leap year calculation

Question

In order to find leap years, why must the year be indivisible by 100 and divisible by 400? I understand why it must be divisible by 4. Please explain the algorithm.

Answer 1

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

Answer 2

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

Answer 3

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).

Answer 4

PHP:

// is number of days in the year 366?  (php days of year is 0 based)
return ((int)date('z', strtotime('Dec 31')) === 365);

Answer 5

In Java Below code calculates leap year count between two given year. Determine starting and ending point of the loop.

Then if parameter modulo 4 is equal 0 and parameter modulo 100 not equal 0 or parameter modulo 400 equal zero then it is leap year and increase counter.

static int calculateLeapYearCount(int year, int startingYear) {
        int min = Math.min(year, startingYear);
        int max = Math.max(year, startingYear);
        int counter = 0;
        for (int i = min; i < max; i++) {
            if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0) {
                counter = counter + 1;
            }
        }
        return counter;
    }

Answer 6

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

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