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README
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Proceso que permite calcular el dia anterior a cuando se esta ejecutando.
Principalmente para los sistemas unix en que la utileria date no esta actualizada y no tiene dicha funcionalidad.
Este script lo desarrollo Carlos Aquiles
Documentacion bastante buena para calcular corrimientos de fechas
Background.
In order to do date calculations we must translate any given date specified by the three components
year, month, and day, into an integer day number. If we choose a base date for day 0 wisely,
calculations will be simplified. For this method the base date will be March 1, year 0.
Note that this year 0 is not the Gregorian year 0, it is a reference date only. However,
calculations involving dates after the establishment of the Gregorian calendar will be correct.
Year calculations.
To four decimal places, the length of a year has been measured as 365.2424 days.
Various calendars have tried to accommodate the extra fraction, which is just less than a quarter
of a day.
In October of 1582, the Gregorian calendar was instituted. It takes the length of a year to be
365.2425 days, which is one part in ten thousand (0.0001) too big, but is accurate enough to last
through the next few millenia. This approximation 365.2425 can be written as:
365.0000 + 0.2500 - 0.0100 + 0.0025
or
365 + 1/4 - 1/100 + 1/400
Now the number of days in y years is:
d = 365y + y/4 - y/100 + y/400 (1)
If we now make this a discrete function by considering integer d's and disregarding remainders, we have:
d = 365y + int(y/4) - int(y/100) + int(y/400) (2)
Every four years the second term y/4 will add an extra day, but that will be neutralized
every century by the third term y/100. However, every four centuries, the leap day will
be added after all, as the last term y/400 kicks in.
This discrete function is the source of the Gregorian leap year rules. Function (2)
approximates the real function (1) very closely, and the two are equal for years
divisible by 400. In the intervening years of this 400 year cycle, the error maxima
are 1.4775 days in year 303 of the cycle, and -0.72 days in year 96 of the cycle.
Because of this error, finding the year given a number of days (the inverse function)
is not exact, but we can find a very close approximation.
Given d days, the year number can be approximated as
y = d / 365.2425 (3)
This will be correct for nearly all dates, except in some cases where d is within the
error maxima from a year boundary. When this happens, we can test for it and adjust
the year accordingly.
Month calculations.
Because the lengths of the months are not the same, and especially because the length of
February is not fixed, it would seem at first to be impossible to calculate day offsets
in a year without looking up tables and checking for leap years. However, we can get around
the problem by defining the start of our calendar year with March. Numbering the months in
this way has the special advantage that leap days are always added at the end of the year,
and do not change the day offsets for the beginnings of the months. Using basic linear
regression on the month indexes and the day offsets, we can find functions to map each to
the other.
Given the month index, we can calculate the day number for the first of the month from the
integer part of the function
d = f(m) = (306*m + 5)/10:
d = f(m) = (306*m + 5)/10
m f(m) int(f)
Mar 0 0.5 0
Apr 1 31.1 31
May 2 61.7 61
Jun 3 92.3 92
Jul. 4 122.9 122
Aug 5 153.5 153
Sep 6 184.1 184
Oct 7 214.7 214
Nov 8 245.3 245
Dec 9 275.9 275
Jan 10 306.5 306
Feb 11 337.1 337
Now, the inverse function, given the day index into the year, we can calculate the month
index by taking the integer part of the function m = f(d) = (100*d + 52)/3060.
The following table shows the values of the function for the first and last days of each month:
m = f(d) = (100*d + 52)/3060
month length index s=first day e=last day f(s) f(e)
Mar 31 0 0 30 0.017 0.997
Apr 30 1 31 60 1.030 1.978
May 31 2 61 91 2.010 2.991
Jun 30 3 92 121 3.024 3.971
Jul 31 4 122 152 4.004 4.984
Aug 31 5 153 183 5.017 5.997
Sep 30 6 184 213 6.030 6.978
Oct 31 7 214 244 7.010 7.991
Nov 30 8 245 274 8.024 8.971
Dec 31 9 275 305 9.004 9.984
Jan 30 10 306 336 10.017 10.997
Feb 28/29 11 337 365/366 11.030 11.945
Gracias a Gary Katch (http://alcor.concordia.ca/~gpkatch/)
# vim: set ts=3 et sw=3 sts=3 si ai tw=100 lbr: