Computus 2011

Rough Maths 5

Last year I posted a rough versification of the computus, the algorithm used to calculate the date of Easter Sunday for a given year. Possibly influenced by the date of the post, or an underestimation of how far I’ll take a daft idea, at least one person thought I’d made it all up, when in fact I’d used Augustus De Morgan’s Easter chapter in his Budget of Paradoxes.

The computus is one of the earliest examples in European culture of an algorithm: a step-by-step set of instructions for performing a calculation. It’s a massive hack, by which I mean that it’s an inelegant and hairy solution to a fairly ridiculous problem posed by the untidy arrangements of the Earth, the Sun and the Moon.

Here is the table, with workings in the third column: this spoils the layout, which is clearer in the 2010 version.

In calculating the values for 2011, I found two errors in my first version. Programming is an exercise in humility.

I Add one to the year you are given. 2011 + 1 = 2012
II Divide the year by four, rounded down. 2011 / 4 = 502
III From the centuries, take sixteen (if you can) 20 – 16 = 4
IV And divide that by four, rounded down. 4 / 4 = 1
V Add I, II and IV, then take away III, 2011 + 502 + 1 – 4 = 2511
VI Then take that value modulo seven 2551 mod 7 = 5
(Divide by seven and keep the remainder)
Subtracting from seven again: 7 – 5 = 2
VII The year’s dominical letter. 2 = B
(We’ll use it as if it’s a number.)
VIII Take I mod nineteen (if it’s zero, nineteen) 2012 mod 19 = 17
This is the year’s golden number.
Now take seventeen from the centuries, 20 – 17 = 3
IX Over twenty-five (chuck the remainder) 3 / 5 = 0
Take IX and 15 from the centuries 20 – 0 – 15 = 5
X Over three (and chuck the remainder) 5 / 3 = 1
To VIII, add ten times (VIII minus one) 17 + 10 * 16 = 177
XI Take that sum modulo thirty 177 mod 30 = 27
Add XI, X and IV and then take away III, (27 + 1 + 1 – 4) = 25
(If it’s large enough, modulo thirty) 25 mod 30 = 25
If it be twenty-four, make it twenty-five;
If twenty-five, and if VIII is more than eleven (17 > 11)
Make it twenty-six instead; 25 = 26
If it’s zero, set it to thirty.
XII The result is the epact; a good Scrabble word, 26
The age of the moon on New Years’ Day.
If the epact is less than twenty-four, (no)
XIII(b) Subtract it from forty-five (write that down)
Then subtract the epact from twenty-seven
Divide that by seven and keep the remainder:
XIV(b) If it’s zero, change it to seven.
If the epact is higher than twenty-three, (26 > 23)
XIII(b) Subtract it from seventy-five instead 75 – 26 = 49
Then subtract the epact from fifty-seven, 57 – 26 = 31
Divide that by seven and keep the remainder: 31 mod 7 = 3
XIV(b) If it’s zero, change it to seven. = 3
Then add XIII to VII (the dominical number). 49 + 2 = 51
If XIV’s more than VII, add seven more. (3 > 2) 51 + 7 = 58
XV And then take away what you got for XIV. 58 – 3 = 55
If the result is below thirty-two, (no)
Easter Sunday’s in March, and that’s the date,
Otherwise, it’s in April – subtract thirty-one. 55 – 31 = 24 April

PS This one got in the way of the promised next post about division. It’s coming.

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