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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.

  • 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 5x8 rectangle with area 40

  • and a 3x8 with area 24,

  • for a score of 16.

  • That's pretty bad.

  • 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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B1 US TED-Ed canvas score x4 rectangle area

Can you solve the Mondrian squares riddle? - Gordon Hamilton

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    ally.chang posted on 2020/02/25
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