Forbidden numbers
Forbidden numbers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Lagrange's theorem tells us that every positive integer can be written as a sum of at most four squares.
Many, (square numbers) can be written as a "sum" of a single square. E.G.: 4 == 22.
Some numbers require at least two squares to be summed. 5 == 22 + 12
Others require at least three squares. 6 == 22 + 12 + 12
Finally, some, (the focus of this task) require the sum at least four squares. 7 == 22 + 12 + 12 + 12. There is no way to reach 7 other than summing four squares.
These numbers show up in crystallography; x-ray diffraction patterns of cubic crystals depend on a length squared plus a width squared plus a height squared. Some configurations that require at least four squares are impossible to index and are colloquially known as forbidden numbers.
Note that some numbers can be made from the sum of four squares: 16 == 22 + 22 + 22 + 22, but since it can also be formed from less than four, 16 == 42, it does not count as a forbidden number.
- Task
- Find and show the first fifty forbidden numbers.
- Find and show the count of forbidden numbers up to 500, 5,000.
- Stretch
- Find and show the count of forbidden numbers up to 50,000, 500,000.
- See also
Go
A translation of the Python code in the OEIS link. Runtime around 7 seconds.
package main
import (
"fmt"
"math"
"rcu"
)
func isForbidden(n int) bool {
m := n
v := 0
for m > 1 && m%4 == 0 {
m /= 4
v++
}
pow := int(math.Pow(4, float64(v)))
return n/pow%8 == 7
}
func main() {
forbidden := make([]int, 50)
for i, count := 0, 0; count < 50; i++ {
if isForbidden(i) {
forbidden[count] = i
count++
}
}
fmt.Println("The first 50 forbidden numbers are:")
rcu.PrintTable(forbidden, 10, 3, false)
fmt.Println()
limit := 500
count := 0
for i := 1; ; i++ {
if isForbidden(i) {
count++
}
if i == limit {
slimit := rcu.Commatize(limit)
scount := rcu.Commatize(count)
fmt.Printf("Forbidden number count <= %11s: %10s\n", slimit, scount)
if limit == 500_000_000 {
return
}
limit *= 10
}
}
}
- Output:
The first 50 forbidden numbers are: 7 15 23 28 31 39 47 55 60 63 71 79 87 92 95 103 111 112 119 124 127 135 143 151 156 159 167 175 183 188 191 199 207 215 220 223 231 239 240 247 252 255 263 271 279 284 287 295 303 311 Forbidden number count <= 500: 82 Forbidden number count <= 5,000: 831 Forbidden number count <= 50,000: 8,330 Forbidden number count <= 500,000: 83,331 Forbidden number count <= 5,000,000: 833,329 Forbidden number count <= 50,000,000: 8,333,330 Forbidden number count <= 500,000,000: 83,333,328
jq
The following also works with gojq, the Go implementation of jq, except that beyond forbidden(5000), gojq's speed and memory requirements might become a problem.
def count(s): reduce s as $x (0; .+1);
def lpad($len): tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
# The def of _nwise can be omitted if using the C implementation of jq:
def _nwise($n):
def n: if length <= $n then . else .[0:$n] , (.[$n:] | n) end;
n;
def forbidden($max):
def ub($a;$b):
if $b < 0 then 0 else [$a, ($b|sqrt)] | min end;
[false, range(1; 1 + $max)]
| reduce range(1; 1 + ($max|sqrt)) as $i (.;
($i*$i) as $s1
| .[$s1] = false
| reduce range(1; 1 + ub($i; ($max - $s1))) as $j (.;
($s1 + $j*$j) as $s2
| .[$s2] = false
| reduce range(1; 1 + ub($j; ($max - $s2))) as $k (.;
.[$s2 + $k*$k] = false ) ) )
| map( select(.) ) ;
forbidden(500) as $f
| "First fifty forbidden numbers:",
( $f[:50] | _nwise(10) | map(lpad(3)) | join(" ") ),
"\nForbidden number count up to 500: \(count($f[]))",
((5000, 50000, 500000) | "\nForbidden number count up to \(.): \(count(forbidden(.)[])) ")- Output:
First fifty forbidden numbers: 7 15 23 28 31 39 47 55 60 63 71 79 87 92 95 103 111 112 119 124 127 135 143 151 156 159 167 175 183 188 191 199 207 215 220 223 231 239 240 247 252 255 263 271 279 284 287 295 303 311 Forbidden number count up to 500: 82 Forbidden number count up to 5000: 831 Forbidden number count up to 50000: 8330 Forbidden number count up to 500000: 83331
Pascal
Free Pascal
modified formula to calc count. Using count as gag to get next forbidden number.
No runtime.
program ForbiddenNumbers;
{$IFDEF FPC}{$MODE DELPHI}{$OPTIMIZATION ON,ALL}{$ENDIF}
{$IFDEF WINDOWS}{$APPTYPE CONSOLE}{$ENDIF}
uses
sysutils,strutils;
function CntForbiddenTilLimit(lmt:NativeUint):NativeUint;
//n = 4^i * (8*j + 7) | i,j >= 0
var
Power4,Delta,n : NativeUint;
begin
result := 0;
power4 := 1;
repeat
delta := Power4*8;
n := Power4*7;
if n > lmt then
Break;
inc(result,(lmt-n) DIV delta+1);
Power4 *= 4;
until false;
end;
var
lmt,n: NativeUint;
BEGIN
writeln('First fifty forbidden numbers:');
n := 1;
lmt := 1;
repeat
repeat
inc(n);
until CntForbiddenTilLimit(n) >= lmt;
write(n:4);
if LMT MOD 20 = 0 THEN
writeln;
lmt += 1;
until lmt > 50;
writeln;
writeln;
writeln('count of forbidden numbers below ');
lmt := 5;
repeat
writeln(Numb2USA(IntToStr(lmt)):30,Numb2USA(IntToStr(CntForbiddenTilLimit(lmt))):25);
lmt *= 10;
until lmt > High(lmt) DIV 4;
END.
- Output:
First fifty forbidden numbers:
7 15 23 28 31 39 47 55 60 63 71 79 87 92 95 103 111 112 119 124
127 135 143 151 156 159 167 175 183 188 191 199 207 215 220 223 231 239 240 247
252 255 263 271 279 284 287 295 303 311
count of forbidden numbers below
5 0
50 7
500 82
5,000 831
50,000 8,330
500,000 83,331
5,000,000 833,329
50,000,000 8,333,330
500,000,000 83,333,328
5,000,000,000 833,333,330
50,000,000,000 8,333,333,327
500,000,000,000 83,333,333,328
5,000,000,000,000 833,333,333,327
50,000,000,000,000 8,333,333,333,327
500,000,000,000,000 83,333,333,333,326
5,000,000,000,000,000 833,333,333,333,327
50,000,000,000,000,000 8,333,333,333,333,325
500,000,000,000,000,000 83,333,333,333,333,323
Python
""" rosettacode.org/wiki/Forbidden_numbers """
def isforbidden(num):
""" true if num is a forbidden number """
fours, pow4 = num, 0
while fours > 1 and fours % 4 == 0:
fours //= 4
pow4 += 1
return (num // 4**pow4) % 8 == 7
f500k = list(filter(isforbidden, range(500_001)))
for idx, fbd in enumerate(f500k[:50]):
print(f'{fbd: 4}', end='\n' if (idx + 1) % 10 == 0 else '')
for fbmax in [500, 5000, 50_000, 500_000]:
print(
f'\nThere are {sum(x <= fbmax for x in f500k):,} forbidden numbers <= {fbmax:,}.')
- Output:
7 15 23 28 31 39 47 55 60 63 71 79 87 92 95 103 111 112 119 124 127 135 143 151 156 159 167 175 183 188 191 199 207 215 220 223 231 239 240 247 252 255 263 271 279 284 287 295 303 311 There are 82 forbidden numbers <= 500. There are 831 forbidden numbers <= 5,000. There are 8,330 forbidden numbers <= 50,000. There are 83,331 forbidden numbers <= 500,000.
Raku
use Lingua::EN::Numbers;
use List::Divvy;
my @f = (1..*).map: *×8-1;
my @forbidden = lazy flat @f[0], gather for ^∞ {
state ($p0, $p1) = 1, 0;
take (@f[$p0] < @forbidden[$p1]×4) ?? @f[$p0++] !! @forbidden[$p1++]×4;
}
put "First fifty forbidden numbers: \n" ~
@forbidden[^50].batch(10)».fmt("%3d").join: "\n";
put "\nForbidden number count up to {.Int.&cardinal}: " ~
comma +@forbidden.&upto: $_ for 5e2, 5e3, 5e4, 5e5, 5e6;
- Output:
First fifty forbidden numbers: 7 15 23 28 31 39 47 55 60 63 71 79 87 92 95 103 111 112 119 124 127 135 143 151 156 159 167 175 183 188 191 199 207 215 220 223 231 239 240 247 252 255 263 271 279 284 287 295 303 311 Forbidden number count up to five hundred: 82 Forbidden number count up to five thousand: 831 Forbidden number count up to fifty thousand: 8,330 Forbidden number count up to five hundred thousand: 83,331 Forbidden number count up to five million: 833,329
Wren
Version 1
This uses a sieve to filter out those numbers which are the sums of one, two or three squares. Works but very slow (c. 52 seconds).
import "./fmt" for Fmt
var forbidden = Fn.new { |limit, countOnly|
var sieve = List.filled(limit+1, false)
var ones
var twos
var threes
var i = 0
while((ones = i*i) <= limit) {
sieve[ones] = true
var j = i
while ((twos = ones + j*j) <= limit) {
sieve[twos] = true
var k = j
while ((threes = twos + k*k) <= limit) {
sieve[threes] = true
k = k + 1
}
j = j + 1
}
i = i + 1
}
if (countOnly) return sieve.count { |b| !b }
var forbidden = []
for (i in 1..limit) {
if (!sieve[i]) forbidden.add(i)
}
return forbidden
}
System.print("The first 50 forbidden numbers are:")
Fmt.tprint("$3d", forbidden.call(400, false).take(50), 10)
System.print()
for (limit in [500, 5000, 50000, 500000, 5000000]) {
var count = forbidden.call(limit, true)
Fmt.print("Forbidden number count <= $,9d: $,7d", limit, count)
}- Output:
The first 50 forbidden numbers are: 7 15 23 28 31 39 47 55 60 63 71 79 87 92 95 103 111 112 119 124 127 135 143 151 156 159 167 175 183 188 191 199 207 215 220 223 231 239 240 247 252 255 263 271 279 284 287 295 303 311 Forbidden number count <= 500: 82 Forbidden number count <= 5,000: 831 Forbidden number count <= 50,000: 8,330 Forbidden number count <= 500,000: 83,331 Forbidden number count <= 5,000,000: 833,329
Version 2
This is a translation of the formula-based Python code in the OEIS link which at around 1.1 seconds is almost 50 times faster than Version 1 and is also about 3 times faster than the PARI code in that link.
import "./fmt" for Fmt
var isForbidden = Fn.new { |n|
var m = n
var v = 0
while (m > 1 && m % 4 == 0) {
m = (m/4).floor
v = v + 1
}
return (n/4.pow(v)).floor % 8 == 7
}
var f400 = (1..400).where { |i| isForbidden.call(i) }
System.print("The first 50 forbidden numbers are:")
Fmt.tprint("$3d", f400.take(50), 10)
System.print()
for (limit in [500, 5000, 50000, 500000, 5000000]) {
var count = (1..limit).count { |i| isForbidden.call(i) }
Fmt.print("Forbidden number count <= $,9d: $,7d", limit, count)
}- Output:
Same as Version 1