Source code for timelinelib.calendar.gregorian.julian2gregorian
# -*- coding: utf-8 -*-
# Copyright (C) 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018 Rickard Lindberg, Roger Lindberg
#
# This file is part of Timeline.
#
# Timeline is free software: you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
#
# Timeline is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with Timeline. If not, see <http://www.gnu.org/licenses/>.
"""
Contains the GregorianDateTime class and static functions related
to the class.
:doc:`Tests are found here <unit_calendar_gregorian_gregorian>`.
"""
from timelinelib.calendar.gregorian.time import GregorianTime
[docs]def julian_day_to_gregorian_ymd(julian_day):
"""
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).
"""
if julian_day < GregorianTime.MIN_JULIAN_DAY:
raise ValueError("julian_day_to_gregorian_ymd only works for julian days >= %d, but was %d" % (
GregorianTime.MIN_JULIAN_DAY, julian_day))
a = julian_day + 32044
b = ((4 * a) + 3) // 146097
c = a - ((b * 146097) // 4)
d = ((4 * c) + 3) // 1461
e = c - ((1461 * d) // 4)
m = ((5 * e) + 2) // 153
day = e - (((153 * m) + 2) // 5) + 1
month = m + 3 - (12 * (m // 10))
year = (b * 100) + d - 4800 + (m // 10)
return year, month, day