The Mondrian Squares Riddle: A Mathematical Art Challenge

Inspired by the abstract, rectangular paintings of Dutch artist Piet Mondrian, mathematicians have devised an intriguing challenge: covering a square canvas with unique, non-overlapping rectangles and achieving the lowest possible score. This puzzle blends geometry, creativity, and a bit of strategic thinking. Let's dive into how to solve it and minimize your score.

Understanding the Mondrian Squares Puzzle

The challenge involves two key steps:

Tiling the Canvas: Completely cover a square canvas with non-overlapping rectangles. Each rectangle must be unique; if you use a 1x4 rectangle, you can't use a 4x1 elsewhere. However, using a 2x2 rectangle is perfectly acceptable.
Calculating the Score: Find the area of the largest rectangle and subtract the area of the smallest rectangle. The result is your score. The goal is to achieve the lowest possible score.

Solving the 4x4 Canvas Puzzle

Let's start with a 4x4 canvas to illustrate the process.

Initial Attempt: Dividing the canvas into a 3x4 and a 1x4 rectangle yields areas of 12 and 4, respectively. This gives us a score of 12 - 4 = 8.
Optimization: To improve, we can further divide the 3x4 rectangle into a 3x3 and a 3x1. Now we have rectangles of sizes 1x4, 3x3, and 3x1, with areas 4, 9, and 3. The score is now 9 - 3 = 6.

With a small canvas like this, the options are limited, but this demonstrates the basic strategy.

Tackling the 8x8 Canvas Puzzle

Now, let's scale up to an 8x8 canvas. This is where the puzzle becomes more interesting.

First Attempt: Dividing the canvas into a 5x8 and a 3x8 rectangle gives areas of 40 and 24, resulting in a score of 16. Not ideal.
Further Division: Splitting the 5x8 into a 5x5 and a 5x3 reduces the score to 25 - 15 = 10. Still not optimal.

The Key Strategy: Minimizing the Range

The trick is to avoid creating increasingly tiny rectangles, which widens the gap between the largest and smallest areas. Instead, aim for rectangles with areas that fall within a small range.

Since the total area of the 8x8 canvas is 64, the areas of your rectangles must add up to 64. Consider the possible rectangle sizes and their areas, keeping in mind that they must fit within the 8x8 canvas.

Finding the Optimal Solution

Area Considerations: Notice that if you use a rectangle with an odd area (e.g., 5, 9, or 15), you'll need another odd-area rectangle to achieve an even sum.
Trial and Error: Experiment with different combinations. Starting with areas of 20 or more quickly exceeds the total area of 64.
A Solution: Rectangles in the 14-18 range (excluding 15) might seem promising, but they won't fit together properly.
The Best Range: The range of 8 to 14, excluding the 3x3 square, works. This can be achieved with a 2x7, 1x8, 2x6, and 2x4. This gives a score of 14-8 = 6.

Alternative Solution

Another combination that yields the same score of 6 involves using a 3x3, 1x7, and 1x6 instead of the 2x7 and 1x8. The key is to maintain a narrow range of area values.

Beyond the 8x8 Canvas

As the canvas size increases (e.g., 10x10 or 32x32), finding the absolute lowest score becomes increasingly challenging. Expert mathematicians aren't even sure if they've found the true minimums for larger grids.

Tips and Tricks

Intuition over Formula: There's no magic formula for solving this puzzle. It requires a blend of intuition, spatial reasoning, and trial and error.
Strategic Division: Start by dividing the canvas into larger rectangles and then refine the divisions to minimize the area range.
Odd vs. Even: Keep in mind the need to balance odd and even area values to reach the total canvas area.

Your Turn!

How would you divide a 4x4, 10x10, or 32x32 canvas to achieve the lowest possible score? Share your solutions and strategies in the comments below!