On 2020-05-14 08:13, Bruce Gray wrote:
On May 14, 2020, at 7:27 AM, ToddAndMargo via perl6-users
<perl6-us...@perl.org> wrote:
Hi All,
1) how do I get 40 or more digits out of sqrt?
—snip—
Use an Integer Root algorithm on ($number-you-want-the-root-of * 100 **
$number-of-digits-you-want), then shift the decimal point of the result.
Exact code here:
https://rosettacode.org/wiki/Integer_roots#Raku
—
Hope this helps,
Bruce Gray (“Util” on RosettaCode, too)
sub integer_root ( Int $p where * >= 2, Int $n --> Int ) {
my Int $d = $p - 1;
my $guess = 10**($n.chars div $p);
my $iterator = { ( $d * $^x + $n div ($^x ** $d) ) div $p };
my $endpoint = { $^x ** $p <= $n
and ($^x + 1) ** $p > $n };
min (+$guess, $iterator ... $endpoint)[*-1, *-2];
}
say integer_root( 2, 2 * 100 ** 2000 );
It does help! I can reproduce noise to my heart's content!
Questions:
what is $p in the sub declaration? Is it always a 2?
what is `$n` in the sub declaration? Looks like
the third set of numbers is the digits I want.
what is the second number (100)?
And what is `2 * 100 ** 2000 `? Is that `(2 x 100)^ 2000`
((2 times 100) to the 2000 power?
> say integer_root( 2, 2 * 100 ** 4 );
14142
> say 3.sqrt
1.7320508075688772
> say integer_root( 2, 3 * 100 ** 16 );
17320508075688772
-T
