Talk:Ham sandwich theorem

message: The two-dimensional proof seems to have a fairly large hole in its logic:
* At the first step, we determine an angle and a corresponding translation offset for that angle, such that pancake #1 will be bisected.  This is in fact a function--let's call it t(\alpha)=x , where x is the translation offset for the line with rotation \alpha.
* In step two, we run through all the angles 0-180 ("every angle \alpha\in[0,180^\circ]") and use the intermediate value theorem to conclude that there is some angle that bisects pancake #2.  That is our angle and we're done.
So here is the problem: In step two, we can't just take a line through the origin and rotate it continuously through angles 0-180 degrees.  At step one, each rotation angle ALSO had a translation amount--our t(\alpha). So as you work your way from \alpha=0 to \alpha=180 you have to rotate the AND also apply the relevant translation t(\alpha) to the line.
And here is the problem: We haven't shown that t(\alpha) is continuous.  If t(\alpha) is not continuous, then we cannot apply the intermediate value theorem.  
Even though we are changing \alpha continuously, t(\alpha) might have some *very* significant discontinuities and if it does, p(\alpha) is going to be discontinuous as well.
In fact, in some situations, t(\alpha) is not continuous at all.  Here is one example:
* Pancake #1 consists of union of two unit discs, one centered at (100,0) and the other centered at (200,0).  So for any line oriented within, say, 45 degrees of vertical, you can choose t(\alpha) to be anything between, say, 102 and 198 and it will split pancake #1 very nicely into two equal parts. So in this very large range, t(\alpha) is not guaranteed to be continuous at all. 
Perhaps t(\alpha) can be constructed so as to be continuous.  But if so, a construction must be given for cases such as this, an argument must be made that it can be done for every such pancake, and so on. It can't just be assumed to be continuous.
The typical hypothesis for the ham sandwich theorem is something like "Given n compact sets in \mathbb{R}^n". There is no requirement that the "pancakes" be connected.  (The example above takes advantage of the possibility that pancakes could be split into two equal & separated parts.) But even if the pancakes were required to be compact and connected, you still would have to make some sort of argument that function t is continuous.
Obviously, the theorem is true so there must be some way to fix it up.  But I can't see a simple way at all.  Can anyone?
At minimum, the article should state that there must be a proof that t is continuous but it is beyond the scope of this article or such.
Bhugh (talk) 06:19, 4 April 2019 (UTC)
replies: text: Your objection to the given proof was correct; the proof is now fixed. AxelBoldt (talk) 16:46, 19 January 2022 (UTC)

text: AxelBoldt (talk) 16:46, 19 January 2022 (UTC)
thread_title: Major hole in the proof of "Two-dimensional variant: proof using a rotating-knife"

message: In this proof, there is the flat statement, "This function f is continuous."  In a typical more rigorous proof (e.g. Brian Libgober) it takes maybe a page of dense argument to establish this.  In Libgober it is half of page 14 and all of page 15. 
We don't need to duplicate that kind of argument here but we can at least acknowledge that an argument is required rather than simply stating the conclusion baldly--as though it is either easy or obvious.
Bhugh (talk) 06:19, 4 April 2019 (UTC)
replies: text: AxelBoldt (talk) 16:46, 19 January 2022 (UTC)
thread_title: And, the same issue applies to the n-dimensional variant: proof using the Borsuk–Ulam theorem

message: I have cut the following part from the page since it contains a false proof of the discrete case. The proof could possibly be transformed into a proof of the continuous case (the opposite of discrete) and inserted back:
The discrete two-dimensional form of the ham-sandwich theorem can be proved directly using continuity as follows.
First we show that, for every angle θ, there is an oriented line of angle θ that bisects the red set of points.  To see this, just slide an oriented line of angle θ perpendicularly; at one extreme, all the red points are on the left side of the line, and at the other extreme, all the red points are on the right side of the line, so by continuity there is a position in between where they are equal.  More precisely, we can consider the continuous function f mapping the perpendicular offset of the line to the difference between the number of red points on the left and the number of red points on the right; at one extreme, f is r (the number of red points), and at the other extreme f is −r, so by the intermediate value theorem, somewhere in between, f must be zero, corresponding to a red bisector.
Second we argue the existence of a red bisector that is also a blue bisector.  Look at the red bisector of angle 0 obtained from the previous step.  This line has some number b1 of blue points on its left, and some number b2 of blue points on its right.  Now consider a red bisector of angle 180°, namely, the reverse of the previous line.  This line has b2 blue points on its left and b1 points on its right.  If we define g(θ) to be the difference between the number of blue points on the left and right of the red bisector of angle θ, then g(0) = b1 − b2 and g(180°) = b2 − b1.  By the intermediate value theorem, there must be an angle θ for which g(θ) = 0, corresponding to a simultaneous red and blue bisector.  (Here we omit the proof that g is in fact continuous.)
Here f is not continuous in the discrete case: it 'jumps' at every point. In particular, if θ is vertical and we have one point with x-coordinate 0, two points with x-coordinate 1 and two point with x-coordinate 2, then any vertical line not going through the leftmost point has an even number of points on one side and an odd number of points on the other side. The line through the leftmost point has more points on the right hand side than on the left.
Marco Streng 18:09, 20 August 2007 (UTC)
replies: text: The usual technique for handling this complication is to approximate the discrete distribution by a sufficiently similar continuous distribution (e.g. one that assigns uniform weight to a sufficiently small disk around each discrete point), apply the proof above for this continuous case, and then argue that any solution for the continuous approximation also solves the discrete case. One could also use a similar idea based on the fact that the function takes integer values and changes by one at each step. But I agree that without some such argument this is not a good enough proof for the discrete case. —David Eppstein 20:05, 20 August 2007 (UTC)
thread_title: Proof of discrete case

message: *Citation: For each point p on the surface of the sphere S, we can define a continuum of oriented hyperplanes perpendicular to the (normal) vector from the origin to p, [...]
**Why is there a continuum of oriented hyperplanes? For example in R3, with a 2-sphere, I can only find one unique hyperplane (that is, a plane touching the unit sphere at point p). Thanks for any corrections to my misconception. --Abdull (talk) 13:38, 9 August 2008 (UTC)
replies: text: If you have one such hyperplane, you can create a continuum of such hyperplanes by translating (moving parallelly) the given one. Note that a hyperplane does not need to pass through the origin. AxelBoldt (talk) 16:46, 19 January 2022 (UTC)
thread_title:  Continuum 

message: I removed 'The ham sandwich theorem is also sometimes referred to as the "ham and cheese sandwich theorem"' - none of the cited sources seems to mention it, and it appears to have been called the "ham sandwich theorem" for many decades.--mcld (talk) 13:37, 25 February 2013 (UTC)
thread_title:  No cheese 

message: The reduction to the Borsuk-Ulam theorem is incorrect as stated, because the function defined is not obviously continuous in the case where the hyperplane choice is ambiguous.  I think continuity actually does hold, but an argument to this effect seems warranted. GIrving (talk) 21:38, 12 March 2013 (UTC)
replies: text: The standard method for this is to replace Ai by a volume distribution that is within ε of its original distribution, but nowhere zero (e.g. take (1 − ε) times the original distribution plus ε times a Gaussian), giving a unique and continuous choice of the splitting hyperplane, and then apply a limiting argument as ε → 0. —David Eppstein (talk) 21:56, 12 March 2013 (UTC)
thread_title:  Hole in proof 

message: Seems to me that even the illustration on the page is a counter example for the phrasing. I see nothing that indicates that the number of dots of each color is equivalent so I assume it should read that the bisect equally divides each set such that the number of red and blue dots matches the number of red and blue dots on the other side of the line respectively.  As an example 8 red 2 blue would need to be bisected to 4 red and 1 blue on each side of the line.
For a finite set of points in the plane, each colored "red" or "blue", there is a line that simultaneously bisects the red points and bisects the blue points, that is, the number of red points on either side of the line is equal and the number of blue points on either side of the line is equal.(198.135.124.85 (talk) 20:23, 5 May 2015 (UTC)
thread_title:  Definition Correctness 

message: Substituted at 02:10, 5 May 2016 (UTC)
thread_title: Assessment comment

message: There is no mention of legal values for the number of sets, , in the article.  Clearly it works for .  It seems to work trivially for , where it the solution is the median.  It looks like it even works for algebraic curves of higher order when ; for example, I think we can always find two points on a line that bisect two sets by virtue of containing half of each set between the two points.  I could try to put that all in, but in my hands it is original research.  Does anyone have an authoritative citation or so much familiarity with the material that it is obvious to them?  Also, Is there a reasonable interpretation for ?  𝕃eegrc (talk) 12:07, 15 June 2016 (UTC)
replies: text: It is in the first sentence: "n measurable "objects" in n-dimensional space". That is, the number of sets equals the dimension of the space they are in. —David Eppstein (talk) 16:10, 15 June 2016 (UTC)
thread_title:  Legal values for the number of sets, {{mvar|n}} 

message: I went ahead and amended the citation to "Steinhaus 1938", it's certainly not titled "A Note on the Ham Sandwich Theorem". Rather there isn't really a title per se, but it's in the journal's regular "Notatki z tolologii" [Topology notes] section. Steinhaus also isn't listed as an author, per se. Rather the author is presumably one of the journal's editors (Stanisław Warhaftman with the participation of Edward Stenz & Kazimierz Zarankiewicz). The note relays information from Steinhaus, but it seems inaccurate to cite this as if he wrote this himself. The volume is also clearly labeled "Tom XI." -- is there a reason it was cited as being volume 9? Umimmak (talk) 20:54, 28 August 2020 (UTC)
replies: text: I changed the reference; it shouldn't list Steinhaus as author. AxelBoldt (talk) 16:46, 19 January 2022 (UTC)
thread_title:  "Steinhaus 1938" 

message: Can someone please confirm that there are cases when 2 objects are only bisectable mutually by 1 (magic line) hyperplane when found or placed in a particular spatial relationship? (no particular spatial relationship being required by the theorem as I understand it and thus not any spatial relationship verboten).
If the answer is yes in an absolute sense (somebody does) and adherents wish to claim the mathematical proof of the theorem is irrefutable we have a paradox; metaphysical: in the mathematical or linguistic realm or both.
82.29.184.92 (talk) 11:17, 23 November 2020 (UTC)
thread_title:  Why is position of third object guaranteed to support theorem? 

message: https://www.youtube.com/watch?v=YCXmUi56rao </Numberphile The Ham Sanwich Problem>
It is not clear that Centroid is taken universally by the mathematical community to be the point, such that only hyperplanes it lies on can be equal volume bisectors of a compound object and no hyperplane it lies on cannot be.
:It is not true that a hyperplane that bisects an object's volume must pass through the object's centroid. Take for instance, in the plane, a large disk, connected with a thin long rod to another, smaller disk (like in the figure O--o). If the rod is long enough, the centroid will be somewhere between the two disks, on the rod. If the volume of the rod is small enough, any volume-bisecting line will pass through the larger disk, and there will be many volume-bisecting lines that don't pass through the centroid.
:It is true that there will always be at least one volume-bisecting hyperplane that passes through the object's centroid, but the same is true for any other point, not just the centroid: just rotate the hyperplane until there's equal volume on both sides. AxelBoldt (talk) 16:28, 19 January 2022 (UTC)
replies: text: Your crankery is off-topic here. This talk page should only be about improvements to the article, based on reliable sources, not on personal reflections. —David Eppstein (talk) 16:44, 19 January 2022 (UTC)
thread_title:  Can Objects Be Treated As Points For The Purpose of Bisection? 

message: Without bothering implies that the original authors being discussed here considered the generalisation too obvious or trivial to devote much attention to. Is this really what was intended? If so, then the formulation can stand as it is (but omitting automatically which certainly cannot mean in context what the likely non-native speaker who put this in wanted to convey). If not, without taking the trouble (or similar) would be clearer. 2A01:CB0C:CD:D800:CD80:544B:3131:E36A (talk) 12:51, 29 April 2022 (UTC)
replies: text: . Thanks! AxelBoldt (talk) 19:07, 2 July 2022 (UTC)
thread_title:  "without bothering to automatically state the theorem in the n-dimensional case"