Giuga numbers

From Rosetta Code
Revision as of 06:53, 7 July 2022 by rosettacode>Horsth (→‎{{header|Free Pascal}}: simplifiying, no performance improvement.)
Giuga 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.
Definition

A Giuga number is a composite number n which is such that each of its distinct prime factors f divide (n/f - 1) exactly.

All known Giuga numbers are even though it is not known for certain that there are no odd examples.

Example

30 is a Giuga number because its distinct prime factors are 2, 3 and 5 and:

  • 30/2 - 1 = 14 is divisible by 2
  • 30/3 - 1 = 9 is divisible by 3
  • 30/5 - 1 = 5 is divisible by 5


Task

Determine and show here the first four Giuga numbers.

Stretch

Determine the fifth Giuga number and any more you have the patience for.

References




C++

Based on the Go solution. Takes 26 minutes on my system (Intel Core i5 3.2GHz). <lang cpp>#include <iostream>

// Assumes n is even with exactly one factor of 2. bool is_giuga(unsigned int n) {

   unsigned int m = n / 2;
   auto test_factor = [&m, n](unsigned int p) -> bool {
       if (m % p != 0)
           return true;
       m /= p;
       return m % p != 0 && (n / p - 1) % p == 0;
   };
   if (!test_factor(3) || !test_factor(5))
       return false;
   static constexpr unsigned int wheel[] = {4, 2, 4, 2, 4, 6, 2, 6};
   for (unsigned int p = 7, i = 0; p * p <= m; ++i) {
       if (!test_factor(p))
           return false;
       p += wheel[i & 7];
   }
   return m == 1 || (n / m - 1) % m == 0;

}

int main() {

   std::cout << "First 5 Giuga numbers:\n";
   // n can't be 2 or divisible by 4
   for (unsigned int i = 0, n = 6; i < 5; n += 4) {
       if (is_giuga(n)) {
           std::cout << n << '\n';
           ++i;
       }
   }

}</lang>

Output:
First 5 Giuga numbers:
30
858
1722
66198
2214408306

FreeBASIC

<lang freebasic>Function isGiuga(m As Uinteger) As Boolean

   Dim As Uinteger n = m
   Dim As Uinteger f = 2, l = Sqr(n)
   Do
       If n Mod f = 0 Then
           If ((m / f) - 1) Mod f <> 0 Then Return False
           n /= f
           If f > n Then Return True
       Else    
           f += 1
           If f > l Then Return False
       End If
   Loop

End Function

Dim As Uinteger n = 3, c = 0, limit = 4 Print "The first "; limit; " Giuga numbers are: "; Do

   If isGiuga(n) Then c += 1: Print n; "  ";
   n += 1

Loop Until c = limit</lang>

Output:
The first 4 Giuga numbers are: 30  858  1722  66198

Go

Translation of: Wren

I thought I'd see how long it would take to 'brute force' the fifth Giuga number and the answer (without using parallelization, Core i7) is about 1 hour 38 minutes. <lang go>package main

import "fmt"

var factors []int var inc = []int{4, 2, 4, 2, 4, 6, 2, 6}

// Assumes n is even with exactly one factor of 2. // Empties 'factors' if any other prime factor is repeated. func primeFactors(n int) {

   factors = factors[:0]
   factors = append(factors, 2)
   last := 2
   n /= 2
   for n%3 == 0 {
       if last == 3 {
           factors = factors[:0]
           return
       }
       last = 3
       factors = append(factors, 3)
       n /= 3
   }
   for n%5 == 0 {
       if last == 5 {
           factors = factors[:0]
           return
       }
       last = 5
       factors = append(factors, 5)
       n /= 5
   }
   for k, i := 7, 0; k*k <= n; {
       if n%k == 0 {
           if last == k {
               factors = factors[:0]
               return
           }
           last = k
           factors = append(factors, k)
           n /= k
       } else {
           k += inc[i]
           i = (i + 1) % 8
       }
   }
   if n > 1 {
       factors = append(factors, n)
   }

}

func main() {

   const limit = 5
   var giuga []int
   // n can't be 2 or divisible by 4
   for n := 6; len(giuga) < limit; n += 4 {
       primeFactors(n)
       // can't be prime or semi-prime
       if len(factors) > 2 {
           isGiuga := true
           for _, f := range factors {
               if (n/f-1)%f != 0 {
                   isGiuga = false
                   break
               }
           }
           if isGiuga {
               giuga = append(giuga, n)
           }
       }
   }
   fmt.Println("The first", limit, "Giuga numbers are:")
   fmt.Println(giuga)

}</lang>

Output:
The first 5 Giuga numbers are:
[30 858 1722 66198 2214408306]

J

We can brute force this task building a test for giuga numbers and checking the first hundred thousand integers (which takes a small fraction of a second):

<lang J>giguaP=: {{ (1<y)*(-.1 p:y)**/(=<.) y ((_1+%)%]) q: y }}"0</lang>

<lang J> 1+I.giguaP 1+i.1e5 30 858 1722 66198</lang>

These numbers have some interesting properties but there's an issue with guaranteeing correctness of more sophisticated approaches.

Julia

<lang ruby>using Primes

isGiuga(n) = all(f -> f != n && rem(n ÷ f - 1, f) == 0, factor(Vector, n))

function getGiuga(N)

   gcount = 0
   for i in 4:typemax(Int)
       if isGiuga(i)
           println(i)
           (gcount += 1) >= N && break
       end
   end

end

getGiuga(4)

</lang>

Output:
30      
858 
1722
66198

Ad hoc faster version

<lang ruby>using Primes

function getgiugas(numberwanted, verbose = true)

   n, found, nfound = 6, Int[], 0
   starttime = time()
   while nfound < numberwanted
       if n % 5 == 0 || n % 7 == 0 || n % 11 == 0
           for (p, e) in eachfactor(n)
               (e != 1 || rem(n ÷ p - 1, p) != 0) && @goto nextnumber
           end
           verbose && println(n, "  (elapsed: ", time() - starttime, ")")
           push!(found, n)
           nfound += 1
       end
       @label nextnumber
       n += 6 # all mult of 6
   end
   return found

end

@time getgiugas(2, false) @time getgiugas(6)

</lang>

Output:
30  (elapsed: 0.0)
858  (elapsed: 0.0)
1722  (elapsed: 0.0)
66198  (elapsed: 0.0009999275207519531)
2214408306  (elapsed: 18.97099995613098)
24423128562  (elapsed: 432.06500005722046)
432.066249 seconds (235 allocations: 12.523 KiB)

Pascal

Free Pascal

OK.Cheating to find square free numbers like julia in distance 6
That means always factors 2,3 and minimum one of 5,7,11.

<lang pascal>program Giuga;

{$IFDEF FPC}

 {$MODE DELPHI}  {$OPTIMIZATION ON,ALL}  {$COPERATORS ON}

{$ELSE}

 {$APPTYPE CONSOLE}

{$ENDIF} uses

 sysutils

{$IFDEF WINDOWS},Windows{$ENDIF}

 ;

//###################################################################### //prime decomposition only squarefree and multiple of 6

type

 tprimeFac = packed record
                pfpotPrimIdx : array[0..9] of Uint64;
                pfMaxIdx  : Uint32;
              end;
 tpPrimeFac = ^tprimeFac;
 tPrimes = array[0..65535] of Uint32;

var

 {$ALIGN 8}
 SmallPrimes: tPrimes;
 {$ALIGN 32}

procedure InitSmallPrimes; //get primes. #0..65535.Sieving only odd numbers const

 MAXLIMIT = (821641-1) shr 1;

var

 pr : array[0..MAXLIMIT] of byte;
 p,j,d,flipflop :NativeUInt;

Begin

 SmallPrimes[0] := 2;
 fillchar(pr[0],SizeOf(pr),#0);
 p := 0;
 repeat
   repeat
     p +=1
   until pr[p]= 0;
   j := (p+1)*p*2;
   if j>MAXLIMIT then
     BREAK;
   d := 2*p+1;
   repeat
     pr[j] := 1;
     j += d;
   until j>MAXLIMIT;
 until false;
 SmallPrimes[1] := 3;
 SmallPrimes[2] := 5;
 j := 3;
 d := 7;
 flipflop := (2+1)-1;//7+2*2,11+2*1,13,17,19,23
 p := 3;
 repeat
   if pr[p] = 0 then
   begin
     SmallPrimes[j] := d;
     inc(j);
   end;
   d += 2*flipflop;
   p+=flipflop;
   flipflop := 3-flipflop;
 until (p > MAXLIMIT) OR (j>High(SmallPrimes));

end;

function OutPots(pD:tpPrimeFac;n:NativeInt):Ansistring; var

 s: String[31];
 chk,p: NativeInt;

Begin

 str(n,s);
 result := s+' :';
 with pd^ do
 begin
   chk := 1;
   For n := 0 to pfMaxIdx-1 do
   Begin
     if n>0 then
       result += '*';
     p := pfpotPrimIdx[n];
     chk *= p;
     str(p,s);
     result += s;
   end;
   str(chk,s);
   result += '_chk_'+s+'<';
 end;

end;

function IsSquarefreeDecomp6(var res:tPrimeFac;n:Uint64):boolean;inline; //factorize only not prime/semiprime and squarefree n= n div 6 var

 pr,i,q,idx :NativeUInt;

Begin

 with res do
 Begin
   Idx := 2;
   q := n DIV 5;
   if n = 5*q then
   Begin
     pfpotPrimIdx[2] := 5;
     n := q;
     q := q div 5;
     if q*5=n then
       EXIT(false);
     inc(Idx);
   end;
   q := n DIV 7;
   if n = 7*q then
   Begin
     pfpotPrimIdx[Idx] := 7;
     n := q;
     q := q div 7;
     if q*7=n then
       EXIT(false);
     inc(Idx);
   end;
   q := n DIV 11;
   if n = 11*q then
   Begin
     pfpotPrimIdx[Idx] := 11;
     n := q;
     q := q div 11;
     if q*11=n then
       EXIT(false);
     inc(Idx);
   end;
   if Idx < 3 then
     Exit(false);
   i := 5;
   while i < High(SmallPrimes) do
   begin
     pr := SmallPrimes[i];
     q := n DIV pr;
     //if n < pr*pr
     if pr > q then
       BREAK;
     if n = pr*q then
     Begin
       pfpotPrimIdx[Idx] := pr;
       n := q;
       q := n div pr;
       if pr*q = n then
         EXIT(false);
       inc(Idx);
     end;
     inc(i);
    end;
    if n <> 1 then
    begin
      pfpotPrimIdx[Idx] := n;
      inc(Idx);
    end;
    pfMaxIdx := idx;
 end;
 exit(true);

end;

function ChkGiuga(n:Uint64;pPrimeDecomp :tpPrimeFac):boolean;inline; var

 p : Uint64;
 idx: NativeInt;

begin

 with pPrimeDecomp^ do
 Begin
   idx := pfMaxIdx-1;
   repeat
     p := pfpotPrimIdx[idx];
     result := (((n DIV p)-1)MOD p) = 0;
     if not(result) then
       EXIT;
     dec(idx);
   until idx<0;
 end;

end;

const

 LMT = 24423128562;//2214408306;//

var

 PrimeDecomp :tPrimeFac;
 T0:Int64;
 n,n6 : UInt64;
 cnt:Uint32;

Begin

 InitSmallPrimes;
 T0 := GetTickCount64;
 with PrimeDecomp do
 begin
   pfpotPrimIdx[0]:= 2;
   pfpotPrimIdx[1]:= 3;
 end;
 n := 0;
 n6 := 0;
 cnt := 0;
 repeat
   //only multibles of 6
   inc(n,6);
   inc(n6);
   //no square factor of 2
   if n AND 3 = 0 then
     continue;
   //no square factor of 3
   if n MOD 9 = 0 then
     continue;
   if IsSquarefreeDecomp6(PrimeDecomp,n6)then
     if ChkGiuga(n,@PrimeDecomp) then
     begin
       inc(cnt);
       writeln(cnt:3,'..',OutPots(@PrimeDecomp,n),'  ',(GettickCount64-T0)/1000:6:3,' s');
     end;
 until n >= LMT;
 T0 := GetTickCount64-T0;
 writeln('Found ',cnt);
 writeln('Tested til ',n,' runtime ',T0/1000:0:3,' s');
 writeln;
 writeln(OutPots(@PrimeDecomp,n));

end.</lang> {{out|@home AMD 5600G ( 4,4 Ghz ) fpc3.2.2 -O4 -Xs}

  1..30 :2*3*5_chk_30<   0.000 s
  2..858 :2*3*11*13_chk_858<   0.000 s
  3..1722 :2*3*7*41_chk_1722<   0.000 s
  4..66198 :2*3*11*17*59_chk_66198<   0.000 s
  5..2214408306 :2*3*11*23*31*47057_chk_2214408306<  17.120 s 
  6..24423128562 :2*3*7*43*3041*4447_chk_24423128562<  450.180 s
Found 6
Tested til 24423128562 runtime 450.180 s

24423128562 :2*3*7*43*3041*4447_chk_24423128562
TIO.RUN (~2.3 Ghz )takes ~4x runtime ? ( 2214408306 DIV 2 ) in 36 secs :-(

Perl

<lang perl>#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Giuga_numbers use warnings; use ntheory qw( factor forcomposites ); use List::Util qw( all );

forcomposites

 {
 my $n = $_;
 all { ($n / $_ - 1) % $_ == 0 } factor $n and print "$n\n";
 } 4, 67000;</lang>
Output:
30
858
1722
66198

Phix

with javascript_semantics
constant limit = 4
sequence giuga = {}
integer n = 4
while length(giuga)<limit do
    sequence pf = prime_factors(n)
    for f in pf do
        if remainder(n/f-1,f) then pf={} exit end if
    end for
    if length(pf) then giuga &= n end if
    n += 2
end while
printf(1,"The first %d Giuga numbers are: %v\n",{limit,giuga})
Output:
The first 4 Giuga numbers are: {30,858,1722,66198}

Python

Translation of: FreeBASIC

<lang python>#!/usr/bin/python

from math import sqrt

def isGiuga(m):

   n = m
   f = 2
   l = sqrt(n)
   while True:
       if n % f == 0:
           if ((m / f) - 1) % f != 0:
               return False
           n /= f
           if f > n:
               return True
       else:
           f += 1
           if f > l:
               return False


if __name__ == '__main__':

   n = 3
   c = 0
   print("The first 4 Giuga numbers are: ")
   while c < 4:
       if isGiuga(n):
           c += 1
           print(n)
       n += 1</lang>


Raku

<lang perl6>my @primes = (3..60).grep: &is-prime;

print 'First four Giuga numbers: ';

put sort flat (2..4).map: -> $c {

   @primes.combinations($c).map: {
       my $n = [×] 2,|$_;
       $n if all .map: { ($n / $_ - 1) %% $_ };
   }

}</lang>

Output:
First 4 Giuga numbers: 30 858 1722 66198

Wren

Simple brute force but assumes all Giuga numbers will be even, must be square-free and can't be semi-prime.

Takes only about 0.05 seconds to find the first four Giuga numbers but finding the fifth would take many hours using this approach, so I haven't bothered. <lang ecmascript>var factors = [] var inc = [4, 2, 4, 2, 4, 6, 2, 6]

// Assumes n is even with exactly one factor of 2. // Empties 'factors' if any other prime factor is repeated. var primeFactors = Fn.new { |n|

   factors.clear()
   var last = 2
   factors.add(2)
   n = (n/2).truncate
   while (n%3 == 0) {
       if (last == 3) {
           factors.clear()
           return
       }
       last = 3
       factors.add(3)
       n = (n/3).truncate     
   }
   while (n%5 == 0) {
       if (last == 5) {
           factors.clear()
           return
       }
       last = 5
       factors.add(5)
       n = (n/5).truncate
   }
   var k = 7
   var i = 0
   while (k * k <= n) {
       if (n%k == 0) {
           if (last == k) {
               factors.clear()
               return
           }
           last = k
           factors.add(k)
           n = (n/k).truncate
       } else {
           k = k + inc[i]
           i = (i + 1) % 8
       }
   }
   if (n > 1) factors.add(n)

}

var limit = 4 var giuga = [] var n = 6 // can't be 2 or 4 while (giuga.count < limit) {

   primeFactors.call(n)
   // can't be prime or semi-prime
   if (factors.count > 2 && factors.all { |f| (n/f - 1) % f == 0 }) {
       giuga.add(n)
   }
   n = n + 4 // can't be divisible by 4

} System.print("The first %(limit) Giuga numbers are:") System.print(giuga)</lang>

Output:
The first 4 Giuga numbers are:
[30, 858, 1722, 66198]

XPL0

<lang XPL0>func Giuga(N0); \Return 'true' if Giuga number int N0; int N, F, Q1, Q2, L; [N:= N0; F:= 2; L:= sqrt(N); loop [Q1:= N/F;

       if rem(0) = 0 then      \found a prime factor
               [Q2:= N0/F;
               if rem((Q2-1)/F) # 0 then return false;
               N:= Q1;
               if F>N then quit;
               ]
       else    [F:= F+1;
               if F>L then return false;
               ];
       ];

return true; ];

int N, C; [N:= 3; C:= 0; loop [if Giuga(N) then

               [IntOut(0, N);  ChOut(0, ^ );
               C:= C+1;
               if C >= 4 then quit;
               ];
       N:= N+1;
       ];

]</lang>

Output:
30 858 1722 66198