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From: cak3g@astsun9.astro.Virginia.EDU (Colin Klipsch)
Newsgroups: comp.lang.c
Subject: Leap Year Checker...even more to it
Summary: More complex than you think
Message-ID: <1990Sep30.013852.8764@murdoch.acc.Virginia.EDU>
Date: 30 Sep 90 01:38:52 GMT
References: <1115.26ff47b6@iccgcc.decnet.ab.com> <242@srchtec.UUCP> <4404@catfish11.UUCP>
Sender: news@murdoch.acc.Virginia.EDU
Organization: University of Virginia
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In article <4404@catfish11.UUCP> yedinak@motcid.UUCP (Mark A. Yedinak) writes:
>johnb@srchtec.UUCP (John Baldwin) writes:
>: (Stan Brown, Oak Road Systems) writes:
>
>:|In article <24700010@sunc1>, mccaugh@sunc1.cs.uiuc.edu writes:
>:|> Ordinarily, a leap-year is a multiple of four, so that--given leap-year y--
>:|> (y%4 == 0) ought to indicate if y designates a leap-year.
>
>:|Bzzzzzt!  Nope, but thanks for playing.  Vanna has lovely gifts for you.

To be nitpicky to the point of nausea, I will point out that the rule
for deciding leap years has FOUR conditions, not three.

A correct algorithm is:

#define FALSE 0
#define TRUE  1

int is_leap_year(int year)
{
   if (year%4000 == 0) return FALSE;
   else if (year%400 == 0) return TRUE;
   else if (year%100 == 0) return FALSE;
   else if (year%4 == 0) return TRUE;
   else return FALSE;
}

With the addition of the divisible-by-4000 condition, the Gregorian
calendar is now accurate to one day in 20,000 years.  (Assuming its
term of use lasts long enough for the rule to be practiced, which I
doubt!)

Source: _Exploration of the Universe_, G. Abell

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