if have multiples object, how to arrange it so that each number of object in row x column will form a near square?
exp:14 objects arrange to something like this:
0 0 0 0
0 0 0 0
0 0 0 0
0 0
+6
A:
Take the ceil of the square root?
in python:
import math
ceil(14**(.5))
Which returns:
>>> from math import ceil
>>> ceil(14**(.5))
4.0
Enrico Carlesso
2010-03-17 14:23:04
+2
A:
Get the square root of the number of items:
n = 14 ^ 0.5 ~ 3.7417
Round up to the nearest integer:
n = ceil(14 ^ 0.5) = 4
Now simply arrange the items in rows of n items until you run out.
Guffa
2010-03-17 14:24:50
+1
A:
Given n objects, the size of the resulting "square" is ceil(sqrt(n)), with special shortcut cases for n = 0 (don't draw anything) or n = 1 (just a 1x1 square).
Also, the question title is a little misleading. You can never arrange N objects in an NxN square, unless N is 1.
John Feminella
2010-03-17 14:25:34
`ceil(sqrt(0)) == 0` and `ceil(sqrt(1)) == 1`; there should be no need for special cases.
Thomas
2010-05-08 12:14:34
@Thomas: Mathematically, they're not special cases. Programmatically, they are; it's very common to have shortcut conditions that execute simpler versions of an algorithm when you only have to deal with zero or one things.
John Feminella
2010-05-08 14:15:14
If special cases are handled just as well by the general case, why add extra code to write, read, debug and maintain? Unless this code is actually a performance bottleneck, and the shortcuts do speed things up; but adding them should be the exception, not the rule.
Thomas
2010-05-08 15:29:24
+2
A:
That depends. Assuming that the number of columns should be equal to or greater than the number of rows then the following calculates the number of columns (in pseudo-code):
Ceiling(Sqrt(n))
where n is the number of items.
Jakob Christensen
2010-03-17 14:26:09