Gnumed.timelinelib.calendar.gregorian.gregorian

Package Gnumed :: Package timelinelib :: Package calendar :: Package gregorian :: Module gregorian |
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# Module gregorian

source code

Classes
| GregorianDateTime

Functions
days_in_month(year, month) | source code
is_leap_year(year) | source code
is_valid_time(hour, minute, second) | source code
is_valid(year, month, day) | source code
julian_day_to_gregorian_ymd(julian_day) This algorithm is described here: | source code
gregorian_ymd_to_julian_day(year, month, day) This algorithm is described here: | source code

Imports: GregorianDelta, GregorianTime

Function Details

### julian_day_to_gregorian_ymd(julian_day) | source code
    This algorithm is described here:

    * http://www.tondering.dk/claus/cal/julperiod.php#formula

    Integer division works differently in C and in Python for negative numbers.
    C truncates towards 0 and Python truncates towards negative infinity:
    http://python-history.blogspot.se/2010/08/why-pythons-integer-division-floors.html

    The above source don't state which to be used. If we can prove that
    division-expressions are always positive, we can be sure this algorithm
    works the same in C and in Python.

    We must prove that:

    1) m             >= 0
    2) ((5 * e) + 2) >= 0  =>  e >= 0
    3) (1461 * d)    >= 0  =>  d >= 0
    4) ((4 * c) + 3) >= 0  =>  c >= 0
    5) (b * 146097)  >= 0  =>  b >= 0
    6) ((4 * a) + 3) >= 0  =>  a >= 0

    Let's work from the top:

    julian_day >= 0                   =>

    a >= 0 + 32044
       = 32044                        =>

    This proves 6).

    b >= ((4 * 32044) + 3) // 146097
       = 0

    This proves 5).

    Let's look at c:

    c = a - ((b * 146097) // 4)
      = a - (((((4 * a) + 3) // 146097) * 146097) // 4)

    For c to be >= 0, then

    (((((4 * a) + 3) // 146097) * 146097) // 4) <= a

    Let's look at this component: ((((4 * a) + 3) // 146097) * 146097)

    This expression can never be larger than (4 * a) + 3. That gives this:

    ((4 * a) + 3) // 4 <= a, which holds.

    This proves 4).

    Now, let's look at d:

    d = ((4 * c) + 3) // 1461

    If c is >= 0, then d is also >= 0.

    This proves 3).

    Let's look at e:

    e = c - ((1461 * d) // 4)
      = c - ((1461 * (((4 * c) + 3) // 1461)) // 4)

    The same resoning as above can be used to conclude that e >= 0.

    This proves 2).

    Now, let's look at m:

    m = ((5 * e) + 2) // 153

    If e >= 0, then m is also >= 0.

    This proves 1).

### gregorian_ymd_to_julian_day(year, month, day) | source code
    This algorithm is described here:

    * http://www.tondering.dk/claus/cal/julperiod.php#formula
    * http://en.wikipedia.org/wiki/Julian_day#Converting_Julian_or_Gregorian_calendar_date_to_Julian_Day_Number

    Integer division works differently in C and in Python for negative numbers.
    C truncates towards 0 and Python truncates towards negative infinity:
    http://python-history.blogspot.se/2010/08/why-pythons-integer-division-floors.html

    The above sources don't state which to be used. If we can prove that
    division-expressions are always positive, we can be sure this algorithm
    works the same in C and in Python.

    We must prove that:

    1) y >= 0
    2) ((153 * m) + 2) >= 0

    Let's prove 1):

    y = year + 4800 - a
      = year + 4800 - ((14 - month) // 12)

    year >= -4713 (gives a julian day of 0)

    so

    year + 4800 >= -4713 + 4800 = 87

    The expression ((14 - month) // 12) varies between 0 and 1 when month
    varies between 1 and 12. Therefore y >= 87 - 1 = 86, and 1) is proved.

    Let's prove 2):

    m = month + (12 * a) - 3
      = month + (12 * ((14 - month) // 12)) - 3

    1 <= month <= 12

    m(1)  = 1  + (12 * ((14 - 1)  // 12)) - 3 = 1  + (12 * 1) - 3 = 10
    m(2)  = 2  + (12 * ((14 - 2)  // 12)) - 3 = 2  + (12 * 1) - 3 = 11
    m(3)  = 3  + (12 * ((14 - 3)  // 12)) - 3 = 3  + (12 * 0) - 3 = 0
    m(4)  = 4  + (12 * ((14 - 4)  // 12)) - 3 = 4  + (12 * 0) - 3 = 1
    m(5)  = 5  + (12 * ((14 - 5)  // 12)) - 3 = 5  + (12 * 0) - 3 = 2
    m(6)  = 6  + (12 * ((14 - 6)  // 12)) - 3 = 6  + (12 * 0) - 3 = 3
    m(7)  = 7  + (12 * ((14 - 7)  // 12)) - 3 = 7  + (12 * 0) - 3 = 4
    m(8)  = 8  + (12 * ((14 - 8)  // 12)) - 3 = 8  + (12 * 0) - 3 = 5
    m(9)  = 9  + (12 * ((14 - 9)  // 12)) - 3 = 9  + (12 * 0) - 3 = 6
    m(10) = 10 + (12 * ((14 - 10) // 12)) - 3 = 10 + (12 * 0) - 3 = 7
    m(11) = 11 + (12 * ((14 - 11) // 12)) - 3 = 11 + (12 * 0) - 3 = 8
    m(12) = 12 + (12 * ((14 - 12) // 12)) - 3 = 12 + (12 * 0) - 3 = 9

    So, m is always > 0. Which also makes the expression ((153 * m) + 2) > 0,
    and 2) is proved.