daysbetweendates Solver equation for 17b

10082018, 10:33 PM
(This post was last modified: 10092018 09:33 PM by Albert Chan.)
Post: #21




RE: daysbetweendates Solver equation for 17b
(10082018 09:13 PM)Dieter Wrote: what's wrong with the shorter and less complicated standard formula for the JulianGregorian difference? That is good ! (above "/" meant floor division. To be safe, for month < 3, set y = y  1) I were using 3rd century calendar match trivia. 2 days for "zero" century is simpler X = 0,1,2,3, 4,5,6,7 ... => days apart = 2,2,1,0, 1,1,2,3 ... = (1,1,2,3, 4,4,5,6 ...)  3 > Days apart = (floor(X/4)  1) * 3 + max(X % 4, 1) For X = 21 = 5 * 4 + 1, Days apart = 4 * 3 + 1 = 13 For X = 47 = 12 * 4 + 1, Days apart = 13 * 3 + 1 = 38 Update: this is faster, Days apart = floor(0.75 X)  2 

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