Can you solve the Mondrian squares riddle? - Gordon Hamilton
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 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 4x4,
10x10,
or 32x32 canvas?
Give it a try
and post your results in the comments.
