

A puzzle: dissect a square into congruent pieces, all touching the center - ColinWright

There's a math problem I've been working on for some time that looks like it's finally been solved.  To explain it to people I usually start with a warm-up:<p><pre><code>    Cut a square into indentical pieces so
    that they all touch the center point.
</code></pre>
As I say, this is a warm up.  It's fairly easy to do, and serves to introduce the ideas involved. So we do that, and we move on.<p>Over the weekend, though, I wondered how many solutions there are to the above problem.  I rapidly came up with a small number, and was reasonably content.<p>Then a friend showed me another.  Then another friend found another.  Now I have rather more than I thought or expected.  Today, a work colleague came up with yet another.<p>How could I miss so many?  What are they all?<p>I invite the HN community to explore this puzzle/problem.  Maybe I still haven't got all the answers.<p>Who will start me off?
======
SHOwnsYou
I am not very interested in puzzles, math, or much else that this question
relates to, but I am incredibly curious-- why is this question being asked?

I don't understand the significance of this problem.

So there are nearly infinitely many ways to accomplish this. What does that
mean though?

What are the implications of there being infinitely many ways to do this? Why
is it interesting?

~~~
ColinWright
Some puzzles don't have "significance", they are simply puzzles. Some people
find some (but not all) puzzles intriguing. Why do people do Sudoku? Or
crosswords?

This particular problem is also an excellent example of learning to think
"outside the box," insofar as almost everyone comes up with the same
solutions, and almost no one comes up with _all_ the solutions.

Including me.

Finally, this is just a warm up to an unanswered question in math/puzzles.
Solving that may turn out to be mundane, but it may lead to the development of
new techniques and insights. You never know. It's like the Collatz Conjecture.
Simple to state, no one knows if it's true, no one know how to prove (or
disprove) it, and who knows what techniques may be developed to answer it.

You say there are "nearly infinitely many ways to accomplish this." Can I ask,
what do you mean by "nearly infinite"? Anything that's not infinite is
infinitely far away from being infinite, so I'm a little confused by your
statement.

And learning to think about these things is an excellent exercise in logic and
reasoning. Explaining solutions is superb training in communication.

And for some, it's fun.

~~~
SHOwnsYou
>> not infinite is infinitely far away from infinite

Nit picky in my opinion, but the reason I didn't assign it a number and rather
opted for the more generic "nearly infinite" is because there are a ton of
possibilities.

Here is my idea for you and will eventually come up with most solutions (might
miss some outliers, this is my <5 minutes of thought on the topic):

Write a program that shows a square and it's center point.

A single line increment out from the center point going at a random starting
angle between 181 and 360 (in other words, they generally tend toward going
left). The line moves the smallest amount possible each iteration (1 pixel for
example).

After the line has moved one pixel, a new angle is picked for the line move,
from 0-360, and it increments 1 more pixel. If the randomly generated line
crosses the center line of the square (if you folded the square in half from
left to right) force the line to randomly pick a new angle before
incrementing.

Do not allow the line to cross over itself.

Now you have your first line. Take this line, replicate it, rotating it 90
degrees, then 180, then 270.

This is your first square. Set the program to keep generating new squares, add
rules as you see fit.

Now you have more solutions to this puzzle than you can count.

Edit: This is pretty badly explained, but it is based on the idea of a single
line leaving the midpoint, moving randomly until it reaches an edge. You then
replicate that single line 3 more times, each rotating by 90 additional
degrees.

~~~
ColinWright
In essence, take any line from the center to the edge, and take three copies
(giving four in total) rotated by 90 degrees. Provided they don't cross,
that's a solution.

And that's one infinite family of solutions, one that I didn't (initially)
find.

There are more.

Initially I had 5 actual solutions, and I thought I had them all. Then someone
produced this infinite family, and I suddenly had my mind expanded. I've since
found what I hope - but have not yet proved - is all solutions.

Can you find any more? You might not care, but that's the challenge. I find it
akin to the best sort of programming, except it doesn't, in the end, actually
_do_ anything.

~~~
SHOwnsYou
I won't find any more solutions. I will let the program find them, which,
given enough time should find every single possible solution.

Any line drawing variant is just the concept I've already outlined expanded.

 _So I guess my question is:_ Is the challenge in this finding other ways to
generate solutions, even though a method that will generate all possible
solutions has already been found?

~~~
ColinWright
I must have mis-understood you. The method you have outlined - as I understand
it - definitely will not find all solutions, and I don't see how you can think
it would.

Perhaps you should explain it again. It seems like you find a squiggly line,
rotate it 4 times by 90 degrees, say that's a solution, do it in all possible
ways, and claim that's everything.

Have I misunderstood?

------
bartonfink
By identical, do you allow for rotations - i.e. two pieces are identical if
one can be oriented to perfectly overlay the other? If so, there are an
infinite number of solutions.

~~~
ColinWright
Yes, but simply saying there are infinitely many solutions is not enough.
There are infinitely many solutions, but there is more you can say.

It's a shame this will sink without making it to the front page - I'd really
like to get the HN input before writing the blog post.

------
spicyj
You can make any pinwheel-type shape with four wheels and can vary the line
from center to edge in any path. This includes a bunch of spirally-looking
things.

(Also you can just cut it in half similarly instead of in four.)

Picture:

[http://cl.ly/3K1i2K3O0p2f1U291636/Screen_shot_2011-05-25_at_...](http://cl.ly/3K1i2K3O0p2f1U291636/Screen_shot_2011-05-25_at_9.35.01_AM.png)

~~~
ColinWright
So you have two infinite families - the generality is what I originally
missed.

There are more ...

------
vanishing
Any straight line through the center point will create 2 identical halves.

Any 2 straight lines through the center point at 90 degrees to each other will
create 4 identical pieces.

In both of the above cases each line segment from the center to the edge can
be distorted in any way which does not intersect the edge of the square or any
of the other lines and the distortion can be rotated 180 degrees in the first
case or 90 degrees 3 times in the second to create new identical shapes.

The only special case seems to be using 4 lines to divide the square into 8
pieces which is, I think, the only configuration of 8 pieces.

I think that's every possible solution.

I'm weak at math, but I think that means there are an infinite number of
configurations for each of an infinite number of configurations for both of
the first 2 solutions, and then there's that 1 extra solution. So does that
mean there are uncountably infinite solutions? I'm not sure how you would
apply the diagonal method to this.

~~~
ColinWright
You've found two infinite families and one sporadic solution.

There is another sporadic solution, and the 8 piece solution you've found is
not, in fact a sporadic.

And yes, the infinities are uncountable.

~~~
pozorvlak
_puzzles_

Oh, right - you can vary the shape of the diagonal lines, provided each "arm"
has rotational symmetry about its mid-point.

OK, so we've got three infinite families and the trivial solution (only one
piece - is that your sporadic solution?). I think that might be all: each
piece can contain 4, 2, 1 or 1/2 of the original square's corners, since
(lacuna) all pieces must contain the same number of corners and further
subdividing the corners (into 1/3s, say) would mean some pieces don't touch
the centre (another lacuna).

~~~
ColinWright
OK, that's now the set of solutions I've got. You've also gone some way to
showing them to be complete.

More to do, though.

And now do it for an equilateral triangle.

------
atakan_gurkan
1) Start from the centre and reach any of the sides following a curve, such
that as you traverse the curve your distance from the centre should increase,
e.g., a circular arc that is less than a semi-circle. (The construction of
this curve gives [at least] an uncountably infinite cubed number of
degeneracies.)

2) Now take the reflection of this curve wrt the centre, and attach it to the
original curve. This will give two-congruent pieces.

3) You can take the resulting curve and rotate it around the centre by pi/2.
The two curves will give 4 congruent pieces.

The restriction on the original curve, ie. its monotonic behaviour, guarantees
that the curves in steps 2 and 3 will not intersect.

~~~
ColinWright
That gives two infinite families. However, those families contain more
examples than your process generates, there is another infinite family, there
is a sporadic, and I have just been shown another example for which I am
unsure if it is a sporadic, or one member of an infinite family.

Not only do I not know everything, I truly don't know very much at all.

------
pozorvlak
Suppose the square can be dissected into n pieces. Then any symmetry of the
square induces a permutation of those n pieces. Hence, the symmetry group of
the square (D_8) must be embeddable as a subgroup of S_n. Hence there is no
dissection into three pieces, since |D_8| = 8, which does not divide |S_3| =
6.

I suspect this line of argument can be taken further, but wanted to post it
before someone else did :-)

 __Edit __: no, that doesn't work, because we already have a dissection into
two pieces, and |S_2| = 2. Dammit. Not all of the symmetries of the square
must map pieces to pieces.

------
pozorvlak
You can divide it into eight isoceles (pi/4, pi/2, pi/4) triangles, by
quartering up/down and then quartering diagonally.

~~~
ColinWright
That's one solution. There are more ...

------
ColinWright
A heart-felt "Thank You" to those of you who have contributed and up-voted.
Alas, as I type, this item has now slipped to the third page of HN and is
unlikely to be seen by "the masses."

So it is only you, the cognoscenti, who have had the privilege (if it be such)
of discussing the dissection.

Again, thank you.

~~~
SHOwnsYou
This is #1 on my "Ask" page.

~~~
ColinWright
Woah! Cool! Thanks ...

I wonder how many people read the "Ask" page.

