## Cutting and rearranging a half-colored square

For the discussion of math. Duh.

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### Cutting and rearranging a half-colored square

**This hypothetical happens in math land, where there are no physical constraints and anything that can be conceived mathematically can be achieved.**

You are given a square piece of paper. The paper is divided down the middle into two regions of equal size. The left region is white, the right region is red.
You do the following:
1. Cut the paper along the boundary of the two regions so that you have two pieces that are identical in all ways except for their color.
2. Cut each piece into two identical pieces so you now have four pieces total laid out in a WWRR pattern.
3. Swap any WR patterns you see so that they become RW. Do this working from left to right until you have an alternating pattern of colors.
4. For every WR pair (again working from left to right), repeat steps 2 and 3.
5. Repeat step 4 ad infinitum.

Assume the pieces of paper are laid out flat on the table and they are placed next to each other in perfect alignment. So at the end of every step, the combined pieces look like a complete square.
Define an iteration as one repetition of step 4. As the number of iterations approach infinity, what will the square look like?

Now suppose the bottom edge of the original paper is mapped uniformly (and bijectively) to the real number line over the interval [0,1]. So the bottom left corner will be 0 and the bottom right corner will be 1.
Notice at the end of every iteration, you will still have a complete square. That square can be mapped to [0,1] in a similar fashion as the original. Thus, let Mj denote the mapping corresponding to the square at the end of the jth iteration.

What is the relationship between M0 and Mn, as n -> infinity?
If a particular number 0 < x < 1 corresponds to some point A under M0 and another point B under Mn (A and B are points on the bottom edge of the square). What is the distance between A and B as a function of x?
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### Re: Cutting and rearranging a half-colored square

I don't think I understand along which axis you are cutting? Are the cuts perpendicular to the x-axis? Could you draw a picture of the first two iterations?

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### Re: Cutting and rearranging a half-colored square

It can converge only at dyadic points. The left and right end-points will always be white and red, respectively, but anywhere in between, on the k’th iteration a point is white if the k’th digit of its binary expansion is 0, and red if it is 1.

To find the color of a dyadic, we must specify which side of the cut it falls on. This is equivalent to choosing either its terminating or nonterminating expansion.
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### Re: Cutting and rearranging a half-colored square

>-) wrote:I don't think I understand along which axis you are cutting? Are the cuts perpendicular to the x-axis? Could you draw a picture of the first two iterations?

Does this help? I've labeled the regions to help you get a better idea of how the swaps occur.

In the first iteration EG and BD are swapped. In the second iteration, B and C swap and G and F swap. It's kind of like a fractal.

Qaanol,
What makes the end points special? After the nth iteration, you can divide the square into 2^n rectangles, each have the exact same color properties as the original square (since the length in the vertical direction doesn't really matter, it's just there to help visualize things).
Consider the point that originally corresponds to 0.25. After one iteration, it becomes the left "end point" of a rectangle with the pattern WR.

(EDITs: fixing so many typos)
Last edited by Cradarc on Sat May 23, 2015 12:59 am UTC, edited 1 time in total.
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### Re: Cutting and rearranging a half-colored square

I'm pretty sure your map is actually undefined on the dyadics at the limit: eventually they will always be between two regions, and then you can't assign them a clear color. Sure, you can say that the left boundary is white and the right one is red, but what about the middle? And what about the other boundaries? You can try and assign then a color in some systematic way, but it will probably be arbitrary to some extent.

Other than that, what Qaanol says applies: a point is white in the kth iteration if the kth digit of its binary expansion is 0(excluding the thing I said with the dyadics: you can choose a coloring for the dyadics by still following that rule, which means that we consider a dyadic white after we cut through it). If a real number ends with an infinite string of 0s(or 1s: both are equivalent), then it is dyadic. Therefore, all the other numbers have an infinite number of 1s and an infinite number of 0s after the kth digit, for any k(otherwise we'd have an infinite string of 0s or 1s and the number would be dyadic): this means that their color swaps infinitely many times, and thus diverges as we let M_n go to infinity.

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### Re: Cutting and rearranging a half-colored square

Color assignment only happens once at the beginning. Everything that happens afterwards simply involves moving the points/pieces around.
No piece is thrown away and no colors are reassigned. The square can't just lose its color.
In my opinion, the square would become pink as the number of iterations approaches infinity. Over time, it becomes "more and more pink" in the sense it becomes more and more difficult to isolate the red and the white. The half-colored square becomes a full colored square of a different color.

As for the second part:
I'm interested in the location of each point on the edge relative to their location in the original square. If you tracked the trajectory of a particular point on the square's edge, would it converge? If so, to where?
For example, the point originally assigned to 0.25 would converge to 0.5. It doesn't matter which side of the cut you place it. On the left side, the convergence happens as n->infinity. On the right side, it reaches 0.5 after the first iteration.
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### Re: Cutting and rearranging a half-colored square

Cradarc wrote:In my opinion, the square would become pink as the number of iterations approaches infinity. Over time, it becomes "more and more pink" in the sense it becomes more and more difficult to isolate the red and the white. The half-colored square becomes a full colored square of a different color.
The problem with that is that it doesn't become pink in math land with no physical constraints, only in the real world. Sufficiently narrow red-and-white stripes in the real world will look pink, but that's only because of the physical limitations of our eyes (we can only see the stripes if they're thick enough, otherwise they blend together and look pink). (That, plus there are additional limitations as the stripes approach the size of the molecules in the paper, and the wavelength of light, etc., but that doesn't change my point.) An abstract mathematical piece of paper with (for instance) a billion red and white stripes still just has a billion red and white stripes, even though we'd never be able to see it that way in the real world.
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### Re: Cutting and rearranging a half-colored square

Cradarc wrote:Over time, it becomes "more and more pink" in the sense it becomes more and more difficult to isolate the red and the white.
Nope.

At every iteration, almost every point in the square (i.e. a set with measure 1) is either red, surrounded by other red points as part of a red vertical strip, or it's white, surrounded by other white points as part of a white vertical strip. You said yourself this happens in math land, where there are no physical constraints. Which means the actual physical width of each strip doesn't matter at all, so long as each strip still has a finite nonzero width (which it does at every finite step right from the beginning).

It's like if you take a square wave:

And at every iteration you double its frequency.

Your claim that it becomes pink is like saying after repeated doubling of the wave's frequency you end up with a constant value of zero. But how can that be? At every finite step there is no point on the wave that's at 0. At every finite step in fact almost all points are at -1 or +1. They may switch between those values but none of them ever get any closer to 0.
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### Re: Cutting and rearranging a half-colored square

Cradarc wrote:Color assignment only happens once at the beginning. Everything that happens afterwards simply involves moving the points/pieces around.
No piece is thrown away and no colors are reassigned. The square can't just lose its color.
In my opinion, the square would become pink as the number of iterations approaches infinity. Over time, it becomes "more and more pink" in the sense it becomes more and more difficult to isolate the red and the white. The half-colored square becomes a full colored square of a different color.

The square gets "pinker" in the sense that in each iteration, red strips and white strips are closer and closer together, and more "mixed": however, at all steps all strips are either red or white, so there is no sense in which we could say that M_∞ is pink(or talk about the colors of M_∞ at all).

The problem about specific points is what I said before: what happens to the points you cut along to(dyadics) is undefined. Consider the point assigned to 1/4(0.25). The first iteration cuts along it, cuts along 3/4, and swaps E&G with B&D. So did the 1/4 point stay with A&C or did it go with E&G? There is no non-arbitrary way of choosing. All dyadics except 0 and 1 suffer from this problem.

(For the rest it should be possible to say where they end up after the nth iteration, and they actually should converge somewhere, but I'll think about that later).

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### Re: Cutting and rearranging a half-colored square

If the interval is [0,1], it seems like each iteration n just swaps the nth digit of the binary expansion with the following digit. So in the first step everything that starts with 0.00... stays, the interval that starts with 0.01... moves to 0.10..., numbers that start with 0.10... move to 0.01..., and everything starting with 0.11... stays. (Proof by induction left for the reader.)

In the limit, this amounts to just taking the first digit and getting rid of it by sliding it down to infinity. Everything in [0,1/2) gets doubled, everything in (1/2,1] is reduced by 1/2 and then doubled. And this is true regardless of which expansion we choose. For example, as already pointed out, 1/4 goes to 1/2 either way,
because 0.00111... -> 0.00111... -> 0.01011... -> 0.01101... -> 0.01110... -> 0.01111... and so on,
while 0.01000... -> 0.10000... -> 0.10000... -> 0.1000... and so on.

The only point that depends on expansion is 1/2, because if it's 0.10000... (i.e. it stays on the red side of the slice) then it goes eventually to 0, whereas if it's 0.0111111... (i.e. it stays on the white side of the slice) then it goes eventually to 1.
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### Re: Cutting and rearranging a half-colored square

I understand that it never truly becomes pink because we never defined pink in the original construction (and anything not explicitly constructed in math land does not exist). However, I think it can be thought of as a better and better approximation to pink.

Let's say you pick and random point on the edge of the n->infinity square, and ask "what color is this point?". As previously established, there is no definitive answer to this question. However, we know for certain that the color is either red or white. It makes sense (although I don't have a proof) that the probability of the point being one color or the other is 0.5. The expected value of the color would be the average of red and white, which we call pink.
This is further supported by Gmalivuk's formula. The range corresponding to the final location of [0,1/2) and (1/2,1] overlaps perfectly. Any given point on the edge of the final square will correspond to both red and white. There exists a red and white piece that converges to any given location (with the exception of the endpoints). The superposition is not merely statistical.
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### Re: Cutting and rearranging a half-colored square

Cradarc wrote:Let's say you pick and random point on the edge of the n->infinity square, and ask "what color is this point?". As previously established, there is no definitive answer to this question. However, we know for certain that the color is either red or white. It makes sense (although I don't have a proof) that the probability of the point being one color or the other is 0.5. The expected value of the color would be the average of red and white, which we call pink.
No, we don't know this for certain and it doesn't make sense, because the sequence of whole squares doesn't converge to any limit, just like the frequency-doubling square wave doesn't converge. So it doesn't really make much sense to talk about the end result unless you first describe and define a new notion of convergence.

If you pick either the terminating or the non-terminating expansion for the dyadic numbers, then those points and those points only will strictly be either red or white. But the set of those points are of measure zero, so almost all points in the square do not have a definite color.
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### Re: Cutting and rearranging a half-colored square

Consider a starting position P0 = 50-d, where d is a very small positive value.
This point is colored white.
After one iteration, P0 -> P1 = 75-d
After two iterations, P1 -> P2 = 87.5-d
etc.
If we take the limit as d->0, Clearly this will converge to 1 from below.

We tracked the location of a single white point and discovered it moves to location 1 from below.

Now consider Q0 = 1.
This point is colored red.
With each iteration, the location of this point does not change.

We tracked the location of a single red point and discovered it stays at location 1.

Saying the trajectories of P and Q converge to two distinct locations is equivalent to saying 0.9999.... and 1 are two distinct values.
The idea is that I am not terminating the process, but taking the limit. It never truly becomes pink, but it can become arbitrarily close to pink.
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### Re: Cutting and rearranging a half-colored square

The points converge. I'm the one who first said what they converge to, remember?

But that is not the same as saying the colors converge.

The points converge because at each step a point can only move at most half as far as any points moved on the previous step. But at every step half of all points "move" the full "distance" between red and white, and no point ever moves any other distance.

You are not arguing that 0.999... = 1, you are arguing that the sequence 0,1,0,1,0,1... converges to 1/2.
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### Re: Cutting and rearranging a half-colored square

Color is a property of each point, agreed? So if the points converge to the same location, then both colors must exist at that location.
If you have a sequence of 0,1,0,1,0..., then the points didn't really converge. If they did you would not be able to parse them out into such a sequence. In actuality, every point on the square's edge could contain such a sequence. So they are effectively superposed.

You can play the same trick on 0.999... = 1. Just define zero to be "even", the next number larger than zero to be "odd", the next number larger than that to be even, etc. The problem is there's no such thing as "the next number larger than zero". To define such a number, you will need to define an increment, and any increment can be made smaller.
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### Re: Cutting and rearranging a half-colored square

Cradarc wrote:Color is a property of each point, agreed? So if the points converge to the same location, then both colors must exist at that location.

If you keep cutting the pieces of paper, you correctly noticed that the right border of the red side stays at the right border, while the right border of the white side moves over and gets arbitrarily close to the right border. That does not mean that they occupy the same location; indeed they don't on any finite iteration. They just get arbitrarily close to each other, but always maintain a non-zero distance.

And you can't just jump to the conclusion that on some ill defined "infinite iteration" everything assumes the value of its limit. That's not how limits work.

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### Re: Cutting and rearranging a half-colored square

Cradarc wrote:Color is a property of each point, agreed? So if the points converge to the same location, then both colors must exist at that location.
If you have a sequence of 0,1,0,1,0..., then the points didn't really converge. If they did you would not be able to parse them out into such a sequence. In actuality, every point on the square's edge could contain such a sequence. So they are effectively superposed.

The point 1/3 is *exactly* such a sequence. Its binary expansion is 0.01010101..., and the strip of square over it goes from white to red to white to red... with every iteration. The fact that those strips are getting narrower and narrower doesn't matter, because you're only considering a single point and you already declared that physical restrictions don't apply.

So yes, you are still claiming that the limit of 0,1,0,1,0,1,0,1,... is 1/2. You're arguing something like, because it spends less time on each number than it did on the last, it must converge. But the speed of going through a sequence doesn't affect convergence, just as the width of a color strip doesn't for the square.

Edit: The thing you don't seem to understand, Cradarc, is that there are two different things going on here. One is the movement of points from the original square to new locations on the number line. That process has a limit, because the distance each point moves on iteration n is at most 1/2^n. The other process is the recoloring process. Specifically the process of assigning a color to each point on the number line. This process does *not* have a pointwise limit, because almost all points have an infinite number of color changes to look forward to.
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### Re: Cutting and rearranging a half-colored square

Just to try to short circuit this loop we seem stuck in, what we are discussing is known as a supertask.

For instance, if I turn on a light a minute before midnight, and then turn it off half a minute before midnight, then on again at time -1/2^(2n) minutes and off again at -1/2^(2n+1). What is the state of the light at midnight?

Some people like assigning a state to what happens at the end of such processes, others don't. Other's have complicated rules about when we can and can't. But, in some cases, reasonable people can differ. In the case of the lamp, there is no sensible "half" position for the light, but there is a sensible half position for these colours. But, reasonable people can differ. I tend to think that the "no colour" position makes more sense here, but that is just my opinion.
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### Re: Cutting and rearranging a half-colored square

For another way of looking at it, what color would you say the point x=0.5 is in iteration 0? It is unclear from the statement of the problem whether this point should be red or white or something else. As more iterations are made, more of these ambiguous points are created (at 0.25, 0.75, etc.). In the limit, the interval will contain (countably) infinitely many of these region boundary points. The rest of the (uncountably many) points will have alternated between red and white infinitely many times, so they do not have a clear color assignment either.

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### Re: Cutting and rearranging a half-colored square

gmalivuk wrote:The point 1/3 is *exactly* such a sequence. Its binary expansion is 0.01010101..., and the strip of square over it goes from white to red to white to red... with every iteration. The fact that those strips are getting narrower and narrower doesn't matter, because you're only considering a single point and you already declared that physical restrictions don't apply.

The strip of square over 1/3 has zero width.

I know there are two things going on, but IMO they are closely related. The color moves with the point. A point that starts with a color (or any orthogonal dimension if you will) doesn't lose it when you move it around. Saying the color becomes undefined is as absurd as saying it approaches pink.

Jestingrabbit,
The link to supertasks is interesting. I think the Cantor Set and Density may also be relevant.
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### Re: Cutting and rearranging a half-colored square

Cradarc wrote:Saying the color becomes undefined is as absurd as saying it approaches pink.

Not really. It's a bit like asking: Is "this sentence is false" true or false?

One answer would be to say its truth value is undefined.

Compare to: A point can only be red or white by definition - yet there's a reason it can't be red (and the same reason it can't be white).

This happens because you're describing a process that can't actually be carried out in the real world. So when describing what happens in the imaginary, mathematical world, you're free to use any definition you like - so long as the definition is useful and others agree there's not a better one.

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### Re: Cutting and rearranging a half-colored square

Cradarc, you need to define a new version of the word "limit" if you want to keep holding onto your idea that the *points* in the square somehow "become" (or "approach") pink.

Normally, limits are tied with the idea of "getting closer" to something, but there's no way in which the color directly over 1/3 gets closer to pink, because the color directly over 1/3 is always either totally red or totally white, which are both equally far away from pink, and the color changes with every iteration.

Until you *define* a new type of limit that doesn't require any "getting closer", we have to conclude that the limit over 1/3 is *undefined*. (Or if you don't want to develop a new definition of "limit", at least take the time to define "pink" in a way that includes something broader than just the solid color halfway between red and white.)

In the same way, you have to redefine at least one of "limit" or "0.5" before you can consistently claim that the limit of the sequence (0,1,0,1,0,1,0,1,...) is 0.5.
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### Re: Cutting and rearranging a half-colored square

Gmalivuk, I understand why you think that.
However, that is only one way to think about the problem.

Suppose locations on the edge of the square are slots. The points are colored balls inside the slots. The slots and balls are arbitrarily small.

How (I think) you are thinking about it is:
Select a particular slot and look at the color of the ball inside that slot as the algorithm is iterated ad infinitum.
When you do this, you encounter the problem that balls are being shifted in and out of the slot at every step. So if the process never stops, you can't say with certainty that a ball of a particular color exists in the slot. In fact, you probably can't say definitively if a ball exists in the slot at all! This is because the system never stabilizes if the operation doesn't stop.

How I am thinking about it is:
In the original configuration, select a particular ball. Allow the algorithm to iterate ad infinitum and track the movement of the ball.
Because the process never stops, the ball is constantly moving. However, the movement of the ball is constrained to a smaller and smaller block* of the total number of slots. The size of the block* can be made arbitrarily small as the number of iterations increase.

Your previous formula shows that there exists a white ball which will eventually be confined to any block in the range [0,1):
Everything in [0,1/2) gets doubled

In addition, there exists a red ball which will eventually be confined to any block in the range (0,1]
everything in (1/2,1] is reduced by 1/2 and then doubled

So if you pick any block in (0,1), there will exist a red ball and a white ball which will eventually be confined to it.

*A block is a set of adjacent slots.

The million dollar question:
What does it mean to pick out a "single slot"?
Any slot can always be divided into a number of smaller slots, and can therefore be considered a block.

To me, slots and blocks are identical. Whenever we identify a specific location, we are actually identifying an abstract region. This region has no fixed size, but is allowed to become as small as it is needed for whatever purpose we have in mind for it.
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### Re: Cutting and rearranging a half-colored square

I am aware that there are two things going on, as I said before:
gmalivuk wrote:Edit: The thing you don't seem to understand, Cradarc, is that there are two different things going on here. One is the movement of points from the original square to new locations on the number line. That process has a limit, because the distance each point moves on iteration n is at most 1/2^n. The other process is the recoloring process. Specifically the process of assigning a color to each point on the number line. This process does *not* have a pointwise limit, because almost all points have an infinite number of color changes to look forward to.
The first process I talk about is where you pick one ball and follow it. The second process is where you pick one slot and follow it.

The problem with saying slots and balls are identical is that the two processes behave totally differently in the limit.

Talking about adjacent slots that can nevertheless be subdivided any number of times seems a bit inconsistent, but I do get what you're trying to say in a mathematical sense: any interval in [0,1] will eventually contain both red and white strips.

As I said in my last post, if you redefine "pink" you can get your "pink square". It's just not a single solid color any more, but rather a dense intermixing of red and white.
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### Re: Cutting and rearranging a half-colored square

I don't think anyone is arguing that the informal statement "the square turns pink in the limit" is incorrect. Clearly, if you weave together very very thin red and white strips, the eye perceives the total square as pink.

The issue is that you cannot even state "the square turns pink in the limit" with mathematical precision without saying exactly what you mean by "the square turns pink". Certainly you can't say that the individual points turn pink. You haven't even defined what it means for a point to be pink, only red and white. You don't want to define "the point turns pink" to mean "it swaps between red and white infinitely often", since we could have a process for which the whole square is red, then white, then red, then white, etc. and that limit's color is ambiguous, at best. Your notion of having colored balls moving about and abstract regions with no fixed size is not a formal definition.
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### Re: Cutting and rearranging a half-colored square

For another way of looking at it, what color would you say the point x=0.5 is in iteration 0? It is unclear from the statement of the problem whether this point should be red or white or something else.

I think that measure hit the nail on the head. You don't even have to start subdividing. Given that you have defined the left region as white and the right region as red, what color is at 0.50000000...?

Whatever definition you choose to apply will affect the answer.

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### Re: Cutting and rearranging a half-colored square

Leaving aside that Cradarc still can’t spell my name even a month after I corrected her or him on the exact same mistake, I think there’s a lot of talking past one another going on here.

Define “point x has color y” to be the ordered pair (x, y). In other words, treat color as a full-fledged dimension in its own right. Now do all the coloring in half-open intervals, eg. [0, ½). And while we’re at it, let “white” = 0 and “red = 1”.

Initially, we have the set S0 consisting of the points on the line y=0 with 0≤x<½ and the points on the line y=1 with ½≤x<1. After one iteration, we have the set S1 consisting of the points on the line y=0 with 0≤x<¼ ⋃ ½≤x<¾ and the points on the line y=1 with ¼≤x<½ ⋃ ¾≤x<1. Subsequent iterations yield Sk as expected.

If we take the liminf of these sets, meaning the union of the intersections of the tails, we end up with the dyadic x-values at y=0 and nothing else.

If we take the limsup of these sets, meaning the intersection of the unions of the tails, we end up with the entire interval 0≤x<1 at y=0 and that same interval except for the dyadics at y=1.

However, we can do better. Instead of having a discrete sequence of sets, we can make a continuous transition from one to the next. In other words, let the subscript of the sets become a real parameter, represented by a full-fledged dimension of its own. We can call it z.

Now we don’t have a sequence at all, just a set of points in 3 dimension. At z=0 we have the points (x, y, 0) where (x, y) is in S0, and so on: for each natural number k we have the points (x, y, k) where (x, y) is in Sk.

There are, of course, uncountably-many ways to connect the “slices” at each z=k to one another. Perhaps the simplest way is to hold each x constant and let y go linearly with z to make a straight line from (x, y, k) to (x, y, k+1). If we let z rise without bound then the y-values will keep oscillating between 0 and 1 everywhere except dyadic x-values, so the limit will only be points of the form (d, 0, ∞) where d is dyadic.

Another simple way to connect the “slices” is to hold each y constant and let the x’s from the nearer half of each interval go linearly with z where necessary to connect with the next slice up. If we let z rise without bound then the x-values will move less and less, so the limit will be points of the form (x, 0, ∞) for 0≤x<1 and points of the form (x, 1, ∞) for non-dyadic x in the same interval.

Even though each dyadic x-value d only has (d, 1, z) in the set when z is less than some finite bound (which depends on d), there is nonetheless for any given dyadic d some other x-value, call it v, such that the path beginning at (v, 1, ∞) and following the continuous mapping from the prior paragraph approaches (d, 1, ∞). In this “path-limit” sense, the limit consists of all points (x, y, ∞) for 0≤x<1 and y∈{0, 1}.

As it happens, the specific description of the situation given in the original post makes it abundantly clear that the pairs (x, y) can change their x-value, but not their y-value. And the mechanism by which the x-values change is consistent with the continuous piece-wise linear paths I described two paragraphs up. Moreover, the end-state is defined by the position that each point (x, y, 0) approaches along its path as z→∞.

With this understanding, the problem yields immediately to elementary analysis. Each point (x, y, 0) maintains its initial y-value, and its x-value changes by excising the first digit after the decimal point of its binary representation, as Gmalivuk described. Algebraically, when 0≤x<½ the limit is 2x, and when ½≤x<1 the limit is 2x−1. In short, we have (x, y, 0) → (2x%1, y, ∞).

Therefore the set of points that exist in the limit contains precisely (x, 0, ∞) ⋃ (x, 1, ∞), for 0≤x<1. That is two horizontal line segments. In the language of the original problem, every physical location along the edge of the square ends up with two slips of paper: one red, and one white. No point is pink, instead every point is both white and red.

This is one of the things we have to get used to when treating color as a dimension: just as a shape can contain points with multiple y-values for the same x, so too can an object have more than one color at the same location. In fact, the case at hand is quite simple, what with there being two discrete colors. But we must be prepared to deal with things which are a whole spectrum of colors all at once. In other words, objects that extend along the color dimension for some interval—or even forever!
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### Re: Cutting and rearranging a half-colored square

Apologies for botching your name again, Qaanol. My fingers have a natural tendency to type "u" after every "q" that is part of a word.

I don't disagree with any of your logic. I just disagree with how the result is interpreted.
For me, "pink" is an infinitely dense mixture of red and white. I think of red and white existing at any location rather than just saying the color is undefined at the location.
Cradarc wrote:Over time, it becomes "more and more pink" in the sense that it becomes more and more difficult to isolate the red and the white.
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### Re: Cutting and rearranging a half-colored square

Cradarc wrote:For me, "pink" is an infinitely dense mixture of red and white.
As I said before.

Cradarc wrote:Over time, it becomes "more and more pink" in the sense that it becomes more and more difficult to isolate the red and the white.
But at every finite step, it is as "easy" as ever to isolate them (remember we have no physical restrictions). Pick any point in the square, and it is unambiguously red or unambiguously white. Almost every point is furthermore surrounded by an entire disc of all red or all white.
Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
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