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#### 你能解决这个蒙德里安方块谜题吗？

Can you solve the Mondrian squares riddle? - Gordon Hamilton

Stick around after this episode for a bonus riddle.

Dutch artist Piet Mondrian’s abstract,rectangular paintings

inspired mathematicians to create a two-fold challenge

First, we must completely cover a squarecanvas 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 usidentical 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 areaof 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 12and 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 happenswhen the canvas gets bigger.

Try out an 8×8;what’s the lowest score you can get?

Pause here if you wantto figure it out yourself.

To get our bearings,we can start as before:

dividing the canvas roughly in two.

That gives us a 5×8 rectanglewith area 40

and a 3×8 with area 24,

for a score of 16.

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 dividingthe biggest rectangle.

But that would leave uswith increasingly tiny rectangles,

which would increase the rangebetween 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 areaof the canvas is 64,

the areas need to add up to that.

Let’s make a listof 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 valuesare left out

because rectangles like 1x13or 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 wayto 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 rangethat 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 themwith 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?

Stacey