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

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.

Answer 2

You could just check if the Year number is divisible by both 4 and 400. You dont really need to check if it is indivisible by 100. The reason 400 comes into question is because according to the Gregorian Calendar, our "day length" is slightly off, and thus to compensate that, we have 303 regular years (365 days each) and 97 leap years (366 days each). The difference of those 3 extra years that are not leap years is to stay in cycle with the Gregorian calendar, which repeats every 400 years. Look up Christian Zeller's congruence equation. It will help understanding the real reason. Hope this helps :)

Answer 3

just wrote this in Coffee-Script:

is_leap_year = ( year ) ->
  assert isa_integer year
  return true   if year % 400 == 0
  return false  if year % 100 == 0
  return true   if year %   4 == 0
  return false

# parseInt? that's not even a word. 
# Let's rewrite that using real language:
integer = parseInt 

isa_number = ( x ) ->
  return Object.prototype.toString.call( x ) == '[object Number]' and not isNaN( x )

isa_integer = ( x ) ->
  return ( isa_number x ) and ( x == integer( x ) )

of course, the validity checking done here goes a little further than what was asked for, but i find it a necessary thing to do in good programming.

note that the return values of this function indicate leap years in the so-called proleptic gregorian calendar, so for the year 1400 it indicates false, whereas in fact that year was a leap year, according to the then-used julian calendar. i will still leave it as such in the datetime library i'm writing because writing correct code to deal with dates quickly gets surprisingly involved, so i will only ever support the gregorian calendar (or get paid for another one).

Answer 4

Python 3.5

def is_leap_baby(year):
    if ((year % 4 is 0) and (year % 100 is not 0)) or (year % 400 is 0):
        return "{0}, {1} is a leap year".format(True, year)
    return "{0} is not a leap year".format(year)

print(is_leap_baby(2014))
print(is_leap_baby(2012))

Answer 5

this is enough to check if a year is a leap year.

if( (year%400==0 || year%100!=0) &&(year%4==0))
    cout<<"It is a leap year";
else
    cout<<"It is not a leap year";

Answer 6

a) The year is 365.242199 days.

b) If every year was 365 days, in 100 years we would lose 24.2199 days. That's why we add 24 days per century (every 4 years EXCEPT when divisible by 100)

c) But still we lose 0.21299 days/century. So in 4 centuries we lose 0.8796 days. That's why we add 1 day per 4 centuries (every fourth century we DO count a leap year).

d) But that means we lose -0.1204 days (we go forward) per quadricentennial (4 centuries). So in 8 quadricentennial (3200 years) we DO NOT count a leap year.

e) But that means we lose 0.0368 days per 3200 years. So in 24x3200 years (=76800years) we lose 0.8832 days. That's why we DO count a leap year.

and so on... (by then we will have destroyed the planet, so it doesn't matter)

What I cannot understand though, is why we don't count a leap year every 500 years instead of 400. In that way we would converge more rapidly to the correct time (we would lose 2.3 hours/500 years).