Re: [MLUG - DISCUSSION] Computing the Date of Easter

[Mark Rages <EMAIL:PROTECTED>]

On Thu, Apr 08, 2004 at 02:27:00PM -0500, Mike Miller wrote:
> On Thu, 8 Apr 2004, Woodsmall, Ryan (IATS) wrote:
> 
> > The Perl time stuff does NOT like dates before 1902 or after 2037.  It
> > craps out.  The first bit of code should work better for dates.  I knew
> > there was a reason that I didn't just use the timelocal() stuff; I just
> > didn't know why...  I promise.
> 
> I believe you!
> 


Because it's fun to compare languages, here's the Python version.  Note I just 
cut 'n' pasted your original math.  Requires Python 2 or newer.  (In RedHat 7.X 
systems, this is installed at /usr/bin/python2)

#!/usr/bin/python
import time,sys,re

def compute_easter(y):
    c = y / 100
    n = y - 19 * ( y / 19 )
    k = ( c - 17 ) / 25
    i = c - c / 4 - ( c - k ) / 3 + 19 * n + 15
    i = i - 30 * ( i / 30 )
    i = i - ( i / 28 ) * ( 1 - ( i / 28 ) * ( 29 / ( i + 1 ) )
                           * ( ( 21 - n ) / 11 ) )
    j = y + y / 4 + i + 2 - c + c / 4
    j = j - 7 * ( j / 7 )
    l = i - j
    m = 3 + ( l + 40 ) / 44
    d = l + 28 - 31 * ( m / 4 )

    return (y,m,d,0,0,0,0,0,0)

def date_format(etime):
    return time.strftime("%A, %b %d, %Y",time.localtime(time.mktime(etime)))

years=[]
args=sys.argv[1:]

if not args:
    years.append(int(time.strftime('%Y')))
else:
    for arg in args:
        for regex in [re.compile(r'(\d+)\.\.(\d+)'),
                      re.compile(r'(\d+)-(\d+)'),
                      re.compile(r'(\d+)')]:
            m=regex.match(arg)
            if m:
                years=years+range(int(m.groups()[0]),
                                  1+int(m.groups()[-1]))
                break

print "Easter falls on the following day%s:"%('s'*(not len(years)==1))

print '\n'.join(map(date_format,map(compute_easter,years)))

# This is the end of the program

-- 
To invent, you need a good imagination and a pile of junk. -Thomas Edison
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