On 2006-12-09, Clive D.W. Feather challenged, and I couldn't resist:
> For something more challenging, try the 8 Bank Holidays in England:
>
> ...
> (8) Second weekday after 24th December.
second weekday after 24th December in Gregorian year( Y )
= Gregorian calendar( Y, December, 26 )
+ 2·(floor( D / 5 )) d - (floor( D / 7 )) d
where
D = day of the week number of Gregorian calendar( Y, December, 25 )
(as in ISO 8601 with values in {1..7} and 1 for Monday)
= 1 + ((floor(Y/100) mod 4)·5 + (Y mod 100)·3 - (Y mod 4)·2) mod 7
Sorry for being off-topic.
Michael Deckers
