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• 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 1x4, we can't use a 4x1 in another spot,

• but a 2x2 rectangle would be fine.

• Let's try that.

• Say we have a canvas measuring 4x4.

• We can't chop it directly in half,

• since that would give us identical rectangles of 2x4.

• But the next closest option - 3x4 and 1x4 - 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 1x4

• while breaking the 3x4 into a 3x3 and a 3x1.

• 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 8x8; what's the lowest score you can get?

• Pause here if you want to figure it out yourself.

• To get our bearings, we can start as before:

• dividing the canvas roughly in two.

• That gives us a 5x8 rectangle with area 40

• and a 3x8 with area 24,

• for a score of 16.

• Dividing that 5x8 into a 5x5 and a 5x3 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 1x13 or 2x9 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 2x7 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 3x3 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 2x7 and 1x8

• and replacing them with a 3x3, 1x7, and 1x6.

• 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 herejust 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 4x4,

• 10x10,

• or 32x32 canvas?

• Give it a try and post your results in the comments.

Dutch artist Piet Mondrian's abstract, rectangular paintings

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# Can you solve the Mondrian squares riddle? - Gordon Hamilton

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ally.chang posted on 2020/02/25
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