O'Halloran numbers
You are encouraged to solve this task according to the task description, using any language you may know.
For this task, for our purposes, a cuboid is a 3 dimensional object, with six rectangular faces, where all angles are right angles, opposite faces of the cuboid are equal, and where each dimension is a positive integer unit length. It will subsequently be referred to simply as a cuboid; but be aware that it references the above definition.
The surface area of a cuboid is two times the length times the width, plus two times the length times the height, plus two times the width times the height. A cuboid will always have an even integer surface area. The minimum surface area a cuboid may have is 6; one where the l, w, and h measurements are all 1.
2 × ( l × w + w × h + h × l ) 2 × ( 1 × 1 + 1 × 1 + 1 × 1 ) = 6
Different cuboid configurations (may) yield different surface areas, but the surface area is always an integer and is always even.
A cuboid with l = 2, w = 1 h = 1 has a surface area of 10
2 × ( 2 × 1 + 1 × 1 + 1 × 2 ) = 10
There is no configuration which will yield a surface area of 8.
There are 16 known even integer values below 1000 which can not be a surface area for any integer cuboid. It is conjectured, though not rigorously proved, that no others exist.
- Task
- Find and display the sixteen even integer values larger than 6 (the minimum cuboid area) and less than 1000
that can not be the surface area of a cuboid.
- See also
ALGOL 68
As in the Phix and other samples, we only need to consider one arrangement of each l, b and h values.
BEGIN # find O'Halloran numbers - numbers that cannot be the surface area of #
# a cuboid with integer dimensions #
INT count := 0;
INT max area = 1 000;
INT half max area = max area OVER 2;
FOR n FROM 8 BY 2 TO max area DO
BOOL ohalloran := TRUE;
FOR l TO half max area WHILE ohalloran DO
FOR b FROM l TO half max area WHILE INT lb = l * b;
lb < n AND ohalloran
DO
FOR h FROM b TO half max area WHILE INT bh = b * h, lh = l * h;
INT sum = 2 * ( lb + bh + lh );
sum <= n AND ( ohalloran := sum /= n )
DO SKIP OD
OD
OD;
IF ohalloran THEN
print( ( " ", whole( n, 0 ) ) );
count +:= 1
FI
OD
END- Output:
8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
AppleScript
on OHalloranNumbers(max)
script o
property evens : {missing value, missing value}
end script
repeat with n from 6 to max by 2
set end of o's evens to n
end repeat
set halfMax to max div 2
set sixthMax to halfMax div 3
repeat with x from 1 to sixthMax
repeat with y from x to (sixthMax div x)
repeat with halfArea from ((x + x + y) * y) to halfMax by (x + y)
set o's evens's item halfArea to missing value
end repeat
end repeat
end repeat
return o's evens's integers
end OHalloranNumbers
OHalloranNumbers(1000 - 1)
(* Repeat logic condensed from:
repeat with x from 1 to sixthMax
repeat with y from x to sixthMax
set xy to x * y
if (xy > sixthMax) then exit repeat
repeat with z from y to sixthMax
set halfArea to xy + (x + y) * z
if (halfArea > halfMax) then exit repeat
set o's evens's item halfArea to missing value
end repeat
end repeat
end repeat
*)
- Output:
{8, 12, 20, 36, 44, 60, 84, 116, 140, 156, 204, 260, 380, 420, 660, 924}
Arturo
found: select 6..998 => even?
loop 1..497 'l ->
loop 1..l 'w [
lw: l * w
if lw >= 498 -> break
loop 1..w 'h [
sa: 2 * (lw + (w*h) + h*l)
(sa < 1000)? -> 'found -- sa
-> break
]
]
print found
- Output:
8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
BASIC
Rosetta Code problem: https://rosettacode.org/wiki/O%27Halloran_numbers
by Jjuanhdez, 02/2023
BASIC256
maxArea = 1000
halfMaxArea = maxArea/2
dim T[halfMaxArea] #table of half areas
for i = 0 to halfMaxArea-1
T[i] = True #assume all are O'kalloran numbers
next i
for l = 1 to maxArea
for w = 1 to halfMaxArea
for h = 1 to halfMaxArea
suma = l*w + l*h + w*h
if suma < halfMaxArea then T[suma] = False #not an O'kalloran number
next h
next w
next l
print "All known O'Halloran numbers:"
for l = 6/2 to halfMaxArea-1
if T[l] then print int(l*2); " ";
next l- Output:
All known O'Halloran numbers: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Chipmunk Basic
100 cls
110 maxarea = 1000
120 halfmaxarea = maxarea/2
130 dim t(halfmaxarea)'table of half areas
140 for i = 0 to halfmaxarea-1
150 t(i) = 1 'assume all are O'kalloran numbers
160 next i
170 for l = 1 to maxarea
180 for w = 1 to halfmaxarea
190 for h = 1 to halfmaxarea
200 suma = l*w+l*h+w*h
210 if suma < halfmaxarea then t(suma) = 0 'not an O'kalloran number
220 next h
230 next w
240 next l
250 print "All known O'Halloran numbers:"
260 print "[ ";
270 for l = 6/2 to halfmaxarea-1
280 if t(l) then print int(l*2);" ";
290 next l
300 print chr$(8);"]"
310 end
FreeBASIC
Const maxArea = 1000, halfMaxArea = maxArea/2
Dim As Integer i, l, w, h, suma
Dim As Boolean T(halfMaxArea) 'table of half areas
For i = 0 To halfMaxArea-1
T(i) = True 'assume all are O'kalloran numbers
Next i
For l = 1 To maxArea
For w = 1 To halfMaxArea
For h = 1 To halfMaxArea
suma = l*w + l*h + w*h
If suma < halfMaxArea Then T(suma) = False 'not an O'kalloran number
Next h, w, l
Print "All known O'Halloran numbers:"
Print "[";
For l = 6/2 To halfMaxArea-1
If T(l) Then Print l*2; ",";
Next l
Print Chr(8); " ]"
Sleep- Output:
All known O'Halloran numbers: [ 8, 12, 20, 36, 44, 60, 84, 116, 140, 156, 204, 260, 380, 420, 660, 924 ]
QBasic
The Chipmunk Basic solution works without any changes.
True BASIC
LET maxarea = 1000
LET halfmaxarea = maxarea/2
DIM t(1)
MAT REDIM t(0 TO halfmaxarea) !Table of half areas
FOR i = 0 TO halfmaxarea-1
LET t(i) = 1 !assume all are O'kalloran numbers
NEXT i
FOR l = 1 TO maxarea
FOR w = 1 TO halfmaxarea
FOR h = 1 TO halfmaxarea
LET suma = l*w+l*h+w*h
IF suma < halfmaxarea THEN LET t(suma) = 0 !not an O'kalloran number
NEXT h
NEXT w
NEXT l
PRINT "All known O'Halloran numbers:"
FOR l = 6/2 TO halfmaxarea-1
IF t(l) = 1 THEN PRINT INT(l*2);" ";
NEXT l
END
Yabasic
maxArea = 1000
halfMaxArea = maxArea/2
dim T(halfMaxArea) //table of half areas
for i = 0 to halfMaxArea-1
T(i) = True //assume all are O'kalloran numbers
next i
for l = 1 to maxArea
for w = 1 to halfMaxArea
for h = 1 to halfMaxArea
suma = l*w + l*h + w*h
if suma < halfMaxArea T(suma) = False //not an O'kalloran number
next h
next w
next l
print "All known O'Halloran numbers:"
print "[";
for l = 6/2 to halfMaxArea-1
if T(l) print l*2, ",";
next l
print chr$(8), "]"- Output:
All known O'Halloran numbers: [8,12,20,36,44,60,84,116,140,156,204,260,380,420,660,924]
C
More or less.
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
int main() {
bool *found = calloc(1000, sizeof(bool)); // all false initially
int i, l, w, h, lw, sa;
for (l = 1; l < 498; ++l) {
for (w = 1; w <= l; ++w) {
lw = l * w;
if (lw >= 498) break;
for (h = 1; h <= w; ++h) {
sa = (lw + w*h + h*l) * 2;
if (sa < 1000) found[sa] = true;
}
}
}
printf("All known O'Halloran numbers:\n[");
for (i = 6; i < 1000; i += 2) {
if (!found[i]) printf("%d, ", i);
}
printf("\b\b]\n");
free(found);
return 0;
}
- Output:
All known O'Halloran numbers: [8, 12, 20, 36, 44, 60, 84, 116, 140, 156, 204, 260, 380, 420, 660, 924]
C#
using System;
using System.Linq;
class OHalloranNumbers
{
static void Main(string[] args)
{
int maximumArea = 1000;
int halfMaximumArea = maximumArea / 2;
bool[] ohalloranNumbers = Enumerable.Repeat(true, halfMaximumArea).ToArray();
for (int length = 1; length < maximumArea; length++)
{
for (int width = 1; width < halfMaximumArea; width++)
{
for (int height = 1; height < halfMaximumArea; height++)
{
int halfArea = length * width + length * height + width * height;
if (halfArea < halfMaximumArea)
{
ohalloranNumbers[halfArea] = false;
}
}
}
}
Console.WriteLine("Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid:");
for (int i = 3; i < halfMaximumArea; i++)
{
if (ohalloranNumbers[i])
{
Console.Write(i * 2 + " ");
}
}
Console.WriteLine();
}
}
- Output:
Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
C++
#include <cstdint>
#include <iostream>
#include <vector>
int main() {
const uint32_t maximum_area = 1'000;
const uint32_t half_maximum_area = maximum_area / 2;
std::vector<uint32_t> ohalloran_numbers(half_maximum_area, true);
for ( uint32_t length = 1; length < maximum_area; ++length ) {
for ( uint32_t width = 1; width < half_maximum_area; ++width ) {
for ( uint32_t height = 1; height < half_maximum_area; ++height ) {
uint32_t half_area = length * width + length * height + width * height;
if ( half_area < half_maximum_area ) {
ohalloran_numbers[half_area] = false;
}
}
}
}
std::cout << "Values larger than 6 and less than 1_000 which cannot be the surface area of a cuboid:"
<< std::endl;
for ( uint32_t i = 3; i < half_maximum_area; ++i ) {
if ( ohalloran_numbers[i] ) {
std::cout << i * 2 << " ";
}
}
std::cout << std::endl;
}
- Output:
Values larger than 6 and less than 1_000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Dart
import "dart:io";
void main() {
const int maximumArea = 1000;
int halfMaximumArea = maximumArea ~/ 2;
List<bool> ohalloranNumbers = List<bool>.filled(halfMaximumArea, true);
for (int length = 1; length < maximumArea; length++) {
for (int width = 1; width < halfMaximumArea; width++) {
for (int height = 1; height < halfMaximumArea; height++) {
int halfArea = length * width + length * height + width * height;
if (halfArea < halfMaximumArea) {
ohalloranNumbers[halfArea] = false;
}
}
}
}
print("Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid:");
for (int i = 3; i < halfMaximumArea; i++) {
if (ohalloranNumbers[i]) {
stdout.write('${i * 2} ');
}
}
}
- Output:
Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Delphi
procedure ShowOHalloranNumbers(Memo: TMemo);
var L, W, H: integer;
var CubeArea, I, Cnt: integer;
const Limit = 1000;
const Limit2 = Limit div 2;
const Limit4 = Limit div 4;
var ValidOH: array [0..Limit2] of boolean;
var S: string;
begin
{Since we are using CuboidArea/2, use half the space }
for I:= 0 to Limit2-1 do ValidOH[I]:= true;
for L:= 1 to Limit4 do
for W:= 1 to L do
for H:= 1 to L do
begin
{Calculate 1/2 Cuboid value}
CubeArea:=L * W + L * H + W * H;
{Make sure number doesn't overrun array}
if CubeArea>=Length(ValidOH) then continue;
{Mark it as not an OHalloran number}
ValidOH[CubeArea]:= false;
end;
S:=''; Cnt:=0;
{Since we are using half the space}
{everything has to be doubled}
for I:=2 * 3 to Limit2-1 do
if ValidOH[I] then
begin
Inc(Cnt);
S:=S+Format('%8D',[I*2]);
If (Cnt mod 5)=0 then S:=S+CRLF;
end;
Memo.Lines.Add(S);
Memo.Lines.Add('Count='+IntToStr(Cnt));
end;
- Output:
12 20 36 44 60
84 116 140 156 204
260 380 420 660 924
Count=15
Elapsed Time: 6.353 ms.
EasyLang
len found[] 1000
for l = 1 to 497
for w = 1 to l
lw = l * w
if lw >= 498
break 1
.
for h = 1 to w
sa = (lw + w * h + h * l) * 2
if sa <= 1000
found[sa] = 1
.
.
.
.
for i = 6 step 2 to 998
if found[i] = 0
write i & " "
.
.Fortran
program ohalloran_numbers
implicit none
integer, parameter :: max_area = 1000
integer :: half_max, i, j, k, area
logical, dimension(max_area) :: areas
half_max = max_area / 2
! Initialize areas with .TRUE.
areas = .TRUE.
do i = 1, max_area
do j = 1, half_max
if (i * j > half_max) exit
do k = 1, half_max
area = 2 * (i * j + i * k + j * k)
if (area > max_area) exit
areas(area) = .FALSE. ! Mark as not an O'Halloran number
end do
end do
end do
! Print the O'Halloran numbers
print *, "Even surface areas < NOT ", max_area, " achievable by any regular integer-valued cuboid:"
do i = 1, max_area
if (mod(i, 2) == 0 .AND. areas(i)) then
write(*,'(I4,1X)', advance='no') i
endif
end do
print *, "" ! To add a final newline after the list
end program ohalloran_numbers
- Output:
Even surface areas < NOT 1000 achievable by any regular integer-valued cuboid: 2 4 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
FutureBasic
_maxArea = 1000
_halfMaxArea = 500
NSUInteger i, l, w, h, sum
BOOL table( _halfMaxArea )
// Populate boolean table
for i = 0 to _halfMaxArea - 1
table( i ) = YES
next
for l = 1 to _maxArea
for w = 1 to _maxArea
for h = 1 to _halfMaxArea
sum = l * w + l * h + w * h
if sum < _halfMaxArea then table( sum ) = NO
next
next
next
printf @"All known O'Halloran numbers:"
for l = 6/2 to _halfMaxArea - 1
if table( l ) then print l * 2,
next
print chr$(8),
HandleEvents- Output:
All known O'Halloran numbers: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Go
package main
import "fmt"
func main() {
found := make([]bool, 1000) // all false initially
for l := 1; l < 498; l++ {
for w := 1; w <= l; w++ {
lw := l * w
if lw >= 498 {
break
}
for h := 1; h <= w; h++ {
sa := (lw + w*h + h*l) * 2
if sa < 1000 {
found[sa] = true
}
}
}
}
fmt.Printf("All known O'Halloran numbers:\n[")
for i := 6; i < 1000; i += 2 {
if !found[i] {
fmt.Printf("%d, ", i)
}
}
fmt.Printf("\b\b]\n")
}
- Output:
All known O'Halloran numbers: [8, 12, 20, 36, 44, 60, 84, 116, 140, 156, 204, 260, 380, 420, 660, 924]
J
require'stats'
2*(3}.i.501)-.+/1 */\.(|:3 comb 42)-i:1
8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Here, we use combinations with repetitions to generate the various relevant cuboid side lengths. Then we multiply all three pairs of these side length combinations and sum the pairs. Then we remove these sums from the sequence 3..500, and finally we multiply the remaining 16 values by 2.
Java
import java.util.Arrays;
public final class OHalloranNumbers {
public static void main(String[] args) {
final int maximumArea = 1_000;
final int halfMaximumArea = maximumArea / 2;
boolean[] ohalloranNumbers = new boolean[halfMaximumArea];
Arrays.fill(ohalloranNumbers, true);
for ( int length = 1; length < maximumArea; length++ ) {
for ( int width = 1; width < halfMaximumArea; width++ ) {
for ( int height = 1; height < halfMaximumArea; height++ ) {
int halfArea = length * width + length * height + width * height;
if ( halfArea < halfMaximumArea ) {
ohalloranNumbers[halfArea] = false;
}
}
}
}
System.out.println("Values larger than 6 and less tha 1_000 which cannot be the surface area of a cuboid:");
for ( int i = 3; i < halfMaximumArea; i++ ) {
if ( ohalloranNumbers[i] ) {
System.out.print(i * 2 + " ");
}
}
System.out.println();
}
}
- Output:
Values larger than 6 and less than 1_000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
JavaScript
// find O'Halloran numbers - numbers that cannot be the surface area of
// a cuboid with integer dimensions
'use strict'
const maxArea = 1000
const halfMaxArea = Math.trunc( maxArea / 2 )
let list = ""
for( let n = 8; n <= maxArea; n += 2 )
{
let oHalloran = true
for( let l = 1; l <= halfMaxArea && oHalloran; l ++ )
{
for( let b = l; b <= halfMaxArea; b ++ )
{
const lb = l * b
if( lb >= n || ! oHalloran )break
for( let h = b; h <= halfMaxArea; h ++ )
{
const bh = b * h, lh = l * h
const sum = 2 * ( lb + bh + lh )
oHalloran = sum != n
if( sum > n || ! oHalloran )break;
}
}
}
if( oHalloran )
{
list = list + " " + n
}
}
console.log( list )
- Output:
8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
jq
Also works with jaq and with gojq, the Go implementation of jq
# Emit a stream of possible cuboid areas less than or equal to the specified limit,
# $maxarea, which should be an even integer.
def cuboid_areas(maxarea):
maxarea as $maxarea
# min area per face is 1 so if total area is $maxarea, the max dimension is ($maxarea-4)/2
| ($maxarea/2) as $halfmax
| ($halfmax - 2) as $max
| foreach range(1; 1+$max) as $i (null;
label $loopj
# By symmetry, we can assume i <= j <= k
| foreach range($i; 1+$max) as $j (.;
($i*$j) as $ij
| if $ij + 2 > $halfmax then break $loopj else . end
| label $loopk
| foreach range($j; 1+$max) as $k (.;
($ij + $i*$k + $j*$k) as $total
| if $total > $halfmax then break $loopk
else 2 * $total
end) ) );
1000 as $n
| "Even integers greater than 6 but less than \($n) that cannot be cuboid surface areas:",
[range(6;$n;2)] - [cuboid_areas($n-2)]Even integers greater than 6 but less than 1000 that cannot be cuboid surface areas: [8,12,20,36,44,60,84,116,140,156,204,260,380,420,660,924]
Julia
""" Rosetta code task: rosettacode.org/wiki/O%27Halloran_numbers """
const max_area, half_max = 1000, 500
const areas = trues(max_area)
areas[1:2:max_area] .= false
for i in 1:max_area
for j in 1:half_max
i * j > half_max && break
for k in 1:half_max
area = 2 * (i * j + i * k + j * k)
area > max_area && break
areas[area] = false
end
end
end
println("Even surface areas < $max_area NOT achievable by any regular integer-valued cuboid:\n",
[n for n in eachindex(areas) if areas[n]])
- Output:
Even surface areas < 1000 NOT achievable by any regular integer-valued cuboid: [2, 4, 8, 12, 20, 36, 44, 60, 84, 116, 140, 156, 204, 260, 380, 420, 660, 924]
Kotlin
object OHalloranNumbers {
@JvmStatic
fun main(args: Array<String>) {
val maximumArea = 1_000
val halfMaximumArea = maximumArea / 2
var ohalloranNumbers = BooleanArray(halfMaximumArea) { true }
for (length in 1 until maximumArea) {
for (width in 1 until halfMaximumArea) {
for (height in 1 until halfMaximumArea) {
val halfArea = length * width + length * height + width * height
if (halfArea < halfMaximumArea) {
ohalloranNumbers[halfArea] = false
}
}
}
}
println("Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid:")
for (i in 3 until halfMaximumArea) {
if (ohalloranNumbers[i]) {
print("${i * 2} ")
}
}
println()
}
}
fun main() {
OHalloranNumbers.main(arrayOf())
}
- Output:
Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Lua
function calculateOHalloranNumbers(maximumArea)
local halfMaximumArea = maximumArea / 2
local ohalloranNumbers = {}
-- Initialize the ohalloranNumbers table with all true values
for i = 1, halfMaximumArea do
ohalloranNumbers[i] = true
end
-- Calculate surface areas of possible cuboids and exclude them
for length = 1, maximumArea - 1 do
for width = 1, halfMaximumArea - 1 do
for height = width, halfMaximumArea - 1 do
local halfArea = length * width + length * height + width * height
if halfArea < halfMaximumArea then
ohalloranNumbers[halfArea] = false
else
break
end
end
if length * width * 2 >= halfMaximumArea then
break
end
end
end
-- Print out the O'Halloran numbers
print("Values larger than 6 and less than " .. maximumArea .. " which cannot be the surface area of a cuboid:")
for i = 3, halfMaximumArea-1 do
if ohalloranNumbers[i] then
io.write((2 * (i)) .. " ")
end
end
print()
end
-- Call the function with the maximum area of 1000
calculateOHalloranNumbers(1000)
- Output:
Values larger than 6 and less than 1000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Mathematica/Wolfram Language
(* Rosetta code task: rosettacode.org/wiki/O%27Halloran_numbers *)
maxArea = 1000;
halfMax = 500;
areas = Table[True, {maxArea}];
areas[[1 ;; maxArea ;; 2]] = False;
For[i = 1, i <= maxArea, i++,
For[j = 1, j <= halfMax,
j++, If[i*j > halfMax, Break[]];
For[k = 1, k <= halfMax,
k++, area = 2 (i*j + i*k + j*k);
If[area > maxArea, Break[]];
areas[[area]] = False;
];
];
];
Print["Even surface areas < ", maxArea, " NOT achievable by any regular integer-valued cuboid:\n",
Select[Range[maxArea], areas[[#]] &]]
- Output:
Even surface areas < 1000 NOT achievable by any regular integer-valued cuboid:
{2, 4, 8, 12, 20, 36, 44, 60, 84, 116, 140, 156, 204, 260, 380, 420, 660, 924}
MATLAB/Octave
max_area = 1000;
half_max = max_area / 2;
areas = ones(1, max_area); % Initialize areas with 1's
for i = 1:max_area
for j = 1:half_max
if (i * j > half_max)
break;
endif
for k = 1:half_max
area = 2 * (i * j + i * k + j * k);
if (area > max_area)
break;
endif
areas(area) = 0; % Mark as not an O'Halloran number
endfor
endfor
endfor
% Print the O'Halloran numbers
printf("Even surface areas < NOT %d achievable by any regular integer-valued cuboid:\n", max_area);
for n = 1:max_area/2
if (areas(2*n))
printf("%d ", 2*n);
endif
endfor
printf("\n");
- Output:
Even surface areas < NOT 1000 achievable by any regular integer-valued cuboid: 2 4 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Nim
import std/algorithm
const
M = 1000 # Maximum area.
N = M div 2 # Maximum half area.
var isOHalloran: array[3..N, bool]
isOHalloran.fill(true)
for length in 1..N:
for width in 1..length:
let plw = length * width
if plw > N: break
let slw = length + width
for height in 1..width:
let halfArea = plw + height * slw
if halfArea > N: break
isOHalloran[halfArea] = false
echo "All known O'Halloran numbers:"
for n in 3..N:
if isOHalloran[n]:
stdout.write 2 * n, ' '
echo()
- Output:
All known O'Halloran numbers: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
OCaml
let maximum_area = 1000;;
let half_maximum_area = maximum_area / 2;;
let ohalloran_numbers = Array.make half_maximum_area true;;
for length = 1 to maximum_area - 1 do
for width = 1 to half_maximum_area - 1 do
for height = 1 to half_maximum_area - 1 do
let half_area = length * width + length * height + width * height in
if half_area < half_maximum_area then
ohalloran_numbers.(half_area) <- false
done;
done;
done;;
Printf.printf "Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid:\n";
for i = 3 to half_maximum_area - 1 do
if ohalloran_numbers.(i) then
Printf.printf "%d " (i * 2)
done;
- Output:
Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
PARI/GP
{
max_area = 1000;
half_max = max_area / 2;
areas = vector(max_area, i, 1); /* Initialize areas with 1's */
for (i = 1, max_area,
for (j = 1, half_max,
if (i * j > half_max, break);
for (k = 1, half_max,
area = 2 * (i * j + i * k + j * k);
if (area > max_area, break);
areas[area] = 0; /* Mark as not an O'Halloran number */
)
)
);
/* Print the O'Halloran numbers */
print("Even surface areas < NOT ", Str(max_area) , " achievable by any regular integer-valued cuboid:");
for (n = 2/2, max_area/2,
if (areas[2*n], print1(2*n, " "));
);
}- Output:
Even surface areas < NOT 1000 achievable by any regular integer-valued cuboid: 2 4 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Perl
use v5.36;
my @A;
my $threshold = 10_000; # redonkulous overkill
for my $x (1..$threshold) {
X: for my $y (1..$x) {
last X if $x*$y > $threshold;
Y: for my $z (1..$y) {
last Y if (my $area = 2 * ($x*$y + $y*$z + $z*$x)) > $threshold;
$A[$area] = "$x,$y,$z";
}
}
say 'Even integer surface areas NOT achievable by any regular, integer dimensioned cuboid';
for (0..$#A) {
print "$_ " if $_ < $threshold and $_ > 6 and 0 == $_ % 2 and not $A[$_];
}
- Output:
Even integer surface areas NOT achievable by any regular, integer dimensioned cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Phix
Since we are going to check {1,1,2}, there is no point checking {1,2,1} or {2,1,1} etc.
with javascript_semantics constant max_area = 1000, half_max = max_area/2 sequence areas = repeat(0,5)&tagset(max_area,6) for x=1 to max_area do if odd(x) then areas[x] = 0 end if for y=x to floor(half_max/x) do for z=y to half_max do atom area = 2 * (x * y + x * z + y * z) if area > max_area then exit end if areas[area] = 0 end for end for end for printf(1,"Even surface areas < %d NOT achievable by any regular integer-valued cuboid:\n%s\n", {max_area,join(filter(areas,"!=",0),fmt:="%d")})
- Output:
You can also set max_area to 1,000,000 and get no more results.
Even surface areas < 1000 NOT achievable by any regular integer-valued cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
PHP
<?php
function main() {
$maximumArea = 1000;
$halfMaximumArea = $maximumArea / 2;
$ohalloranNumbers = array_fill(0, $halfMaximumArea, true);
for ($length = 1; $length < $maximumArea; $length++) {
for ($width = 1; $width < $halfMaximumArea; $width++) {
for ($height = 1; $height < $halfMaximumArea; $height++) {
$halfArea = $length * $width + $length * $height + $width * $height;
if ($halfArea < $halfMaximumArea) {
$ohalloranNumbers[$halfArea] = false;
}
}
}
}
echo "Values larger than 6 and less than $maximumArea which cannot be the surface area of a cuboid:\n";
for ($i = 3; $i < $halfMaximumArea; $i++) {
if ($ohalloranNumbers[$i]) {
echo ($i * 2) . " ";
}
}
echo "\n";
}
main();
?>
- Output:
Values larger than 6 and less than 1000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Python
#!/usr/bin/python
maxArea = 1000
halfMaxArea = 500
T = [] #table of half areas
for i in range(halfMaxArea):
T.append(True) #assume all are O'kalloran numbers
for l in range(1,maxArea):
for w in range(1, halfMaxArea):
for h in range(1, halfMaxArea):
suma = l*w + l*h + w*h
if suma < halfMaxArea:
T[suma] = False #not an O'kalloran number
print("All known O'Halloran numbers:")
for l in range(3,halfMaxArea):
if T[l]:
print(l*2, end=" ");
- Output:
All known O'Halloran numbers: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
PROMAL
;;; find O'Halloran numbers - numbers that cannot be the surface area of
;;; a cuboid with integer dimensions
PROGRAM OHalloran
INCLUDE LIBRARY
CON WORD maxArea = 1000
WORD count
WORD halfMaxArea
WORD n
WORD l
WORD b
WORD h
WORD lb
WORD lh
WORD bh
WORD sum
BYTE isOHalloran
BEGIN
count = 0
halfMaxArea = maxArea / 2
n = 8
WHILE n <= maxArea
isOHalloran = 1
l = 1
REPEAT
b = l
lb = l * b
REPEAT
h = b
REPEAT
bh = b * h
lh = l * h
sum = 2 * ( lb + ( h * ( b + l ) ) )
IF sum = n
isOHalloran = 0
h = h + 1
UNTIL h > halfMaxArea OR sum > n OR NOT isOHalloran
b = b + 1
lb = l * b
UNTIL b > halfMaxArea OR lb >= n OR NOT isOHalloran
l = l + 1
UNTIL l > halfMaxArea OR NOT isOHalloran
IF isOHalloran
OUTPUT " #W", n
count = count + 1
n = n + 2
END- Output:
8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
R
max_area <- 1000
half_max <- max_area / 2
areas <- rep(1, max_area) # Initialize areas with 1's
for (i in 1:max_area) {
for (j in 1:half_max) {
if (i * j > half_max) {
break
}
for (k in 1:half_max) {
area <- 2 * (i * j + i * k + j * k)
if (area > max_area) {
break
}
areas[area] <- 0 # Mark as not an O'Halloran number
}
}
}
# Print the O'Halloran numbers
cat("Even surface areas < NOT", max_area, "achievable by any regular integer-valued cuboid:\n")
for (n in 3:(max_area/2)) {
if (areas[2*n] == 1) {
cat(2*n, " ")
}
}
cat("\n") # To ensure the output ends with a newline
- Output:
Even surface areas < NOT 1000 achievable by any regular integer-valued cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Racket
#lang racket
(define (main)
(let ([maximum-area 1000]
[ohalloran-numbers (make-hash)])
;; Initialize the hash table
(for ([i (in-range (quotient maximum-area 2))])
(hash-set! ohalloran-numbers i #t))
;; Calculate cuboid surface areas and update the hash table
(for* ([length (in-range 1 maximum-area)]
[width (in-range 1 (quotient maximum-area 2))]
[height (in-range 1 (quotient maximum-area 2))])
(let* ([half-area (+ (* length width) (* length height) (* width height))])
(when (< half-area (quotient maximum-area 2))
(hash-set! ohalloran-numbers half-area #f))))
;; Print the results
(printf "Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid:\n")
(for ([i (in-range 3 (quotient maximum-area 2))])
(when (hash-ref ohalloran-numbers i)
(printf "~a " (* i 2))))
(printf "\n")))
(main)
- Output:
Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Raku
my @Area;
my $threshold = 1000; # a little overboard to make sure we get them all
for 1..$threshold -> $x {
for 1..$x -> $y {
last if $x × $y > $threshold;
for 1..$y -> $z {
push @Area[my $area = ($x × $y + $y × $z + $z × $x) × 2], "$x,$y,$z";
last if $area > $threshold;
}
}
}
say "Even integer surface areas NOT achievable by any regular, integer dimensioned cuboid:\n" ~
@Area[^$threshold].kv.grep( { $^key > 6 and $key %% 2 and !$^value } )»[0];
- Output:
Even integer surface areas NOT achievable by any regular, integer dimensioned cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Ruby
class OHalloranNumbers
def self.main
maximum_area = 1_000
half_maximum_area = maximum_area / 2
ohalloran_numbers = Array.new(half_maximum_area, true)
(1...maximum_area).each do |length|
(1...half_maximum_area).each do |width|
(1...half_maximum_area).each do |height|
half_area = length * width + length * height + width * height
ohalloran_numbers[half_area] = false if half_area < half_maximum_area
end
end
end
puts "Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid:"
(3...half_maximum_area).each do |i|
print "#{i * 2} " if ohalloran_numbers[i]
end
puts
end
end
OHalloranNumbers.main
- Output:
Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Rust
fn main() {
const MAXIMUM_AREA: u32 = 1_000;
const HALF_MAXIMUM_AREA: u32 = MAXIMUM_AREA / 2;
let mut ohalloran_numbers = vec![true; HALF_MAXIMUM_AREA as usize];
for length in 1..MAXIMUM_AREA {
for width in 1..HALF_MAXIMUM_AREA {
for height in 1..HALF_MAXIMUM_AREA {
let half_area = length * width + length * height + width * height;
if half_area < HALF_MAXIMUM_AREA {
ohalloran_numbers[half_area as usize] = false;
}
}
}
}
println!("Values larger than 6 and less than 1_000 which cannot be the surface area of a cuboid:");
for i in 3..HALF_MAXIMUM_AREA {
if ohalloran_numbers[i as usize] {
print!("{} ", i * 2);
}
}
println!();
}
- Output:
Values larger than 6 and less than 1_000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
RPL
« 1000 DUP 2 / FLOOR 1 → maxarea halfmax ohalloran
« { }
8 maxarea FOR n
1 ohalloran SF
WHILE DUP halfmax ≤ ohalloran FS? AND REPEAT
1 halfmax FOR b
DUP b *
IF DUP n ≥ ohalloran FC? OR THEN DROP halfmax 'b' STO
ELSE
b halfmax FOR h
OVER h * OVER + h b * + DUP +
IF DUP n == THEN ohalloran CF END
IF n > ohalloran FC? OR THEN halfmax 'h' STO END
NEXT DROP
END
NEXT
1 +
END DROP
IF ohalloran FS? THEN n + END
2 STEP
» » 'OHALL' STO
- Output:
1: { 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924 }
Scala
object OHalloranNumbers {
def main(args: Array[String]): Unit = {
val maximumArea = 1_000
val halfMaximumArea = maximumArea / 2
var ohalloranNumbers = Array.fill[Boolean](halfMaximumArea)(true)
for {
length <- 1 until maximumArea
width <- 1 until halfMaximumArea
height <- 1 until halfMaximumArea
} {
val halfArea = length * width + length * height + width * height
if (halfArea < halfMaximumArea) {
ohalloranNumbers(halfArea) = false
}
}
println("Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid:")
for {
i <- 3 until halfMaximumArea
if ohalloranNumbers(i)
} {
print(i * 2 + " ")
}
println()
}
}
- Output:
Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Swift
import Foundation
func main() {
let maximumArea = 1_000
let halfMaximumArea = maximumArea / 2
var ohalloranNumbers = Array(repeating: true, count: halfMaximumArea)
for length in 1..<maximumArea {
for width in 1..<halfMaximumArea {
for height in 1..<halfMaximumArea {
let halfArea = length * width + length * height + width * height
if halfArea < halfMaximumArea {
ohalloranNumbers[halfArea] = false
}
}
}
}
print("Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid:")
for i in 3..<halfMaximumArea where ohalloranNumbers[i] {
print(i * 2, terminator: " ")
}
print()
}
// Running the main function
main()
- Output:
Values larger than 6 and less than 1,000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Tcl
proc calculateOHalloranNumbers { maximumArea } {
set halfMaximumArea [expr {$maximumArea / 2}]
set ohalloranNumbers [list]
# Initialize the ohalloranNumbers list with all true values
for {set i 0} {$i < $halfMaximumArea} {incr i} {
lappend ohalloranNumbers true
}
# Calculate surface areas of possible cuboids and exclude them
for {set length 1} {$length < $maximumArea} {incr length} {
for {set width 1} {$width < $halfMaximumArea} {incr width} {
for {set height $width} {$height < $halfMaximumArea} {incr height} {
set halfArea [expr {$length * $width + $length * $height + $width * $height}]
if {$halfArea < $halfMaximumArea} {
lset ohalloranNumbers $halfArea false
} else {
break
}
}
if { $length * $width * 2 >= $halfMaximumArea } {
break
}
}
}
# Print out the O'Halloran numbers
puts "Values larger than 6 and less than $maximumArea which cannot be the surface area of a cuboid:"
for {set i 3} {$i < $halfMaximumArea} {incr i} {
if {[lindex $ohalloranNumbers $i]} {
puts -nonewline "[expr {2 * $i}] "
}
}
puts ""
}
# Call the procedure with the maximum area of 1000
calculateOHalloranNumbers 1000
- Output:
Values larger than 6 and less than 1000 which cannot be the surface area of a cuboid: 8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
Wren
import "./seq" for Lst
var found = []
for (l in 1..497) {
for (w in 1..l) {
var lw = l * w
if (lw >= 498) break
for (h in 1..w) {
var sa = (lw + w*h + h*l) * 2
if (sa < 1000) found.add(sa) else break
}
}
}
var allEven = (6..998).where { |i| i%2 == 0 }.toList
System.print("All known O'Halloran numbers:")
System.print(Lst.except(allEven, found))
- Output:
All known O'Halloran numbers: [8, 12, 20, 36, 44, 60, 84, 116, 140, 156, 204, 260, 380, 420, 660, 924]
XPL0
int L, W, H, HA, I;
char T(1000/2); \table of half areas
[for I:= 0 to 1000/2-1 do
T(I):= true; \assume all are O'Halloran numbers
for L:= 1 to 250 do
for W:= 1 to 250/L do
for H:= 1 to 250/L do
[HA:= L*W + L*H + W*H;
if HA < 500 then \not an O'Halloran number
T(HA):= false;
];
for I:= 6/2 to 1000/2-1 do
if T(I) then
[IntOut(0, I*2); ChOut(0, ^ )];
]- Output:
8 12 20 36 44 60 84 116 140 156 204 260 380 420 660 924
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