# 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 ${\displaystyle (a,b,c)}$ where ${\displaystyle a, and ${\displaystyle a^{2}+b^{2}=c^{2}.}$

They are called primitive triples if ${\displaystyle a,b,c}$ are co-prime, that is, if their pairwise greatest common divisors ${\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}$.

Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (${\displaystyle {\rm {gcd}}(a,b)=1}$).

Each triple forms the length of the sides of a right triangle, whose perimeter is ${\displaystyle P=a+b+c}$.

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 maximum 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.

## 360 Assembly

Translation of: Perl
*        Pythagorean triples -     12/06/2018
PYTHTRI CSECT
USING PYTHTRI,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
SAVE (14,12) save previous context
MVC PMAX,=F'1' pmax=1
LA R6,1 i=1
DO WHILE=(C,R6,LE,=F'6') do i=1 to 6
L R5,PMAX pmax
MH R5,=H'10' *10
ST R5,PMAX pmax=pmax*10
MVC PRIM,=F'0' prim=0
MVC COUNT,=F'0' count=0
L R1,PMAX pmax
BAL R14,ISQRT isqrt(pmax)
SRA R0,1 /2
ST R0,NMAX nmax=isqrt(pmax)/2
LA R7,1 n=1
DO WHILE=(C,R7,LE,NMAX) do n=1 to nmax
LA R9,1(R7) m=n+1
LR R5,R9 m
AR R5,R7 +n
MR R4,R9 *m
SLA R5,1 *2
LR R8,R5 p=2*m*(m+n)
DO WHILE=(C,R8,LE,PMAX) do while p<=pmax
LR R1,R9 m
LR R2,R7 n
BAL R14,GCD gcd(m,n)
IF C,R0,EQ,=F'1' THEN if gcd(m,n)=1 then
L R2,PRIM prim
LA R2,1(R2) +1
ST R2,PRIM prim=prim+1
L R4,PMAX pmax
SRDA R4,32 ~
DR R4,R8 /p
A R5,COUNT +count
ST R5,COUNT count=count+pmax/p
ENDIF , endif
LA R9,2(R9) m=m+2
LR R5,R9 m
AR R5,R7 +n
MR R4,R9 *m
SLA R5,1 *2
LR R8,R5 p=2*m*(m+n)
ENDDO , enddo n
LA R7,1(R7) n++
ENDDO , enddo n
L R1,PMAX pmax
XDECO R1,XDEC edit pmax
MVC PG+15(9),XDEC+3 output pmax
L R1,COUNT count
XDECO R1,XDEC edit count
MVC PG+33(9),XDEC+3 output count
L R1,PRIM prim
XDECO R1,XDEC edit prim
MVC PG+55(9),XDEC+3 output prim
XPRNT PG,L'PG print
LA R6,1(R6) i++
ENDDO , enddo i
L R13,4(0,R13) restore previous savearea pointer
RETURN (14,12),RC=0 restore registers from calling sav
NMAX DS F nmax
PMAX DS F pmax
COUNT DS F count
PRIM DS F prim
PG DC CL80'Max Perimeter: ........., Total: ........., Primitive:'
XDEC DS CL12
GCD EQU * --------------- function gcd(a,b)
STM R2,R7,GCDSA save context
LR R3,R1 c=a
LR R4,R2 d=b
GCDLOOP LR R6,R3 c
SRDA R6,32 ~
DR R6,R4 /d
LTR R6,R6 if c mod d=0
BZ GCDELOOP then leave loop
LR R5,R6 e=c mod d
LR R3,R4 c=d
LR R4,R5 d=e
B GCDLOOP loop
GCDELOOP LR R0,R4 return(d)
LM R2,R7,GCDSA restore context
BR R14 return
GCDSA DS 6A context store
ISQRT EQU * --------------- function isqrt(n)
STM R3,R10,ISQRTSA save context
LR R6,R1 n=r1
LR R10,R6 sqrtn=n
SRA R10,1 sqrtn=n/2
IF LTR,R10,Z,R10 THEN if sqrtn=0 then
LA R10,1 sqrtn=1
ELSE , else
LA R9,0 snm2=0
LA R8,0 snm1=0
LA R7,0 sn=0
LA R3,0 okexit=0
DO UNTIL=(C,R3,EQ,=A(1)) do until okexit=1
AR R10,R7 sqrtn=sqrtn+sn
LR R9,R8 snm2=snm1
LR R8,R7 snm1=sn
LR R4,R6 n
SRDA R4,32 ~
DR R4,R10 /sqrtn
SR R5,R10 -sqrtn
SRA R5,1 /2
LR R7,R5 sn=(n/sqrtn-sqrtn)/2
IF C,R7,EQ,=F'0',OR,CR,R7,EQ,R9 THEN if sn=0 or sn=snm2 then
LA R3,1 okexit=1
ENDIF , endif
ENDDO , enddo until
ENDIF , endif
LR R5,R10 sqrtn
MR R4,R10 *sqrtn
IF CR,R5,GT,R6 THEN if sqrtn*sqrtn>n then
BCTR R10,0 sqrtn=sqrtn-1
ENDIF , endif
LR R0,R10 return(sqrtn)
LM R3,R10,ISQRTSA restore context
BR R14 return
ISQRTSA DS 8A context store
YREGS
END PYTHTRI
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

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

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

## ANSI Standard BASIC

Translation of BBC BASIC Program

100 DECLARE EXTERNAL SUB tri
110 !
120 PUBLIC NUMERIC U0(3,3), U1(3,3), U2(3,3), all, prim
130 DIM seed(3)
140 MAT READ U0, U1, U2
150 DATA 1, -2, 2, 2, -1, 2, 2, -2, 3
160 DATA 1, 2, 2, 2, 1, 2, 2, 2, 3
170 DATA -1, 2, 2, -2, 1, 2, -2, 2, 3
180 !
200 DATA 3, 4, 5
210 FOR power = 1 TO 7
220 LET all = 0
230 LET prim = 0
240 CALL tri(seed, 10^power , all , prim)
250 PRINT "Up to 10^";power,
260 PRINT USING "######### triples ######### primitives":all,prim
270 NEXT power
280 END
290 !
300 EXTERNAL SUB tri(i(), mp, all, prim)
310 DECLARE EXTERNAL FUNCTION SUM
320 DECLARE NUMERIC t(3)
330 !
340 IF SUM(i) > mp THEN EXIT SUB
350 LET prim = prim + 1
360 LET all = all + INT(mp / SUM(i))
370 !
380 MAT t = U0 * i
390 CALL tri(t, mp , all , prim)
400 MAT t = U1 * i
410 CALL tri(t, mp , all , prim)
420 MAT t = U2 * i
430 CALL tri(t, mp , all , prim)
440 END SUB
450 !
460 EXTERNAL FUNCTION SUM(a())
470 LET temp = 0
480 FOR i=LBOUND(a) TO UBOUND(a)
490 LET temp = temp + a(i)
500 NEXT i
510 LET SUM = temp
520 END FUNCTION

## AutoHotkey

#NoEnv
SetBatchLines, -1
#SingleInstance, Force

; Greatest common divisor, from http://rosettacode.org/wiki/Greatest_common_divisor#AutoHotkey
gcd(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 minute count_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 all 1000: 70 prim, 325 all 10000: 703 prim, 4858 all 100000: 7026 prim, 64741 all 1000000: 70229 prim, 808950 all 10000000: 702309 prim, 9706567 all ... ## D ### Lazy Functional Version With hints from the Haskell solution. void main() @safe { import std.stdio, std.range, std.algorithm, std.typecons, std.numeric; enum triples = (in uint n) pure nothrow @safe /*@nogc*/ => iota(1, n + 1) .map!(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[].only.sum <= n) .map!(t => tuple(t[0 .. 2].gcd == 1, t[])); auto xs = triples(100); writeln("Up to 100 there are ", xs.count, " triples, ", xs.filter!q{ a[0] }.count, " are primitive."); } Output: Up to 100 there are 17 triples, 7 are primitive. ### Shorter Version ulong[2] tri(ulong lim, ulong a=3, ulong b=4, ulong c=5) pure nothrow @safe @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() /*@safe*/ { import std.stdio; foreach (immutable p; 1 .. 9) 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] 100000000 [7023027, 113236940] Run-time (32 bit system): about 0.80 seconds with ldc2. ### 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 @safe @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() /*@safe*/ { foreach (immutable p; 1 .. 9) writeln(10 ^^ p, ' ', tri(10 ^^ p).array); } The output is the same. Run-time (32 bit system): about 0.67 seconds with ldc2. ### 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. Run-time: about 0.27 seconds with ldc2. 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. ## Eiffel class APPLICATION create make feature make local perimeter: INTEGER do perimeter := 100 from until perimeter > 1000000 loop total := 0 primitive_triples := 0 count_pythagorean_triples (3, 4, 5, perimeter) io.put_string ("There are " + total.out + " triples, below " + perimeter.out + ". Of which " + primitive_triples.out + " are primitives.%N") perimeter := perimeter * 10 end end count_pythagorean_triples (a, b, c, perimeter: INTEGER) -- Total count of pythagorean triples and total count of primitve triples below perimeter. local p: INTEGER do p := a + b + c if p <= perimeter then primitive_triples := primitive_triples + 1 total := total + perimeter // p count_pythagorean_triples (a + 2 * (- b + c), 2 * (a + c) - b, 2 * (a - b + c) + c, perimeter) count_pythagorean_triples (a + 2 * (b + c), 2 * (a + c) + b, 2 * (a + b + c) + c, perimeter) count_pythagorean_triples (- a + 2 * (b + c), 2 * (- a + c) + b, 2 * (- a + b + c) + c, perimeter) end end feature {NONE} primitive_triples: INTEGER total: INTEGER end Output: There are 17 triples, below 100. Of which 7 are primitives. There are 325 triples, below 1000. Of which 70 are primitives. There are 4858 triples, below 10000. Of which 703 are primitives. There are 64741 triples, below 100000. Of which 7026 are primitives. There are 808950 triples, below 1000000. Of which 70229 are primitives. ## Elixir Translation of: Ruby defmodule RC do def count_triples(limit), do: count_triples(limit,3,4,5) defp count_triples(limit, a, b, c) when limit<(a+b+c), do: {0,0} defp count_triples(limit, a, b, c) do {p1, t1} = count_triples(limit, a-2*b+2*c, 2*a-b+2*c, 2*a-2*b+3*c) {p2, t2} = count_triples(limit, a+2*b+2*c, 2*a+b+2*c, 2*a+2*b+3*c) {p3, t3} = count_triples(limit,-a+2*b+2*c,-2*a+b+2*c,-2*a+2*b+3*c) {1+p1+p2+p3, div(limit, a+b+c)+t1+t2+t3} end end list = for n <- 1..8, do: Enum.reduce(1..n, 1, fn(_,acc)->10*acc end) Enum.each(list, fn n -> IO.inspect {n, RC.count_triples(n)} end) 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}} ## 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 triples count_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 ## ERRE PROGRAM PIT BEGIN PRINT(CHR$(12);) !CLS
PRINT(TIME$) FOR POWER=1 TO 7 DO PLIMIT=10#^POWER UPPERBOUND=INT(1+PLIMIT^0.5) PRIMITIVES=0 TRIPLES=0 EXTRAS=0 ! will count the in-range multiples of any primitive FOR M=2 TO UPPERBOUND DO FOR N=1+(M MOD 2=1) TO M-1 STEP 2 DO TERM1=2*M*N TERM2=M*M-N*N TERM3=M*M+N*N PERIMETER=TERM1+TERM2+TERM3 IF PERIMETER<=PLIMIT THEN TRIPLES=TRIPLES+1 A=TERM1 B=TERM2 REPEAT R=A-B*INT(A/B) A=B B=R UNTIL R<=0 ! we've found a primitive triple if a = 1, since hcf =1. ! and it is inside perimeter range. Save it in an array IF (A=1) AND (PERIMETER<=PLIMIT) THEN PRIMITIVES=PRIMITIVES+1 !----------------------------------------------- !swap so in increasing order of side length !----------------------------------------------- IF TERM1>TERM2 THEN SWAP(TERM1,TERM2) !----------------------------------------------- !we have the primitive & removed any multiples. !Now calculate ALL the multiples in range. !----------------------------------------------- NEX=INT(PLIMIT/PERIMETER) EXTRAS=EXTRAS+NEX END IF !scan END FOR END FOR PRINT("Primit. with perimeter <=";10#^power;"is";primitives;"&";extras;"non-prim.triples.") PRINT(TIME$)
END FOR

PRINT PRINT("** End **")
END PROGRAM
Output:
16:08:39
Primit. with perimeter <= 10 is 0 & 0 non-prim.triples.
16:08:39
Primit. with perimeter <= 100 is 7 & 17 non-prim.triples.
16:08:39
Primit. with perimeter <= 1000 is 70 & 325 non-prim.triples.
16:08:39
Primit. with perimeter <= 10000 is 703 & 4858 non-prim.triples.
16:08:39
Primit. with perimeter <= 100000 is 7026 & 64741 non-prim.triples.
16:08:41
Primit. with perimeter <= 1000000 is 70229 & 808950 non-prim.triples.
16:09:07
Primit. with perimeter <= 10000000 is 702309 & 9706567 non-prim.triples.
16:13:10

** End **

## 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 r
end function

atom max_peri
max_peri = 10
while max_peri <= 100000000 do
printf(1,"%d: ", max_peri)
? tri(max_peri, {3, 4, 5})
max_peri *= 10
end 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 math
math.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 ok FLOAD 'C:\Forth\ancien_fibo_pythag.F' ok ok ok ok 3 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 ok 10 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 do end subroutine new_tri end 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 do end 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 ## FreeBASIC The upper limit is set to 12(10^12), this will take about 3-4 hours. If you can't wait that long better lower it to 11(10^11). ### Version 1 Normal version ' version 30-05-2016 ' compile with: fbc -s console ' primitive pythagoras triples ' a = m^2 - n^2, b = 2mn, c = m^2 + n^2 ' m, n are positive integers and m > n ' m - n = odd and GCD(m, n) = 1 ' p = a + b + c ' max m for give perimeter ' p = m^2 - n^2 + 2mn + m^2 + n^2 ' p = 2mn + m^2 + m^2 + n^2 - n^2 = 2mn + 2m^2 ' m >> n and n = 1 ==> p = 2m + 2m^2 = 2m(1 + m) ' m >> 1 ==> p = 2m(m) = 2m^2 ' max m for given perimeter = sqr(p / 2) Function gcd(x As UInteger, y As UInteger) As UInteger Dim As UInteger t While y t = y y = x Mod y x = t Wend Return x End Function Sub pyth_trip(limit As ULongInt, ByRef trip As ULongInt, ByRef prim As ULongInt) Dim As ULongInt perimeter, lby2 = limit Shr 1 Dim As UInteger m, n Dim As ULongInt a, b, c For m = 2 To Sqr(limit / 2) For n = 1 + (m And 1) To (m - 1) Step 2 ' common divisor, try next n If (gcd(m, n) > 1) Then Continue For a = CULngInt(m) * m - n * n b = CULngInt(m) * n * 2 c = CULngInt(m) * m + n * n perimeter = a + b + c ' perimeter > limit, since n goes up try next m If perimeter >= limit Then Continue For, For prim += 1 If perimeter < lby2 Then trip += limit \ perimeter Else trip += 1 End If Next n Next m End Sub ' ------=< MAIN >=------ Dim As String str1, buffer = Space(14) Dim As ULongInt limit, trip, prim Dim As Double t, t1 = Timer Print "below triples primitive time" Print For x As UInteger = 1 To 12 t = Timer limit = 10 ^ x : trip = 0 : prim = 0 pyth_trip(limit, trip, prim) LSet buffer, Str(prim) : str1 = buffer Print Using "10^## ################ "; x; trip; If x > 7 Then Print str1; Print Using " ######.## sec."; Timer - t Else Print str1 End If Next x Print : Print Print Using "Total time needed #######.## sec."; Timer - t1 ' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End Output: below triples primitive time 10^ 1 0 0 10^ 2 17 7 10^ 3 325 70 10^ 4 4858 703 10^ 5 64741 7026 10^ 6 808950 70229 10^ 7 9706567 702309 10^ 8 113236940 7023027 0.94 sec. 10^ 9 1294080089 70230484 10.13 sec. 10^10 14557915466 702304875 109.75 sec. 10^11 161750315680 7023049293 1204.56 sec. 10^12 1779214833461 70230492763 13031.84 sec. Total time needed 14357.31 sec. ### Version 2 Attempt to make a faster version (about 20% faster) ' version 30-05-2016 ' compile with: fbc -s console ' max m for give perimeter ' p = m^2 - n^2 + 2mn + m^2 + n^2 ' p = 2mn + m^2 + m^2 + n^2 - n^2 = 2mn + 2m^2 ' m >> n and n = 1 ==> p = 2m + 2m^2 = 2m(1 + m) ' m >> 1 ==> p = 2m(m) = 2m^2 ' max m for given perimeter = sqr(p / 2) Function gcd(x As UInteger, y As UInteger) As UInteger Dim As UInteger t While y t = y y = x Mod y x = t Wend Return x End Function Sub pyth_trip_fast(limit As ULongInt, ByRef trip As ULongInt, ByRef prim As ULongInt) Dim As ULongInt perimeter, lby2 = limit Shr 1 Dim As UInteger mx2 = 4 For m As UInteger = 2 To Sqr(limit / 2) perimeter = (CULngInt(m) * m * 2) - IIf(m And 1, 0, m * 2) mx2 = mx2 + 4 For n As UInteger = 1 + (m And 1) To (m - 1) Step 2 perimeter += mx2 ' common divisor, try next n If (gcd(m, n) > 1) Then Continue For ' perimeter > limit, since n goes up try next m If perimeter >= limit Then Continue For, For prim += 1 If perimeter < lby2 Then trip += limit \ perimeter Else trip += 1 End If Next n Next m End Sub ' ------=< MAIN >=------ Dim As String str1, buffer = Space(14) Dim As ULongInt limit, trip, prim Dim As Double t, t1 = Timer Print "below triples primitive time" Print For x As UInteger = 1 To 12 t = Timer limit = 10 ^ x : trip = 0 : prim = 0 pyth_trip_fast(limit, trip, prim) LSet buffer, Str(prim) : str1 = buffer Print Using "10^## ################ "; x; trip; If x > 7 Then Print str1; Print Using " ######.## sec."; Timer - t Else Print str1 End If Next x Print : Print Print Using "Total time needed #######.## sec."; Timer - t1 ' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End Output: below triples primitive time 10^ 1 0 0 10^ 2 17 7 10^ 3 325 70 10^ 4 4858 703 10^ 5 64741 7026 10^ 6 808950 70229 10^ 7 9706567 702309 10^ 8 113236940 7023027 0.66 sec. 10^ 9 1294080089 70230484 7.48 sec. 10^10 14557915466 702304875 83.92 sec. 10^11 161750315680 7023049293 945.95 sec. 10^12 1779214833461 70230492763 10467.94 sec. Total time needed 11506.01 sec. The time needed is about 11 times the time needed for the previous limit. To calculate 10^12 with the fast version that would take about 32 hours. The variable that holds the number of triples will eventual overflow at 10^18 - 10^19. To reach that stage you need the program to run for a few years. ## 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 ## Haskell pytr :: Int -> [(Bool, Int, Int, Int)] pytr n = filter (\(_, a, b, c) -> a + b + c <= n) [ (prim a b c, a, b, c) | a <- xs , b <- drop a xs , c <- drop b xs , a ^ 2 + b ^ 2 == c ^ 2 ] where xs = [1 .. n] prim a b _ = gcd a b == 1 main :: IO () main = putStrLn$
"Up to 100 there are " ++
show (length xs) ++
" triples, of which " ++
show (length $filter (\(x, _, _, _) -> x) xs) ++ " are primitive." where xs = pytr 100 Output: Up to 100 there are 17 triples, of which 7 are primitive. Or equivalently (desugaring the list comprehension down to nested concatMaps, and pruning back the search space a little): pythagoreanTriplesBelow :: Int -> [[Int]] pythagoreanTriplesBelow n = let m = quot n 2 in concatMap (\x -> concatMap (\y -> concatMap (\z -> if x + y + z <= n && x ^ 2 + y ^ 2 == z ^ 2 then [[x, y, z]] else []) [y + 1 .. m]) [x + 1 .. m]) [1 .. m] -- TEST ------------------------------------------------------------------------- main :: IO () main = mapM_ (print . length) ([id, filter (\[x, y, _] -> gcd x y == 1)] <*> [pythagoreanTriplesBelow 100]) Output: 17 7 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]]
]

triangleCount max_p = (length t, sum $map ((max_p div) . sum) t) where t = triangles max_p main :: IO () main = mapM_ ((putStrLn . (\n -> show n ++ " " ++ show (triangleCount n))) . (10 ^)) [1 .. 7] Output: 10 (0,0) 100 (7,17) 1000 (70,325) 10000 (703,4858) 100000 (7026,64741) 1000000 (70229,808950) 10000000 (702309,9706567) ## Icon and Unicon This uses the elegant formula (#IV) from Formulas for generating Pythagorean triples link numbers link 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 order every (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 ## 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 100 1 3 4 5 1 5 12 13 0 6 8 10 1 7 24 25 1 8 15 17 0 9 12 15 1 9 40 41 0 10 24 26 0 12 16 20 1 12 35 37 0 15 20 25 0 15 36 39 0 16 30 34 0 18 24 30 1 20 21 29 0 21 28 35 0 24 32 40 (# , [: {. +/) pytr 10 0 0 (# , [: {. +/) pytr 100 17 7 (# , [: {. +/) pytr 1000 325 70 (# , [: {. +/) pytr 10000 4858 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 [email protected]<[email protected]% +/)"1 prim ) usage for trips is the same as for pytr. Thus: (# , 1 {. +/) trips 10 0 0 (# , 1 {. +/) trips 100 17 7 (# , 1 {. +/) trips 1000 325 70 (# , 1 {. +/) trips 10000 4858 703 (# , 1 {. +/) trips 100000 64741 7026 (# , 1 {. +/) trips 1000000 808950 70229 (# , 1 {. +/) trips 10000000 9706567 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 1e8 7023027 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. ### Brute force primitives with scaling ### 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. ## JavaScript ### ES6 Exhaustive search of a full cartesian product. Not scalable. (() => { 'use strict'; // Arguments: predicate, maximum perimeter // pythTripleCount :: ((Int, Int, Int) -> Bool) -> Int -> Int const pythTripleCount = (p, maxPerim) => { const xs = enumFromTo(1, Math.floor(maxPerim / 2)); return concatMap(x => concatMap(y => concatMap(z => ((x + y + z <= maxPerim) && (x * x + y * y === z * z) && p(x, y, z)) ? [ [x, y, z] ] : [], // (Empty lists disappear under concatenation) xs.slice(y)), xs.slice(x)), xs ) .length; }; // GENERIC FUNCTIONS -------------------------------------- // concatMap :: (a -> [b]) -> [a] -> [b] const concatMap = (f, xs) => xs.length > 0 ? [].concat.apply([], xs.map(f)) : []; // enumFromTo :: Enum a => a -> a -> [a] const enumFromTo = (m, n) => (typeof m !== 'number' ? ( enumFromToChar ) : enumFromToInt) .apply(null, [m, n]); // enumFromToInt :: Int -> Int -> [Int] const enumFromToInt = (m, n) => n >= m ? Array.from({ length: Math.floor(n - m) + 1 }, (_, i) => m + i) : []; // gcd :: Int -> Int -> Int const gcd = (x, y) => { const _gcd = (a, b) => (b === 0 ? a : _gcd(b, a % b)); return _gcd(Math.abs(x), Math.abs(y)); }; // MAIN --------------------------------------------------- return [10, 100, 1000] .map(n => ({ maxPerimeter: n, triples: pythTripleCount(x => true, n), primitives: pythTripleCount((x, y, _) => gcd(x, y) === 1, n) })); })(); Output: [{"maxPerimeter":10, "triples":0, "primitives":0}, {"maxPerimeter":100, "triples":17, "primitives":7}, {"maxPerimeter":1000, "triples":325, "primitives":70}] ## 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''':
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.jq 10: [0,0] 100: [17,7] 1000: [325,70] 10000: [4858,703] 100000: [64741,7026] 1000000: [808950,70229] 10000000: [9706567,702309] 100000000: [113236940,7023027] ## Julia This solution uses the the Euclidian concept of m and n as generators of Pythagorean triplets. When m and n are coprime and have opposite parity, the generated triplets are primitive. It works reasonably well up to a limit of 10^10. function primitiven{T<:Integer}(m::T) 1 < m || return T[] m != 2 || return T[1] !isprime(m) || return T[2:2:m-1] rp = trues(m-1) if isodd(m) rp[1:2:m-1] = false end for p in keys(factor(m)) rp[p:p:m-1] = false end T[1:m-1][rp] end function pythagoreantripcount{T<:Integer}(plim::T) primcnt = 0 fullcnt = 0 11 < plim || return (primcnt, fullcnt) for m in 2:plim p = 2m^2 p+2m <= plim || break for n in primitiven(m) q = p + 2m*n q <= plim || break primcnt += 1 fullcnt += div(plim, q) end end return (primcnt, fullcnt) end println("Counting Pythagorian Triplets within perimeter limits:") println(" Limit All Primitive") for om in 1:10 (pcnt, fcnt) = pythagoreantripcount(10^om) println(@sprintf " 10^%02d %11d %9d" om fcnt pcnt) end Output: Counting Pythagorian Triplets within perimeter limits: Limit All Primitive 10^01 0 0 10^02 17 7 10^03 325 70 10^04 4858 703 10^05 64741 7026 10^06 808950 70229 10^07 9706567 702309 10^08 113236940 7023027 10^09 1294080089 70230484 10^10 14557915466 702304875 ## Kotlin Translation of: Go Due to deep recursion, I needed to increase the stack size to 4MB to get up to a maximum perimeter of 10 billion. Expect a run time of around 30 seconds on a typical laptop. // version 1.1.2 var total = 0L var prim = 0L var maxPeri = 0L fun newTri(s0: Long, s1: Long, s2: Long) { val p = s0 + s1 + s2 if (p <= maxPeri) { prim++ total += maxPeri / p newTri( s0 - 2 * s1 + 2 * s2, 2 * s0 - s1 + 2 * s2, 2 * s0 - 2 * s1 + 3 * s2) newTri( s0 + 2 * s1 + 2 * s2, 2 * s0 + s1 + 2 * s2, 2 * s0 + 2 * s1 + 3 * s2) newTri(-s0 + 2 * s1 + 2 * s2, -2 * s0 + s1 + 2 * s2, -2 * s0 + 2 * s1 + 3 * s2) } } fun main(args: Array<String>) { maxPeri = 100 while (maxPeri <= 10_000_000_000L) { prim = 0 total = 0 newTri(3, 4, 5) println("Up to$maxPeri: $total triples,$prim primatives")
maxPeri *= 10
}
}
Output:
Up to 100: 17 triples, 7 primatives
Up to 1000: 325 triples, 70 primatives
Up to 10000: 4858 triples, 703 primatives
Up to 100000: 64741 triples, 7026 primatives
Up to 1000000: 808950 triples, 70229 primatives
Up to 10000000: 9706567 triples, 702309 primatives
Up to 100000000: 113236940 triples, 7023027 primatives
Up to 1000000000: 1294080089 triples, 70230484 primatives
Up to 10000000000: 14557915466 triples, 702304875 primatives

## Lasso

// Brute Force: Too slow for large numbers
define 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.

## Mercury

From List comprehensions:

:- module comprehension.
:- interface.
:- import_module io.
:- import_module int.

:- type triple ---> triple(int, int, int).

:- pred pythTrip(int::in,triple::out) is nondet.
:- pred main(io::di, io::uo) is det.

:- implementation.
:- import_module solutions.

pythTrip(Limit,triple(X,Y,Z)) :-
nondet_int_in_range(1,Limit,X),
nondet_int_in_range(X,Limit,Y),
nondet_int_in_range(Y,Limit,Z),
pow(Z,2) = pow(X,2) + pow(Y,2).

main(!IO) :-
solutions((pred(Triple::out) is nondet :- pythTrip(20,Triple)),Result),
write(Result,!IO).

## 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

## Nim

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: array[0..2, int]) =
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

## Ol

; triples generator based on Euclid's formula, creates lazy list
(define (euclid-formula max)
(let loop ((a 3) (b 4) (c 5) (tail #null))
(if (<= (+ a b c) max)
(cons (tuple a b c) (lambda ()
(let ((d (- b)) (z (- a)))
(loop (+ a d d c c) (+ a a d c c) (+ a a d d c c c) (lambda ()
(loop (+ a b b c c) (+ a a b c c) (+ a a b b c c c) (lambda ()
(loop (+ z b b c c) (+ z z b c c) (+ z z b b c c c) tail))))))))
tail)))

; let's do calculations
(define (calculate max)
(let loop ((p 0) (t 0) (ll (euclid-formula max)))
(cond
((null? ll)
(cons p t))
((function? ll)
(loop p t (ll)))
(else
(let ((triple (car ll)))
(loop (+ p 1) (+ t (div max (apply + triple)))
(cdr ll)))))))

; print values for 10..100000
(for-each (lambda (max)
(print max ": " (calculate max)))
(map (lambda (n) (expt 10 n)) (iota 6 1)))

Output:
10: (0 . 0)
100: (7 . 17)
1000: (70 . 325)
10000: (703 . 4858)
100000: (7026 . 64741)
1000000: (70229 . 808950)

## 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

Works with: Rakudo version 2018.09

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

constant limit = 100;

for [X] [^limit] xx 3 -> (\a, \b, \c) {
say [a, b, c] if a < b < c and a**2 + b**2 == c**2
}
Output:
3 4 5
5 12 13
6 8 10
7 24 25
8 15 17
9 12 15
9 40 41
10 24 26
11 60 61
12 16 20
12 35 37
13 84 85
14 48 50
15 20 25
15 36 39
16 30 34
16 63 65
18 80 82
20 21 29
20 48 52
21 28 35
21 72 75
24 32 40
24 45 51
24 70 74
25 60 65
27 36 45
28 45 53
30 40 50
30 72 78
32 60 68
33 44 55
33 56 65
35 84 91
36 48 60
36 77 85
39 52 65
39 80 89
40 42 58
40 75 85
42 56 70
45 60 75
48 55 73
48 64 80
51 68 85
54 72 90
57 76 95
60 63 87
65 72 97

Here is a slightly less naive brute force implementation, but still not practical for large perimeter limits.

my $limit = 10000; my atomicint$i = 0;
my @triples[$limit/2]; (3 ..$limit/2).race.map: -> $c { for 1 ..$c -> $a { my$b = ($c *$c - $a *$a).sqrt;
last if $c +$a + $b >$limit;
last if $a >$b;
@triples[$i++] = ([gcd]$c, $a,$b) > 1 ?? 0 !! 1 if $b ==$b.Int;
}
}

say my $result = "There are {[email protected]triples.grep:{$_ !eqv Any}} Pythagorean Triples with a perimeter <= $limit," ~"\nof which {[+] @triples.grep: so *} are primitive."; Output: There are 4858 Pythagorean Triples with a perimeter <= 10000, of which 703 are primitive. Here's a much faster version. Hint, "oyako" is Japanese for "parent/child". :-) sub triples($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 $limit; } Output: 10 => (0 0) 100 => (7 17) 1000 => (70 325) 10000 => (703 4858) 100000 => (7026 64741) 1000000 => (70229 808950) 10000000 => (702309 9706567) 100000000 => (7023027 113236940) 1000000000 => (70230484 1294080089) ^C The geometric sequence of limits will continue on forever, so eventually when you get tired of waiting (about a billion on my computer), you can just stop it. Another efficiency trick of note: we avoid passing the limit in as a parameter to the inner helper routine, but instead supply the limit via the lexical scope. Likewise, the accumulators are referenced lexically, so only the triples themselves need to be passed downward, and nothing needs to be returned. Here is an alternate version that avoids naming any scalars that can be handled by vector processing instead. Using vectorized ops allows a bit more potential for parallelization in theory, but techniques like the use of complex numbers to add two numbers in parallel, and the use of gather/take generate so much overhead that this version runs 70-fold slower than the previous one. constant @coeff = [[+1, -2, +2], [+2, -1, +2], [+2, -2, +3]], [[+1, +2, +2], [+2, +1, +2], [+2, +2, +3]], [[-1, +2, +2], [-2, +1, +2], [-2, +2, +3]]; sub triples($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, $complex.im})"; } for 10, 100, 1000, 10000 ->$limit {
say triples $limit; } Output: 10 => (0 0) 100 => (7 17) 1000 => (70 325) 10000 => (703 4858) ## Phix Translation of: Pascal atom total, prim, maxPeri = 10 procedure tri(atom s0, s1, s2) atom p = s0 + s1 + s2 if p<=maxPeri then prim += 1 total += floor(maxPeri/p) tri( s0+2*(-s1+s2), 2*( s0+s2)-s1, 2*( s0-s1+s2)+s2); tri( s0+2*( s1+s2), 2*( s0+s2)+s1, 2*( s0+s1+s2)+s2); tri(-s0+2*( s1+s2), 2*(-s0+s2)+s1, 2*(-s0+s1+s2)+s2); end if end procedure while maxPeri<=1e8 do prim := 0; total := 0; tri(3, 4, 5); printf(1,"Up to %d: %d triples, %d primitives.\n", {maxPeri,total,prim}) maxPeri *= 10; end while 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. ## 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; ## PowerShell function triples($p) {
if($p -gt 4) { # ai + bi + ci = pi <= p # ai < bi < ci --> 3ai < pi <= p and ai + 2bi < pi <= p$pa = [Math]::Floor($p/3) 1..$pa | foreach {
$ai =$_
$pb = [Math]::Floor(($p-$ai)/2) ($ai+1)..$pb | foreach {$bi = $_$pc = $p-$ai-$bi ($bi+1)..$pc | where {$ci = $_$pi = $ai +$bi + $ci$ci*$ci -eq$ai*$ai +$bi*$bi } | foreach { [pscustomobject]@{ a = "$ai"
b = "$bi" c = "$ci"
p = "$pi" } } } } } else { Write-Error "$p is not greater than 4"
}
}
function gcd ($a,$b) {
function pgcd ($n,$m) {
if($n -le$m) {
if($n -eq 0) {$m}
else{pgcd $n ($m%$n)} } else {pgcd$m $n} }$n = [Math]::Abs($a)$m = [Math]::Abs($b) (pgcd$n $m) }$triples = (triples 100)

$coprime =$triples |
where {((gcd $_.a$_.b) -eq 1) -and ((gcd $_.a$_.c) -eq 1) -and ((gcd $_.b$_.c) -eq 1)}

"There are $(($triples).Count) Pythagorean triples with perimeter no larger than 100
and $(($coprime).Count) of them are coprime."

Output:

There are 17 Pythagorean triples with perimeter no larger than 100 and 7 of them are coprime.

## PureBasic

${\displaystyle (a=m^{2}-n^{2})\land (b=2mn)\land (c=m^{2}+n^{2})\land (p=a+b+c)\rightarrow }$

${\displaystyle p=2m^{2}+2mn\leq 100,000,000\rightarrow }$

${\displaystyle m^{2}+mn\leq 50,000,000\rightarrow }$

${\displaystyle m^{2}+mn-50,000,000\leq 0\rightarrow }$

${\displaystyle n\leq 10,000}$

Procedure.i ConsoleWrite(t.s) ; compile using /CONSOLE option
OpenConsole()
PrintN (t.s)
CloseConsole()
ProcedureReturn 1
EndProcedure

Procedure.i StdOut(t.s) ; compile using /CONSOLE option
OpenConsole()
Print(t.s)
CloseConsole()
ProcedureReturn 1
EndProcedure

Procedure.i gcDiv(n,m) ; greatest common divisor
if n=0:ProcedureReturn m:endif
while m <> 0
if n > m
n - m
else
m - n
endif
wend
ProcedureReturn n
EndProcedure

st=ElapsedMilliseconds()

nmax =10000
power =8

dim primitiveA(power)
dim alltripleA(power)
dim pmaxA(power)

x=1
for i=1 to power
x*10:pmaxA(i)=x/2
next

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
next
next

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)
next
ConsoleWrite("")
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:
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
from sys import setrecursionlimit
setrecursionlimit(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

### using GCD for determinacy

/*REXX program counts the number of  Pythagorean triples  that exist given a maximum    */
/*──────────────────── perimeter of N, and also counts how many of them are primitives.*/
parse arg N . /*obtain optional argument from the CL.*/
if N=='' | N=="," then N= 100 /*Not specified? Then use the default.*/
do j=1 for N; @.j= j*j; end /*pre-compute some squares. */
N66= N * 2%3 /*calculate 2/3 of N (for a+b). */
T= 0; P= 0 /*set the number of Triples, Primitives*/
do a=3 to N%3 /*limit side to 1/3 of the perimeter.*/
do b= a+1 /*the triangle can't be isosceles. */
ab= a + b /*compute a partial perimeter (2 sides)*/
if ab>=N66 then iterate a /*is a+b≥66% perimeter? Try different A*/
aabb= @.a + @.b /*compute the sum of a²+b² (shortcut)*/
do c=b+1 /*compute the value of the third side. */
if ab+c > N then iterate a /*is a+b+c>perimeter ? Try different A.*/
if @.c >aabb then iterate b /*is c² > a²+b² ? Try " B.*/
if @.c\==aabb then iterate /*is c² ¬= a²+b² ? Try " C.*/
T= T + 1 /*eureka. We found a Pythagorean triple*/
P= P + (gcd(a, b)==1) /*is this triple a primitive triple? */
end /*c*/
end /*b*/
end /*a*/
_= left('', 7) /*for padding the output with 7 blanks.*/
say 'max perimeter =' N _ "Pythagorean triples =" T _ 'primitives =' P
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
gcd: procedure; parse arg x,y; do until y==0; parse value x//y y with y x; end; return x
output   when using the default input of:     100
max perimeter = 100         Pythagorean triples = 17         primitives = 7
output   when using the default input of:     1000
max perimeter = 1000         Pythagorean triples = 325         primitives = 70

### using single evenness for determinacy

This REXX version takes advantage that primitive Pythagorean triples must have one and only one   even   number.

This REXX version is about   10%   faster than the 1st REXX version.

Non-primitive Pythagorean triples are generated after a primitive triple is found.

/*REXX program counts the number of  Pythagorean triples  that exist given a maximum    */
/*──────────────────── perimeter of N, and also counts how many of them are primitives.*/
parse arg N . /*obtain optional argument from the CL.*/
if N=='' | N=="," then N= 100 /*Not specified? Then use the default.*/
@.= 0; do j=1 for N; @.j= j*j; end /*pre-compute some squares. */
N66= N * 2%3 /*calculate 2/3 of N (for a+b). */
P= 0; T= 0; do a=3 to N%3 /*limit side to 1/3 of the perimeter.*/
aEven= a//2==0 /*set variable to 1 if A is even. */
do b=a+1 by 1+aEven; ab= a + b /*the triangle can't be isosceles. */
if ab>=N66 then iterate a /*is a+b≥66% perimeter? Try different A*/
aabb= @.a + @.b /*compute the sum of a²+b² (shortcut)*/
do c=b + 1 /*compute the value of the third side. */
if aEven then if c//2==0 then iterate /*both A&C even? Skip it*/
if ab+c>n then iterate a /*a+b+c > perimeter? Try different A. */
if @.c > aabb then iterate b /*is c² > a²+b² ? " " B. */
if @.c\==aabb then iterate /*is c² ¬= a²+b² ? " " C. */
if @.a.b.c then iterate /*Is this a duplicate? Then try again.*/
T= T + 1 /*Eureka! We found a Pythagorean triple*/
P= P + 1 /*count this also as a primitive triple*/
do m=2 while a*m+b*m+c*m<=N /*generate non-primitives Pythagoreans.*/
T= T + 1 /*Eureka! We found a Pythagorean triple*/
am= a*m; bm= b*m; cm= c*m /*create some short-cut variable names.*/
@.am.bm.cm= 1 /*mark Pythagorean triangle as a triple*/
end /*m*/
end /*c*/
end /*b*/
end /*a*/ /*stick a fork in it, we're all done. */
_= left('', 7) /*for padding the output with 7 blanks.*/
say 'max perimeter =' N _ "Pythagorean triples =" T _ 'primitives =' P
output   is identical to the 1st REXX version.

## Ring

size = 100
sum = 0
prime = 0
for i = 1 to size
for j = i + 1 to size
for k = 1 to size
if pow(i,2) + pow(j,2) = pow(k,2) and (i+j+k) < 101
if gcd(i,j) = 1 prime += 1 ok
sum += 1
see "" + i + " " + j + " " + k + nl ok
next
next
next
see "Total : " + sum + nl
see "Primitives : " + prime + nl

func gcd gcd, b
while b
c = gcd
gcd = b
b = c % b
end
return gcd

Output:

3 4 5
5 12 13
6 8 10
7 24 25
8 15 17
9 12 15
9 40 41
10 24 26
12 16 20
12 35 37
15 20 25
15 36 39
16 30 34
18 24 30
20 21 29
21 28 35
24 32 40
Total : 17
Primitives : 7

## Ruby

Translation of: Java
class PythagoranTriplesCounter
def initialize(limit)
@limit = limit
@total = 0
@primitives = 0
generate_triples(3, 4, 5)
end

private
def generate_triples(a, b, c)
perim = a + b + c
return if perim > @limit

@primitives += 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)
end
end

perim = 10
while perim <= 100_000_000
c = PythagoranTriplesCounter.new perim
p [perim, c.total, c.primitives]
perim *= 10
end

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]

## Rust

fn f1 (a : u64, b : u64, c : u64, d : u64) -> u64 {
let mut primitive_count = 0;
for triangle in [[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],
[2*b + 2*c - a, b + 2*c - 2*a, 2*b + 3*c - 2*a]] .iter() {
let l = triangle[0] + triangle[1] + triangle[2];
if l > d { continue; }
primitive_count += 1 + f1(triangle[0], triangle[1], triangle[2], d);
}
primitive_count
}

fn f2 (a : u64, b : u64, c : u64, d : u64) -> u64 {
let mut triplet_count = 0;
for triangle in [[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],
[2*b + 2*c - a, b + 2*c - 2*a, 2*b + 3*c - 2*a]] .iter() {
let l = triangle[0] + triangle[1] + triangle[2];
if l > d { continue; }
triplet_count += (d/l) + f2(triangle[0], triangle[1], triangle[2], d);
}
triplet_count
}

fn main () {
let new_th_1 = thread::Builder::new().stack_size(32 * 1024 * 1024).spawn (move || {
let mut i = 100;
while i <= 100_000_000_000 {
println!(" Primitive triples below {} : {}", i, f1(3, 4, 5, i) + 1);
i *= 10;
}
}).unwrap();

let new_th_2 =thread::Builder::new().stack_size(32 * 1024 * 1024).spawn (move || {
let mut i = 100;
while i <= 100_000_000_000 {
println!(" Triples below {} : {}", i, f2(3, 4, 5, i) + i/12);
i *= 10;
}
}).unwrap();

new_th_1.join().unwrap();
new_th_2.join().unwrap();
}
Output:
Primitive triples below 100 : 7
Triples below 100 : 17
Primitive triples below 1000 : 70
Triples below 1000 : 325
Primitive triples below 10000 : 703
Triples below 10000 : 4858
Primitive triples below 100000 : 7026
Triples below 100000 : 64741
Primitive triples below 1000000 : 70229
Triples below 1000000 : 808950
Primitive triples below 10000000 : 702309
Triples below 10000000 : 9706567
Primitive triples below 100000000 : 7023027
Triples below 100000000 : 113236940
Primitive triples below 1000000000 : 70230484
Triples below 1000000000 : 1294080089
Primitive triples below 10000000000 : 702304875
Triples below 10000000000 : 14557915466
Primitive triples below 100000000000 : 7023049293
Triples below 100000000000 : 161750315680

real	2m22.676s
user	3m39.239s
sys	0m0.024s

## Scala

Output:
Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
object PythagoreanTriples extends App {

println(" Limit Primatives All")

for {e <- 2 to 7
limit = math.pow(10, e).longValue()
} {
var primCount, tripCount = 0

def parChild(a: BigInt, b: BigInt, c: BigInt): Unit = {
val perim = a + b + c
val (a2, b2, c2, c3) = (2 * a, 2 * b, 2 * c, 3 * c)
if (limit >= perim) {
primCount += 1
tripCount += (limit / perim).toInt
parChild(a - b2 + c2, a2 - b + c2, a2 - b2 + c3)
parChild(a + b2 + c2, a2 + b + c2, a2 + b2 + c3)
parChild(-a + b2 + c2, -a2 + b + c2, -a2 + b2 + c3)
}
}

parChild(BigInt(3), BigInt(4), BigInt(5))
println(f"a + b + c <= ${limit.toFloat}%3.1e$primCount%9d $tripCount%12d") } } ## Scheme Works with: Gauche Scheme (use srfi-42) (define (py perim) (define prim 0) (values (sum-ec (: c perim) (: b c) (: a b) (if (and (<= (+ a b c) perim) (= (square c) (+ (square b) (square a))))) (begin (when (= 1 (gcd a b)) (inc! prim))) 1) prim)) Testing: gosh> (py 100) 17 7 ## Scratch Scratch is a visual programming language. Click the link, then "see inside" to see the code. Output: 17 Pythagorean triples with a perimeter less than 100, 7 of which are primitive. ## Seed7 Translation of: C efficient method The example below uses bigInteger numbers:$ 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.

## Sidef

Translation of: Perl 6
func triples(limit) {
var primitive = 0
var civilized = 0

func oyako(a, b, c) {
(var perim = a+b+c) > limit || (
primitive++
civilized += int(limit / 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 n (1..Inf) {
say triples(10**n)
}
Output:
10 => (0 0)
100 => (7 17)
1000 => (70 325)
10000 => (703 4858)
100000 => (7026 64741)
1000000 => (70229 808950)
^C

## Swift

Translation of: Pascal
var total = 0
var prim = 0
var maxPeri = 100

func newTri(s0:Int, _ s1:Int, _ s2: Int) -> () {

let p = s0 + s1 + s2
if p <= maxPeri {
prim += 1
total += maxPeri / 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)
}
}

while maxPeri <= 100_000_000 {
prim = 0
total = 0
newTri(3, 4, 5)
print("Up to \(maxPeri) : \(total) triples \( prim) primitives.")
maxPeri *= 10
}
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.

## 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

## VBA

Translation of: Pascal
Dim total As Variant, prim As Variant, maxPeri As Variant
Private Sub newTri(s0 As Variant, s1 As Variant, s2 As Variant)
Dim p As Variant
p = CDec(s0) + CDec(s1) + CDec(s2)
If p <= maxPeri Then
prim = prim + 1
total = total + maxPeri \ 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 If
End Sub
Public Sub Program_PythagoreanTriples()
maxPeri = CDec(100)
Do While maxPeri <= 10000000#
prim = CDec(0)
total = CDec(0)
newTri 3, 4, 5
Debug.Print "Up to "; maxPeri; ": "; total; " triples, "; prim; " primitives."
maxPeri = maxPeri * 10
Loop
End Sub
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.

## VBScript

Translation of: Perl

For i=1 To 8
WScript.StdOut.WriteLine triples(10^i)
Next

Function triples(pmax)
prim=0 : count=0 : nmax=Sqr(pmax)/2 : n=1
Do While n <= nmax
m=n+1 : p=2*m*(m+n)
Do While p <= pmax
If gcd(m,n)=1 Then
prim=prim+1
count=count+Int(pmax/p)
End If
m=m+2
p=2*m*(m+n)
Loop
n=n+1
Loop
triples = "Max Perimeter: " & pmax &_
", Total: " & count &_
", Primitive: " & prim
End Function

Function gcd(a,b)
c = a : d = b
Do
If c Mod d > 0 Then
e = c Mod d
c = d
d = e
Else
gcd = d
Exit Do
End If
Loop
End Function

## Visual Basic

Translation of: VBA
Works with: Visual Basic version 5
Works with: Visual Basic version 6
Works with: VBA version Access 97
Works with: VBA version 6.5
Works with: VBA version 7.1
Option Explicit

Dim total As Long, prim As Long, maxPeri As Long

Public Sub NewTri(ByVal s0 As Long, ByVal s1 As Long, ByVal s2 As Long)
Dim p As Long, x1 As Long, x2 As Long
p = s0 + s1 + s2
If p <= maxPeri Then
prim = prim + 1
total = total + maxPeri \ p
x1 = s0 + s2
x2 = s1 + s2
NewTri s0 + 2 * (-s1 + s2), 2 * x1 - s1, 2 * (x1 - s1) + s2
NewTri s0 + 2 * x2, 2 * x1 + s1, 2 * (x1 + s1) + s2
NewTri -s0 + 2 * x2, 2 * (-s0 + s2) + s1, 2 * (-s0 + x2) + s2
End If
End Sub

Public Sub Main()
maxPeri = 100
Do While maxPeri <= 10& ^ 8
prim = 0
total = 0
NewTri 3, 4, 5
Debug.Print "Up to "; maxPeri; ": "; total; " triples, "; prim; " primitives."
maxPeri = maxPeri * 10
Loop
End Sub
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.

## zkl

Translation of: D
fcn tri(lim,a=3,b=4,c=5){
p:=a + b + c;
if(p>lim) return(0,0);
T(1,lim/p).zipWith('+,
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.

## ZX Spectrum Basic

ZX Spectrum: 8 bit microprocessor 3.5 Mhz doing all the work. In an effort to get some decent speed the program is made to be as fast as it can.

It takes about 90 seconds for limit = 10 000 and 17 minutes for limit=100 000 and 3.5 hours for limit = 1000 000.

To set the limits. Set in line nr: 1 L to the starting limit. Set in line nr: 11 IF L<=(last limit to calculate)

Ex. start as limit 100 and end on limit 1000. Set in line nr: 1 LET L=100. Set in line nr: 11 IF L<=1000 THEN GO TO 2

1 LET Y=0: LET X=0: LET Z=0: LET V=0: LET U=0: LET L=10: LET T=0: LET P=0: LET N=4: LET M=0: PRINT "limit   trip.   prim."
2 FOR U=2 TO INT (SQR (L/2)): LET Y=U-INT (U/2)*2: LET N=N+4: LET M=U*U*2: IF Y=0 THEN LET M=M-U-U
3 FOR V=1+Y TO U-1 STEP 2: LET M=M+N: LET X=U: LET Y=V
4 LET Z=Y: LET Y=X-INT (X/Y)*Y: LET X=Z: IF Y<>0 THEN GO TO 4
5 IF X>1 THEN GO TO 8
6 IF M>L THEN GO TO 9
7 LET P=P+1: LET T=T+INT (L/M)
8 NEXT V
9 NEXT U
10 PRINT L;TAB 8;T;TAB 16;P
11 LET N=4: LET T=0: LET P=0: LET L=L*10: IF L<=100000 THEN GO TO 2
Output:
limit   trip.   prim.
10      0       0
100     17      7
1000    325     70
10000   4858    703
100000  64741   7026
1000000 808950  70229