Get Easter Date

 

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Returns teh date for Easter in any given Year

--put easterdate(2000) -- [#day: 23, #month: 4, #year: 2000] --As an added bonus, since Easter always falls on a Sunday, you could theoretically combine this function with a leap year function --and get the day of the week for any date between 1583 and 4099.  But you would also have to be pretty much insane. --Thanks to Greg Mallen for the original algorithm and Ronald W. Mallen of Adelaide Australia for the research.  You can check out his web page at http://www.assa.org.au/edm.html --Any suggestions on how to make the algorithm above even smaller, gratefully appreciated. on GetEasterDate y    --"all arguments are integer types    --    --"INPUTS    --"y is the year (1583 to 4099)    --    --"OUTPUTS    --"d day of month    --"m month    --    --"==========================================================    --    --"This algorithm is adapted from a faq document by Claus Tondering    --"URL: http://www.pip.dknet.dk/~pip10160/calendar.faq2.txt    --"E-mail: c-t@pip.dknet.dk.    --    --"The FAQ algorithm is based in part on the algorithm of Oudin (1940)    --"as quoted in "Explanatory Supplement to the Astronomical Almanac",    --"P. Kenneth Seidelmann, editor.    --    --"Additions by Greg Mallen:    --"(1) adds method 2 calculations (original calculation, converted to    --Gregorian date)    --"(2) modified d and m calculations to account for later dates produced by    --method 2    --    --"==========================================================    --    --Dim g   "golden year - 1    --Dim c   "century    --Dim h   " = (23 - Epact) mod 30    --Dim i   "no of days from March 21 to Paschal Full Moon    --Dim p   "no of days from March 21 to Sunday on or before PFM    --            "(-6 to 28 methods 1 & 3, to 56 for method 2)    --Dim e   "extra days to add for method 2 (converting Julian date to Gregorian    --date)    --    --"calc intermediate vars    --"~~~~~~~~~~~~~~~~~~~~~~    g = y Mod 19    c = y / 100    h = (c - c / 4 - (8 * c + 13) / 25 + 19 * g + 15) Mod 30    i = h - (h / 28) * (1 - (h / 28) * (29 / (h + 1)) * ((21 - g) / 11))    --"return day and month    --"~~~~~~~~~~~~~~~~~~~~    p = i - ((y + y / 4 + i + 2 - c + c / 4) Mod 7)    --" p can be from -6 to 56 (for all valid years 326 to 4099, all methods)    --" corresponds to dates 22 March to 23 May    --" (later dates apply to method 2, although 23 May never actually occurs)    return([#year:y, #day:1 + (p + 27 + (p + 6) / 40) Mod 31, #month:3 + (p + 26) / 30]) end