A liturgical one-liner

It started with a colleague showing me a little “Easter Egg” (excuse the pun) in the calendar program ncal. The code and the result of running it was as follows:

15 April 2001 31 March 2002 20 April 2003 11 April 2004 27 March 2005 16 April 2006 8 April 2007 23 March 2008 12 April 2009 4 April 2010 (Again I am resorting to the Perl mode for syntax highlighting my bash command). I was asked what this command or its output represented, and it was not too difficult to infer that these were (the western) dates of Pasch for the first 10 years of this century.

The problem arose when I considered this too easy, as it is a feature built into the calendar program. “What about other liturgical dates?” I wondered. In the western calendar, the only ones which move (apart from transference of a minor feast because of coincidence with a more important moveable feast that year) do so in line with Pasch. Thus the feast days for which it is an interesting question to ask what date they are on in a given year, can be found by adding days to the date of Pasch. As an example, I chose to find the date of Pentecost for the year 2010 with a one-liner. As the old saying goes “Well, it is a one-liner, but it depends how long your lines are”. For me the rule is that one does not use semi-colons (to separate what are effectively several consecutive one-liners) or custom scripts. I usually also add the rule of “No backticks”, but I break that rule here.

First, the one-liner which I used and its output:

2010-05-23 00:00:00 UTC and now, more verbosely, an explanation of how and why this works:

ncal -y 2010 -e

Find the date of Pasch in 2010

sed ‘s/^ //’

Remove any initial space

sed ‘s/ /-/g’

Turn any remaining spaces into hyphens

xargs date +%j -d

Use xargs to give this date as the final argument to “date +%j -d” which outputs a given date as a day of the year

sed ‘s/.*/60*60*24*(48+&)/’

Turn the number from the previous step into a sum, such as “60*60*24*(48+094)”

bc

Output the answer to the sum, which is the number of seconds from the start of that year until the start of the day of Pentecost

xargs echo `date -d 2010-01-01 +%s` +

The backticked section produces the number of seconds from the Epoch until the start of the year we are interested in, so this command creates a sum with that number, then a plus sign, then the number of seconds from the previous step (thanks to xargs)

bc

Calculate that sum, which is the number of seconds since the Epoch until the start of the day of Pentecost

sed ‘s/^/@/’

Put a commercial “at” sign at the beginning of the number, to put it in the form needed for the undocumented feature in the next step

xargs date -u +”%F %T %Z” -d

Output in ISO format the date (and time) of the start of Pentecost that year, found from its timestamp

There are a few things to say about this calculation, apart from just how the individual steps work. For instance, even if Pasch could fall before 03-01 (instead of being limited to between 03-22 and 04-25 inclusive), the calculation would not fall foul of leap years as it does not rely on there being a certain number of days in any month. The other point to mention is due to the use of “48” as the number days to be added after Pasch to reach Pentecost. To explain this, I first have to point out that date +%j gives the number of the day of the year (DoY) on which Pasch falls, with 01-01 being day 1, thus: DoY * 24 * 60 * 60 includes the seconds up until the end of that day of the year. This means one day must be subtracted to get the seconds until the start of that day. Then note that there are 49 days in 7 weeks, and Pentecost is the 50th day after 7 weeks of Pasch (starting on and including the day of Pasch). So the period is 49 days, starting from the beginning of Pasch or 48 days starting from the end of Pasch.

Wikipedia says About 2.1 billion years from now, the continual increase of the Sun’s radiation will cause the Earth’s oceans to boil away, removing the bulk of the tidal friction and acceleration. Even without this, the slowdown to a month-long day would still not have been completed by 4.5 billion years from now when the Sun will evolve into a red giant and likely destroy both the Earth and Moon. What if we built a shield for the Earth to stop the oceans boiling, so that before the Sun becomes a red giant the Earth has, for example, a 20-day-long day and a 40-day-long month? Surely while this was true, every point on the Earth could only ever see half of the moon’s phases (in two each 10-day phases with 10-day gaps of daytime between them). If those countries never had a full moon, could they have Pasch those years?