Leap year calculation

The length of a year is (more or less) 365.242196 days. So we have to subtract, more or less, a quarter of a day to make it fit :

365.242196 - 0.25 = 364.992196 (by adding 1 day in 4 years) : but oops, now it's too small!! lets add a hundreth of a day (by not adding that day once in a hundred year :-))

364.992196 + 0,01 = 365.002196 (oops, a bit too big, let's add that day anyway one time in about 400 years)

365.002196 - 1/400 = 364.999696

Almost there now, just play with leapseconds now and then, and you're set.

(Note : the reason no more corrections are applied after this step is because a year also CHANGES IN LENGTH!!, that's why leapseconds are the most flexible solution, see for examlple here)

That's why i guess

There's an algorithm on wikipedia to determine leap years:

function isLeapYear (year):
    if ((year modulo 4 is 0) and (year modulo 100 is not 0))
    or (year modulo 400 is 0)
        then true
    else false

There's a lot of information about this topic on the wikipedia page about leap years, inclusive information about different calendars.

In general terms the algorithm for calculating a leap year is as follows...

A year will be a leap year if it is divisible by 4 but not by 100. If a year is divisible by 4 and by 100, it is not a leap year unless it is also divisible by 400.

Thus years such as 1996, 1992, 1988 and so on are leap years because they are divisible by 4 but not by 100. For century years, the 400 rule is important. Thus, century years 1900, 1800 and 1700 while all still divisible by 4 are also exactly divisible by 100. As they are not further divisible by 400, they are not leap years

I'm sure Wikipedia can explain it better than I can, but it is basically to do with the fact that if you added an extra day every four years we'd get ahead of the sun as its time to orbit the sun is less than 365.25 days so we compensate for this by not adding leap days on years that are not divisible by 400 eg 1900.

Hope that helps

Here is a simple implementation of the wikipedia algorithm, using the javascript ternary operator:

isLeapYear = (year % 100 === 0) ? (year % 400 === 0) : (year % 4 === 0);

You really should try to google first.

Wikipedia has a explanation of leap years. The algorithm your describing is for the Proleptic Gregorian calendar.

More about the math around it can be found in the article Calendar Algorithms (PDF).

Leap years are arbitrary, and the system used to describe them is a man made construct. There is no why.

What I mean is there could have been a leap year every 28 years and we would have an extra week in those leap years ... but the powers that be decided to make it a day every 4 years to catch up.

It also has to do with the earth taking a pesky 365.25 days to go round the sun etc. Of course it isn't really 365.25 is it slightly less (365.242222...), so to correct for this discrepancy they decided drop the leap years that are divisible by 100.

If you're interested in the reasons for these rules, it's because the time it takes the earth to make exactly one orbit around the sun is a long imprecise decimal value. It's not exactly 365.25. It's slightly less than 365.25, so every 100 years, one leap day must be eliminated (365.25 - 0.01 = 365.24). But that's not exactly correct either. The value is slightly larger than 365.24. So only 3 out of 4 times will the 100 year rule apply (or in other words, add back in 1 day every 400 years; 365.25 - 0.01 + 0.0025 = 365.2425).

There are on average, roughly 365.2425 days in a year at the moment (the Earth is slowing down but let's ignore that for now).

The reason we have leap years every 4 years is because that gets us to 365.25 on average [(365+365+365+366) / 4 = 365.25, 1461 days in 4 years].

The reason we don't have leap years on the 100-multiples is to get us to 365.24 `[(1461 x 25 - 1) / 100 = 365.24, 36,524 days in 100 years.

Then the reason we once again have a leap year on 400-multiples is to get us to 365.2425 [(36,524 x 4 + 1) / 400 = 365.2425, 146,097 days in 400 years].

I believe there may be another rule at 3600-multiples but I've never coded for it (Y2K was one thing but planning for one and a half thousand years into the future is not necessary in my opinion - keep in mind I've been wrong before).

So, the rules are, in decreasing priority:

• multiple of 400 is a leap year.
• multiple of 100 is not a leap year.
• multiple of 4 is a leap year.
• anything else is not a leap year.

Simply Because year 2000 is a leap year and it is divisible by 100 and dividable by 4. SO to guarantee it is correct leap we need to ensure it is divisible by 400. 2000 % 4 = 0 2000 % 100 = 0 According to algorithm it's not leap, but it is dividable by 400 2000 % 400 = 0 so it is leap.

I found this problem in the book "Illustrated Guide to Python 3". It was in a very early chapter that only discussed the math operations, no loops, no comparisons, no conditionals. How can you tell if a given year is a leap year?

Below is what I came up with:

y = y % 400
a = y % 4
b = y % 100
c = y // 100
ly = (0**a) * ((1-(0**b)) + 0**c)   # ly is not zero for leap years, else 0

This is the most efficient way, I think.

Python:

def leap(n):
    if n % 100 == 0:
        n = n / 100
    return n % 4 == 0

C# implementation

public bool LeapYear()
{
     int year = 2016;

     return year % 4 == 0 && year % 100 != 0 || year % 400 == 0 ;
}

From 1700 to 1917, official calendar was the Julian calendar. Since then they we use the Gregorian calendar system. The transition from the Julian to Gregorian calendar system occurred in 1918, when the next day after January 31st was February 14th. This means that 32nd day in 1918, was the February 14th.

In both calendar systems, February is the only month with a variable amount of days, it has 29 days during a leap year, and 28 days during all other years. In the Julian calendar, leap years are divisible by 4 while in the Gregorian calendar, leap years are either of the following:

Divisible by 400.

Divisible by 4 and not divisible by 100.

So the program for leap year will be:

Python:

def leap_notleap(year):

    yr = ''
    if year <= 1917:
        if year % 4 == 0:
            yr = 'leap'
        else:
            yr = 'not leap'
    elif year >= 1919:
        if (year % 400 == 0) or (year % 4 == 0 and year % 100 != 0):
            yr = 'leap'
        else:
            yr = 'not leap'
    else:
        yr = 'none actually, since feb had only 14 days'

    return yr

In shell you can use cal -j YYYY which prints the julian day of the year, If the last julian day is 366, then it is a leap year.

$ function check_leap_year
 {
  year=$1
   if [ `cal -j $year | awk 'NF>0' | awk 'END { print $NF } '` -eq 366 ];
   then
      echo "$year -> Leap Year";
   else
      echo "$year -> Normal Year" ;
   fi
 }
$ check_leap_year 1900
1900 -> Normal Year
$ check_leap_year 2000
2000 -> Leap Year
$ check_leap_year 2001
2001 -> Normal Year
$ check_leap_year 2020
2020 -> Leap Year
$

Using awk, you can do

$ awk -v year=1900 ' BEGIN { jul=strftime("%j",mktime(year " 12 31 0 0 0 ")); print jul } '
365
$ awk -v year=2000 ' BEGIN { jul=strftime("%j",mktime(year " 12 31 0 0 0 ")); print jul } '
366
$ awk -v year=2001 ' BEGIN { jul=strftime("%j",mktime(year " 12 31 0 0 0 ")); print jul } '
365
$ awk -v year=2020 ' BEGIN { jul=strftime("%j",mktime(year " 12 31 0 0 0 ")); print jul } '
366
$

BIS will be 1 if the year is leap, otherwise 0 in this boolean logic:

BIS = A MOD 4=0 - (A MOD 100=0 AND A>1600) + (A MOD 400=0 AND A>1600)