*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.

Admirable post…