⚠️ Warning: This is a draft ⚠️
This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.
{{task}}
One form of '''Pythagorean quadruples''' is (for positive integers '''a''', '''b''', '''c''', and '''d'''):
:::::::: a2 + b2 + c2 = d2
An example:
:::::::: 22 + 32 + 62 = 72
::::: which is:
:::::::: 4 + 9 + 36 = 49
;Task:
For positive integers up '''2,200''' (inclusive), for all values of '''a''', '''b''', '''c''', and '''d''',
find (and show here) those values of '''d''' that ''can't'' be represented.
Show the values of '''d''' on one line of output (optionally with a title).
;Related tasks:
- [[Euler's sum of powers conjecture]].
- [[Pythagorean triples]].
;Reference: :* the Wikipedia article: [https://en.wikipedia.org/wiki/Pythagorean_quadruple Pythagorean quadruple].
ALGOL 68
As with the optimised REXX solution, we find the values of d for which there are no a^2 + b^2 = d^2 - c^2.
BEGIN
# find values of d where d^2 =/= a^2 + b^2 + c^2 for any integers a, b, c #
# where d in [1..2200], a, b, c =/= 0 #
# max number to check #
INT max number = 2200;
INT max square = max number * max number;
# table of numbers that can be the sum of two squares #
[ 1 : max square ]BOOL sum of two squares; FOR n TO max square DO sum of two squares[ n ] := FALSE OD;
FOR a TO max number DO
INT a2 = a * a;
FOR b FROM a TO max number WHILE INT sum2 = ( b * b ) + a2;
sum2 <= max square DO
sum of two squares[ sum2 ] := TRUE
OD
OD;
# now find d such that d^2 - c^2 is in sum of two squares #
[ 1 : max number ]BOOL solution; FOR n TO max number DO solution[ n ] := FALSE OD;
FOR d TO max number DO
INT d2 = d * d;
FOR c TO d - 1 WHILE NOT solution[ d ] DO
INT diff2 = d2 - ( c * c );
IF sum of two squares[ diff2 ] THEN
solution[ d ] := TRUE
FI
OD
OD;
# print the numbers whose squares are not the sum of three squares #
FOR d TO max number DO
IF NOT solution[ d ] THEN
print( ( " ", whole( d, 0 ) ) )
FI
OD;
print( ( newline ) )
END
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
AppleScript
-- double :: Num -> Num
on double(x)
x + x
end double
-- powersOfTwo :: Generator [Int]
on powersOfTwo()
iterate(double, 1)
end powersOfTwo
on run
-- Two infinite lists, from each of which we can draw an arbitrary number of initial terms
set xs to powersOfTwo() -- {1, 2, 4, 8, 16, 32 ...
set ys to fmapGen(timesFive, powersOfTwo()) -- {5, 10, 20, 40, 80, 160 ...
-- Another infinite list, derived from the first two (sorted in rising value)
set zs to mergeInOrder(xs, ys) -- {1, 2, 4, 5, 8, 10 ...
-- Taking terms from the derived list while their value is below 2200 ...
takeWhileGen(le2200, zs)
--> {1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048}
end run
-- le2200 :: Num -> Bool
on le2200(x)
x ≤ 2200
end le2200
-- timesFive :: Num -> Num
on timesFive(x)
5 * x
end timesFive
-- mergeInOrder :: Generator [Int] -> Generator [Int] -> Generator [Int]
on mergeInOrder(ga, gb)
script
property a : uncons(ga)
property b : uncons(gb)
on |λ|()
if (Nothing of a or Nothing of b) then
missing value
else
set ta to Just of a
set tb to Just of b
if |1| of ta < |1| of tb then
set a to uncons(|2| of ta)
return |1| of ta
else
set b to uncons(|2| of tb)
return |1| of tb
end if
end if
end |λ|
end script
end mergeInOrder
-- GENERIC -----------------------------------------------------------------
-- fmapGen <$> :: (a -> b) -> Gen [a] -> Gen [b]
on fmapGen(f, gen)
script
property g : gen
property mf : mReturn(f)'s |λ|
on |λ|()
set v to g's |λ|()
if v is missing value then
v
else
mf(v)
end if
end |λ|
end script
end fmapGen
-- iterate :: (a -> a) -> a -> Gen [a]
on iterate(f, x)
script
property v : missing value
property g : mReturn(f)'s |λ|
on |λ|()
if missing value is v then
set v to x
else
set v to g(v)
end if
return v
end |λ|
end script
end iterate
-- Just :: a -> Maybe a
on Just(x)
{type:"Maybe", Nothing:false, Just:x}
end Just
-- length :: [a] -> Int
on |length|(xs)
set c to class of xs
if list is c or string is c then
length of xs
else
(2 ^ 29 - 1) -- (maxInt - simple proxy for non-finite)
end if
end |length|
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- Nothing :: Maybe a
on Nothing()
{type:"Maybe", Nothing:true}
end Nothing
-- take :: Int -> [a] -> [a]
-- take :: Int -> String -> String
on take(n, xs)
set c to class of xs
if list is c then
if 0 < n then
items 1 thru min(n, length of xs) of xs
else
{}
end if
else if string is c then
if 0 < n then
text 1 thru min(n, length of xs) of xs
else
""
end if
else if script is c then
set ys to {}
repeat with i from 1 to n
set v to xs's |λ|()
if missing value is v then
return ys
else
set end of ys to v
end if
end repeat
return ys
else
missing value
end if
end take
-- takeWhileGen :: (a -> Bool) -> Gen [a] -> [a]
on takeWhileGen(p, xs)
set ys to {}
set v to |λ|() of xs
tell mReturn(p)
repeat while (|λ|(v))
set end of ys to v
set v to xs's |λ|()
end repeat
end tell
return ys
end takeWhileGen
-- Tuple (,) :: a -> b -> (a, b)
on Tuple(a, b)
{type:"Tuple", |1|:a, |2|:b, length:2}
end Tuple
-- uncons :: [a] -> Maybe (a, [a])
on uncons(xs)
set lng to |length|(xs)
if 0 = lng then
Nothing()
else
if (2 ^ 29 - 1) as integer > lng then
if class of xs is string then
set cs to text items of xs
Just(Tuple(item 1 of cs, rest of cs))
else
Just(Tuple(item 1 of xs, rest of xs))
end if
else
Just(Tuple(item 1 of take(1, xs), xs))
end if
end if
end uncons
{{Out}}
{1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048}
AWK
# syntax: GAWK -f PYTHAGOREAN_QUADRUPLES.AWK
# converted from Go
BEGIN {
n = 2200
s = 3
for (a=1; a<=n; a++) {
a2 = a * a
for (b=a; b<=n; b++) {
ab[a2 + b * b] = 1
}
}
for (c=1; c<=n; c++) {
s1 = s
s += 2
s2 = s
for (d=c+1; d<=n; d++) {
if (ab[s1]) {
r[d] = 1
}
s1 += s2
s2 += 2
}
}
for (d=1; d<=n; d++) {
if (!r[d]) {
printf("%d ",d)
}
}
printf("\n")
exit(0)
}
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
C
Version 1
Five seconds on my Intel Linux box.
#include <stdio.h>
#include <math.h>
#include <string.h>
#define N 2200
int main(int argc, char **argv){
int a,b,c,d;
int r[N+1];
memset(r,0,sizeof(r)); // zero solution array
for(a=1; a<=N; a++){
for(b=a; b<=N; b++){
int aabb;
if(a&1 && b&1) continue; // for positive odd a and b, no solution.
aabb=a*a + b*b;
for(c=b; c<=N; c++){
int aabbcc=aabb + c*c;
d=(int)sqrt((float)aabbcc);
if(aabbcc == d*d && d<=N) r[d]=1; // solution
}
}
}
for(a=1; a<=N; a++)
if(!r[a]) printf("%d ",a); // print non solution
printf("\n");
}
{{out}}
$ clang -O3 foo.c -lm
$ ./a.out
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
===Version 2 (much faster)=== Translation of second version of FreeBASIC entry. Runs in about 0.15 seconds.
#include <iostream>
#include <stdio.h>
#include <string.h>
#define N 2200
#define N2 2200 * 2200 * 2
int main(int argc, char **argv) {
int a, b, c, d, a2, s = 3, s1, s2;
int r[N + 1];
memset(r, 0, sizeof(r));
int *ab = calloc(N2 + 1, sizeof(int)); // allocate on heap, zero filled
for (a = 1; a <= N; a++) {
a2 = a * a;
for (b = a; b <= N; b++) ab[a2 + b * b] = 1;
}
for (c = 1; c <= N; c++) {
s1 = s;
s += 2;
s2 = s;
for (d = c + 1; d <= N; d++) {
if (ab[s1]) r[d] = 1;
s1 += s2;
s2 += 2;
}
}
for (d = 1; d <= N; d++) {
if (!r[d]) printf("%d ", d);
}
printf("\n");
free(ab);
return 0;
}
{{out}}
Same as first version.
C++
{{trans|D}}
#include <iostream>
#include <vector>
constexpr int N = 2200;
constexpr int N2 = 2 * N * N;
int main() {
using namespace std;
vector<bool> found(N + 1);
vector<bool> aabb(N2 + 1);
int s = 3;
for (int a = 1; a < N; ++a) {
int aa = a * a;
for (int b = 1; b < N; ++b) {
aabb[aa + b * b] = true;
}
}
for (int c = 1; c <= N; ++c) {
int s1 = s;
s += 2;
int s2 = s;
for (int d = c + 1; d <= N; ++d) {
if (aabb[s1]) {
found[d] = true;
}
s1 += s2;
s2 += 2;
}
}
cout << "The values of d <= " << N << " which can't be represented:" << endl;
for (int d = 1; d <= N; ++d) {
if (!found[d]) {
cout << d << " ";
}
}
cout << endl;
return 0;
}
{{out}}
The values of d <= 2200 which can't be represented:
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
C#
{{trans|Java}}
using System;
namespace PythagoreanQuadruples {
class Program {
const int MAX = 2200;
const int MAX2 = MAX * MAX * 2;
static void Main(string[] args) {
bool[] found = new bool[MAX + 1]; // all false by default
bool[] a2b2 = new bool[MAX2 + 1]; // ditto
int s = 3;
for(int a = 1; a <= MAX; a++) {
int a2 = a * a;
for (int b=a; b<=MAX; b++) {
a2b2[a2 + b * b] = true;
}
}
for (int c = 1; c <= MAX; c++) {
int s1 = s;
s += 2;
int s2 = s;
for (int d = c + 1; d <= MAX; d++) {
if (a2b2[s1]) found[d] = true;
s1 += s2;
s2 += 2;
}
}
Console.WriteLine("The values of d <= {0} which can't be represented:", MAX);
for (int d = 1; d < MAX; d++) {
if (!found[d]) Console.Write("{0} ", d);
}
Console.WriteLine();
}
}
}
{{out}}
The values of d <= 2200 which can't be represented:
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Crystal
{{trans|Ruby}}
n = 2200
l_add, l = Hash(Int32, Bool).new(false), Hash(Int32, Bool).new(false)
(1..n).each do |x|
x2 = x * x
(x..n).each { |y| l_add[x2 + y * y] = true }
end
s = 3
(1..n).each do |x|
s1 = s
s += 2
s2 = s
((x+1)..n).each do |y|
l[y] = true if l_add[s1]
s1 += s2
s2 += 2
end
end
puts (1..n).reject{ |x| l[x] }.join(" ")
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
D
{{trans|C}}
import std.bitmanip : BitArray;
import std.stdio;
enum N = 2_200;
enum N2 = 2*N*N;
void main() {
BitArray found;
found.length = N+1;
BitArray aabb;
aabb.length = N2+1;
uint s=3;
for (uint a=1; a<=N; ++a) {
uint aa = a*a;
for (uint b=1; b<N; ++b) {
aabb[aa + b*b] = true;
}
}
for (uint c=1; c<=N; ++c) {
uint s1 = s;
s += 2;
uint s2 = s;
for (uint d=c+1; d<=N; ++d) {
if (aabb[s1]) {
found[d] = true;
}
s1 += s2;
s2 += 2;
}
}
writeln("The values of d <= ", N, " which can't be represented:");
for (uint d=1; d<=N; ++d) {
if (!found[d]) {
write(d, ' ');
}
}
writeln;
}
{{out}}
The values of d <= 2200 which can't be represented:
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
FreeBASIC
From the Wikipedia page
[https://en.wikipedia.org/wiki/Pythagorean_quadruple Alternate parametrization, second version both A and B even.]Time just less then 0.7 second on a AMD Athlon II X4 645 3.34GHz win7 64bit. Program uses one core. When the limit is set to 576 (abs. minimum for 2200), the time is about 0.85 sec.
' version 12-08-2017
' compile with: fbc -s console
#Define max 2200
Dim As UInteger l, m, n, l2, l2m2
Dim As UInteger limit = max * 4 \ 15
Dim As UInteger max2 = limit * limit * 2
ReDim As Ubyte list_1(max2), list_2(max2 +1)
' prime sieve, list_2(l) contains a 0 if l = prime
For l = 4 To max2 Step 2
list_1(l) = 1
Next
For l = 3 To max2 Step 2
If list_1(l) = 0 Then
For m = l * l To max2 Step l * 2
list_1(m) = 1
Next
End If
Next
' we do not need a and b (a and b are even, l = a \ 2, m = b \ 2)
' we only need to find d
For l = 1 To limit
l2 = l * l
For m = l To limit
l2m2 = l2 + m * m
list_2(l2m2 +1) = 1
' if l2m2 is a prime, no other factors exits
If list_1(l2m2) = 0 Then Continue For
' find possible factors of l2m2
' if l2m2 is odd, we need only to check the odd divisors
For n = 2 + (l2m2 And 1) To Fix(Sqr(l2m2 -1)) Step 1 + (l2m2 And 1)
If l2m2 Mod n = 0 Then
' set list_2(x) to 1 if solution is found
list_2(l2m2 \ n + n) = 1
End If
Next
Next
Next
For l = 1 To max
If list_2(l) = 0 Then Print l; " ";
Next
Print
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Brute force
Based on the second REXX version: A^2 + B^2 = D^2 - C^2. Faster then the first version, about 0.2 second
' version 14-08-2017
' compile with: fbc -s console
#Define n 2200
Dim As UInteger s = 3, s1, s2, x, x2, y
ReDim As Ubyte l(n), l_add(n * n * 2)
For x = 1 To n
x2 = x * x
For y = x To n
l_add(x2 + y * y) = 1
Next
Next
For x = 1 To n
s1 = s
s += 2
s2 = s
For y = x +1 To n
If l_add(s1) = 1 Then l(y) = 1
s1 += s2
s2 += 2
Next
Next
For x = 1 To n
If l(x) = 0 Then Print Str(x); " ";
Next
Print
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Go
{{trans|FreeBASIC}}
package main
import "fmt"
const (
N = 2200
N2 = N * N * 2
)
func main() {
s := 3
var s1, s2 int
var r [N + 1]bool
var ab [N2 + 1]bool
for a := 1; a <= N; a++ {
a2 := a * a
for b := a; b <= N; b++ {
ab[a2 + b * b] = true
}
}
for c := 1; c <= N; c++ {
s1 = s
s += 2
s2 = s
for d := c + 1; d <= N; d++ {
if ab[s1] {
r[d] = true
}
s1 += s2
s2 += 2
}
}
for d := 1; d <= N; d++ {
if !r[d] {
fmt.Printf("%d ", d)
}
}
fmt.Println()
}
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Haskell
powersOfTwo :: [Int]
powersOfTwo = iterate (2 *) 1
unrepresentable :: [Int]
unrepresentable = merge powersOfTwo ((5 *) <$> powersOfTwo)
merge :: [Int] -> [Int] -> [Int]
merge xxs@(x:xs) yys@(y:ys)
| x < y = x : merge xs yys
| otherwise = y : merge xxs ys
main :: IO ()
main = do
putStrLn "The values of d <= 2200 which can't be represented."
print $ takeWhile (<= 2200) unrepresentable
{{out}}
The values of d <= 2200 which can't be represented.
[1,2,4,5,8,10,16,20,32,40,64,80,128,160,256,320,512,640,1024,1280,2048]
J
Approach: generate the set of all triple sums of squares, then select the legs for which there aren't any squared "d"s. The solution is straightforward interactive play.
Filter =: (#~`)(`:6)
B =: *: A =: i. >: i. 2200
S1 =: , B +/ B NB. S1 is a raveled table of the sums of squares
S1 =: <:&({:B)Filter S1 NB. remove sums of squares exceeding bound
S1 =: ~. S1 NB. remove duplicate entries
S2 =: , B +/ S1
S2 =: <:&({:B)Filter S2
S2 =: ~. S2
RESULT =: (B -.@:e. S2) # A
RESULT
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Java
{{trans|Kotlin}}
public class PythagoreanQuadruple {
static final int MAX = 2200;
static final int MAX2 = MAX * MAX * 2;
public static void main(String[] args) {
boolean[] found = new boolean[MAX + 1]; // all false by default
boolean[] a2b2 = new boolean[MAX2 + 1]; // ditto
int s = 3;
for (int a = 1; a <= MAX; a++) {
int a2 = a * a;
for (int b = a; b <= MAX; b++) a2b2[a2 + b * b] = true;
}
for (int c = 1; c <= MAX; c++) {
int s1 = s;
s += 2;
int s2 = s;
for (int d = c + 1; d <= MAX; d++) {
if (a2b2[s1]) found[d] = true;
s1 += s2;
s2 += 2;
}
}
System.out.printf("The values of d <= %d which can't be represented:\n", MAX);
for (int d = 1; d <= MAX; d++) {
if (!found[d]) System.out.printf("%d ", d);
}
System.out.println();
}
}
{{out}}
The values of d <= 2200 which can't be represented:
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
JavaScript
{{Trans|Haskell}}
(() => {
'use strict';
// main :: IO ()
const main = () => {
const xs = takeWhileGen(
x => 2200 >= x,
mergeInOrder(
powersOfTwo(),
fmapGen(x => 5 * x, powersOfTwo())
)
);
return (
console.log(JSON.stringify(xs)),
xs
);
}
// powersOfTwo :: Gen [Int]
const powersOfTwo = () =>
iterate(x => 2 * x, 1);
// mergeInOrder :: Gen [Int] -> Gen [Int] -> Gen [Int]
const mergeInOrder = (ga, gb) => {
function* go(ma, mb) {
let
a = ma,
b = mb;
while (!a.Nothing && !b.Nothing) {
let
ta = a.Just,
tb = b.Just;
if (fst(ta) < fst(tb)) {
yield(fst(ta));
a = uncons(snd(ta))
} else {
yield(fst(tb));
b = uncons(snd(tb))
}
}
}
return go(uncons(ga), uncons(gb))
};
// GENERIC FUNCTIONS ----------------------------
// fmapGen <$> :: (a -> b) -> Gen [a] -> Gen [b]
function* fmapGen(f, gen) {
const g = gen;
let v = take(1, g);
while (0 < v.length) {
yield(f(v))
v = take(1, g)
}
}
// fst :: (a, b) -> a
const fst = tpl => tpl[0];
// iterate :: (a -> a) -> a -> Generator [a]
function* iterate(f, x) {
let v = x;
while (true) {
yield(v);
v = f(v);
}
}
// Just :: a -> Maybe a
const Just = x => ({
type: 'Maybe',
Nothing: false,
Just: x
});
// Returns Infinity over objects without finite length
// this enables zip and zipWith to choose the shorter
// argument when one is non-finite, like cycle, repeat etc
// length :: [a] -> Int
const length = xs => xs.length || Infinity;
// Nothing :: Maybe a
const Nothing = () => ({
type: 'Maybe',
Nothing: true,
});
// snd :: (a, b) -> b
const snd = tpl => tpl[1];
// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = (n, xs) =>
xs.constructor.constructor.name !== 'GeneratorFunction' ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));
// takeWhileGen :: (a -> Bool) -> Generator [a] -> [a]
const takeWhileGen = (p, xs) => {
const ys = [];
let
nxt = xs.next(),
v = nxt.value;
while (!nxt.done && p(v)) {
ys.push(v);
nxt = xs.next();
v = nxt.value
}
return ys;
};
// Tuple (,) :: a -> b -> (a, b)
const Tuple = (a, b) => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});
// uncons :: [a] -> Maybe (a, [a])
const uncons = xs => {
const lng = length(xs);
return (0 < lng) ? (
lng < Infinity ? (
Just(Tuple(xs[0], xs.slice(1))) // Finite list
) : (() => {
const nxt = take(1, xs);
return 0 < nxt.length ? (
Just(Tuple(nxt[0], xs))
) : Nothing();
})() // Lazy generator
) : Nothing();
};
// MAIN ---
return main();
})();
{{Out}}
[1,2,4,5,8,10,16,20,32,40,64,80,128,160,256,320,512,640,1024,1280,2048]
jq
The following is a direct solution but with some obvious optimizations.
Its main value may be to illustrate how looping with breaks can
be accomplished in jq without foreach. Notice also how
first/1 is used in is_pythagorean_quad/0 to avoid unnecessary computation.
# Emit a proof that the input is a pythagorean quad, or else false
def is_pythagorean_quad:
. as $d
| (.*.) as $d2
| first(
label $continue_a | range(1; $d) | . as $a | (.*.) as $a2
| if 3*$a2 > $d2 then break $continue_a else . end
| label $continue_b | range($a; $d) | . as $b | (.*.) as $b2
| if $a2 + 2 * $b2 > $d2 then break $continue_b else . end
| (($d2-($a2+$b2)) | sqrt) as $c
| if ($c | floor) == $c then [$a, $b, $c] else empty end )
// false;
# The specific task:
[range(1; 2201) | select( is_pythagorean_quad | not )] | join(" ")
'''Invocation and Output'''
jq -r -n -f program.jq
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Julia
{{works with|Julia|0.6}} {{trans|C}}
function quadruples(N::Int=2200)
r = falses(N)
ab = falses(2N ^ 2)
for a in 1:N, b in a:N
ab[a ^ 2 + b ^ 2] = true
end
s = 3
for c in 1:N
s1, s, s2 = s, s + 2, s + 2
for d in c+1:N
if ab[s1] r[d] = true end
s1 += s2
s2 += 2
end
end
return find(.! r)
end
println("Pythagorean quadruples up to 2200: ", join(quadruples(), ", "))
{{out}}
Pythagorean quadruples up to 2200: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048
Kotlin
Version 1
This uses a similar approach to the REXX optimized version. It also takes advantage of a hint in the C entry that there is no solution if both a and b are odd (confirmed by Wikipedia article). Runs in about 7 seconds on my modest laptop which is more than 4 times faster than the brute force version would have been:
// version 1.1.3
const val MAX = 2200
const val MAX2 = MAX * MAX - 1
fun main(args: Array<String>) {
val found = BooleanArray(MAX + 1) // all false by default
val p2 = IntArray(MAX + 1) { it * it } // pre-compute squares
// compute all possible positive values of d * d - c * c and map them back to d
val dc = mutableMapOf<Int, MutableList<Int>>()
for (d in 1..MAX) {
for (c in 1 until d) {
val diff = p2[d] - p2[c]
val v = dc[diff]
if (v == null)
dc.put(diff, mutableListOf(d))
else if (d !in v)
v.add(d)
}
}
for (a in 1..MAX) {
for (b in 1..a) {
if ((a and 1) != 0 && (b and 1) != 0) continue
val sum = p2[a] + p2[b]
if (sum > MAX2) continue
val v = dc[sum]
if (v != null) v.forEach { found[it] = true }
}
}
println("The values of d <= $MAX which can't be represented:")
for (i in 1..MAX) if (!found[i]) print("$i ")
println()
}
{{out}}
The values of d <= 2200 which can't be represented:
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
===Version 2 (much faster)=== This is a translation of the second FreeBASIC version and runs in about the same time (0.2 seconds).
One thing I've noticed about the resulting sequence is that it appears to be an interleaving of the two series 2 ^ n and 5 * (2 ^ n) for n >= 0 though whether it's possible to prove this mathematically I don't know.
// version 1.1.3
const val MAX = 2200
const val MAX2 = MAX * MAX * 2
fun main(args: Array<String>) {
val found = BooleanArray(MAX + 1) // all false by default
val a2b2 = BooleanArray(MAX2 + 1) // ditto
var s = 3
for (a in 1..MAX) {
val a2 = a * a
for (b in a..MAX) a2b2[a2 + b * b] = true
}
for (c in 1..MAX) {
var s1 = s
s += 2
var s2 = s
for (d in (c + 1)..MAX) {
if (a2b2[s1]) found[d] = true
s1 += s2
s2 += 2
}
}
println("The values of d <= $MAX which can't be represented:")
for (d in 1..MAX) if (!found[d]) print("$d ")
println()
}
{{out}}
Same as Version 1.
Lua
-- initialize
local N = 2200
local ar = {}
for i=1,N do
ar[i] = false
end
-- process
for a=1,N do
for b=a,N do
if (a % 2 ~= 1) or (b % 2 ~= 1) then
local aabb = a * a + b * b
for c=b,N do
local aabbcc = aabb + c * c
local d = math.floor(math.sqrt(aabbcc))
if (aabbcc == d * d) and (d <= N) then
ar[d] = true
end
end
end
end
-- print('done with a='..a)
end
-- print
for i=1,N do
if not ar[i] then
io.write(i.." ")
end
end
print()
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
=={{header|Modula-2}}== {{trans|C}}
MODULE PythagoreanQuadruples;
FROM FormatString IMPORT FormatString;
FROM RealMath IMPORT sqrt;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE WriteInteger(i : INTEGER);
VAR buffer : ARRAY[0..16] OF CHAR;
BEGIN
FormatString("%i", buffer, i);
WriteString(buffer)
END WriteInteger;
(* Main *)
CONST N = 2200;
VAR
r : ARRAY[0..N] OF BOOLEAN;
a,b,c,d : INTEGER;
aabb,aabbcc : INTEGER;
BEGIN
(* Initialize *)
FOR a:=0 TO HIGH(r) DO
r[a] := FALSE
END;
(* Process *)
FOR a:=1 TO N DO
FOR b:=a TO N DO
IF (a MOD 2 = 1) AND (b MOD 2 = 1) THEN
(* For positive odd a and b, no solution *)
CONTINUE
END;
aabb := a*a + b*b;
FOR c:=b TO N DO
aabbcc := aabb + c*c;
d := INT(sqrt(FLOAT(aabbcc)));
IF (aabbcc = d*d) AND (d <= N) THEN
(* solution *)
r[d] := TRUE
END
END
END
END;
FOR a:=1 TO N DO
IF NOT r[a] THEN
(* pritn non-solution *)
WriteInteger(a);
WriteString(" ")
END
END;
WriteLn;
ReadChar
END PythagoreanQuadruples.
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Nim
{{trans|FreeBasic}}
Version 1
const N = 2_200
var d, b = 0 r = newSeq[bool](N + 1)
for a in 1..N: for b in a..N: var aabb = 0 if (a and 1).bool and (b and 1).bool: continue aabb = a * a + b * b for c in b..N: var aabbcc = 0 aabbcc = aabb + c * c d = sqrt(aabbcc.float).int if aabbcc == d * d and d <= N: r[d] = true for i in 1..N: if not r[I]: stdout.write i, " "
{{out}}
```txt
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Version 2
const N = 2_200 const N2 = N * N * 2
var a2, d, s1, s2 = 0 s = 3 r = newSeq[bool](N + 1) ab = newSeq[bool](N2 + 1)
for a in 1..N: a2 = a * a for b in a..N: ab[a2 + b * b] = true
for c in 1..N: s1 = s s += 2 s2 = s for d in c+1..N: if ab[s1]: r[d] = true s1 += s2 s2 += 2
for d in 1..N: if not r[d]: stdout.write d, " "
{{out}}
```txt
Same as Version 1.
Pascal
{{works with|Free Pascal}} compiled with fpc 3.2.0 ( 2019.01.10 ) -O4 -Xs
version 1
Brute froce, but not as brute as [http://rosettacode.org/mw/index.php?title=Pythagorean_quadruples#Ring Ring].Did it ever run?
Stopping search if limit is reached
program pythQuad;
//find phythagorean Quadrupel up to a,b,c,d <= 2200
//a^2 + b^2 +c^2 = d^2
//find all values of d which are not possible
//brute force
//split in two procedure to reduce register pressure for CPU32
const
MaxFactor =2200;
limit = MaxFactor*MaxFactor;
type
tIdx = NativeUint;
tSum = NativeUint;
var
check : array[0..MaxFactor] of boolean;
checkCnt : LongWord;
procedure Find2(s:tSum;idx:tSum);
//second sum (a*a+b*b) +c*c =?= d*d
var
s1 : tSum;
d : tSum;
begin
d := trunc(sqrt(s+idx*idx));// calculate first sqrt
For idx := idx to MaxFactor do
Begin
s1 := s+idx*idx;
If s1 <= limit then
Begin
while s1 > d*d do //adjust sqrt
inc(d);
inc(checkCnt);
IF s1=d*d then
check[d] := true;
end
else
Break;
end;
end;
procedure Find1;
//first sum a*a+b*b
var
a,b : tIdx;
s : tSum;
begin
For a := 1 to MaxFactor do
For b := a to MaxFactor do
Begin
s := a*a+b*b;
if s < limit then
Find1(s,b)
else
break;
end;
end;
var
i : NativeUint;
begin
Find1;
For i := 1 to MaxFactor do
If Not(Check[i]) then
write(i,' ');
writeln;
writeln(CheckCnt,' checks were done');
end.
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
929605937 checks were done
real 0m2.323s -> 9 cpu-cycles per check on Ryzen 5 1600 3,7 Ghz ( Turbo )
version 2
Using a variant of [http://rosettacode.org/wiki/Pythagorean_quadruples#optimized REXX optimized] optimized
As I now see the same as [http://rosettacode.org/wiki/Pythagorean_quadruples#ALGOL_68 Algol68]
Quite fast.
program pythQuad_2;
//find phythagorean Quadrupel up to a,b,c,d <= 2200
//a^2 + b^2 +c^2 = d^2
//a^2 + b^2 = d^2-c^2
const
MaxFactor =2200;
limit = MaxFactor*MaxFactor;
type
tIdx = NativeUint;
tSum = NativeUint;
var
// global variables are initiated with 0 at startUp
sumA2B2 :array[0..limit] of byte;
check : array[0..MaxFactor] of byte;
procedure BuildSumA2B2;
var
a,b : tIdx;
s : tSum;
begin
For a := 1 to MaxFactor do
For b := 1 to a do
Begin
s := a*a+b*b;
if s < limit then
sumA2B2[s] := 1
else
break;
end;
end;
procedure CheckDifD2C2;
var
d,c : tIdx;
s : tSum;
begin
For d := 1 to MaxFactor do
//c < d => (d*d-c*c) > 0
For c := d-1 downto 1 do
Begin
s := d*d-c*c;
if sumA2B2[s] <> 0 then
Check[d] := 1;
end;
end;
var
i : NativeUint;
begin
BuildSumA2B2;
CheckDifD2C2;
//FindHoles;
For i := 1 to MaxFactor do
If Check[i] = 0 then
write(i,' ');
writeln;
end.
{{Out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
real 0m0.018s //4.8 Mb -> Level 3 cache 16 Mb ( Ryzen 5 1600 )
//MaxFactor =22000;484 Mb -> no level X Cache
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048 2560 4096 5120 8192 10240 16384 20480
real 0m4.184s
Perl
{{trans|Perl 6}}
my $N = 2200;
push @sq, $_**2 for 0 .. $N;
my @not = (0) x $N;
@not[0] = 1;
for my $d (1 .. $N) {
my $last = 0;
for my $a (reverse ceiling($d/3) .. $d) {
for my $b (1 .. ceiling($a/2)) {
my $ab = $sq[$a] + $sq[$b];
last if $ab > $sq[$d];
my $x = sqrt($sq[$d] - $ab);
if ($x == int $x) {
$not[$d] = 1;
$last = 1;
last
}
}
last if $last;
}
}
sub ceiling { int $_[0] + 1 - 1e-15 }
for (0 .. $#not) {
$result .= "$_ " unless $not[$_]
}
print "$result\n"
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Perl 6
{{works with|Rakudo|2018.09}}
my \N = 2200;
my @sq = (0 .. N)»²;
my @not = False xx N;
@not[0] = True;
(1 .. N).race.map: -> $d {
my $last = 0;
for $d ... ($d/3).ceiling -> $a {
for 1 .. ($a/2).ceiling -> $b {
last if (my $ab = @sq[$a] + @sq[$b]) > @sq[$d];
if (@sq[$d] - $ab).sqrt.narrow ~~ Int {
@not[$d] = True;
$last = 1;
last
}
}
last if $last;
}
}
say @not.grep( *.not, :k );
{{out}}
(1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048)
Phix
constant N = 2200,
N2 = N*N*2
sequence found = repeat(false,N),
squares = repeat(false,N2)
-- first mark all numbers that can be the sum of two squares
for a=1 to N do
integer a2 = a*a
for b=a to N do
squares[a2+b*b] = true
end for
end for
-- now find all d such that d^2 - c^2 is in squares
for d=1 to N do
integer d2 = d*d
for c=1 to d-1 do
if squares[d2-c*c] then
found[d] = true
exit
end if
end for
end for
sequence res = {}
for i=1 to N do
if not found[i] then res &= i end if
end for
?res
{{out}}
{1,2,4,5,8,10,16,20,32,40,64,80,128,160,256,320,512,640,1024,1280,2048}
PicoLisp
{{trans|C}}
(de quadruples (N)
(let (AB NIL S 3 R)
(for A N
(for (B A (>= N B) (inc B))
(idx
'AB
(+ (* A A) (* B B))
T ) ) )
(for C N
(let (S1 S S2)
(inc 'S 2)
(setq S2 S)
(for (D (+ C 1) (>= N D) (inc D))
(and (idx 'AB S1) (idx 'R D T))
(inc 'S1 S2)
(inc 'S2 2) ) ) )
(make
(for A N
(or (idx 'R A) (link A)) ) ) ) )
(println (quadruples 2200))
{{out}}
(1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048)
Python
Search
{{trans|Julia}}
def quad(top=2200):
r = [False] * top
ab = [False] * (top * 2)**2
for a in range(1, top):
for b in range(a, top):
ab[a * a + b * b] = True
s = 3
for c in range(1, top):
s1, s, s2 = s, s + 2, s + 2
for d in range(c + 1, top):
if ab[s1]:
r[d] = True
s1 += s2
s2 += 2
return [i for i, val in enumerate(r) if not val and i]
if __name__ == '__main__':
n = 2200
print(f"Those values of d in 1..{n} that can't be represented: {quad(n)}")
{{out}}
Those values of d in 1..2200 that can't be represented: [1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]
Composition of simpler generators
Or, as an alternative to search – a generative solution (defining the generator we need as a composition of simpler generators): {{Trans|Haskell}} {{Trans|JavaScript}} {{Trans|AppleScript}}
from itertools import (islice)
# main :: IO ()
def main():
print (
takeWhileGen(lambda x: 2200 > x)(
mergeInOrder(powersOfTwo())(
map(lambda x: 5 * x, powersOfTwo())
)
)
)
# powersOfTwo :: Gen [Int]
def powersOfTwo():
return iterate(lambda x: 2 * x)(1)
# mergeInOrder :: Gen [Int] -> Gen [Int] -> Gen [Int]
def mergeInOrder(ga):
def go(ma, mb):
a = ma
b = mb
while not a['Nothing'] and not b['Nothing']:
ta = a['Just']
tb = b['Just']
if ta[0] < tb[0]:
yield(ta[0])
a = uncons(ta[1])
else:
yield(tb[0])
b = uncons(tb[1])
return lambda gb: go(uncons(ga), uncons(gb))
# GENERIC ABSTRACTIONS ------------------------------------
# iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
def go(x):
v = x
while True:
yield(v)
v = f(v)
return lambda x: go(x)
# Just :: a -> Maybe a
def Just(x):
return {'type': 'Maybe', 'Nothing': False, 'Just': x}
# Nothing :: Maybe a
def Nothing():
return {'type': 'Maybe', 'Nothing': True}
# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else list(islice(xs, n))
)
# takeWhileGen :: (a -> Bool) -> Gen [a] -> [a]
def takeWhileGen(p):
def go(xs):
vs = []
v = next(xs)
while (None is not v and p(v)):
vs.append(v)
v = next(xs)
return vs
return lambda xs: go(xs)
# uncons :: [a] -> Maybe (a, [a])
def uncons(xs):
if isinstance(xs, list):
return Just((xs[0], xs[1:])) if 0 < len(xs) else Nothing()
else:
nxt = take(1)(xs)
return Just((nxt[0], xs)) if 0 < len(nxt) else Nothing()
# MAIN ---
main()
{{Out}}
[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]
Racket
{{trans|Python}}
#lang racket
(require data/bit-vector)
(define (quadruples top)
(define top+1 (add1 top))
(define 1..top (in-range 1 top+1))
(define r (make-bit-vector top+1))
(define ab (make-bit-vector (add1 (sqr (* top 2)))))
(for* ((a 1..top) (b (in-range a top+1))) (bit-vector-set! ab (+ (sqr a) (sqr b)) #t))
(for/fold ((s 3))
((c 1..top))
(for/fold ((s1 s) (s2 (+ s 2)))
((d (in-range (add1 c) top+1)))
(when (bit-vector-ref ab s1)
(bit-vector-set! r d #t))
(values (+ s1 s2) (+ s2 2)))
(+ 2 s))
(for/list ((i (in-naturals 1)) (v (in-bit-vector r 1)) #:unless v) i))
(define (report n)
(printf "Those values of d in 1..~a that can't be represented: ~a~%" n (quadruples n)))
(report 2200)
{{out}}
Those values of d in 1..2200 that can't be represented: (1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048)
REXX
brute force
This version is a brute force algorithm, with some optimization (to save compute time)
which pre-computes some of the squares of the positive integers used in the search.
/*REXX pgm computes/shows (integers), D that aren't possible for: a² + b² + c² = d² */
parse arg hi . /*obtain optional argument from the CL.*/
if hi=='' | hi=="," then hi=2200; high= 3 * hi /*Not specified? Then use the default.*/
@.=. /*array of integers to be squared. */
!.=. /* " " " squared. */
do j=1 for high /*precompute possible squares (to max).*/
_= j*j; !._= j; if j<=hi then @.j= _ /*define a square; D value; squared # */
end /*j*/
d.=. /*array of possible solutions (D) */
do a=1 for hi-2; aodd= a//2 /*go hunting for solutions to equation.*/
do b=a to hi-1;
if aodd then if b//2 then iterate /*Are A and B both odd? Then skip.*/
ab = @.a + @.b /*calculate sum of 2 (A,B) squares.*/
do c=b to hi; abc= ab + @.c /* " " " 3 (A,B,C) " */
if !.abc==. then iterate /*Not a square? Then skip it*/
s=!.abc; d.s= /*define this D solution as being found*/
end /*c*/
end /*b*/
end /*a*/
say
say 'Not possible positive integers for d ≤' hi " using equation: a² + b² + c² = d²"
say
$= /* [↓] find all the "not possibles". */
do p=1 for hi; if d.p==. then $=$ p /*Not possible? Then add it to the list*/
end /*p*/ /* [↓] display list of not-possibles. */
say substr($, 2) /*stick a fork in it, we're all done. */
{{out|output|text= when using the default input:}}
Not possible positive integers for d ≤ 2200 using equation: a² + b² + c² = d²
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
optimized
This REXX version is an optimized version, it solves the formula:
:::::::: a2 + b2 = d2 - c2
This REXX version is around '''60''' times faster then the previous version.
Programming note: testing for '''a''' and '''b''' both being odd (lines '''15''' and '''16''' that each contain a '''do''' loop) as
being a case that won't produce any solutions actually slows up the calculations and makes the program execute slower.
/*REXX pgm computes/shows (integers), D that aren't possible for: a² + b² + c² = d² */
parse arg hi . /*obtain optional argument from the CL.*/
if hi=='' | hi=="," then hi=2200 /*Not specified? Then use the default.*/
high= hi * 3 /*D can be three times the HI (max).*/
@.= . /*array of integers (≤ hi) squared.*/
do s=1 for high; _= s*s; r._= s; @.s=_ /*precompute squares and square roots. */
end /*s*/
!.= /*array of differences between squares.*/
do c=1 for high; cc = @.c /*precompute possible differences. */
do d=c+1 to high; dif= @.d - cc /*process D squared; calc differences*/
!.dif= !.dif cc /*add CC to the !.DIF list. */
end /*d*/
end /*c*/
d.=. /*array of the possible solutions (D). */
do a=1 for hi-2 /*go hunting for solutions to equation.*/
do b=a to hi-1; ab= @.a + @.b /*calculate sum of two (A,B) squares.*/
if !.ab=='' then iterate /*Not a difference? Then ignore it. */
do n=1 for words(!.ab) /*handle all ints that satisfy equation*/
abc= ab + word(!.ab, n) /*add the C² integer to A² + B² */
_= r.abc /*retrieve the square root of C² */
d._= /*mark the D integer as being found. */
end /*n*/
end /*b*/
end /*a*/
say
say 'Not possible positive integers for d ≤' hi " using equation: a² + b² + c² = d²"
say
$= /* [↓] find all the "not possibles". */
do p=1 for hi; if d.p==. then $= $ p /*Not possible? Then add it to the list*/
end /*p*/ /* [↓] display list of not-possibles. */
say substr($, 2) /*stick a fork in it, we're all done. */
{{out|output|text= is the same as the 1st REXX version.}}
Ring
# Project : Pythagorean quadruples
limit = 2200
pq = list(limit)
for n = 1 to limit
for m = 1 to limit
for p = 1 to limit
for x = 1 to limit
if pow(x,2) = pow(n,2) + pow(m,2) + pow(p,2)
pq[x] = 1
ok
next
next
next
next
pqstr = ""
for d = 1 to limit
if pq[d] = 0
pqstr = pqstr + d + " "
ok
next
see pqstr + nl
{{Out}} 1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Ruby
{{trans|VBA}}
n = 2200
l_add, l = {}, {}
1.step(n) do |x|
x2 = x*x
x.step(n) {|y| l_add[x2 + y*y] = true}
end
s = 3
1.step(n) do |x|
s1 = s
s += 2
s2 = s
(x+1).step(n) do |y|
l[y] = true if l_add[s1]
s1 += s2
s2 += 2
end
end
puts (1..n).reject{|x| l[x]}.join(" ")
{{Out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
Considering the observations in the Rust and Sidef sections and toying with Enumerators :
squares = Enumerator.new{|y| (0..).each{|n| y << 2**n} }
squares5 = Enumerator.new{|y| (0..).each{|n| y << 2**n*5} }
pyth_quad = Enumerator.new do |y|
n = squares.next
m = squares5.next
loop do
if n < m
y << n
n = squares.next
else
y << m
m = squares5.next
end
end
end
# this takes less than a millisecond
puts pyth_quad.take_while{|n| n <= 1000000000}.join(" ")
{{Out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048 2560 4096 5120 8192 10240 16384 20480 32768 40960 65536 81920 131072 163840 262144 327680 524288 655360 1048576 1310720 2097152 2621440 4194304 5242880 8388608 10485760 16777216 20971520 33554432 41943040 67108864 83886080 134217728 167772160 268435456 335544320 536870912 671088640
Rust
{{output?}}
This is equivalent to https://oeis.org/A094958 which simply contains positive integers of the form 2^n or 5*2^n. Multiple implementations are provided.
use std::collections::BinaryHeap;
fn a094958_iter() -> Vec<u16> {
(0..12)
.map(|n| vec![1 << n, 5 * (1 << n)])
.flatten()
.filter(|x| x < &2200)
.collect::<BinaryHeap<u16>>()
.into_sorted_vec()
}
fn a094958_filter() -> Vec<u16> {
(1..2200) // ported from Sidef
.filter(|n| ((n & (n - 1) == 0) || (n % 5 == 0 && ((n / 5) & (n / 5 - 1) == 0))))
.collect()
}
fn a094958_loop() -> Vec<u16> {
let mut v = vec![];
for n in 0..12 {
v.push(1 << n);
if 5 * (1 << n) < 2200 {
v.push(5 * (1 << n));
}
}
v.sort();
return v;
}
fn main() {
println!("{:?}", a094958_iter());
println!("{:?}", a094958_loop());
println!("{:?}", a094958_filter());
}
#[cfg(test)]
mod tests {
use super::*;
static HAPPY: &str = "[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]";
#[test]
fn test_a094958_iter() {
assert!(format!("{:?}", a094958_iter()) == HAPPY);
}
#[test]
fn test_a094958_loop() {
assert!(format!("{:?}", a094958_loop()) == HAPPY);
}
#[test]
fn test_a094958_filter() {
assert!(format!("{:?}", a094958_filter()) == HAPPY);
}
}
Scala
{{Out}}Best seen running in your browser either by [https://scalafiddle.io/sf/drfij1d/0 ScalaFiddle (ES aka JavaScript, non JVM)] or [https://scastie.scala-lang.org/6AHn7YXSRbKHzmOY5rWAwg Scastie (remote JVM)].
object PythagoreanQuadruple extends App {
val MAX = 2200
val MAX2: Int = MAX * MAX * 2
val found = Array.ofDim[Boolean](MAX + 1)
val a2b2 = Array.ofDim[Boolean](MAX2 + 1)
var s = 3
for (a <- 1 to MAX) {
val a2 = a * a
for (b <- a to MAX) a2b2(a2 + b * b) = true
}
for (c <- 1 to MAX) {
var s1 = s
s += 2
var s2 = s
for (d <- (c + 1) to MAX) {
if (a2b2(s1)) found(d) = true
s1 += s2
s2 += 2
}
}
println(f"The values of d <= ${MAX}%d which can't be represented:")
val notRepresented = (1 to MAX).filterNot(d => found(d) )
println(notRepresented.mkString(" "))
}
Sidef
# Finds all solutions (a,b) such that: a^2 + b^2 = n^2
func sum_of_two_squares(n) is cached {
n == 0 && return [[0, 0]]
var prod1 = 1
var prod2 = 1
var prime_powers = []
for p,e in (n.factor_exp) {
if (p % 4 == 3) { # p = 3 (mod 4)
e.is_even || return [] # power must be even
prod2 *= p**(e >> 1)
}
elsif (p == 2) { # p = 2
if (e.is_even) { # power is even
prod2 *= p**(e >> 1)
}
else { # power is odd
prod1 *= p
prod2 *= p**((e - 1) >> 1)
prime_powers.append([p, 1])
}
}
else { # p = 1 (mod 4)
prod1 *= p**e
prime_powers.append([p, e])
}
}
prod1 == 1 && return [[prod2, 0]]
prod1 == 2 && return [[prod2, prod2]]
# All the solutions to the congruence: x^2 = -1 (mod prod1)
var square_roots = gather {
gather {
for p,e in (prime_powers) {
var pp = p**e
var r = sqrtmod(-1, pp)
take([[r, pp], [pp - r, pp]])
}
}.cartesian { |*a|
take(Math.chinese(a...))
}
}
var solutions = []
for r in (square_roots) {
var s = r
var q = prod1
while (s*s > prod1) {
(s, q) = (q % s, s)
}
solutions.append([prod2 * s, prod2 * (q % s)])
}
for p,e in (prime_powers) {
for (var i = e%2; i < e; i += 2) {
var sq = p**((e - i) >> 1)
var pp = p**(e - i)
solutions += (
__FUNC__(prod1 / pp).map { |pair|
pair.map {|r| sq * prod2 * r }
}
)
}
}
solutions.map {|pair| pair.sort } \
.uniq_by {|pair| pair[0] } \
.sort_by {|pair| pair[0] }
}
# Finds all solutions (a,b,c) such that: a^2 + b^2 + c^2 = n^2
func sum_of_three_squares(n) {
gather {
for k in (1 .. n//3) {
var t = sum_of_two_squares(n**2 - k**2) || next
take(t.map { [k, _...] }...)
}
}
}
say gather {
for n in (1..2200) {
sum_of_three_squares(n) || take(n)
}
}
{{out}}
[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]
Numbers d that cannot be expressed as a^2 + b^2 + c^2 = d^2, are numbers of the form 2^n or 5*2^n:
say gather {
for n in (1..2200) {
if ((n & (n-1) == 0) || (n%%5 && ((n/5) & (n/5 - 1) == 0))) {
take(n)
}
}
}
{{out}}
[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]
Swift
{{trans|C}}
func missingD(upTo n: Int) -> [Int] {
var a2 = 0, s = 3, s1 = 0, s2 = 0
var res = [Int](repeating: 0, count: n + 1)
var ab = [Int](repeating: 0, count: n * n * 2 + 1)
for a in 1...n {
a2 = a * a
for b in a...n {
ab[a2 + b * b] = 1
}
}
for c in 1..<n {
s1 = s
s += 2
s2 = s
for d in c+1...n {
if ab[s1] != 0 {
res[d] = 1
}
s1 += s2
s2 += 2
}
}
return (1...n).filter({ res[$0] == 0 })
}
print(missingD(upTo: 2200))
{{out}}
[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]
VBA
{{trans|FreeBasic}}
Const n = 2200
Public Sub pq()
Dim s As Long, s1 As Long, s2 As Long, x As Long, x2 As Long, y As Long: s = 3
Dim l(n) As Boolean, l_add(9680000) As Boolean '9680000=n * n * 2
For x = 1 To n
x2 = x * x
For y = x To n
l_add(x2 + y * y) = True
Next y
Next x
For x = 1 To n
s1 = s
s = s + 2
s2 = s
For y = x + 1 To n
If l_add(s1) Then l(y) = True
s1 = s1 + s2
s2 = s2 + 2
Next
Next
For x = 1 To n
If Not l(x) Then Debug.Print x;
Next
Debug.Print
End Sub
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048
zkl
{{trans|ALGOL 68}}
# find values of d where d^2 =/= a^2 + b^2 + c^2 for any integers a, b, c #
# where d in [1..2200], a, b, c =/= 0 #
# max number to check #
const max_number = 2200;
const max_square = max_number * max_number;
# table of numbers that can be the sum of two squares #
sum_of_two_squares:=Data(max_square+1,Int).fill(0); # 4 meg byte array
foreach a in ([1..max_number]){
a2 := a * a;
foreach b in ([a..max_number]){
sum2 := ( b * b ) + a2;
if(sum2 <= max_square) sum_of_two_squares[ sum2 ] = True; # True-->1
}
}
# now find d such that d^2 - c^2 is in sum of two squares #
solution:=Data(max_number+1,Int).fill(0); # another byte array
foreach d in ([1..max_number]){
d2 := d * d;
foreach c in ([1..d-1]){
diff2 := d2 - ( c * c );
if(sum_of_two_squares[ diff2 ]){ solution[ d ] = True; break; }
}
}
# print the numbers whose squares are not the sum of three squares #
foreach d in ([1..max_number]){
if(not solution[ d ]) print(d, " ");
}
println();
{{out}}
1 2 4 5 8 10 16 20 32 40 64 80 128 160 256 320 512 640 1024 1280 2048