Understanding the Polygon Shape Calculator get amount inside function

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Describe your issue in detail here.

Basically I am trying to understand what the below line in the project rubric is asking of me. Please if someone could just break it down:

get_amount_inside: Takes another shape (square or rectangle) as an argument. Returns the number of times the passed in shape could fit inside the shape (with no rotations). For instance, a rectangle with a width of 4 and a height of 8 could fit in two squares with sides of 4.

Your code so far

class Rectangle:

name = 'Rectangle'

def __init__(self, width, height):
    self.width = width
    self.height = height

def __str__(self):
    return f"{self.name}(width={self.width}, height={self.height})"

def set_width(self, width) -> int:
    self.width = width
    return self.width

def set_height(self, height) -> int:
    self.height = height
    return self.height

def get_area(self) -> int:
    return self.width * self.height

def get_perimeter(self) -> int:
    return (2 * self.set_width(self.width)) + (2 * self.set_height(self.height))

def get_diagonal(self) -> int:
    return ((self.set_width(self.width) ** 2) + (self.set_height(self.height) ** 2)) ** .5

def get_picture(self) -> str:
    display = ''
    if self.width > 50 or self.height > 50:
        return "Too big for picture."
    for i in range(0, self.height):
        display += '*' * self.width
        display += '\n'
    return display

def get_amount_inside(self, shape):
    # Get the Greatest Common Denominator (GCD)
    odd = []
    even = []
    width = self.width
    height = self.height

    if width % 2 == 0:
        for i in range(1, width + 1):
            if width % i == 0:
        for j in range(1, width + 1):
            if width % j == 0:
    if height % 2 == 0:
        for i in range(1, height + 1):
            if height % i == 0:
        for j in range(1, height + 1):
            if height % j != 0:

    # Union of both list
    # and finding the common factors bw them
    union = list(set(odd) & set(even))
    greatest = max(union)
    return greatest
    # if shape is self:
    #     return 'rectangle'
    # return "square"

class Square(Rectangle):

name = 'Square'

def __init__(self, side_length):
    self.side_length = side_length
    super().__init__(width=side_length, height=side_length)

def __str__(self):
    return f"{self.name}(side={self.side_length})"

def set_side(self, side) -> int:
    self.side_length = side
    return self.side_length

def set_height(self, height) -> int:
    return self.set_side(height)

def set_width(self, width) -> int:
    return self.set_side(width)

def get_picture(self) -> str:
    display = ''
    if self.side_length > 50:
        return "Too big for picture."
    for i in range(0, self.side_length):
        display += '*' * self.side_length
        display += '\n'
    return display

Your browser information:

User Agent is: Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/93.0.4577.82 Safari/537.36

Challenge: Polygon Area Calculator

Link to the challenge:

You got a shape A and a shape B and have to calculate how often you can fit B into A.
Take a pencil and a sheet of paper and draw a 3x3 square. Then draw a 2x2 square into the 3x3 square. How many can you draw inside? One.
Now take a 3x4 rectangle - how many can you draw inside? Two.
4x4? Here you can fit a total of four 2x2 squares inside.

1 Like

Thanks I was able to figure out the solution. I knew it had to do with comparing two shapes but only one dimension was shown and I didn’t realize that I had to compare that to another set of dimensions. Soon as I figured this out I was able to solve the challenge. Thank you.

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