From: ikastan@sol.uucp (ilias kastanas 08-14-90) Subject: Re: Cutting a Bagel Challenge Date: 17 Mar 1999 08:25:51 GMT Newsgroups: sci.math In article , Donald T. Davis, Jr. wrote: @In article <7cmtlh$7m9@cocoa.brown.edu>, "A" wrote: @ @> How many pieces can a perfect bagel be sliced into with 3 straight cuts? @ @ten. the trick here is to describe the cuts clearly in words. @first, cut the bagel across the hole, as for making a sandwich. @then, lie the bagel flat on the counter, and make two vertical @cuts, each tangent to the hole, with the two cuts crossing @anywhere inside the bagel. these two cuts separate the bagel @into two nearly semicircular arcs. one arc is cut into four @parts by the crossing vertical cuts, the other arc is intact. @each of these 5 parts is bisected by the 1st, horizontal cut. @ @i don't know how to prove that this is maximal. Twelve is possible. (Eleven, too!). Hint: get the most out of the first two cuts. Ilias ============================================================================== From: horst.kraemer@snafu.de (Horst Kraemer) Subject: Re: Cutting a Bagel Challenge Date: Thu, 18 Mar 1999 01:05:35 GMT Newsgroups: sci.math On 17 Mar 1999 08:25:51 GMT, ikastan@sol.uucp (ilias kastanas 08-14-90) wrote: > > Twelve is possible. (Eleven, too!). > Too easy. Even thirteen pieces are possible by three plane cuts through of a torus. Definition of the cutting planes upon request. But I'm sure you will find it out yourself... Regards Horst ============================================================================== From: ol3@webtv.net (Oscar Lanzi III) Subject: Re: Cutting a Bagel Challenge Date: Wed, 17 Mar 1999 21:24:59 -0600 (CST) Newsgroups: sci.math How old is this problem? I saw it in an old Scientific American puzzle book and so I know the answer. I'm also informed from this source that the technical problem of implementing this solution and getting the indicated number of pieces is a bit tricky, as it inolves cutting a pair of slender pyramids to very close tolerance. So I guess you could say I cheated, or I looked it up in the literature! I am being a bit evasive, but I do not want to spoil the fun for all you mathematics buffs out there. But if you look at this posting closely, you'll see a subtle hint that will give away the answer. --OL ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Cutting a Bagel Challenge Date: 21 Mar 1999 09:39:29 GMT Newsgroups: sci.math In article <7cmtlh$7m9@cocoa.brown.edu>, "A" wrote: > How many pieces can a perfect bagel be sliced into with 3 straight cuts? Donald T. Davis, Jr. wrote: > i don't know how to prove that this is maximal. Well, I can go the other way: I know _how_ to prove maximality, but I'm too lazy to complete the proof. In fact, I'm sure I've seen this particular question answered before, but what I rather do is describe a method of proof which would answer the (to me) more interesting question: how many different homeomorphism classes of spaces are there of the form X = (torus) - (N planes) ? The original question concerned N=3 and asked only for the invariant pi_0(X) (the number of connected components) but with no extra effort we can handle the more refined problem. I'll describe the process in some detail, just in case there's someone who wants to carry the computations through to completion. Let me begin with N=1: what are the possible sets of homeomorphism types of pieces which can result from a single planar slice? This is a question of Morse theory, and the answer is: the homeomorphism types can only change when the intersection of the plane and the torus is singular, i.e. the plane is tangent to the torus. For an example, consider a family of vertical planes slicing the torus (assumed to have a horizontal central circle). The set of pieces varies from: torus union empty-set torus union ball cylinder union cylinder (homeomorphic to two balls) ball union torus empty-set union torus (up to homeomorphism), and the transitions occur when the plane is tangent along the outside or along an inside "saddle point" (where the surface of the torus meets the plane in a figure-eight). So what we need to do is consider the family of all planes in R^3, remove the ones which are tangent to the torus, and then look at the components of the remaining set. The homeomorphism type of X doesn't change on each component, and although it's not obviously necessary that it _do_ change when we pass from component to component, that turns out to be the case here. In case you've never thought about "the set of all planes" before, let me observe that this set P is almost the same as R^3 itself. The non-vertical planes pair off nicely with the triples [a,b,c], such a triple corresponding to the plane z=ax+by+c. To include the vertical planes too (and to capitalize on the compactness of the torus) it's probably better to use some other local homeomorphisms with R^3. For example, to any plane we may associate the point (r+1)*v, where v is the outward-pointing unit vector and r is the distance of the plane to the origin; this pairs off the set of planes with the vectors outside the closed unit ball in R^3, except that (since "outward" makes no sense for planes containing the origin) we must identify antipodal points on the boundary of this ball. Or we could more naturally pair off the same plane with the point r*v in R^3, except that all planes through the origin would get paired off with the origin itself this way. (Note that r*v is simply the point in the plane closest to the origin.) It turns out this last representation is good enough for our purposes, if for example we use a torus like (x^2+y^2+z^2-5/4)^2+4z^2-1 = 0, which rings around the origin: it's easy to examine the ways these planes can partition the torus. So let us use this representation for planes, and look to see which planes are tangent to the torus somewhere. It's easy enough to parameterize the torus as [cos(u)*(1 + cos(v)/2), sin(u)*(1 + cos(v)/2), sin(v)/2] and then retain the linear part of the Taylor series of this mapping to describe the tangent plane, which I can then map (for each u and v) to a point in R^3 as indicated above. For example, when u=0, I get the tangent plane to correspond to the point [(1/2)* cos(v)* | (1+2*cos(v)) | , 0 , (1/2)* sin(v)* | (1+2*cos(v)) | ] if I've done the algebra right. You get the points for other u > 0 by rotating this answer around the z-axis. It looks rather like a pinched torus from the outside, but there are chambers on the inside, too. Removing these points corresponding to tangent planes (call this the subset Q of P ) then leaves three regions in R^3 : the unbounded region on the outside corresponds to those planes which are too far from the origin to touch the torus; the main volume of the shape consists of the planes which chop off a nibble of the bagle (without destroying its fundamental group), and one inner chamber corresponds to the planes which slice the torus into two "C" like pieces. There are two other chambers in this representation, which correspond to planes which are roughly horizontal but a little above or below the central plane. Both of them correspond to planes which slice the bagel into two toaster-ready pieces. In my representation they appear to be two disjoint components in R^3 but that's an artifact of the failure to distinguish planes through the origin; in reality this is a single connected component of plane-space P . So we have discovered the conclusion we thought we knew -- that one plane can meet the torus in three fundamentally different ways -- by looking for components of the open set P - Q in "plane-space" P . If one plane meets the torus transversally (i.e. is not tangent), then we can investigate how a second plane will further chop up the pieces. As in the previous section, we expect the homeomorphism type of the complements of the planes not to change as this second plane varies, unless it passes through a tangent plane of the torus. However, that misses another special case: we should apply Morse theory also to the intersection of the first plane with the boundary of the torus (a curve). When the second plane is tangent to this, it can happen that the homeomorphism type will change. So here's how you can describe the fundamentally distinct ways of slicing the bagel with two planes. Begin with P x P , the space of all ordered pairs of planes in R^3. You can remove the diagonal of PxP to obtain the set of pairs of _distinct_ planes; indeed, you can then mod out by the action of Sym(2) to get a 6-dimensional manifold which parameterizes the set of unordered pairs of distinct planes in R^3. From this set, remove the sets Q x P and P x Q ; what remains are pairs of non-tangent planes. Further remove the set of pairs whose intersection is tangent to the torus somewhere; this again is a hypersurface in P x P. The net result is the complement in P x P of a union of closed submanifolds, hence open. On each of the components of this set, the homeomorphism types of the pieces of torus stay fixed. Thus we can find the fundamentally distinct slicings of a bagel by two planes simply by testing with one point from each of these components. Finally, we describe the solution to the original problem in the same way: in the 9-dimensional manifold P x P x P we locate the points corresponding to triples of planes in which * Two of the planes coincide * One of the planes is tangent to the torus * The intersection of some pair of planes is tangent to the torus * The intersection of all three planes lies on the torus. What remains is an open set. Pick representative points in the components of that open set, slice the bagel accordingly, and observe the homeomorphism types of (or just count) the pieces which remain. Select the slicing with the maximum number of components, if that's your goal. Let me close with some comments about identifying the components. Since P is 3-dimensional, we can "draw" it and be very explicit about the components corresponding to a single planar slice. In the other cases, we have to do something more algebraic. For example, to locate the components of the open set in P x P x P, use one of the near-identifications of P with R^3. Then use Morse theory _again_ to observe that the open set meets each "chunk" { (x1, ..., x9) ; a < x9 < b } in a space which is a product of the slice x9=(a+b)/2, as long as a and b are consecutive critical values of x9 on the hypersurface. (identifying components of adjacent "chunks" is probably harder but unnecessary for our problem). So we analyze the open set in R^9 by analyzing a collection of open sets in R^8. Repeat to lower the dimension to 3 or lower, then draw a picture. I don't know if there's an easier way to count components of an open set in a high-dimensional space; certainly this looked too much like work for me to attempt it, but it's also clearly a finite problem. dave ============================================================================== From http://mathworld.wolfram.com/TorusCutting.html : With cuts of a torus of genus 1, the maximum number of pieces which can be obtained is N(n) = n(n^2+3n+8)/6 The first few terms are 2, 6, 13, 24, 40, 62, 91, 128, 174, 230, ... (Sloane's A003600).