I have n elements as a input and a function make_grid(n) which will calculate the dimensions of a grid that will contain the elements. Let's suppose that n = 12, then the function must calculate that the width is 4 and the height is 3 and not 1 and 12 or so. Similarly, n = 24 should return 6 and 4.
I tried to use ceil(sqrt(n)) to get one dimension, but is not a general case at all, and playing with cases (n even, sqrt(n) == ceil(sqrt(n))) hasn't worked.
Edit: Finding the optimum column and row size for a table with n elements and a given range for its proportion I already see this question but the coding throws me for n = 24 dimensions 5 and 5. Any help?
The approach is as follows:
Take the integer n as the input of the function. The goal is to obtain the "squarest" table. As @John suggested we must calculate sqrt(n) to get an idea of the dimensions. On the other hand we must calculate all the divisors of n in order to do choose the nearest divisors to sqrt(n).
How do we choose the nearest low value? we can use this tip (Python): Finding index of an item closest to the value in a list that's not entirely sorted and get the index of the nearest value in the list of divisors, let's say hIndex.
Then the other dimension can be calculated dividing n by divisors[hIndex] or using a new index wIndex = hIndex + 1 and get divisors[wIndex].
The Python code is this (note that I used al lazy evaluation to find the divisors):
import numbers
from math import sqrt
def get_dimensions(n):
tempSqrt = sqrt(n)
divisors = []
currentDiv = 1
for currentDiv in range(n):
if n % float(currentDiv + 1) == 0:
divisors.append(currentDiv+1)
#print divisors this is to ensure that we're choosing well
hIndex = min(range(len(divisors)), key=lambda i: abs(divisors[i]-sqrt(n)))
if divisors[hIndex]*divisors[hIndex] == n:
return divisors[hIndex], divisors[hIndex]
else:
wIndex = hIndex + 1
return divisors[hIndex], divisors[wIndex]