## Advent Calendar - December 9, 2019

Monday, Dec 9, 2019| Tags: Perl

### | Day 8 | Day 9 | Day 10 |

The gift is presented by Dave Cross. Today he is talking about his solutions to Task #1: Find 5 Weekends of “The Weekly Challenge - 019”.

#### Write a script to display months from the year 1900 to 2019 where you find 5 weekends i.e. 5 Friday, 5 Saturday and 5 Sunday.

This would be simple enough to just brute-force. But when I started to think about it, I realised there’s a bit of a trick we can use which can cut down our search space quite significantly.

If we’re looking for a month with five Fridays, Saturdays and Sundays then we need a month with 31 days (as four weeks is twenty-eight days and we need three extra days). Only seven months ever have 31 daysJanuary, March, May, July, August, October and December. There is no point at all in ever looking in any other month. You might also realise that those three extra days need to be Friday 29th, Saturday 30th and Sunday 31st. And that means that the first day of the month also needs to be a Friday.

So, the problem simplifies to “Find months with 31 days where the 1st is a Friday”. And here’s the code I wrote to do that:

``````#!/usr/bin/perl

use strict;
use warnings;
use feature 'say';

use Time::Piece;

# Array of months with 31 days
my @months = (qw[Jan Mar May Jul Aug Oct Dec]);
my \$format = '%Y-%b';

for my \$y (1900 .. 2019) {
for my \$m (@months) {
# Get the first day of the month as a Time::Piece object
my \$first = Time::Piece->strptime("\$y-\$m", \$format);

# Print the date if the 1st is a Friday
say \$first->strftime('%b %Y') if \$first->day eq 'Fri';
}
}
``````

If you have any suggestion then please do share with us perlweeklychallenge@yahoo.com.