How to resolve the algorithm Day of the week step by step in the UNIX Shell programming language

Published on 12 May 2024 09:40 PM
#Unix shell

How to resolve the algorithm Day of the week step by step in the UNIX Shell programming language

Table of Contents

Problem Statement

A company decides that whenever Xmas falls on a Sunday they will give their workers all extra paid holidays so that, together with any public holidays, workers will not have to work the following week (between the 25th of December and the first of January).

In what years between 2008 and 2121 will the 25th of December be a Sunday? Using any standard date handling libraries of your programming language; compare the dates calculated with the output of other languages to discover any anomalies in the handling of dates which may be due to, for example, overflow in types used to represent dates/times similar to   y2k   type problems.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Day of the week step by step in the UNIX Shell programming language

Source code in the unix programming language

#! /bin/bash

for (( i=2008; i<=2121; ++i ))
do
 date -d "$i-12-25"
done  |grep Sun

exit 0

Sun Dec 25 00:00:00 CET 2011
Sun Dec 25 00:00:00 CET 2016
Sun Dec 25 00:00:00 CET 2022
Sun Dec 25 00:00:00 CET 2033
date: invalid date `2038-12-25'

bash-3.00$ date --version
date (coreutils) 5.2.1
Written by David MacKenzie.

Copyright (C) 2004 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
bash-3.00$ uname -a
Linux brslln01 2.6.9-67.ELsmp #1 SMP Wed Nov 7 13:56:44 EST 2007 x86_64 x86_64 x86_64 GNU/Linux
bash-3.00$ for((i=2009; i <= 2121; i++)); do  date -d "$i-12-25" |egrep Sun; done
Sun Dec 25 00:00:00 GMT 2011
Sun Dec 25 00:00:00 GMT 2016
Sun Dec 25 00:00:00 GMT 2022
Sun Dec 25 00:00:00 GMT 2033
Sun Dec 25 00:00:00 GMT 2039
Sun Dec 25 00:00:00 GMT 2044
Sun Dec 25 00:00:00 GMT 2050
Sun Dec 25 00:00:00 GMT 2061
Sun Dec 25 00:00:00 GMT 2067
Sun Dec 25 00:00:00 GMT 2072
Sun Dec 25 00:00:00 GMT 2078
Sun Dec 25 00:00:00 GMT 2089
Sun Dec 25 00:00:00 GMT 2095
Sun Dec 25 00:00:00 GMT 2101
Sun Dec 25 00:00:00 GMT 2107
Sun Dec 25 00:00:00 GMT 2112
Sun Dec 25 00:00:00 GMT 2118
bash-3.00$

seq 2008 2121 | xargs -IYEAR -n 1 date +%c -d 'Dec 25 YEAR' | grep Sun

y=2008
while test $y -lt 2122; do
	cal 12 $y | tail +3 | cut -c1-2 | grep -Fq 25 && echo 25 Dec $y
	y=`expr $y + 1`
done

zmodload zsh/datetime
for (( year = 2010; year <= 2121; year++ ));
  if [[ $(strftime '%A' $(strftime -r '%F' $year-12-25)) == Sunday ]] print $year

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