module timelinelib.calendar.gregorian.julian2gregorian¶
Contains the GregorianDateTime class and static functions related to the class.
-
timelinelib.calendar.gregorian.julian2gregorian.
julian_day_to_gregorian_ymd
(julian_day)[source]¶ This algorithm is described here:
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:
m >= 0
((5 * e) + 2) >= 0 => e >= 0
(1461 * d) >= 0 => d >= 0
((4 * c) + 3) >= 0 => c >= 0
(b * 146097) >= 0 => b >= 0
((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).