] [^limit] xx 3 -> \a, \b, \c {    say [a, b, c] if a < b < c and a**2 + b**2 == c**2}

Here is a slightly less naive brute force implementation that is not really practical for large perimeter limits. 100 is no problem. 1000 is slow. 10000 is glacial.

my %triples;my $limit = 100; for 3 ..$limit/2 -> $c { for 1 ..$c -> $a { my$b = ($c *$c - $a *$a).sqrt;       last if $c +$a + $b >$limit;       last if $a >$b;       if $b ==$b.Int {          my $key = "$a $b$c";          %triples{$key} = ([gcd]$c, $a,$b.Int) > 1 ?? 0  !! 1;          say $key, %triples{$key} ?? ' - primitive' !! '';       }   } } say "There are {+%triples.keys} Pythagorean Triples with a perimeter <= $limit," ~"\nof which {[+] %triples.values} are primitive."; 3 4 5 - primitive 6 8 10 5 12 13 - primitive 9 12 15 8 15 17 - primitive 12 16 20 7 24 25 - primitive 15 20 25 10 24 26 20 21 29 - primitive 18 24 30 16 30 34 21 28 35 12 35 37 - primitive 15 36 39 24 32 40 9 40 41 - primitive There are 17 Pythagorean Triples with a perimeter <= 100, of which 7 are primitive.  Here's a much faster version. Hint, "oyako" is Japanese for "parent/child". :-) Works with: niecza sub triples($limit) {    my $primitive = 0; my$civilized = 0;     sub oyako($a,$b, $c) { my$perim = $a +$b + $c; return if$perim > $limit; ++$primitive; $civilized +=$limit div $perim; oyako($a - 2*$b + 2*$c,  2*$a -$b + 2*$c, 2*$a - 2*$b + 3*$c);        oyako( $a + 2*$b + 2*$c, 2*$a + $b + 2*$c,  2*$a + 2*$b + 3*$c); oyako(-$a + 2*$b + 2*$c, -2*$a +$b + 2*$c, -2*$a + 2*$b + 3*$c);    }     oyako(3,4,5);    "$limit => ($primitive $civilized)";} for 10,100,1000 ... * ->$limit {    say triples $limit;} Output: 10 => (0 0) 100 => (7 17) 1000 => (70 325) 10000 => (703 4858) 100000 => (7026 64741) 1000000 => (70229 808950) 10000000 => (702309 9706567) 100000000 => (7023027 113236940) 1000000000 => (70230484 1294080089) ^C The geometric sequence of limits will continue on forever, so eventually when you get tired of waiting (about a billion on my computer), you can just stop it. Another efficiency trick of note: we avoid passing the limit in as a parameter to the inner helper routine, but instead supply the limit via the lexical scope. Likewise, the accumulators are referenced lexically, so only the triples themselves need to be passed downward, and nothing needs to be returned. Here is an alternate version that avoids naming any scalars that can be handled by vector processing instead: constant @coeff = [[+1, -2, +2], [+2, -1, +2], [+2, -2, +3]], [[+1, +2, +2], [+2, +1, +2], [+2, +2, +3]], [[-1, +2, +2], [-2, +1, +2], [-2, +2, +3]]; sub triples($limit) {     sub oyako(@trippy) {        my $perim = [+] @trippy; return if$perim > $limit; take (1 + ($limit div $perim)i); for @coeff -> @nine { oyako (map -> @three { [+] @three »*« @trippy }, @nine); } return; } my$complex = 0i + [+] gather oyako([3,4,5]);    "$limit => ({$complex.re, $complex.im})";} for 10,100,1000 ... * ->$limit {    say triples $limit;} Using vectorized ops allows a bit more potential for parallelization, though this is probably not as big a win in this case, especially since we do a certain amount of multiplying by 1 that the scalar version doesn't need to do. Note the cute trick of adding complex numbers to add two numbers in parallel. The use of gather/take allows the summation to run in a different thread than the helper function, at least in theory... In practice, this solution runs considerably slower than the previous one, due primarily to passing gather/take values up many levels of dynamic scope. Eventually this may be optimized. Also, this solution currently chews up gigabytes of memory, while the previous solution stays at a quarter gig or so. This likely indicates a memory leak somewhere in niecza. ## PicoLisp Translation of: C (for (Max 10 (>= 100000000 Max) (* Max 10)) (let (Total 0 Prim 0 In (3 4 5)) (recur (In) (let P (apply + In) (when (>= Max P) (inc 'Prim) (inc 'Total (/ Max P)) (for Row (quote (( 1 -2 2) ( 2 -1 2) ( 2 -2 3)) (( 1 2 2) ( 2 1 2) ( 2 2 3)) ((-1 2 2) (-2 1 2) (-2 2 3)) ) (recurse (mapcar '((U) (sum * U In)) Row) ) ) ) ) ) (prinl "Up to " Max ": " Total " triples, " Prim " primitives.") ) ) Output: Up to 10: 0 triples, 0 primitives. Up to 100: 17 triples, 7 primitives. Up to 1000: 325 triples, 70 primitives. Up to 10000: 4858 triples, 703 primitives. Up to 100000: 64741 triples, 7026 primitives. Up to 1000000: 808950 triples, 70229 primitives. Up to 10000000: 9706567 triples, 702309 primitives. Up to 100000000: 113236940 triples, 7023027 primitives. ## PL/I Version 1 *process source attributes xref or(!); /********************************************************************* * REXX pgm counts number of Pythagorean triples * that exist given a max perimeter of N, * and also counts how many of them are primatives. * 05.05.2013 Walter Pachl translated from REXX version 2 *********************************************************************/ pyt: Proc Options(main); Dcl sysprint Print; Dcl (addr,mod,right) Builtin; Dcl memn Bin Fixed(31) Init(0); Dcl mabca(300) Char(12); Dcl 1 mabc, 2 ma Dec fixed(7), 2 mb Dec fixed(7), 2 mc Dec fixed(7); Dcl mabce Char(12) Based(addr(mabc)); Dcl 1 abc, 2 a Dec fixed(7), 2 b Dec fixed(7), 2 c Dec fixed(7); Dcl abce Char(12) Based(addr(abc)); Dcl (prims,trips,m,n,aa,aabb,cc,aeven,ab) Dec Fixed(7); mabca=''; trips=0; prims=0; n=100; la: Do a=3 To n/3; aa=a*a; /* limit side to 1/3 of perimeter.*/ aeven=mod(a,2)=0; lb:Do b=a+1 By 1+aeven; /* triangle can't be isosceles. */ ab=a+b; /* compute partial perimeter. */ If ab>=n Then Iterate la; /* a+b>perimeter? Try different A*/ aabb=aa+b*b; /* compute sum of a² + b² (cheat)*/ Do c=b+1 By 1; cc=c*c; /* 3rd side: also compute c² */ If aeven Then If mod(c,2)=0 Then Iterate; If ab+c>n Then Iterate la; /* a+b+c > perimeter? Try diff A.*/ If cc>aabb Then Iterate lb; /* c² > a²+b² ? Try different B.*/ If cc^=aabb Then Iterate; /* c² ¬= a²+b² ? Try different C.*/ If mema(abce) Then Iterate; trips=trips+1; /* eureka. */ prims=prims+1; /* count this primitive triple. */ Put Edit(a,b,c,' ',right(a**2+b**2,5),right(c**2,5),a+b+c) (Skip,f(4),2(f(5)),a,2(f(6)),f(9)); Do m=2 By 1; ma=a*m; mb=b*m; mc=c*m; /* gen non-primitives. */ If ma+mb+mc>n Then Leave; /* is this multiple a triple ? */ trips=trips+1; /* yuppers, then we found another.*/ If mod(m,2)=1 Then /* store as even multiple. */ call mems(mabce); Put Edit(ma,mb,mc,' * ', right(ma**2+mb**2,5),right(mc**2,5),ma+mb+mc) (Skip,f(4),2(f(5)),a,2(f(6)),f(9)); End; /* m */ End; /* c */ End; /* b */ End; /* a */ Put Edit('max perimeter = ',n, /* show a single line of output. */ 'Pythagorean triples =',trips, 'primitives =',prims) (Skip,a,f(5),2(x(9),a,f(4))); mems: Proc(e); Dcl e Char(12); memn+=1; mabca(memn)=e; End; mema: Proc(e) Returns(bit(1)); Dcl e Char(12); Do memi=1 To memn; If mabca(memi)=e Then Return('1'b); End; Return('0'b); End; End; Version 2  pythagorean: procedure options (main, reorder); /* 23 January 2014 */ declare (a, b, c) fixed (3); declare (asquared, bsquared) fixed; declare (triples, primitives) fixed binary(31) initial (0); do a = 1 to 100; asquared = a*a; do b = a+1 to 100; bsquared = b*b; do c = b+1 to 100; if a+b+c <= 100 then if asquared + bsquared = c*c then do; triples = triples + 1; if GCD(a,b) = 1 then primitives = primitives + 1; end; end; end; end; put skip data (triples, primitives); GCD: procedure (a, b) returns (fixed binary (31)) recursive; declare (a, b) fixed binary (31); if b = 0 then return (a); return (GCD (b, mod(a, b)) ); end GCD; end pythagorean; Output: TRIPLES= 17 PRIMITIVES= 7;  Output for P=1000: TRIPLES= 325 PRIMITIVES= 70;  ## PureBasic $(a=m^2-n^2) \and (b=2mn) \and (c=m^2+n^2) \and (p=a+b+c) \rightarrow$ $p=2m^2+2mn \le 100,000,000 \rightarrow$ $m^2+mn \le 50,000,000 \rightarrow$ $m^2+mn -50,000,000 \le 0 \rightarrow$ $n \le 10,000$  Procedure.i ConsoleWrite(t.s) ; compile using /CONSOLE option OpenConsole() PrintN (t.s) CloseConsole() ProcedureReturn 1EndProcedure Procedure.i StdOut(t.s) ; compile using /CONSOLE option OpenConsole() Print(t.s) CloseConsole() ProcedureReturn 1EndProcedure Procedure.i gcDiv(n,m) ; greatest common divisorif n=0:ProcedureReturn m:endifwhile m <> 0 if n > m n - m else m - n endifwendProcedureReturn nEndProcedure st=ElapsedMilliseconds() nmax =10000power =8 dim primitiveA(power)dim alltripleA(power)dim pmaxA(power) x=1for i=1 to power x*10:pmaxA(i)=x/2next for n=1 to nmax for m=(n+1) to (nmax+1) step 2 ; assure m-n is odd d=gcDiv(n,m) p=m*m+m*n for i=1 to power if p<=pmaxA(i) if d =1 primitiveA(i)+1 ; right here we have the primitive perimeter "seed" 'p' k=1:q=p*k ; set k to one to include p : use q as the 'p*k' while q<=pmaxA(i) alltripleA(i)+1 ; accumulate multiples of this perimeter while q <= pmaxA(i) k+1:q=p*k wend endif endif next nextnext for i=1 to power t.s="Up to "+str(pmaxA(i)*2)+": " t.s+str(alltripleA(i))+" triples, " t.s+str(primitiveA(i))+" primitives." ConsoleWrite(t.s)nextConsoleWrite("")et=ElapsedMilliseconds()-st:ConsoleWrite("Elapsed time = "+str(et)+" milliseconds")  Output >cmd /c "C:\_sys\temp\PythagoreanTriples.exe" Up to 10: 0 triples, 0 primitives. Up to 100: 17 triples, 7 primitives. Up to 1000: 325 triples, 70 primitives. Up to 10000: 4858 triples, 703 primitives. Up to 100000: 64741 triples, 7026 primitives. Up to 1000000: 808950 triples, 70229 primitives. Up to 10000000: 9706567 triples, 702309 primitives. Up to 100000000: 113236940 triples, 7023027 primitives. Elapsed time = 5163 milliseconds >Exit code: 0  ## Python Two methods, the second of which is much faster from fractions import gcd def pt1(maxperimeter=100): '''# Naive method ''' trips = [] for a in range(1, maxperimeter): aa = a*a for b in range(a, maxperimeter-a+1): bb = b*b for c in range(b, maxperimeter-b-a+1): cc = c*c if a+b+c > maxperimeter or cc > aa + bb: break if aa + bb == cc: trips.append((a,b,c, gcd(a, b) == 1)) return trips def pytrip(trip=(3,4,5),perim=100, prim=1): a0, b0, c0 = a, b, c = sorted(trip) t, firstprim = set(), prim>0 while a + b + c <= perim: t.add((a, b, c, firstprim>0)) a, b, c, firstprim = a+a0, b+b0, c+c0, False # t2 = set() for a, b, c, firstprim in t: a2, a5, b2, b5, c2, c3, c7 = a*2, a*5, b*2, b*5, c*2, c*3, c*7 if a5 - b5 + c7 <= perim: t2 |= pytrip(( a - b2 + c2, a2 - b + c2, a2 - b2 + c3), perim, firstprim) if a5 + b5 + c7 <= perim: t2 |= pytrip(( a + b2 + c2, a2 + b + c2, a2 + b2 + c3), perim, firstprim) if -a5 + b5 + c7 <= perim: t2 |= pytrip((-a + b2 + c2, -a2 + b + c2, -a2 + b2 + c3), perim, firstprim) return t | t2 def pt2(maxperimeter=100): '''# Parent/child relationship method:# http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples#XI. ''' trips = pytrip((3,4,5), maxperimeter, 1) return trips def printit(maxperimeter=100, pt=pt1): trips = pt(maxperimeter) print(" Up to a perimeter of %i there are %i triples, of which %i are primitive" % (maxperimeter, len(trips), len([prim for a,b,c,prim in trips if prim]))) for algo, mn, mx in ((pt1, 250, 2500), (pt2, 500, 20000)): print(algo.__doc__) for maxperimeter in range(mn, mx+1, mn): printit(maxperimeter, algo)  Output # Naive method Up to a perimeter of 250 there are 56 triples, of which 18 are primitive Up to a perimeter of 500 there are 137 triples, of which 35 are primitive Up to a perimeter of 750 there are 227 triples, of which 52 are primitive Up to a perimeter of 1000 there are 325 triples, of which 70 are primitive Up to a perimeter of 1250 there are 425 triples, of which 88 are primitive Up to a perimeter of 1500 there are 527 triples, of which 104 are primitive Up to a perimeter of 1750 there are 637 triples, of which 123 are primitive Up to a perimeter of 2000 there are 744 triples, of which 140 are primitive Up to a perimeter of 2250 there are 858 triples, of which 156 are primitive Up to a perimeter of 2500 there are 969 triples, of which 175 are primitive # Parent/child relationship method: # http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples#XI. Up to a perimeter of 500 there are 137 triples, of which 35 are primitive Up to a perimeter of 1000 there are 325 triples, of which 70 are primitive Up to a perimeter of 1500 there are 527 triples, of which 104 are primitive Up to a perimeter of 2000 there are 744 triples, of which 140 are primitive Up to a perimeter of 2500 there are 969 triples, of which 175 are primitive Up to a perimeter of 3000 there are 1204 triples, of which 211 are primitive Up to a perimeter of 3500 there are 1443 triples, of which 245 are primitive Up to a perimeter of 4000 there are 1687 triples, of which 280 are primitive Up to a perimeter of 4500 there are 1931 triples, of which 314 are primitive Up to a perimeter of 5000 there are 2184 triples, of which 349 are primitive Up to a perimeter of 5500 there are 2442 triples, of which 385 are primitive Up to a perimeter of 6000 there are 2701 triples, of which 422 are primitive Up to a perimeter of 6500 there are 2963 triples, of which 457 are primitive Up to a perimeter of 7000 there are 3224 triples, of which 492 are primitive Up to a perimeter of 7500 there are 3491 triples, of which 527 are primitive Up to a perimeter of 8000 there are 3763 triples, of which 560 are primitive Up to a perimeter of 8500 there are 4029 triples, of which 597 are primitive Up to a perimeter of 9000 there are 4304 triples, of which 631 are primitive Up to a perimeter of 9500 there are 4577 triples, of which 667 are primitive Up to a perimeter of 10000 there are 4858 triples, of which 703 are primitive Up to a perimeter of 10500 there are 5138 triples, of which 736 are primitive Up to a perimeter of 11000 there are 5414 triples, of which 770 are primitive Up to a perimeter of 11500 there are 5699 triples, of which 804 are primitive Up to a perimeter of 12000 there are 5980 triples, of which 839 are primitive Up to a perimeter of 12500 there are 6263 triples, of which 877 are primitive Up to a perimeter of 13000 there are 6559 triples, of which 913 are primitive Up to a perimeter of 13500 there are 6843 triples, of which 949 are primitive Up to a perimeter of 14000 there are 7129 triples, of which 983 are primitive Up to a perimeter of 14500 there are 7420 triples, of which 1019 are primitive Up to a perimeter of 15000 there are 7714 triples, of which 1055 are primitive Up to a perimeter of 15500 there are 8004 triples, of which 1089 are primitive Up to a perimeter of 16000 there are 8304 triples, of which 1127 are primitive Up to a perimeter of 16500 there are 8595 triples, of which 1159 are primitive Up to a perimeter of 17000 there are 8884 triples, of which 1192 are primitive Up to a perimeter of 17500 there are 9189 triples, of which 1228 are primitive Up to a perimeter of 18000 there are 9484 triples, of which 1264 are primitive Up to a perimeter of 18500 there are 9791 triples, of which 1301 are primitive Up to a perimeter of 19000 there are 10088 triples, of which 1336 are primitive Up to a perimeter of 19500 there are 10388 triples, of which 1373 are primitive Up to a perimeter of 20000 there are 10689 triples, of which 1408 are primitive Barebone minimum for this task: from sys import setrecursionlimitsetrecursionlimit(2000) # 2000 ought to be big enough for everybody def triples(lim, a = 3, b = 4, c = 5): l = a + b + c if l > lim: return (0, 0) return reduce(lambda x, y: (x[0] + y[0], x[1] + y[1]), [ (1, lim / l), triples(lim, a - 2*b + 2*c, 2*a - b + 2*c, 2*a - 2*b + 3*c), triples(lim, a + 2*b + 2*c, 2*a + b + 2*c, 2*a + 2*b + 3*c), triples(lim, -a + 2*b + 2*c, -2*a + b + 2*c, -2*a + 2*b + 3*c) ]) for peri in [10 ** e for e in range(1, 8)]: print peri, triples(peri) Output: 10 (0, 0)100 (7, 17)1000 (70, 325)10000 (703, 4858)100000 (7026, 64741)1000000 (70229, 808950)10000000 (702309, 9706567) ## Racket #lang racket #| Euclid's enumeration formula and counting is fast enough for extra credit. For maximum perimeter P₀, the primitive triples are enumerated by n,m with: 1 ≤ n < m perimeter P(n, m) ≤ P₀ where P(n, m) = (m² - n²) + 2mn + (m² + n²) = 2m(m+n) m and n of different parity and coprime. Since n < m, a simple close non-tight bound on n is P(n, n) < P₀. For each of these the exact set of m's can be enumerated. Each primitive triple with perimeter p represents one triple for each kp ≤ P₀, of which there are floor(P₀/p) k's. |# (define (P n m) (* 2 m (+ m n)))(define (number-of-triples P₀) (for/fold ([primitive 0] [all 0]) ([n (in-naturals 1)] #:break (>= (P n n) P₀)) (for*/fold ([primitive′ primitive] [all′ all]) ([m (in-naturals (+ n 1))] #:break (> (P n m) P₀) #:when (and (odd? (- m n)) (coprime? m n))) (values (+ primitive′ 1) (+ all′ (quotient P₀ (P n m))))))) (define (print-results P₀) (define-values (primitive all) (number-of-triples P₀)) (printf "~a ~a:\n ~a, ~a.\n" "Number of Pythagorean triples and primitive triples with perimeter ≤" P₀ all primitive))(print-results 100)(time (print-results (* 100 1000 1000))) #| Number of Pythagorean triples and primitive triples with perimeter ≤ 100: 17, 7. Number of Pythagorean triples and primitive triples with perimeter ≤ 100000000: 113236940, 7023027. cpu time: 11976 real time: 12215 gc time: 2381|# ## REXX ### version 1 A brute force approach. /*REXX pgm counts number of Pythagorean triples that exist given a max *//* perimeter of N, and also counts how many of them are primitives.*/trips=0; prims=0 /*zero # of triples, primatives. */parse arg N .; if N=='' then n=100 /*get "N". If none, then assume.*/ do a=3 to N%3; aa=a*a /*limit side to 1/3 of perimeter.*/ do b=a+1 /*triangle can't be isosceles. */ ab=a+b /*compute partial perimeter. */ if ab>=N then iterate a /*a+b≥perimeter? Try different A*/ aabb=aa+b*b /*compute sum of a² + b² (cheat)*/ do c=b+1 /*3rd side: also compute c² */ if ab+c>N then iterate a /*a+b+c > perimeter? Try diff A.*/ cc=c*c /*compute C². */ if cc > aabb then iterate b /*c² > a²+b² ? Try different B.*/ if cc\==aabb then iterate /*c² ¬= a²+b² ? Try different C.*/ trips=trips+1 /*eureka. We found a prim triple*/ prims=prims+(gcd(a,b)==1) /*is this triple a primitive? */ end /*a*/ end /*b*/ end /*c*/ say 'max perimeter =' N, /*show a single line of output. */ left('',7) "Pythagorean triples =" trips, /*left('',7)≡7 blanks.*/ left('',7) 'primitives =' primsexit /*stick a fork in it, we're done.*//*──────────────────────────────────GCD subroutine──────────────────────*/gcd: procedure; arg x,y; do until _==0; _=x//y; x=y; y=_; end; return x output when using the default input of: 100 max perimeter = 100 Pythagorean triples = 17 primitives = 7  output when using the input of: 1000 max perimeter = 1000 Pythagorean triples = 325 primitives = 70  ### version 2 Also a brute force approach. This REXX version takes advantage that primitive Pythagorean triples must have one and only one even number. Non-primitive Pythagorean triples are generated after a primitive triple is found. /*REXX pgm counts number of Pythagorean triples that exist given a max *//* perimeter of N, and also counts how many of them are primitives.*/@.=0; trips=0; prims=0 /*zero some REXX variables. */parse arg N .; if N=='' then N=100 /*get "N". If none, then assume.*/ do a=3 to N%3; aa=a*a /*limit side to 1/3 of perimeter.*/ aEven= a//2==0 /*set var to 1 if A is even.*/ do b=a+1 by 1+aEven /*triangle can't be isosceles. */ ab=a+b /*compute partial perimeter. */ if ab>=N then iterate a /*a+b≥perimeter? Try different A*/ aabb=aa+b*b /*compute sum of a² + b² (cheat)*/ do c=b+1 /*C is the third side of triangle*/ if aEven then if c//2==0 then iterate if ab+c>n then iterate a /*a+b+c > perimeter? Try diff A.*/ cc=c*c /*compute C². */ if cc > aabb then iterate b /*c² > a²+b² ? Try different B.*/ if cc\==aabb then iterate /*c² ¬= a²+b² ? Try different C.*/ if @.a.b.c then iterate /*Is this a duplicate? Try again*/ trips=trips+1 /*eureka. We found a prim triple*/ prims=prims+1 /*count this primitive triple. */ do m=2; am=a*m; bm=b*m; cm=c*m /*gen non-primitives.*/ if am+bm+cm>N then leave /*is this multiple a triple ? */ trips=trips+1 /*yuppers, then we found another.*/ if m//2 then @.am.bm.cm=1 /*don't mark if an even multiple.*/ end /*m*/ end /*d*/ end /*b*/ end /*a*/ say 'max perimeter =' N, /*show a single line of output. */ left('',7) "Pythagorean triples =" trips, /*left('',7)≡7 blanks.*/ left('',7) 'primitives =' prims /*stick a fork in it, we're done.*/ output is identical to version 1. ## Ruby Translation of: Java class PythagoranTriplesCounter def initialize(limit) @limit = limit @total = 0 @primatives = 0 generate_triples(3, 4, 5) end attr_reader :total, :primatives private def generate_triples(a, b, c) perim = a + b + c return if perim > @limit @primatives += 1 @total += @limit / perim generate_triples( a-2*b+2*c, 2*a-b+2*c, 2*a-2*b+3*c) generate_triples( a+2*b+2*c, 2*a+b+2*c, 2*a+2*b+3*c) generate_triples(-a+2*b+2*c,-2*a+b+2*c,-2*a+2*b+3*c) endend perim = 10while perim <= 100_000_000 c = PythagoranTriplesCounter.new perim p [perim, c.total, c.primatives] perim *= 10end output [10, 0, 0] [100, 17, 7] [1000, 325, 70] [10000, 4858, 703] [100000, 64741, 7026] [1000000, 808950, 70229] [10000000, 9706567, 702309] [100000000, 113236940, 7023027] ## Seed7 Translation of: C efficient method The example below uses bigInteger numbers: $ include "seed7_05.s7i";  include "bigint.s7i"; var bigInteger: total is 0_;var bigInteger: prim is 0_;var bigInteger: max_peri is 10_; const proc: new_tri (in bigInteger: a, in bigInteger: b, in bigInteger: c) is func  local    var bigInteger: p is 0_;  begin    p := a + b + c;    if p <= max_peri then      incr(prim);      total +:= max_peri div p;      new_tri( a - 2_*b + 2_*c,  2_*a - b + 2_*c,  2_*a - 2_*b + 3_*c);      new_tri( a + 2_*b + 2_*c,  2_*a + b + 2_*c,  2_*a + 2_*b + 3_*c);      new_tri(-a + 2_*b + 2_*c, -2_*a + b + 2_*c, -2_*a + 2_*b + 3_*c);    end if;  end func; const proc: main is func  begin    while max_peri <= 100000000_ do      total := 0_;      prim := 0_;      new_tri(3_, 4_, 5_);      writeln("Up to " <& max_peri <& ": " <& total <& " triples, " <& prim <& " primitives.");      max_peri *:= 10_;    end while;  end func;

Output:

Up to 10: 0 triples, 0 primitives.
Up to 100: 17 triples, 7 primitives.
Up to 1000: 325 triples, 70 primitives.
Up to 10000: 4858 triples, 703 primitives.
Up to 100000: 64741 triples, 7026 primitives.
Up to 1000000: 808950 triples, 70229 primitives.
Up to 10000000: 9706567 triples, 702309 primitives.
Up to 100000000: 113236940 triples, 7023027 primitives.


## Tcl

Using the efficient method based off the Wikipedia article:

proc countPythagoreanTriples {limit} {    lappend q 3 4 5    set idx [set count [set prim 0]]    while {$idx < [llength$q]} {    set a [lindex $q$idx]    set b [lindex $q [incr idx]] set c [lindex$q [incr idx]]    incr idx    if {$a +$b + $c <=$limit} {        incr prim        for {set i 1} {$i*$a+$i*$b+$i*$c <= $limit} {incr i} { incr count } lappend q \ [expr {$a + 2*($c-$b)}] [expr {2*($a+$c) - $b}] [expr {2*($a-$b) + 3*$c}] \        [expr {$a + 2*($b+$c)}] [expr {2*($a+$c) +$b}] [expr {2*($a+$b) + 3*$c}] \ [expr {2*($b+$c) -$a}] [expr {2*($c-$a) + $b}] [expr {2*($b-$a) + 3*$c}]    }    }    return [list $count$prim]}for {set i 10} {$i <= 10000000} {set i [expr {$i*10}]} {    lassign [countPythagoreanTriples $i] count primitive puts "perimeter limit$i => $count triples,$primitive primitive"}

Output:

perimeter limit 10 => 0 triples, 0 primitive
perimeter limit 100 => 17 triples, 7 primitive
perimeter limit 1000 => 325 triples, 70 primitive
perimeter limit 10000 => 4858 triples, 703 primitive
perimeter limit 100000 => 64741 triples, 7026 primitive
perimeter limit 1000000 => 808950 triples, 70229 primitive
perimeter limit 10000000 => 9706567 triples, 702309 primitive


## zkl

Translation of: D
fcn gcd(a,b){while(b){t:=a; a=b; b=t%b} a.abs()}fcn tri(lim,a=3,b=4,c=5){    p := a + b + c;    if (p > lim) return(0,0);    Utils.zipWith('+,       T(1,lim/p),       tri(lim,  a - 2*b + 2*c,  2*a - b + 2*c,  2*a - 2*b + 3*c),       tri(lim,  a + 2*b + 2*c,  2*a + b + 2*c,  2*a + 2*b + 3*c),       tri(lim, -a + 2*b + 2*c, -2*a + b + 2*c, -2*a + 2*b + 3*c)    );}
n:=10; do(10){println("%,d: %s".fmt(n,tri(n).reverse())); n*=10; }
Output:
10: L(0,0)
100: L(17,7)
1,000: L(325,70)
10,000: L(4858,703)
100,000: L(64741,7026)
1,000,000: L(808950,70229)
10,000,000: L(9706567,702309)
VM#1 caught this unhandled exception:
AssertionError : That is one big stack, infinite recursion?
Stack trace for VM#1 ():
<repeats 3578 times>
...


Max stack size is arbitrary but not adjustable.