# Python: Find the years where 25th of December be a Sunday between 2000 and 2150

## Python Itertools: Exercise-21 with Solution

Write a Python program to find the years between 2000 and 2150 when the 25th of December is a Sunday.

**Sample Solution:**

**Python Code:**

```
'''Days of the week'''
# Source:https://bit.ly/30NoXF8
from datetime import date
from itertools import islice
# xmasIsSunday :: Int -> Bool
def xmasIsSunday(y):
'''True if Dec 25 in the given year is a Sunday.'''
return 6 == date(y, 12, 25).weekday()
# main :: IO ()
def main():
'''Years between 2000 and 2150 with 25 December on a Sunday'''
xs = list(filter(
xmasIsSunday,
enumFromTo(2000)(2150)
))
total = len(xs)
print(
fTable(main.__doc__ + ':\n\n' + '(Total ' + str(total) + ')\n')(
lambda i: str(1 + i)
)(str)(index(xs))(
enumFromTo(0)(total - 1)
)
)
# GENERIC -------------------------------------------------
# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))
# index (!!) :: [a] -> Int -> a
def index(xs):
'''Item at given (zero-based) index.'''
return lambda n: None if 0 > n else (
xs[n] if (
hasattr(xs, "__getitem__")
) else next(islice(xs, n, None))
)
# unlines :: [String] -> String
def unlines(xs):
'''A single string formed by the intercalation
of a list of strings with the newline character.
'''
return '\n'.join(xs)
# FORMATTING ---------------------------------------------
# fTable :: String -> (a -> String) ->
# (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> xs -> tabular string.
'''
def go(xShow, fxShow, f, xs):
ys = [xShow(x) for x in xs]
w = max(map(len, ys))
return s + '\n' + '\n'.join(map(
lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
xs, ys
))
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)
# MAIN --
if __name__ == '__main__':
main()
```

Sample Output:

Years between 2000 and 2150 with 25 December on a Sunday: (Total 22) 1 -> 2005 2 -> 2011 3 -> 2016 4 -> 2022 5 -> 2033 6 -> 2039 7 -> 2044 8 -> 2050 9 -> 2061 10 -> 2067 11 -> 2072 12 -> 2078 13 -> 2089 14 -> 2095 15 -> 2101 16 -> 2107 17 -> 2112 18 -> 2118 19 -> 2129 20 -> 2135 21 -> 2140 22 -> 2146

**Python Code Editor:**

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