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

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Will it not be much better if we make one step further. Assuming every 3200 year as no leap year, the length of the year will come

364.999696 + 1/3200 = 364.999696 + .0003125 = 365.0000085

and after this the adjustment will be required after around 120000 years.

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this is enough to check if a year is a leap year   if   year 400  0    year 100  0   amp  amp  year 4  0       cout lt  lt  It is a leap year   else     cout lt  lt  It is not a leap year

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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

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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

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Here comes a rather obsqure idea  When every year dividable with 100 gets 365 days  what shall be done at this time  In the far future  when even years dividable with 400 only can get 365 days   Then there is a possibility or reason to make corrections in years dividable with 80  Normal years will have 365 day and those dividable with 400 can get 366 days  Or is this a loose-loose situation

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PHP      is number of days in the year 366    php days of year is 0 based  return   int date  z   strtotime  Dec 31        365

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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

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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

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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

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C  implementation  public bool LeapYear          int year   2016        return year   4    0  amp  amp  year   100    0    year   400    0

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Here is a simple implementation of the wikipedia algorithm  using the javascript ternary operator   isLeapYear    year   100     0     year   400     0     year   4     0

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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

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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

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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

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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  lt   1917          if year   4    0              yr    leap          else              yr    not leap      elif year  gt   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

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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

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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

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just wrote this in Coffee-Script   is leap year     year   - gt    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   - gt    return Object prototype toString call  x        object Number   and not isNaN  x    isa integer     x   - gt    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

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Return true if the input year is a leap year  Basic modern day code     If year mod 4   0  then leap year   if year mod 100 then normal year   if year mod 400 then leap year   else normal year   Todays rule started 1582 AD   Julian calendar rule with every 4th year started 46BC but is not coherent before 10 AD as declared by Cesar    They did however add some leap years every 3rd year now and then in the years before    Leap years were therefore 45 BC  42 BC  39 BC  36 BC  33 BC  30 BC  27 BC  24 BC  21 BC  18 BC  15 BC  12 BC  9 BC  8 AD  12 AD   Before year 45BC leap year was not added   The year 0 do not exist as it is    2BC 1BC 1AD 2AD    for some calculation this can be an issue   function isLeapYear year  Integer   Boolean  begin   result    false    if year  gt  1582 then    Todays calendar rule was started in year 1582      result      year mod 4   0  and  not year mod 100   0    or  year mod 400   0    else if year  gt  10 then    Between year 10 and year 1582 every 4th year was a leap year      result    year mod 4   0   else   Between year -45 and year 10 only certain years was leap year  every 3rd year but the entire time     case year of       -45  -42  -39  -36  -33  -30  -27  -24  -21  -18  -15  -12  -9          result    true      end  end

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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

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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

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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 gt 0    awk  END   print  NF      -eq 366       then       echo   year - gt  Leap Year      else       echo   year - gt  Normal Year       fi      check leap year 1900 1900 - gt  Normal Year   check leap year 2000 2000 - gt  Leap Year   check leap year 2001 2001 - gt  Normal Year   check leap year 2020 2020 - gt  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

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This is the most efficient way  I think   Python   def leap n       if n   100    0          n   n   100     return n   4    0

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In the Gregorian calendar 3 criteria must be taken into account to identify leap years    The year is evenly divisible by 4  If the year can be evenly divided by 100  it is NOT a leap year  unless  The year is also evenly divisible by 400  Then it is a leap year  Why the year divided by 100 is not leap year

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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  lt  max  i                  if   i   4    0  amp  amp  i   100    0     i   400    0                    counter   counter   1                                  return counter

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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

[algorithm] Leap year calculation

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