Largest proper divisor of n: Difference between revisions

<pre> 1 1 1 2 1 3 1 4 3 5

1 6 1 7 5 8 1 9 1 10

7 11 1 12 5 13 9 14 1 15

1 16 11 17 7 18 1 19 13 20

1 21 1 22 15 23 1 24 7 25

17 26 1 27 11 28 19 29 1 30

1 31 21 32 13 33 1 34 23 35

1 36 1 37 25 38 11 39 1 40

27 41 1 42 17 43 29 44 1 45

13 46 31 47 19 48 1 49 33 50</pre>

Or, reducing the search space, formatting more flexibly (to allow for experiments with larger ranges) and composing functionally:

<lang python>'''Largest proper divisor of'''

from math import isqrt

# maxProperDivisors :: Int -> Int

def maxProperDivisors(n):

'''The largest proper divisor of n.

'''

secondDivisor = find(

lambda x: 0 == (n % x)

)(

range(2, 1 + isqrt(n))

)

return 1 if None is secondDivisor else (

n // secondDivisor

)

# ------------------------- TEST -------------------------

# main :: IO ()

def main():

'''Test for values of n in [1..100]'''

xs = [

maxProperDivisors(n) for n

in range(1, 1 + 100)

]

colWidth = 1 + len(str(max(xs)))

print(

'\n'.join([

''.join(row) for row in

chunksOf(10)([

str(x).rjust(colWidth, " ") for x in xs

])

])

)

# ----------------------- GENERIC ------------------------

# chunksOf :: Int -> [a] -> [[a]]

def chunksOf(n):

'''A series of lists of length n, subdividing the

contents of xs. Where the length of xs is not evenly

divisible, the final list will be shorter than n.

'''

def go(xs):

return (

xs[i:n + i] for i in range(0, len(xs), n)

) if 0 < n else None

return go

# find :: (a -> Bool) -> [a] -> (a | None)

def find(p):

'''Just the first element in the list that matches p,

or None if no elements match.

'''

def go(xs):

try:

return next(x for x in xs if p(x))

except StopIteration:

return None

return go

# MAIN ---

if __name__ == '__main__':

main()</lang>

{{Out}}