#onmedium

gridcalc.py

def calculate_m():
    # get a list of possible squared numbers, this give us the number of
    # rows and columns to use for the canvas, eg 4 = 2 x 2, 6 = 3 x 3...
    # Return: [] of squared numbers up to 10000 (hopefully we don't get that high!)
    m = []
    for i in range(1,10000):
        m.append(i*i)
    return(m)

def grid(n,m):
    # given the amount of words, work out what grid or M we will need
    # 5 words should use 9 (m = 9) a 3 x 3 grid
    # Parameters:
      # n: a list of elements
      # m: a list of squared numbers
    word_count = len(n)
    for i,m in enumerate(m):
        if m - word_count > 0 or m - word_count == 0:
            # loop stops when the amount of words can fit in a grid
            # eg ((m = 9) - (word_count = 5) = greater than 0 so use 9 (3x3)
            return float(m)
            break

m = grid(convert(), calculate_m())