Giuga numbers: Difference between revisions

 

=={{header|AWK}}==

# syntax: GAWK -f GIUGA_NUMBER.AWK

BEGIN {

}

}

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<pre>

===Brute force===

Based on the Go solution. Takes 26 minutes on my system (Intel Core i5 3.2GHz).

 

// Assumes n is even with exactly one factor of 2.

}

}

 

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{{libheader|Boost}}

{{trans|Wren}}

 

#include <algorithm>

for (uint64_t n = 3; n < 7; ++n)

giuga_numbers(n);

 

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=={{header|FreeBASIC}}==

Dim As Uinteger n = m

Dim As Uinteger f = 2, l = Sqr(n)

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

n += 1

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<pre>The first 4 Giuga numbers are: 30 858 1722 66198</pre>

{{trans|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.

 

import "fmt"

fmt.Println("The first", limit, "Giuga numbers are:")

fmt.Println(giuga)

 

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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):

 

 

 

These numbers have some interesting properties but there's an issue with guaranteeing correctness of more sophisticated approaches. That said, here's a translation of the pari/gp implementation on the talk page:

 

 

giuga=: {{

end.

r

 

 

giuga 2

66198

giuga 6

 

=={{header|Julia}}==

 

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

 

getGiuga(4)

<pre>

30

</pre>

=== Ad hoc faster version ===

 

function getgiugas(numberwanted, verbose = true)

@time getgiugas(2, false)

@time getgiugas(6)

<pre>

30 (elapsed: 0.0)

That means always factors 2,3 and minimum one of 5,7,11.<br>

<br>

 

{$IFDEF FPC}

writeln;

writeln(OutPots(@PrimeDecomp,n));

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

<pre>

Generating recursive squarefree numbers of ascending primes and check those numbers.<BR>

2*3 are set fixed. 2*3*5 followed 2*3*7 than 2*3*11. Therefor the results are unsorted.

{

30 = 2 * 3 * 5.

writeln;

end.

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<pre>

 

=={{header|Perl}}==

 

use strict; # https://rosettacode.org/wiki/Giuga_numbers

my $n = $_;

all { ($n / $_ - 1) % $_ == 0 } factor $n and print "$n\n";

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<pre>

 

=={{header|Phix}}==

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #008080;">constant</span> <span style="color: #000000;">limit</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">4</span>

<span style="color: #008080;">end</span> <span style="color: #008080;">while</span>

<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"The first %d Giuga numbers are: %v\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">limit</span><span style="color: #0000FF;">,</span><span style="color: #000000;">giuga</span><span style="color: #0000FF;">})</span>

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<pre>

===Faster version===

{{trans|Wren}}

<span style="color: #000080;font-style:italic;">--

-- demo\rosetta\Giuga_number.exw

<span style="color: #7060A8;">papply</span><span style="color: #0000FF;">({</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">4</span><span style="color: #0000FF;">,</span><span style="color: #000000;">5</span><span style="color: #0000FF;">,</span><span style="color: #000000;">6</span><span style="color: #0000FF;">},</span><span style="color: #000000;">giuga</span><span style="color: #0000FF;">)</span>

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<pre>

</pre>

You can (almost) push things a little further on 64-bit:

<span style="color: #008080;">if</span> <span style="color: #7060A8;">machine_bits</span><span style="color: #0000FF;">()=</span><span style="color: #000000;">64</span> <span style="color: #008080;">then</span> <span style="color: #000000;">giuga</span><span style="color: #0000FF;">(</span><span style="color: #000000;">7</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>

and get

<pre>

=={{header|Python}}==

{{trans|FreeBASIC}}

 

from math import sqrt

c += 1

print(n)

 

 

=={{header|Raku}}==

 

 

print 'First four Giuga numbers: ';

$n if all .map: { ($n / $_ - 1) %% $_ };

}

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<pre>First 4 Giuga numbers: 30 858 1722 66198</pre>

 

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.

var inc = [4, 2, 4, 2, 4, 6, 2, 6]

 

}

System.print("The first %(limit) Giuga numbers are:")

 

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{{libheader|Wren-rat}}

This is a translation of the very fast Pari-GP code in the talk page. Only takes 0.015 seconds to find the first six Giuga numbers.

import "./rat" for Rat

 

giuga.call(n)

System.print()

 

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=={{header|XPL0}}==

int N0;

int N, F, Q1, Q2, L;

N:= N+1;

];

 

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