Loop colomns in a quarter speedy solution for Counting Tiles by macfreek - python coding challenges

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Loop colomns in a quarter solution in Speedy category for Counting Tiles by macfreek

#!/usr/bin/env python3.2
# -*- coding: utf-8 -*-

from math import floor, ceil, sqrt

def checkio(radius):
    """count tiles"""
    full = 0
    total = 0
    # Count only one quadrant, and multiply by 4.
    # Loop through rows "r"
    leftheight = radius
    for r in range(int(ceil(radius))):
        if radius > r+1:
            rightheight = sqrt(radius**2 - (r+1)**2)
        else:
            rightheight = 0
        full += int(rightheight)
        total += int(ceil(leftheight))
        leftheight = rightheight
    print(radius, "->", [4*full, 4*(total-full)])
    return [4*full, 4*(total-full)]

if __name__ == '__main__':
    assert checkio(2) == [4, 12], "First, N=2"
    assert checkio(3) == [16, 20], "Second, N=3"
    assert checkio(2.1) == [4, 20], "Third, N=2.1"
    assert checkio(2.5) == [12, 20], "Fourth, N=2.5"

Dec. 18, 2012

Comments:

DiZ on May 20, 2014, 4:11 p.m.
Good solution, but not very pythonist. I know it's an old one, but seems interesting to comment on it, to give tips to future readers. <ul> <li>First 2 lines: Python allows multi-initializations -> full, total = 0, 0</li> <li>int(ceil(radius)): ceil already generates an integer, no need to use int() on it. By the way, you don't use your import of floor</li> <li>r+1: you always use r+1, why not initialize it directly on range -> for r in range(1, ceil(radius))</li> <li>rightheight assignation: in my view, branching with 5 lines to initialize just one variable seems too expensive, especially with assignation to a constant on 1 branch. It's syntax-sugar but you can use inline statement for this -> rightheight = sqrt(radius * *2 - r * *2) if radius > r else 0</li> </ul> One more word about rightheight and leftheight: they are variables that evolve along your for loop. Difference between them are just r and r+1, you don't really need them. Just initialize your sums and add floored/ceiled square root directly. Below a piece of code with all these points: <pre class='brush: python'>from math import sqrt, floor, ceil def checkio(R): full, total = -floor(R), 0 for n in range(ceil(R)): i = sqrt(R*R-n*n) full += floor(i) total += ceil(i) return [4*full, 4*(total-full)] </pre> You can also use sum paradigm as in Sim0000's solution, or don't use math module at all as in mine's.

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