<p>This seems to have little to do with Banach-Tarski.
In your setting, f is simply not an isometry due to floating-point errors, and in particular there must be a single piece <span class="math-container">$i$</span> such that <span class="math-container">$\mathrm{Vol}(f(P_i))&gt;\mathrm{Vol}(P_i)$</span>, so no need to cut into several pieces.</p>

<p>Banach-Tarski works because the notion of volume is not well-defined on the pieces. This clashes with your requirement that the pieces are defined by "a finite amount of geometric data", and with your writing of <span class="math-container">$\mathrm{Vol}(P_1)+\dots+\mathrm{Vol}(P_n)$</span>.</p>

CC BY-SA 4.0
2020-05-25T14:01:30.593Z
Yes, I think I understand Banach-Tarski, and you are right, my question has little to do with B-T. But there would be a certain pleasure if what I describe is achievable, a pleasure partially due to the contrast with B-T.
Joseph O'Rourke
CC BY-SA 4.0
2020-05-25T14:35:47.737Z
Unless I missed something, what you describe is a function which is programmed with the intent of being an isometry, but in reality is not because of rounding errors. This will probably be the case of any function doing floating point computation at some point, like a rotation whose angle is represented by a float. So I'm not sure what you mean by "if what I describe is achievable".
Denis
CC BY-SA 4.0
2020-05-25T13:31:11.110Z
2020-05-25T13:31:11.110Z
Denis
<p>The <a href="https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox" rel="nofollow noreferrer">Banach-Tarski paradox</a> says that
a ball in <span class="math-container">$\mathbb{R}^3$</span> can be partitioned into a finite number of pieces, whose rearrangement has a larger volume than the original.
It occurred to me that it might be possible to define the pieces 
<span class="math-container">$P_1,P_2,\ldots,P_n$</span> by a finite amount of geometric data, such that
the applied rearrangement function <span class="math-container">$f(P_i)$</span> actually <em>does</em> increase the volume of <span class="math-container">$P_i$</span>, because 
of floating-point round-off error: <span class="math-container">$f$</span> is an isometry but when implemented
in floating-point hardware it is not an exact isometry. So it might be possible that
<span class="math-container">$$\textrm{Vol}(f(P_1))+\ldots \textrm{Vol}(f(P_n)) &gt; \textrm{Vol}(P_1)+\ldots \textrm{Vol}(P_n) \;.$$</span>
Repeating, this might lead to a type of "realization" of the paradox in terms of floating-point
computations.</p>

<p>Whether you think this would be as interesting as I do is in the eye of the beholder.</p>

CC BY-SA 4.0
2020-05-25T14:34:51.217Z
It’s worth defining what you mean by two pieces being “the same” in this picture.  Do you mean that the one piece can be turned into the other by rounding an isometric and applying it to the first piece?
Geoffrey Irving
<set-cover><partition-problem><na.numerical-analysis>
Can the Banach-Tarski paradox be "realized" by floating-point round-off?
Joseph O'Rourke
# Can the Banach-Tarski paradox be "realized" by floating-point round-off?

(Asked by: username_0 on 2020-05-25)

The [Banach-Tarski paradox](https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox) says that
a ball in $\mathbb{R}^3$ can be partitioned into a finite number of pieces, whose rearrangement has a larger volume than the original.
It occurred to me that it might be possible to define the pieces
$P\_1,P\_2,\ldots,P\_n$ by a finite amount of geometric data, such that
the applied rearrangement function $f(P\_i)$ actually *does* increase the volume of $P\_i$, because
of floating-point round-off error: $f$ is an isometry but when implemented
in floating-point hardware it is not an exact isometry. So it might be possible that
$$\textrm{Vol}(f(P\_1))+\ldots \textrm{Vol}(f(P\_n)) > \textrm{Vol}(P\_1)+\ldots \textrm{Vol}(P\_n) \;.$$
Repeating, this might lead to a type of "realization" of the paradox in terms of floating-point
computations.

Whether you think this would be as interesting as I do is in the eye of the beholder.

**Tags:** set-cover, partition-problem, na.numerical-analysis

> **Comments:**
> **username_1**: It’s worth defining what you mean by two pieces being “the same” in this picture.  Do you mean that the one piece can be turned into the other by rounding an isometric and applying it to the first piece?  

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## 1 answers

### Answer by username_2 on 2020-05-25. Score: 5

This seems to have little to do with Banach-Tarski.
In your setting, f is simply not an isometry due to floating-point errors, and in particular there must be a single piece $i$ such that $\mathrm{Vol}(f(P\_i))>\mathrm{Vol}(P\_i)$, so no need to cut into several pieces.

Banach-Tarski works because the notion of volume is not well-defined on the pieces. This clashes with your requirement that the pieces are defined by "a finite amount of geometric data", and with your writing of $\mathrm{Vol}(P\_1)+\dots+\mathrm{Vol}(P\_n)$.

> **Comments:**
> **username_0**: Yes, I think I understand Banach-Tarski, and you are right, my question has little to do with B-T. But there would be a certain pleasure if what I describe is achievable, a pleasure partially due to the contrast with B-T.  
> **username_2**: Unless I missed something, what you describe is a function which is programmed with the intent of being an isometry, but in reality is not because of rounding errors. This will probably be the case of any function doing floating point computation at some point, like a rotation whose angle is represented by a float. So I'm not sure what you mean by "if what I describe is achievable".  

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