# The Weekly Challenge ‐ Perl and Raku

## CY's Take on The Weekly Challenge #246 ‐ Uniqueness

If you want to challenge yourself on programming, especially on Perl and/or Raku, go to https://theweeklychallenge.org, code the latest challenges, submit codes on-time (by GitHub or email).

Do tell me, if I am wrong or you strongly oppose my statements!

It's time for challenges in Week #246 !

#### Task 1: 6 out of 49

Instead of picking out the balls from a box, we can imagine we line up a set of balls labelled 1, 2, 3 ..., 49, and pick out a ball one by one via randomly selecting the k-th ball on the line to be removed.

use v5.30.0;
use warnings;

my @ans;
my @arr = (1..49);
for my \$i (0..5) {
my \$k = int(rand(49-\$i));
push @ans, splice(@arr, \$k, 1);
}

say join "\n", sort {\$a<=>\$b} @ans;

#### Task 2: Linear Recurrence of Second Order

Let's have a brief analysis for the task first. It is given 5 integers, and they form some linear recurrence; we have three linear (Diophantine) equations to play with.

Becase of the requirement of integral parameter solution(s), I started with extended Euclidean algorithm and Bézout coefficients. Some theory can be found on Wikipedia or here (Linear Diophantine Equations.

The pros of the approach are (1) it identifies non-integral linear equations early on; (2) instead of two degeneracies which may be found in solving the two coefficients through the first two linear equations, it reduces to one degeneracy (a[2]*a[0]-a[1]*a[1] may be zero). And by number theoretic arguments, we know a[2] = k a[1] = k^2 a[0] in case of the degeneracy occurs, then we have p + qk = k^2. a[3] = p a[1] + q a[2] = p k a[0] + q k^2 a[0] = k^3 a[0]. Similarly we can deduce a[4] = k^4 a[0].

The code is full of printing of equations which may not be so relevant to the "solution" but for verifying number theoretic algorithms.

# The Weekly Challenge 246
# Task 2 Linear Recurrence of Second Order
# a[n] = p * a[n-2] + q * a[n-1] with n > 1
# where p and q must be integers.
use v5.30.0;
use warnings;

sub ex_euc_alg {
my @r = (\$_[0], \$_[1]);
my @s = (1, 0);
my @t = (0, 1);
my \$i = 0;
while (\$r[-1] != 0) {
my \$q = int(\$r[\$i-1]/\$r[\$i]);
\$r[\$i+1] = \$r[\$i-1] - \$q * \$r[\$i];
while (\$r[\$i+1] > abs(\$r[\$i])) {
\$r[\$i+1] = \$r[\$i+1] - abs(\$r[\$i]);
\$q = \$q + abs(\$r[\$i])/\$r[\$i]
}
while (\$r[\$i+1] < 0) {
\$r[\$i+1] = \$r[\$i+1] + abs(\$r[\$i]);
\$q = \$q - abs(\$r[\$i])/\$r[\$i]
}
\$s[\$i+1] = \$s[\$i-1] - \$q*\$s[\$i];
\$t[\$i+1] = \$t[\$i-1] - \$q*\$t[\$i];
\$i++;
}
my \$a0 = \$_[0];
my \$b0 = \$_[1];
say "\$r[\$i-1] = \$a0*\$s[\$i-1]+\$b0*\$t[\$i-1]";
my (\$d,\$x,\$y) = (\$r[\$i-1], \$s[\$i-1], \$t[\$i-1]);
# for my \$k (-10..10) {
#     my \$x1 = \$x - \$k*\$b0/\$d;
#     my \$y1 = \$y + \$k*\$a0/\$d;
#     say "\$d = \$a0*\$x1+\$b0*\$y1";
# }

return [\$r[\$i-1], \$s[\$i-1], \$t[\$i-1]];
}

sub check {
my @a = @_;

# consective zeros check
if (\$a[0] == 0 && \$a[1] == 0) {
return (\$a[2] == 0 && \$a[3] == 0 && \$a[4] == 0) ? 1 : 0;
}
if (\$a[2] == 0) {
if (\$a[1] == 0) {
return (\$a[3] == 0 && \$a[4] == 0) ? 1 : 0;
}
if (\$a[3] == 0) {
return (\$a[4] == 0) ? 1 : 0;
}
}

my (\$d0, \$d1, \$x0, \$y0, \$u0, \$v0);
my (\$a0, \$b0);
my (\$a1, \$b1);

(\$d0, \$x0, \$y0) = ex_euc_alg(\$a[0], \$a[1])->@*;
(\$a0, \$b0) = (\$a[0], \$a[1]);
return 0 if \$a[2] % \$d0 != 0;
\$x0 = \$x0 *(\$a[2]/\$d0);
\$y0 = \$y0 *(\$a[2]/\$d0);
say "\$a[2] = \$a0*\$x0+\$b0*\$y0";
# for my \$k (-10..10) {
#     my \$x1 = \$x0 + \$b0/\$d0*\$k ;
#     my \$y1 = \$y0 - \$a0/\$d0*\$k;
#     say "\$a[2] = \$a0*\$x1+\$b0*\$y1";
# }

(\$d1, \$u0, \$v0) = ex_euc_alg(\$a[1], \$a[2])->@*;
(\$a1, \$b1) = (\$a[1], \$a[2]);
return 0 if \$a[3] % \$d1 != 0;
\$u0 = \$u0 *(\$a[3]/\$d1);
\$v0 = \$v0 *(\$a[3]/\$d1);
say "\$a[3] = \$a1*\$u0+\$b1*\$v0";
# for my \$j (-10..10) {
#     my \$x1 = \$u0 + \$b1/\$d1*\$j ;
#     my \$y1 = \$v0 - \$a1/\$d1*\$j;
#     say "\$a[3] = \$a1*\$x1+\$b1*\$y1";
# }

# \$x0 + \$b0/\$d0*\$k == \$u0 + \$b1/\$d1*\$j
# \$y0 - \$a0/\$d0*\$k == \$v0 - \$a1/\$d1*\$j

# \$a0\$x0 + \$a0\$b0/\$d0*\$k == \$a0\$u0 + \$a0\$b1/\$d1*\$j
# \$b0\$y0 - \$b0\$a0/\$d0*\$k == \$b0\$v0 - \$b0\$a1/\$d1*\$j

if (\$a[2]*\$a[0] != \$a[1]*\$a[1]) {
my \$j = (\$a0*\$x0 + \$b0*\$y0 - \$a0*\$u0 - \$b0*\$v0)*\$d1/(\$a0*\$b1 - \$b0*\$a1);
my \$x1 = \$u0 + \$b1/\$d1*\$j;
my \$y1 = \$v0 - \$a1/\$d1*\$j;
say "(\$x1, \$y1)";
return 0 if int(\$x1) != \$x1 || int(\$y1) != \$y1;
return (\$a[4] == \$a[2]*\$x1+\$a[3]*\$y1) ? 1 : 0;
}
else {
return ( \$a[1]*\$a[2] == \$a[3]*\$a[0]
&&
\$a[1]*\$a[3] == \$a[4]*\$a[0]) ? 1 : 0;
# Explanation:
# if a[2] = a[1]*a[1]/\$a[0],
# z d0 = x^2/y d0
# since d0 = gcd(x d0, y d0), x, y coprime
# then y must be 1, i.e. x a[0] = a[1]
# and a[2] = x^2 d0 = x a[1]
# i.e. x a[1] = a[2]
}
}

Test Cases

use Test::More tests=>11;
ok check(1, 1, 2, 3, 5);
ok !check(4, 2, 4, 5, 7);
ok check(4, 1, 2, -3, 8);

ok check(3, 9, 27, 81, 243);
ok check(3, 5, 27, 45, 243);
ok check(1, 1, 0, 0, 0);

ok check(0, 0, 0, 0, 0);
ok check(1, 0, 0, 0, 0);
ok check(0, 3, 0, 0, 0);

ok !check(0, 0, 3, 0, 0);
ok check(2, 4, 8, 16, 32);

Hope for Peace! □

The image of Fibonacci spiral is taken from Wikimedia Commons, released in public domin.

Acknowledgement: Thanks to discussion of the PA&A Discord Channel (maintained by Adam Crussell).

Except from images and codes from other personnels, the content of this blogpost is released under a copyleft spirit. One may share (full or partial) content of this blogpost on other platform if you share it under the free and open content spirit.

link for CY's full codes: ch-1.pl, ch-2.pl