##### Riddle

# Can you solve the Mondrian squares riddle?

Dutch artist Piet Mondrian’s abstract, rectangular paintings inspired mathematicians to create a two-fold challenge. First, we must completely cover a square canvas with non-overlapping rectangles.

All must be unique, so if we use a 1×4, we can’t use a 4×1 in another spot, but a 2×2 rectangle would be fine. Let’s try that. Say we have a canvas measuring 4×4. We can’t chop it directly in half, since that would give us identical rectangles of 2×4. But the next closest option – 3×4 and 1×4 – works.

That was easy, but we’re not done yet. Now take the area of the largest rectangle, and subtract the area of the smallest. The result is our score, and the goal is to get as low a score as possible. Here, the largest area is 12 and the smallest is 4, giving us a score of 8. Since we didn’t try to go for a low score that time,

we can probably do better. Let’s keep our 1×4 while breaking the 3×4 into a 3×3 and a 3×1. Now our score is 9 minus 3, or 6. Still not optimal, but better. With such a small canvas, there are only a few options. But let’s see what happens when the canvas gets bigger.

Try out an 8×8; what’s the lowest score you can get? Pause here if you want to figure it out yourself. Answer in: 3 Answer in: 2 Answer in: 1 To get our bearings, we can start as before: dividing the canvas roughly in two. That gives us a 5×8 rectangle with area 40 and a 3×8 with area 24, for a score of 16. That’s pretty bad. Dividing that 5×8 into a 5×5 and a 5×3 leaves us with a score of 10. Better, but still not great.

We could just keep dividing the biggest rectangle. But that would leave us with increasingly tiny rectangles, which would increase the range between the largest and smallest. What we really want is for all our rectangles to fall within a small range of area values. And since the total area of the canvas is 64, the areas need to add up to that. Let’s make a list of possible rectangles and areas.

To improve on our previous score, we can try to pick a range of values spanning 9 or less and adding up to 64. You’ll notice that some values are left out because rectangles like 1×13 or 2×9 won’t fit on the canvas. You might also realize that if you use one of the rectangles with an odd area like 5, 9, or 15, you need to use another odd-value rectangle to get an even sum. With all that in mind, let’s see what works. Starting with area 20 or more puts us over the limit too quickly. But we can get to 64 using rectangles in the 14-18 range, leaving out 15.

Unfortunately, there’s no way to make them fit. Using the 2×7 leaves a gap that can only be filled by a rectangle with a width of 1. Going lower, the next range that works is 8 to 14, leaving out the 3×3 square. This time, the pieces fit. That’s a score of 6. Can we do even better? No. We can get the same score by throwing out the 2×7 and 1×8 and replacing them with a 3×3, 1×7, and 1×6. But if we go any lower down the list, the numbers become so small that we’d need a wider range of sizes to cover the canvas, which would increase the score.

There’s no trick or formula here – just a bit of intuition. It’s more art than science. And for larger grids, expert mathematicians aren’t sure whether they’ve found the lowest possible scores. So how would you divide a 4×4, 10×10, or 32×32 canvas? Give it a try and post your results in the comments.