WWhy is a soap bubble like a railway?
David Wakeham Department of Physics and AstronomyUniversity of British ColumbiaVancouver, BC V6T 1Z1, Canada
Abstract
At a certain infamous tea party, the Mad Hatter posed the following riddle: why is a raven likea writing-desk? We do not answer this question. Instead, we solve a related nonsense query:why is a soap bubble like a railway? The answer is that both minimize over graphs. We give aself-contained introduction to graphs and minimization, starting with minimal networks on theEuclidean plane and ending with close-packed structures for three-dimensional foams. Along theway, we touch on algorithms and complexity, the physics of computation, curvature, chemistry,space-filling polyhedra, and bees from other dimensions. The only prerequisites are high schoolgeometry, some algebra, and a spirit of adventure. These notes should therefore be suitable forhigh school enrichment and bedside reading. [email protected] a r X i v : . [ phy s i c s . pop - ph ] A ug ontents ◦ rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 A minimal history . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Acknowledgments
These notes were inspired by conversations with Rafael Haenel, Pedro Lopes and Haris Amiri ofDTQC, and the hands-on Steiner tree activity they created for the UBC Physics Circle. It gotme curious about the computing power of bubbles! I would particularly like to thank Rafael andPedro for encouragement, the Physics Circle students for letting me test material on them, andScott Aaronson for detailed comments on the draft. I am supported by an International DoctoralFellowship from UBC.
Introduction
The Hatter opened his eyes very wide on hearing this; but all he said was, “Why is a raven likea writing-desk?” “Come, we shall have some fun now!” thought Alice. “I’m glad they’ve begunasking riddles.—I believe I can guess that,” she added aloud. “Do you mean that you think youcan find out the answer to it?” said the March Hare. “Exactly so,” said Alice.
Lewis Carroll
Why is a soap bubble like a railway? I believe we can guess that. Suppose we are designing a railnetwork which joins three cities. If stations are cheap, our biggest expense will be rail itself, and tominimize cost we should make the network as short as possible. For three cities A , B and C , thecheapest network typically looks like the example below left. In addition to stations at each city, weadd a hub station in the middle to minimize length. For a general triangle of cities, hub placementfollows a simple rule: outgoing rail lines are equally spaced, fanning out at angles of ◦ .A two-dimensional bubble, with walls made of soapy water, solves the same problem. Themolecules in the water are attracted to each other, creating surface tension. Tension pulls thesurface taut, and length is once again minimized, due to the budgetary constraints of Nature itself.Like rail lines, bubble walls converge at junctions three at a time, separated by ◦ . The rule evenworks for the soapy walls of a three-dimensional bubble.Of course, rail networks in the real world connect many cities, and the problem is more compli-cated. But it remains true that for the cheapest network, any time we introduce a hub it must havethree rail lines emerge at angles of ◦ , with the same going for multiple bubbles. This makes theconnection between soap bubbles and railways useful : by drilling screws through plexiglass, we canmake a soap bubble computer, and solve network planning problems with soapy water!While soap bubbles can find small railways almost instantaneously, there is a deep but subtlereason they aren’t useful for finding the best way to connect every city in North America. Inprinciple, we just place a screw at the position of every city, dip into soapy water, and withdraw.1n practice, it will probably take longer than the age of the universe for the bubbles to settle down!The problem is just too hard. Although we know what hub stations look like locally —a tridentof three rail lines—there are many different ways to arrange a given number of hubs. We show afew examples above right. As the number of hubs gets large, there are so many that no physicalmechanism , soap bubbles or quantum computers or positronic brains, can quickly search them allto find the shortest candidate, unless there is a wildly clever algorithm we have overlooked. Thisis called the NP Hardness Assumption . Ultimately, this is a physical hypothesis, because it makespredictions about the behaviour of physical objects which compute, such as soap bubbles.If it takes arbitrarily large amounts of computing power to find the cheapest network, it is nolonger cheap.
Approximate answers are preferable if they can be found quickly, and we will give twomethods for rapid (but suboptimal) rail planning below. But bubbles still hold surprises. Once weremove the plexiglass and screws, genuine bubbles are free to form, each cell enclosing some fixedvolume. The laws of physics will now try to minimize the total area of the cell surfaces, so poeticallyspeaking, the forms flowing out of the bubble blower are conjectures made by Nature about the best(i.e. smallest-area) way to enclose some air pockets.For example, the humble spherical bubble harbours the following conjecture: of all surfaces offixed volume V , the sphere has the smallest area. This is the isoperimetric inequality , a result wewill prove later. But surprisingly little is known about more bubbles. While the symmetric doublebubble shown below is the most economic way to enclose two equal volumes, no one knows if thesymmetric triple bubble is optimal for three equal volumes.Our comparative ignorance of bubbles will not stop us launching, undaunted, into the problemof partitioning not two, not three, but an infinite number of equal volumes. As a warm-up, we canconsider the problem for two-dimensional bubbles. We will show that in a large foam of bubbles,the ◦ rule means that most cells are hexagonal. This helps explain why bees prefer a hexagonallattice for building their hives. They are trying not to waste wax! In fact, the hexagonal tessellation,where each hexagon is identical, provably requires the least amount of wax per equal volume cell.In three dimensions, things are more interesting. In addition to the ◦ rule, we need a fewother rules for bubbles which together make up Plateau’s laws . Unlike two dimensions, these laws2on’t tell us precisely how many faces a bubble has, but they do give some constraints. We can usethese constraints to eliminate all but one space-filling pattern, the
Kelvin structure (above middle),made from pruned octahedra. Surprisingly, this is not the best way to separate an infinite numberof cells of equal volume. There is a mutant tessellation made from weaving together two differentequivoluminous shapes called the
Weaire-Phelan structure , shown above right. Although there areno four-dimensional bees to store their honey in Weaire and Phelan’s cells, Nature uses this structureto make superconductors and trap gas. No one knows if there is a way to beat it.
Let’s outline the contents a little more formally. In §2, we start our study of minimization with thesuprisingly rich problem of minimal networks on the triangle. In §2.1, we analyze the equilateraltriangle using symmetry, and argue that a hub should be placed in the center. We deform thissolution in §2.2, and give some loose arguments that the hub collides with a vertex when an internalangle opens to ◦ . This is generalized in §2.3 to give the ◦ rule for general minimal networks.Finally, in §2.4, we give a brief history of minimal networks and related problems.In §3, we use tools from graph theory to take the ◦ rule, which is a local constraint, andturn it into a global constraint on the structure of the network. Trees and their basic properties areintroduced in §3.1, and exploited in §3.2 to put a bound on the maximum number of hubs. Thisallows us to solve some small but nontrivial networks. In §3.3, the bound is turned into a roughargument for the computational hardness of finding minimal networks, while §3.4 provides someeasily computable alternatives, namely the minimal spanning tree and Steiner insertion heuristic.With §4, we move laterally into the realm of soap bubbles. We build soap bubble computers in§4.1 to solve our minimal network problems, where our computational hardness results resurface aspredictions about physics. In §4.2, we introduce Euler’s formula and apply it to bubble networks,while in §4.3, we make a simple scaling argument that most bubbles in a large foam are hexagonal.This is related to the fact that bees build hexagonal hives, and the honeycomb theorem that beesknow the best way to partition the plane into cells of equal size. The planar minimal bubble problem make its appearance in §4.4, along with a heuristic proof of the isoperimetric inequality.The last section, §5, considers three-dimensional bubbles. After defining mean curvature in§5.1, we state Plateau’s laws in §5.2, motivating them by analogy with bubble networks. With§5.3, we describe the three-dimensional bubble problem and Plateau’s problem for wireframes andbubble blowers. Finally, in §5.4 we generalize Euler’s formula to study network constraints on three-dimensional foams, and conclude with a whirlwind tour of regular tessellations of space, the Kelvinproblem, the Weiare-Phelan surprise, and the chemistry of tetrahedrally closed-packed structures. Prerequisites.
The only prerequisites for these notes are high school algebra, geometry anda little physics. You can do a lot of minimization without calculus! The material should thereforebe suitable for high school enrichment in math or physics, and parts of sections 2–3 have beensuccessfully trialled in a physics outreach program. We often resort to heuristics, pictures, andphysical intuition, which may deter some readers. But the price of admission is lower, and we hopethe rides no less fun!
Exercises.
There are around 40 problems of varying difficulty. Many of these are used sub-sequently in the text. I hope this is not a weakness, but rather than an incentive to solve them!Difficult exercises are labelled with a mountain ( (cid:21) ), or an icy mountain ( (cid:22) ) in the case of greaterabstraction or required background. Mountain ranges ( (cid:23) and (cid:24) ) inflect for length. For solutions,please contact me by email. They will hopefully be included in a future iteration.3
Trains and triangles
Suppose we want to join up three towns A , B and C by rail. Building railways is expensive, sincewe not only need to design and build the rail itself, but acquire the land beneath it. In contrast,stations can be reasonably cheap: we just slap together some sidings, a platform, and a bench ortwo. To minimize cost we should make the total length of the rail network as short as possible.If the railway lets us travel from one town to any other, we say that the rail network is connected .A hub is a station built solely to connect rails. A connected rail network of minimal length is calleda minimal network or Steiner tree . Two possible networks for the three towns are shown in Fig. 1.The “triangle” network is built from two sides of the triangle formed by the three towns, while the“trident” network adds a hub (also called a
Steiner point ) in the middle.Figure 1: Rail networks (triangle and trident) connecting three towns.
Exercise 2.1.
Choosing sides.
Suppose A , B and C are separated by distances AB , AC and BC . A triangularnetwork consists of two sides of the triangle. Which ones should we choose? Exercise 2.2.
Triangle or trident?
In Fig. 1, we have two networks connecting the same towns: two sides of thetriangle, and the trident-shaped network with a hub D in the middle. Check thetrident is shorter. Hint.
Measure lengths with a ruler. Simple but it works!Already, there is a surprise. Although the simplest network consists of two sides of the triangle,this is not minimal, since (to spoil Exercise 2.2) the trident in Fig. 1 is shorter. We can go furtherand optimize the placement of the hub D . The case for a general triangle is tricky, but we can buildmost of the intuition we need by focusing on the special case of an equilateral triangle. Suppose A , B and C sit on the corners of an equilateral triangle of side length d , as in Fig. 2. Thetriangular network has total length L Λ = 2 d . For the trident network on the right, we place the We will see where the term “tree” comes from in §3. D directly in the middle. Let’s trade our engineering for math hats, and find the length of thetrident network using trigonometry.Figure 2: Rail networks on an equilateral triangle. Exercise 2.3.
Equilateral trident.
Show that the length of the trident network is L Y = √ d. Since √ ≈ . < , the trident is shorter than the triangle.Although this beats the triangle network, it’s possible that placing D somewhere other than thecenter could make the network even shorter. But as it turns out, the center is optimal, and we canargue this from symmetry . We draw one of the triangle’s axes of symmetry in red in Fig. 3. Wecan wiggle the hub D left and right along the dark blue line in Fig. 3.Figure 3: Left.
Wiggling the hub.
Right.
Length is an even function of wiggle.Because of symmetry, the total length of the network (light blue lines) is an even function of howfar we have moved D along the dark blue line. On the right in Fig. 3, we depict two possibilitiesfor an even function. Length could either be a minimum on the red line, like the curve on top, or a maximum , like the curve on the bottom. Of course, if we move the hub along the blue line outside This cuts the triangle into two mirror-image halves. on the red line, andfor a minimal network it should lie on that line, as in Fig. 4 (left). But there are two other axes ofsymmetry, associated with B and C . All three intersect at the center of the triangle, as shown inFig. 4 (right). Since D should lie on each of these lines, it must lie at the center!Figure 4: Left.
Length is minimized on the red line.
Right.
Total length is minimized at theintersection of the red lines.
We are now going to take our solution to the equilateral triangle and slowly deform it, sliding thecorners so that the triangle so it is no longer equilateral. What will happen to the optimal positionof the hub D ? Since everything is sliding continuously, the optimal hub should slide continuouslyas well. In Fig. 5, we give an example, with the paths of the corners are depicted in purple, andthe corresponding continuous change of hub in green.Figure 5: Optimal hub position slides as we slide the corners of the triangle.Since the hub position changes continuously, it should stay inside the triangle for small defor-mations of the corners. But for triangles which are far from equilateral, the sliding hub might collide with a corner! In this case, the trident network collapses into a simpler triangular network,formed from two sides of the triangle. In Fig. 6, B remains fixed in position, but A and C lowersymmetrically and open out the angle of the triangle, with the optimal hub D moving verticallydown as they do so. At some critical angle θ crit , it will coincide with B . This does not prove it is a minimum, since there may be other minima we have missed. See Exercise 2.8 for arigorous proof. θ crit , D collides with B .Figure 7: Removing a corner city removes a leg from the equilateral trident.It turns out this critical angle is θ crit = 120 ◦ . Although we won’t provide a watertight proof justyet, we can give a plausibility argument. Let’s return to the equilateral triangle. Instead of addinga hub in the middle, suppose that D is in fact a fourth city fixed in place. Clearly, the solution inFig. 7 (left) is still optimal, since if we could add more hubs to reduce the total length, we couldadd more hubs to improve the network for the equilateral triangle. If we now remove a corner city,such as A , the optimal network removes the corresponding leg of the trident, as in Fig. 7 (right). Exercise 2.4.
Cutting corners.
Suppose that in Fig. 7 (right), we can add a new hub E which reduces the totallength of the network. Explain how adding E could reduce the length of the networkin Fig. 7 (left), and thereby improve our solution for the equilateral triangle.Exercise 2.4 is an example of a proof by contradiction , a favourite proof method among mathemati-cians. To show something is false, we assume it is true and use it to derive a contradiction withknown facts. We then reason backwards to conclude that it cannot be true! The next exercise givesa slightly stronger indication that the critical angle is ◦ . This is the best we can do withoutmore involved math (Exercises 2.9 and 2.8). Exercise 2.5.
Critical isosceles. (cid:21)
The argument above really only establishes that θ crit ≤ ◦ . In principle, thetriangular network might become optimal at some angle θ crit < ◦ . In thisexercise, we will show for an isosceles triangle that this is not the case. We willneed the law of cosines , c = a + b − ab cos θ, ABC , forming an angle of ◦ . We now raise the two nodes A and B symmetrically so that the angle ABC is less than ◦ . You can prove that the green and purple lines are shorter thanthe red lines, so that an interior hub D , making an angle ◦ with green andpurple lines, yields a shorter network.(a) Show using the law of cosines (or otherwise) that c = a + b + ab. (b) From part (a), argue that a + 2 b < c. (c) Conclude that for an isosceles triangle ABC , the critical angle is θ crit = 120 ◦ . ◦ rule Let’s state the general, n -city version of the problem we’ve been studying: Box 2.1.
Minimal networks.
Suppose we have n cities on the plane. The minimal network or Steiner tree is aconfiguration of edges connecting these cities which has minimal total length. Wecan introduce additional hubs in order to minimize this total length.Our work with triangles pays off with a remarkable conclusion about minimal networks connecting any number of cities called the ◦ rule . Readers who are not interested in the proof may simplyinternalize the the contents of the following blue box and move on. Box 2.2.
The ◦ rule. In a minimal network, every hub has three edges separated by angles of ◦ .The argument is ingenious. Our first step is to show that it is impossible for a hub to have edgesseparated by less than ◦ . Suppose we have cities or fixed nodes A , A , . . . , A n connected by aminimal network, and a hub station H with incoming rail lines separated by less than θ crit = 120 ◦ ,as on the left in Fig. 8. There may be other incoming lines, but these will play no role in our proof8nd can be ignored.Figure 8: Left.
A hub with incoming angle less than θ crit . Middle.
Adding two extra stations.
Right.
A shorter network.We can build two new stations on these outgoing legs, h and h , without changing the lengthof track. For simplicity, we take these new stations to be the same distance from H , as in Fig. 8(middle). But from our work in the previous section, we know that the minimal network connecting h , h and H is not the triangle network we have drawn! Instead, it is a trident with another hub h in the middle, Fig. 8 (right). This strictly decreases the length of the network, so our originalnetwork could not be truly minimal.This means that any hub must have spokes separated by at least ◦ . How do we know thatthere are three, separated by exactly ◦ ? Well, suppose two lines enter H , separated by more than ◦ . Then there can only be two incoming edges, joining H to some cities A and B , sinceany additional lines would have to be closer than ◦ to one of these lines. We have the situationdepicted on the left of Fig. 9.Figure 9: Left.
A hub with incoming angle greater than θ crit . Right.
A shorter network.Hopefully you can see what goes wrong: if there is a “kink” in the blue line, then we can obtaina shorter network be deleting H and directly connecting A and B . (Remember that H is a hub,introduced only to shorten the network, and not a city that needs to be connected.) Once again, wehave a contradiction! Strictly speaking, we can have hubs with only two incoming edges, separatedby ◦ . But such a hub is always unnecessary, since all it does is sit on a straight line. If we deletethese useless hubs, we have the general result advertised above, namely that any hub in a minimalnetwork has three equally spaced spokes. Exercise 2.6.
Outer rim.
Our proof applies to hubs only, but similar arguments apply to the cities A , A , . . . , A n . Prove the following:(a) No incoming edges can be separated by less than ◦ .(b) The number of incoming edges is between one and three.9 .4 A minimal history French mathematician
Pierre de Fermat (1607–1665) was the first to ask about minimal networkson the triangle, though he framed it as a geometric problem:
Box 2.3.
Fermat’s problem.
Given three points
A, B, C in the plane, find the point D such that the sum oflengths | DA | + | DB | + | DC | is minimal.He figured out the answer himself, but according to the mathematical custom of the day, sent a letterto Galileo’s student Evangelista Torricelli (1608–1647), challenging him to solve it. Torricellifound the same answer, but using a different method, so the position of the hub is called the
Fermat-Torricelli point in joint honor of its discoverers.
Jakob Steiner (1796–1863) generalized Fermat’squestion to n points on the plane: Box 2.4.
Steiner’s problem.
Given n points A , · · · , A n in the plane, find the point D such that the sum oflengths | DA | + · · · + | DA n | is minimal.Although minimal networks are also called Steiner trees, Steiner’s problem is very different from the n -city problem we’ve been considering. Steiner wanted a single point such that the sum of lengthsto that point is minimal, rather than a connected network of minimal length. Put differently, it isthe minimal network when you are allowed to add at most one hub.Figure 10: A visual history of minimal networks.10n 1836, 200 years later, the great German mathematician Carl Friedrich Gauss (1777–1855)mulled on the design of a minimal rail network between four German cities (Exercise 4.1). Aroundthe same time, the French mathematician
Joseph Diez Gergonne (1771–1859) considered thegeneral n city problem (connecting them via canals rather than railways) and discovered the ◦ rule. The world evidently paid no attention until 1934, when Czech mathematicians VojtěchJarník (1897–1970) and
Miloš Kössler (1884–1961) independently rediscovered Gergonne’s re-sults [22]. The Gergonne-Jarník-Kössler version was popularized under the name
Steiner trees by Richard Courant and
Herbert Robbins in their classic 1941 text,
What is Mathematics? [6].For a more in-depth history, see [4].We finish this section by finding the Steiner point and Steiner tree for regular polygons, atrigonometric construction of the Fermat-Torricelli point for the optimal hub placement (Exercise2.9), and a proof that the ◦ rule does indeed minimize total distance (Exercise 2.8). Exercise 2.7.
Easy polygons.
Consider n cities on the corners of a regular n -sided polygon.(a) Show that for n ≥ , the network formed by removing a single edge from theperimeter satisfies the ◦ rule and requirement (a) from Exercise 2.6. It’sharder to prove, but this is in fact the minimal network! a (b) Use the reasoning in §2.1 to argue that the center of the polygon solvesSteiner’s problem in Box 2.4. a You might wonder why this doesn’t follow immediately. As will explore in §3, and particularlyExercise 3.4, it turns out that satisfying these rules does not guarantee a network is minimal.
Exercise 2.8.
From straight lines to Steiner’s problem. (cid:24)
Here, we give a rigorous proof of the ◦ rule, and immediately extend it findthe analogous rule for Steiner’s problem. The proof makes use of vectors and thedot product, hence the higher difficulty rating. Recall that | v | is the length of thevector v , and ˆ v = v / | v | is the unit vector pointing in the same direction.Choose a point D on the plane, which will act as the “origin”. Consider anotherpoint, A , making a vector a = DA , with unit vector ˆ a .(a) Prove (visually or however you like) that for any other point X , with x = DX , | a | ≤ | a − x | + x · ˆ a , where as in the image above, x · ˆ a is the length of x projected onto a .11b) Consider two points A and B on the plane. Using the previous exercise, showthat for any point X , | DA | + | DB | ≤ | XA | + | XB | + x · (ˆ a + ˆ b ) . (c) Conclude that if we choose D so that ˆ a + ˆ b = , the sum | DA | + | DB | will beminimized. Geometrically, what does correspond to? Does this makes sense?(d) Let’s now introduce three points A, B, C on the plane, with origin D . Gen-eralize (c) to establish that | DA | + | DB | + | DC | is minimized when ˆ a + ˆ b + ˆ c = . Show that this is precisely the ◦ rule.(e) Finally, consider points A , . . . , A n and corresponding vectors a , . . . , a n .Generalize (d) to conclude that if a point D exists such that ˆ a + · · · + ˆ a n = , then it solves Steiner’s problem (Box 2.4).(f) Exploit (e) to solve Steiner’s problem for an arbitrary quadrilateral. Exercise 2.9.
Searching for Fermat-Torricelli. (cid:23)
Here, we give a geometric construction of the Fermat-Torricelli point for any tri-angle. Proceed if you like geometry! So, we’re going to find the interior pointsatisfying the ◦ rule for the blue triangle (below left). Start by attaching equi-lateral triangles (green, red, yellow) on each side, and drawing lines (dark blue)from the outer corners of the equilateral triangles to the opposite corner of ouroriginal triangle, as shown below right.We claim these lines intersect at the point f , and moreover, are separated by anglesof ◦ . To prove this, draw the dotted circles circumscribing each equilateral12riangle. The exercises guide you through a demonstration that the circles intersectat ◦ angles at f , using the inscribed angle theorem.(a) Show that the shaded triangles are congruent. Argue that, in consequence,the three blue lines do interesect at a single point.(b) From part (a), argue that ∠ baf = ∠ bcf .(c) From (b) and the inscribed angle theorem, argue that a, b, c, f lie on a circle.(d) Since the triangle is equilateral, ∠ cab = 60 ◦ . Using the inscribed angletheorem once more, show that ∠ cf b = 120 ◦ . Repeating this argument forthe remaining two triangles gives our result!This construction works provided all angles in the blue triangle are < ◦ .(e) What goes wrong if an angle is ≥ ◦ ?13 Graphs
In a sense, the ◦ rule solves the problem of minimal networks, giving us a mathematical conditionthat hubs must obey. But if I hand you a list of cities and tell you to start designing, you will quicklysee that the ◦ rule is not enough! In this section, we will think more about the layout of networks,including how many hubs we need to consider, the number of network arrangements, the generaldifficulty of finding these networks and methods for approximating them.Studying network layouts is the domain of graph theory . A graph is a bunch of dots connectedby lines, drawn on a page. The technical term for dots is vertices or nodes , and edges for the lines. Ifan edge joins two nodes, we say they are neighbours . Edges must start and end at different vertices,and are allowed to overlap. Vertices can be attached to any number of edges, including zero. Therules are illustrated in Fig. 11. We let E denote the number of edges and N the number of nodes.Figure 11: Left. “Illegal” and “legal” graphs.
Right.
The handshake lemma in action.We will need a simple, general result called the handshake lemma . There are two ways to countedges. The first is simply to count the edges directly, yielding a number E . But our rules tell usthat edges attach to a vertex at each end. So instead, we can go through the vertices and count thenumber of edges which attach to them. This will hit each edge twice , once for the vertex at eitherend, so this way of counting gives E . That’s the handshake lemma! More precisely, suppose thereare N vertices v , v , . . . , v N . If these have n , n , . . . , n N edges attached, the handshake lemmastates that n + n + · · · + n N = 2 E. (1)The name, incidentally, comes from the fact that if vertices v , . . . , v N are people, and edges arehandshakes, we add the number of handshakes each person performs to get twice the total numberof handshakes. Some cities are not joined by rail, say Minsk and Darwin. But in a connected rail network, thereis at least one route between each pair of cities. Anyone who has had the pleasure of exploringTokyo’s subway network will know the dizzying extent to which more than one route from A to B is possible. But in a genuinely minimal train network, A and B will be joined by a unique route.The basic idea is to get rid of routes until one is left. If there is more than one way to get from A to B , the network has unnecessary edges and can be “pruned” to get something shorter. Youmight worry that pruning these unnecessary edges could accidentally disconnect other cities, butthis is never the case! Fig. 12 shows why. Suppose A and B are connected by two paths, labelled p and p , and potentially consisting of more than one edge. The blob to the left is all the verticeswhose paths to B go through A first, and similarly, vertices on the right connect to A through B .14f two nodes are in the same blob, such as C and E , then pruning path p has no effect on whetherthey are connected. If two nodes are in different blobs, like C and D , they can still reach each otherusing path p . We can prune the redundant paths willy nilly!Figure 12: Pruning unnecessary paths. Exercise 3.1.
Pruning along the path.
Generalize the argument above to account for nodes that lie between A and B .(These are nodes which can connect to either A or B without passing through theother, and schematically lie on paths p or p .)Once we have completely pruned the network, there is only a single path connecting any twonodes A and B . Such a network is called a tree because it can be drawn so that edges look likebranches. This finally explains why minimal networks are also called Steiner trees ! An example ofa tree is shown in Fig. 13. Every tree has a special node called a leaf . As the name suggests, thisis at the “end” of the tree’s branches. More formally, a leaf is a node with a single edge, like J , D , G , and I in Fig. 13. It may seem intuitive, but as an exercise in reasoning about trees, let’s prove they must have leaves.Figure 13: A tree network, with a unique path between each node. Exercise 3.2.
Finding leaves. (cid:23)
To begin our proof that each tree has a leaf, we choose a node at random (red,below left) and count the number of steps to each other node.15a) Explain why the number of steps from the red node to any other node iswell-defined in a tree.(b) Consider the node or nodes furthest from the red node (orange, above left).Argue that these must be leaves.
Hint.
If they are not, what is the distancefrom red node to their neighbours?(c) In fact, we can prove something stronger. The previous question tells us howto find a leaf. Repeat the same procedure, but start with the leaf and findthe furthest node. Conclude that every tree (with at least two nodes) has two leaves.(d) Show, using an example, that a tree need not have more than two leaves.
In Fig. 13, you may have noticed the number of edges E = 9 is one less than the number of nodes, N = 10 . This is not a coincidence. For any tree, it turns out that E = N − . We can prove thisfact using the existence of leaves. The idea is simple: keep removing the leaf, and the single edgejoining it to the rest of the tree, until you have a single node left. This requires the removal of N − nodes, and hence N − edges. Since there are no edges now, and we removed one each time, wemust have started with N − edges. Hence, E = N − (2)for trees in general. Equation (2), along with the handshake lemma (1), will allow us to place a capon the maximum number of hubs that can occur in the network.Suppose we are trying to connect n cities, and introduce h hubs in order to do so. The totalnumber of nodes is then N = n + h . The ◦ rule tells that each hub attaches to exactly threeedges. Each of the n cities attaches to at least one edge to ensure it is connected to the rest of thenetwork. Thus, (1) gives E = n + n + · · · + n N ≥ n + 3 h. (3)From (2), we know that E = N − n + h − . Combining this with (3), we find n + h −
1) = 2 n + 2 h − ≥ n + 3 h = ⇒ n − ≥ h. (4)In other words, the number of hubs h is at most n − .16 xercise 3.3. Hubs and nubs.
While hubs always have three attached edges, Exercise 2.6 tells us that cities (fixednodes) have between one and three edges.(a) Show it is always possible to arrange n cities so that h = 0 .(b) At the other end of the spectrum, argue that the maximum number of hubs, h = n − , occurs when the fixed nodes are exactly the leaves of the network.The ◦ rule and hub cap together give us a simple tool for building minimal networks. For n fixed nodes, pick h = n − hubs, with spokes emerging at angles of ◦ , and connect them togetherto form a tree, with the fixed nodes as leaves. Although the angles are fixed, we can extend thespokes and legs, and perform overall rotations of the network. We call this extendable configurationof hubs and spokes a tinkertoy , after the modular children’s toy it vaguely resembles (Fig. 14). Figure 14:
Left.
A real Tinkertoy TM . Right.
A network tinkertoy.We can play with our network tinkertoys, or program a computer to play with them, until they dowhat we want. We give some examples in the following exercises.
Exercise 3.4.
Minimal rectangular network.
Consider four cities on a rectangle of height h and width w ≥ h :(a) Draw the single tinkertoy for n = 4 , and argue from Exercise 2.6 that thisshould describe the minimal network.(b) Fit the tinkertoy to the city, and deduce that the minimal network has length L = w + √ h. (c) Show that the tinkertoy can be oriented in two ways when h < w < √ h .Explain why the horizontal orientation is always minimal. In the mathematics literature, a tinkertoy graph is related to what are called
Steiner topologies . They are slightlydifferent, however, since the Steiner topologies are graphs which care about how they connect to fixed nodes. ◦ rule is not sufficientto guarantee that a network is minimal. Exercise 3.5.
Harder polygons.
Fit a tinkertoy (or three) to the following shapes; no need for exact placement.These networks are minimal, though it take a bit more work to show.
For a small number of hubs, tinkertoys are useful. But are they useful for many hubs? Suppose thatfiddling with tinkertoys is a quick operation, and once a tinkertoy is selected, a human or a computercan quickly check whether the tinkertoy can be extruded to hit our fixed points. If there are manytinkertoys, finding one that fits could still take a while. In Fig. 15, we show a few tinkertoys for h = 6 , suggesting that with more hubs, enumerating them all may turn out to be hard. In fact,as h gets larger, the total number of tinkertoys T h suffers what is called a combinatorial explosion ,growing exponentially as a function of h . A brute force approach, which simply fiddles with eachtinkertoy to see if it can be made to fit the fixed points, will take an exponential amount of time.This is beginning to seem like a hard problem in general! Counting the total number of tinkertoys is difficult. To demonstrate this exponential growth,we are instead going to focus on a subset of tinkertoys we can conveniently enumerate. Trees ingeneral have a complicated structure, so to simplify, we consider only linear tinkertoys. These aretinkertoys where the hubs lie on a “line”, so that no hub has more than two neighbours, for instanceFig. 16 (left). The next problem is that even these linear tinkertoys can be rotated by ◦ . Toavoid counting the same tinkertoy twice, we need some way of knowing which end is which. Asimple method is to start and end with a ∨ -shaped segment, as in Fig. 16 (right). If we rotate ◦ , the tinkertoy is bookended by ∧ -shaped segments, which is clearly distinct. Not every lineartinkertoy has this form, so we call these special tinkertoys oriented .With the notion of oriented tinkertoys, we can immediately find an exponentially growing set!There are h − edges altogether since the hubs form a tree. We fix four (two at each end) to ensure There is a subtlety here. If most tinkertoys can be made to fit, then this brute force approach will run quickly!At least, it runs quickly if fiddling is a quick operation. In reality, the best fiddling algorithms are exponential in n ,so the brute force approach remains exponential, irrespective of how many tinkertoys fit. While I’m not sure howmany fit in general, for the purposes of our heuristic approach, we’ll continue to assume the list is small. h − edges within the grey circle of Fig. 16 (right). As wemove along from the leftmost ∨ , these edges constitute h − turns left or right by ◦ before weexit again to hit the final ∨ . At each point, either a left or a right turn is allowed, so there are h − possible choices altogether. To make this more transparent, we could label left and right turns with s and s respectively, so that a tinkertoy is just a sequence of binary digits, as in Fig. 17. Thus,there are an exponential number of oriented tinkertoys. If you like drawing graphs, you can havea go at findiing the total number T h in the next exercise.Figure 15: A selection of tinkertoys for h = 6 .Figure 16: Left.
A linear tinkertoy.
Right.
An oriented tinkertoy.
Exercise 3.6.
Physicist’s induction. (cid:22)
Calculate the number of tinkertoys T h from h = 0 to h = 6 . You should beable to find the general sequence T h by searching for these numbers in the OnlineEncyclopedia of Integer Sequences. At large h , the OEIS informs us that this You might worry that if we turn too many times, the tinkertoy will collide with itself and no longer be valid. Forinstance, after six right turns, edges of equal length will form a closed hexagon! But we can always adjust the lengthof edges to prevent this from happening, so the count remains valid. T h ≈ h − √ πh / . If we count how the tinkertoys connect to the fixed nodes (“Steiner topologies”),there are dramatically more arrangements: ¯ T h = (2 h )! / h h ! to be precise! a a This can be proved using mathematical induction , rather than the physicist’s induction we’veused here. We leave this as a bonus exercise to the mathematically inducted (ahem).
By now, we should be confident that there are many tinkertoys. If we have to consider even afraction of them at large h , any computer is doomed to failure. For instance, using the countingin Exercise 3.6, suppose a computer can check a billion tinkertoys per second, and wants to designa railway network to connect the ∼ largest cities in North America. If it has to check everytinkertoy, it will take an unimaginably long · − √ π / · s ≈ years . Would a faster computer help? Not likely. If you do more operations per second than there areatoms in the universe, it still takes ∼ years! No realistic improvements in processing speedwill make this problem solvable, unless we find a much much better algorithm. As we’ll discussbelow, most computer scientists think no such algorithm exists, but can’t prove it!Notice that there are two slightly distinct problems here. The first is searching for tinkertoys thatfit; and the second is singling out the truly minimal network from the shortlist of fitting tinkertoys.The two are not the same because, as we saw in Exercise 3.4, just because a tinkertoy fits doesn’tmean it is minimal. The first problem is easier because if somebody hands you a tinkertoy andclaims it fits, you can easily check. In fact, you yourself could make a lucky guess and find atinkertoy which fits immediately. There is an area of computer science called complexity theory which classifies problems according to how hard they are. In the language of complexity theory,finding tinkertoys that fit is called NP , for “ N ondeterministic P olynomial time”. This is a fancy wayof saying you can make a lucky guess and confirm it immediately.In fact, fitting tinkertoys is as hard as any problem in the set NP . “As hard as” is a technical termin complexity theory, meaning that you can transform any algorithm for finding good tinkertoysinto an algorithm for solving any other problem in NP ! It is a key that unlocks the rest of the set.20e call such a task NP-complete , since it gives us “complete” access to every NP problem. Now, ifsomeone hands you a tinkertoy configuration and claims that it’s the minimal network, you mustfirst check that it fits. So finding a minimal network is at least as hard as fitting a tinkertoy. Butyou can’t stop there! You have to keep searching to find all the tinkertoys that fit, checking thelengths, and verifying that the first configuration really is the shortest. The second problem istherefore at least as hard as fitting tinkertoys. This places it in a class called NP-hard [15], which iscomplexity-ese for “as hard as any problem in
NP” . Exercise 3.7.
Tiny tinkertoys.
We’ve been talking about fitting a single tinkertoy, but as we saw in Exercise 3.5,the minimal network is sometimes obtained by cobbling together multiple “tiny”tinkertoys. Argue that including tiny tinkertoys makes finding minimal networksharder, but the problem of fitting potentially easier.
Box 3.1.
Complexity I.
Fitting tinkertoys is
NP-complete . Finding minimal networks is
NP-hard . All these heavy-sounding results about complexity theory make life sound impossible for networkplanners. But while finding the exact minimal network is difficult, approximating is easy! Life,and near-optimal rail travel, go on. We’ll discuss two simple approximation schemes, starting witha generalization of the very first Exercise 2.1. Recall that, for three cities, the triangle networkconsists of the two shortest sides of the triangle. Put differently, we draw an edge between eachcity, and select the two shortest ones, which happen to form a tree which connects everything.For n cities, we do the same thing. Draw an edge between each city, forming what is called the complete graph on n nodes. From these edges, we select a subset which form a tree, connecting eachcity and of minimum total length. This is called a minimum spanning tree (MST) , since it “spans”the cities. We illustrate the construction for n = 4 in Fig. 18.Figure 18: Left.
The complete graph for four cities.
Right.
The minimum spanning tree. Note that if we can quickly fit tinkertoys, we can quickly find the minimal network. So while finding minimalnetworks is hard, it’s only marginally harder than
NP-complete , and if P = NP , finding minimal networks is also in P ! I thank Scott Aaronson for pointing this out. v .1. Add the shortest edge adjacent to v to form a tree T . k ≥ . Add the shortest edge adjacent to T k to form a tree T k +1 .Repeat the last step until we have a tree T n − which spans all the nodes. This algorithm wasdiscovered in 1930 by Jarník [21], but subsequently rediscovered by Robert Prim in 1957 [24], soit is called the
Prim-Jarník algorithm . We implement it for n = 4 in Fig. 19.Figure 19: The Prim-Jarník algorithm for n = 4 . Dark blue edges are added sequentially. Exercise 3.8.
MST is easy.
Here, we will give a very lazy bound on the number of steps required to performthe Prim-Jarník algorithm.(a) Using the handshake lemma (1), show the total number of edges in the com-plete graph on n cities is E complete = n ( n + 1) / .(b) The algorithm has n − steps where it adds an edge. For each step, it mustconsider the available edges. Call this a substep . Give a very lazy argumentthat the total number of substeps for the algorithm is ≤ n .This is a polynomial function of n , rather than an exponential function of n . Exercise 3.9.
Correctness of Prim-Jarník. (cid:22)
Suppose that the Prim-Jarník algorithm produces a tree T which is not minimal,with T (cid:48) (cid:54) = T the genuine MST. Then there must be a step in the constructionwhere we first add an edge e which is not in T (cid:48) . We will show that the algorithmis correct in the sense that this situation cannot occur! There will always be ashorter edge e (cid:48) it should add instead of e . The setup is shown below. By “adjacent to”, we just mean an edge which touches the tree but is not already in it. e , it spans a set ofcities V . The complementary set of cities is ¯ V . Show that e connects a vertex v ∈ V to a vertex ¯ v ∈ ¯ V .(b) Argue that the MST T (cid:48) has an edge e (cid:48) connecting V to ¯ V . Hint.
Use thefact that there is a path from v to ¯ v in T (cid:48) .(c) Explain why removing e (cid:48) from T (cid:48) , and replacing it with e , results in a tree . Hint.
Show there is still exactly one route between any two nodes.(d) From part (c), conclude that the Prim-Jarník algorithm is correct.Finding MSTs is quick. But are they any good, or can they be much longer than the minimalnetwork? Once again, our simple results on triangles provide some insight. Let’s start with anequilateral triangle of side length d . In Exercise 2.3, you found that the minimal network has length L Y = √ d . The MST for the equilateral triangle just consists of any two sides, and therefore haslength L Λ = 2 d . The ratio of these two lengths is ρ = L Λ /L Y = 2 / √ ≈ . , so the MST is about longer than the Steiner tree. This is close enough for many practical purposes.You might wonder, in general, how bad this ratio can get. To start with, let’s see what happenswhen we squeeze or stretch the triangle symmetrically. If we squeeze it, like Fig. 20 (left), the MSTconsists of a long side of length d and the short side which shrinks to zero. Similarly, the minimalnetwork consists of two short sides which approach zero, and a long side which approaches d . Sothe ratio of lengths approaches . Similarly, as we stretch the triangle out like Fig. 20 (right), theMST is the shorter two sides at the top, of total length d , while the hub eventually hits the topvertex, so it coincides with the MST. Once again, the ratio approaches .Figure 20: Stretching and squeezing the equilateral triangle.This hints that the equilateral triangle is the worst-case scenario. In fact, you can show inExercise 3.11 that this ratio is at most / √ for any triangle. In Gilbert and
Pollak ’s magisterial23tudy [16], they conjecture that this holds for any number of cities! In other words, if ρ is the ratioof the length of the MST to the minimal network for any given set of cities, the Gilbert-Pollakconjecture states that ρ ≤ √ . (5)The conjecture remains unproven. The best we can do right now is ρ ≤ . [5].What if we want to do better than ? We can tweak the MST a little to get closer to theoptimal network length. One particularly simple method is the Steiner insertion heuristic [10],which elegantly combines the MST and our work with triangles. The basic observation is that noedges in a minimal network are separated by less than ◦ , since hubs always have edges separatedby exactly ◦ , and edges at fixed nodes must be separated by at least ◦ according to Exercise2.6(a). The idea is to find edges with “bad” angles ( < ◦ ) and replace them with hubs.In more detail, the insertion heuristic works as follows. We first find the MST (using Prim-Jarníkor another quick procedure), and then search for the pair of edges with the smallest angle < ◦ .If no such angle exists, we are done! It such an angle does exist, the two edges connect a vertex,say A , to vertices B and C , as below in Fig. 21. We introduce a hub for these three vertices, whichsatisfies the ◦ rule. And then we do the whole thing again, looking for bad angles to replace,until no more are left. That’s it!Figure 21: Applying the Steiner insertion heuristic to our MST. First, we find the smallest angle < ◦ . Then, add a hub. No more bad angles, so we’re done! Exercise 3.10.
Steiner insertion heuristic.
Let’s explore some general properties of the insertion algorithm.(a) Argue that the insertion of a hub can only decrease length.(b) Give an example showing that the insertion heuristic need not converge tothe globally minimal network.
Hint.
Exercise 3.4(c).(c) Remember our earlier statement that fitting a tinkertoy to a set of fixed nodesis
NP-complete . Explain why the Steiner heuristic can run quickly withoutcontradicting this result.
Hint.
Exercise 3.7.Although Steiner insertion is quick, the optimality varies. A different and less practicalmethod[3] shows that, in principle, you can approximate the minimal network on n cities as closely as youlike , in some number of steps at most polynomial in n . For this reason, minimal networks belong24o a complexity class called PTAS (“ P olynomial T ime A pproximation S cheme”), the problems whichcan be easily approximated. We can update our statement about complexity: Box 3.2.
Complexity II.
Finding minimal networks is
NP-hard but also
PTAS . Exercise 3.11.
Gilbert-Pollak for triangles. (cid:21)
Below, we give a visual proof of the Gilbert-Pollak conjecture for triangles. Thebasic idea is that, in an arbitrary triangle with angles ≤ ◦ , we can attach asmall equilateral triangle to the largest angle (city A below).The lengths L , L , L , L are made up of lengths of coloured lines, but blue lineshave weight , while orange lines have a weight √ / . For instance, L = | DA | + | DB | + | DC | , L = √
32 ( | AC | + | BC | ) . In other words, L is the length of the minimal network, and L is √ / times thelength of the MST.(a) Argue that L ≤ L ≤ L ≤ L .(b) Use this (along with the case where some internal angle is ≥ ◦ ) to establishthe Gilbert-Pollak conjecture for triangles.25 Bubble networks
Humans are not the only players in the minimization game. Nature is also cheap, or rather lazy :it does as little as possible, formally known as the Principle of Least Action. If we play our cardsright, perhaps we can hack the laws of physics to do our minimization for us. In our case, it turnsout we can do network planning with bubbles . Bubbles are formed when a film of liquid separatestwo volumes of air. Surface tension tries to pull the bubble surface taut in all directions, whichresults in the minimization of area . But if there are no constraints, then the surface will shrink untilnothing is left! Really, we mean that bubbles minimize the area of the wall subject to constraints .For building railway networks, we want walls to be one-dimensional, and the contraints to befixed external nodes. We’ll talk about how to do this in a moment, but there is a more naturalconstraint associated with blowing bubbles: they enclose a pocket of air. This explains why soapbubbles are spheres! As we will show in §5.3, a sphere (Fig. 22 (left)) is the smallest surfacecontaining a fixed volume of air. A lone bubble is direct proof of Nature’s laziness.Figure 22: Soap bubbles in three and two dimensions.If we sandwich the bubbles between two plexiglass plates, we will get a two-dimensional networkof bubbles. The vertical walls look like a graph from above, and a single bubble will be a circle (Fig.22 (right)). There are two questions about these networks that immediately present themselves.First, what happens at a junction of bubble walls? And second, what do walls look like away froma junction? The first question is easy to answer. Imagine zooming in on a junction until the walls look straight . Since the bubbles try to minimize wall area, or viewed from above, wall length , theywill obey the ◦ rule, since this is the local rule any length-minimizing network obeys! The situation away from junctions is a little trickier, but as we will see, for both physical(Exercise 4.7) and mathematical reasons (§4.4), a bubble wall can either be straight, or it can curvealong the arc of a circle. Viewing a straight line as the arc of an infinitely large circle, we can justsay that walls are arcs of circles.
Plexiglass gives us two-dimensional bubbles, and length rather than surface area will be minimized.But the constraint will generally be to enclose a fixed area of air per cell. Can we hack this setupto make a soap bubble computer for finding minimal networks? Yes! The key is to give the bubblewalls something to hold onto. If we drill some screws between the plexiglass plates, these will act Zooming in enough means that edges can be reconfigured without having any practical effect on air enclosed. Fig. 23 shows an example withfour screws, and the junctions that can form between bubble walls.Figure 23: A soap bubble computer for finding minimal networks.You can use a soap bubble computer to solve the original railway planning problem.
Exercise 4.1.
Railways and soap bubbles.
As advertised in §2.4, the mathematician Gauss wanted to connect four cities witha minimal rail network. In an 1836 letter to his friend, the astronomer
HeinrichSchuhmacher (1780–1850), Gauss asked:
How does a railway network of minimal length connect the four Germancities of Bremen, Harburg, Hannover, and Braunschweig?
The cities are drawn, along with their GPS coordinates, below:(a) Find the minimum spanning tree using the Prim-Jarník algorithm.(b) Assuming Gilbert-Pollak, lower bound the length of the minimal network.(c) Improve the MST using the Steiner insertion heuristic.(d) Build a soap bubble computer and solve the Gauss’ railway problem. Howdoes this compare to the results of the Steiner insertion heuristic? For a programmable soap bubble computer, you can use suction cups with rods between them. Thanks to PedroLopes for pointing this out.
NP-complete , and finding thegenuine minimal network is
NP-hard . Both problems are at least as hard as everything in NP , theclass of problems where lucky guesses can be checked quickly. But just because a lucky guess can bechecked quickly does not mean your chances of making a lucky guess are good. In fact, computerscientists are almost certain that most problems in NP cannot be solved quickly on a regular digitalcomputer. The set of problems which can be solved quickly is called P , for “ P olynomial time”. Tosummarize, computer scientists believe that P (cid:54) = NP through proving it is the most important openproblem in computer science. But, you might object, a soap bubble is not a regular digital computer; it is built out of the lawsof physics rather than 1s and 0s. Could it do things quickly that would take a digital computer longerthan the age of the universe? The answer is probably no. Computer scientist
Scott Aaronson hypothesized [1] that the problems in
NP-complete (and hence
NP-hard ) cannot be solved quicklyby any computer, digital or analogue. This is called the NP Hardness Assumption .One piece of evidence is that every time we think we have a loophole for quickly solving
NP-complete problems, the loophole disappears on closer examination. The devil is in the details! Butthere is broader philosophical reason for believing NP Hardness: roughly, NP is OP . Many of thehardest problems we know are
NP-complete , and if we could solve them, then as Aaronson says, . . . we would be almost like gods. The NP Hardness Assumption is the belief that suchpower will be forever beyond our reach.
This means we cannot quickly find the minimal rail network for 800 cities using soap bubbles, ablack hole, human DNA, a quantum computer, or any other conceivable mechanism. No one willever know what the network looks like.That raises the question: what do soap bubbles actually do? They cannot quickly find minimalnetworks, since this problem is potentially even harder than NP . But there are several ways for thisto fail. First, they can take a long time to settle down, which Aaronson saw happening in his ownsoap bubble experiments, even for a few screws [1]. Secondly, they could relax into local minima rather than the true minima. Since fitting tinkertoys is NP-complete , even this can take a long time,unless (like the Steiner insertion heuristic) the tinkertoys are small. A final possibility is thatwe simply solve the wrong problem, e.g. by introducing small bubbles which change the networkconfiguration. Based on my own experiments with soap bubble computers, it appears that all ofthese failure modes can be realized!I am not trying to skewer soap bubbles. Indeed, the rest of these notes are really just a love letterto their physico-mathematical beauty. Rather, the moral is that physics and computation interactin interesting ways , with results about computation leading to physical predictions (see Exercise 4.2for some non-bubbly examples). Going in the other direction, physics can lead to new insights intocomputer science, the most spectacular example being quantum computers . These are machinesbased on the laws of quantum mechanics rather than the classical logic of 1s and 0s. Although intheir infancy, thinking about quantum computers has already taught us some remarkable things So important that there is a $1 million bounty on its head! Gamer speak for “overpowered”. See Exercise 3.7 for more on this subtlety.
Exercise 4.2. NP Hardness and the laws of physics.
Here are a few fun ways to solve
NP-complete problems:(a) Create a time machine, and by sending a computer through it again andagain, perform an arbitrary number of computations in finite time.(b) Build a “Zeno hypercomputer”, performing one step in / s, the second stepin / s, the third step in / s, etc., so an infinite steps take second.(c) Store information in infinite precision real numbers, e.g. points on a line, andmanipulate them using basic arithmetic [26].If the NP Hardness Assumption is correct, none of these methods works! In eachcase, what do you think this is telling us about the nature of the universe?
While we can use soap bubbles to learn about minimal networks, we can arguably obtain moreinsight by going in the other direction. What do minimal networks teach us about soap bubbles?In this section, we consider the two-dimensional bubble networks with no screws . We’ll just let thebubbles do their own thing! Fig. 24 shows a real two-dimensional soap foam. We’ve counted thenumber of sides per cell, and surprisingly, most seem to be hexagonal. Is this is a coincidence, or issomething deep going on?Figure 24: Most cells in a bubble network are hexagonal.The answer is something deep. We can actually prove most bubbles are hexagonal using the ◦ Based on a photograph by Klaus-Dieter Keller, Wikimedia Commons.
Euler’s formula , discovered by the prolific Swiss mathematician
Leonhard Euler (1707–1783)in 1735. It states a relationship between the number of nodes N , edges E , and faces F in a graph,proved below in Exercise 4.3: N − E + F = 2 . (6)Importantly, this only holds for connected graphs which can be drawn without any edges crossing,also called planar graphs (Fig. 25). A face is defined as any region enclosed by a loop of edges,including (counterintuitively at first) the exterior of the graph.Figure 25: Left.
A disconnected graph which cannot be drawn without edge crossings.
Right . Aplanar graph. Euler’s formula holds if we count the region outside the graph as a face.One way to obtain a planar graph is to take a three-dimensional polyhedron, remove a singleface, and flatten what remains onto the plane. This flattening process is shown for the cube in Fig.26. The removed face becomes the exterior region of the graph, which is why we count it as a face.Figure 26: Remove the top of the cube and flatten.Let’s check that Euler’s formula works. For the cube, we have F = 6 faces, E = 12 edges, and N = 8 corners, so F − E + N = 2 just as Euler predicts. We can count either using the cube itself,or the flattened graph, provided we count the exterior as a face. Exercise 4.3.
Euler’s formula.
Define the
Euler characteristic χ = N − E + F. Our goal will be to show χ = 2 for a graph without crossings. First, we willestablish Euler’s formula for networks made out of triangles. We can then extendthis to any graph without crossings. Below we depict stages (a), (b), (c) and (e).30a) Show that a lone triangle in the plane obeys Euler’s formula.(b) Suppose a network obeys Euler’s formula. Add a triangle (two edges and anode) to an external edge, and explain why the Euler characteristic doesn’tchange, ∆ χ = 0 . Conclude that the new network obeys Euler’s formula.(c) Explain why a network composed of triangles obeys Euler’s formula.Now we can generalize to any network without crossings.(d) Consider a face, i.e. loop of edges, in such a network. Describe a procedureto add edges so that the face is split into triangles.(e) Show that, after your procedure in part (d), ∆ χ = 0 .(f) Conclude that any network without crossings obeys χ = 2 .When there are no screws, every node in the bubble network is a hub, and therefore obeysthe ◦ rule, with three bubble walls meeting. By the handshake lemma (1), we have E = 3 N .Putting this into Euler’s formula, we can eliminate N and find a relation between the number offaces and number of edges: N − E + F = E − E + F = 2 = ⇒ F − E = 6 . (7)It will be useful to treat the external face a little differently. Let F (cid:48) be the number of internal faces,so that F = F (cid:48) + 1 . Then (7) becomes F (cid:48) − E = 3 .Before proceeding, we need two additional properties of our bubble networks. First of all, anedge cannot dangle into the middle of a face. If it did, the vertex at the end of the dangling edgewould not have three attached edges, only one, which is impossible by the ◦ rule. It follows thatevery edge straddles two distinct faces. Let F s denote the number of internal faces with s sides, and let E b stand for the number ofedges of the outer face of the collection of bubbles. The total number of internal faces is F (cid:48) = F + F + F + · · · . (8)But since each edge is associated with two faces, we can also express edges as E = E b + 1 · F + 2 · F + · · · + s · F s + · · · . (9)If we plug (8) and (9) into F (cid:48) − E = 3 , we finally get E b = 6 F (cid:48) − E + E b = (6 − · F + (6 − · F + · · · + (6 − s ) · F s + · · · . Counterexamples like an edge cutting across two concentric circles are also ruled out by the ◦ rule.
31e will call the RHS the hexagonal difference D hex , since it counts the number of edges which donot belong to a hexagonal face, with a sign depending on whether the face is smaller ( + ) or larger( − ) than a hexagon. So, more simply, we have D hex = 6 + E b . (10)The hexagonal difference is more than the number of boundary edges. We give a few simpleexamples in Fig. 27, with the contribution to D hex indicated in each cell.Figure 27: D hex , the sum of numbers in cells, is always more than E b . Exercise 4.4.
Large and small faces.
Equation (10) already tells us some interesting things about bubble networks.(a) Explain why D hex ≥ .(b) Deduce that · F + 4 · F + · · · + 1 · F ≥ · F + 2 · F + · · · . (c) Suppose a bubble network has two bubbles with four sides and no other smallfaces. What is the maximum number of -sided bubbles?In general, once we count the “small” faces F , . . . , F , we can constrain the numberof “large” faces F , F , . . . . It’s still not clear why most bubbles are hexagonal. At this point, we need to introduce some basicphysical intuition. Suppose the foam has overall size ∼ L . Assuming bubbles have a typical sizeindependent of L , the number of external edges E b ∼ L . The total area of the bubble networkshould scale as A ∼ L . For instance, consider a roughly circular foam of radius L (Fig. 28). Ifbubbles have average edge length (cid:96) , independent of L , then E b ≈ ( π/(cid:96) ) L , while A = πL .32igure 28: As the foam gets large, the number of outer edges scales as L , and the area as L .It follows that, for large L , the hexagonal difference D hex = 6 + E b ∼ L . The “density” ofnon-hexagonal edges d hex is just the total hexagonal difference divided by the area of the foam.Since area scales as L , the density of non-hexagonal edges scales as d hex ∼ D hex L ∼ L . (11)As L becomes larger, edges belonging to non-hexagons become increasingly rare. This explains whya typical cell in a bubble network has six sides, just like Fig. 24. Exercise 4.5.
Bubble blobs. A bubble blob is a set of contiguous bubbles in a bubble network. Let E o denotethe number of edges extending outward from the boundary, and E i the numberextending inward.(a) Explain why the difference from hexagonality is now given by D hex = 6 + E i − E o . (12)(b) Verify that the blob of cells in Fig. 24 satisfies (12).(c) Repeat the scaling argument above, and conclude that in a large blob, de-partures from hexagonality become rare.You may have wondered if the hexagonality of bubbles is related to the fact that bees buildhoneycombs in a hexagonal lattice. It is! Bees have a clear evolutionary reason to minimize theamount of wax used. Charles Darwin (1809–1882) discusses the hive-making instinct and itsrelation to fitness in his
Origin of Species [7]:
That motive power of the process of natural selection having been economy of wax; thatindividual swarm that wasted least honey in the secretion of wax, having succeeded best.
For honeycomb walls to be minimal, they must obey the ◦ rule. If honeycomb cells are equalin size (which bees might prefer for simplicity of construction), then a natural guess at the optimalarrangement is the hexagonal lattice . This is the only regular tessellation satisfying the ◦ rule.33igure 29: The three regular tessellations of the plane: square, triangle, hexagon.The honeycomb conjecture states that the hexagonal lattice is the globally minimal solutionamong tessellations of the plane with equal cell size. It is hard to verify this guess, since youneed to check all possible irregular tilings as well as the regular ones. But in 1999, it turned fromconjecture into theorem after Thomas Hales gave a formal proof [18]. In Exercise 4.6, we explorethe analogous problem for the saddle-shaped hyperbolic plane.
Exercise 4.6.
Hyperbolic honeycomb. (cid:24)
Our scaling argument assumed we were on a regular, Euclidean plane. But we cansee what happens if, instead of working on the Euclidean plane, we work on thestrangely curved hyperbolic plane .Above, we have tiled the hyperbolic plane with heptagons. Each heptagon has thesame area, and sides of equal length, but the curvature means they must be drawnwith different lengths on our flat page!(a) Find A (in heptagon units) and E b for the regions enclosed in (i) green, (ii)purple, (iii) blue. Does the ratio E b /A appear to be decreasing?(b) Argue that, in general, for n > “rings” of heptagons, E b = 4 · n , A = 16 (7 n +1 − ≈ · n . Hint.
Use a geometric sum for A .This shows that on the hyperbolic plane, the scaling E b ∼ L , A ∼ L no longerholds. Instead, the boundary and area scale the same way .(c) Why does the ◦ rule still hold for minimal networks on the hyperbolicplane? Hint.
What happens when you zoom in on a node?34d) Show that for ring n , E o − E i = 2 · n . Using part (c) and similar reasoningto the plane, conclude that a large number of heptagonal rings, D hex = 6 − · n ≈ − · n . (e) Finally, show that our heptagonal tiling has d hex = D hex A ≈ − . (f) Given Exercise 4.4(a), how can D hex be negative?The weird properties of hyperbolic space mean that the optimal tessellation depends on thesize of the cells. The heptagonal tiling is optimal for the cell size pictured above, at least among regular hyperbolic tilings [8]. The “hyperbolic honeycomb conjecture”—that this is optimal among all hyperbolic tessellations with this cell size, including the irregular ones—remains open. Perhapswe should breed some hyperbolic bees, and inspect their honeycomb in a few million years!
The preceding two sections studied two-dimensional bubble foams, assuming there were no fixednodes. The total length is being minimized, but subject to what constraints? The answer issuggested by our earlier discussion of air pockets, and by the honeycomb conjecture. The bees haveno fixed nodes, since they are not trying to connect anything. Instead, they are trying to buildcells to store honey. To simplify the problem, we have considered an infinite number of cells of thesame size, but what if the bees only want six? Or want to vary their serving sizes with cells ofdifferent area? In general, we can ask for the minimal length bubble network enclosing cells of size A , A , . . . , A n .Figure 30: Left.
The standard double bubble.
Middle.
An empty pocket.
Right.
A split bubble.In the same way that we are allowed to add nodes to minimal networks to decrease length, wewill allow empty pockets and bubble splitting (Fig. 30) if it helps us reduces length. We will alwaysask that the bubble network is connected for physical reasons. This leads to...
Box 4.1.
The Planar Minimal Bubble Problem.
Find the connected bubble network of smallest perimeter enclosing cells of area As far as I know, the problem is open for all cell sizes. If the network is not connected, the disconnected parts can “float” relative to each and will soon collide, forminga connected network. , A , . . . , A n , allowing empty pockets and split bubbles.While we derived the ◦ rule directly from minimizing length, the other salient property of bubblenetworks is that walls are straight or arcs of circles. You can see where this comes from using thephysics of surface tension. Exercise 4.7.
Young-Laplace I.
The molecules in a bubble wall are attracted to each other. If you try to bend thesurface, it strains the molecular bonds, which attempt to restore the unstretchedstate. The amount of bending at a point can be quantified by finding a circle whichfits snugly onto the curve (the green circle, below right).If this snug circle has radius R , we say the bend has radius of curvature R . For abubble wall of height L , the restoring force per unit length of curve is f = 2 σL/R ,where σ is the surface tension .(a) Show that if there are no other forces acting on the wall, it must be straight.(b) Now consider the effects of pressure at a point on the wall. If the pressure onone side is P out , and inside is P in , argue the bend will have radius of curvature R = 2 σ ∆ P , (13)where ∆ P = P out − P in . This is called the Young-Laplace law , after
ThomasYoung (1773–1829) and
Pierre-Simon Laplace (1749–1827). Check it isconsistent with (a).(c) Within a cell, pressure differences equalize very quickly, so it is reasonableto assume pressure is constant on a face of the network. Deduce that bubblewalls are either flat or arcs of circles.Although Exercise 4.7 does involve surface tension, it says nothing about minimizing surfacearea or solving the bubble configuration problem. It seems plausible that real bubbles do solve thisproblem, but for the moment, let us view the bubble configurations as physical conjectures aboutminimal-length solutions. In other words, they are guesses made by Nature, awaiting the rubberstamp of mathematical proof.The simplest physical conjecture is for a single bubble of fixed area A . The only smooth wayto draw a single cell is a circle (Fig. 32 (left)), and when Nature is left to its own devices, bubbles36end to assume this form. You can also check (Exercise 4.8) that there is no way to split the singlebubble, or introduce air pockets, while maintaining a connected bubble network. Exercise 4.8.
One bubble to rule them all. (a) Show that, if we split a bubble into parts which contain a total area A , theycannot share any edges. Explain why the same goes for an empty air pocketand the region outside the bubble configuration.(b) Argue that splitting a single bubble, or adding empty pockets, violates (a).Mathematicians as far back as Archimedes (287–212 BC) have suggested that the circle is theshape of smallest perimeter for a fixed area A , a guess called the isoperimetric inequality . Thisguess wasn’t verified until the 19th century, but the modern proof is simple enough to present inoutline. The basic idea is to wobble a line and see how the length and enclosed area change. Tostart with, we consider wobbling the radius of a single circular arc.
Exercise 4.9.
Stretched arcs.
Suppose an arc of length L and radius R is part of a curve enclosing some areaon the plane. Consider extending the radius by a small amount t , where “small”means much smaller than L .(a) Show that the area enclosed changes by ∆ A ≈ Lt. (14)(b) Assuming that the angle subtended by the arc is the same, explain why thelength of the arc changes by ∆ L ≈ LtR . (15)(c) Check that (b) still makes sense for a straight line.For both ∆ A and ∆ L , there are some additional corrections, but these will appearas higher powers of t , starting at t .In general, we can take a curve on the plane and chop it up into k small pieces of length Saying the circle has the least perimeter of all figures of area A = πr is the same as saying it has most area ofall figures with the same perimeter πr . “Isoperimetric” means “same perimeter”. , L , . . . , L k and constant radius of curvature R , R , . . . , R k , setting R i = ∞ for any straightlines. Imagine we wobble the curve by independently changing the radii for each segment, adding t , t , . . . , t k . Using (14), the total change in area is ∆ A = ∆ A + ∆ A + · · · + ∆ A k = L t + L t + · · · + L k t k . (16)From (15), the total change in length is ∆ L = ∆ L + ∆ L + · · · + ∆ L k = L t R + L t R + · · · + L k t k R k . (17)It’s clear that if we deform the curve so as to preserve area, then ∆ A = 0 .But here is the clever part: if the curve is a local minimum of perimeter, then the perimeter lookslike a quadratic function of the wobbling. But we are making t small enough that we can ignorethese quadratic t terms, and keep only the terms proportional to t . Thus, in the approximation wehave used to compute (17), a perimeter-minimizing curve has ∆ L = 0 . You can show in the nextexercise that, given the forms for ∆ A and ∆ L , this is only possible if R = R = · · · = R k = R, i.e. the radius of curvature is constant. Thus, the perimeter-minimizing curve has constant R . Exercise 4.10.
Constant radius of curvature.
If we vary a perimeter-minimizing curve, then ∆ L = 0 . If the wobbles also preservearea, then ∆ A = 0 . We will show that this implies all the radii R , R , . . . , R k arethe same.(a) Suppose that only t and t are nonzero in (16). Show that ∆ A = 0 implies t = − L t L . (b) Now substitute this into (16), and from ∆ L = 0 , argue R = R .(c) Extend this argument to show that R = R = R = · · · = R k .Does constant radius of curvature mean we have a circle? Not necessarily. You could join arcsof the same circle with a “kink”. But we can always approximate a kink as closely as we like bya smooth edge which encloses the same area (Fig. 31). This edge will have a different radius ofcurvature, which contradicts our argument! The only smooth, closed curve we can draw, which hasthe same radius of curvature R at every point, is the circle of radius R itself. This more or lessproves the isoperimetric theorem. So much for a single bubble. The next simplest problems involve two and three bubbles ofequal area A . The standard double bubble (Fig. 32 (middle)) and tripple bubble (Fig. 32 (right)) If we wanted to be rigorous, we would actually chop the line up into an infinite number of pieces using calculus. This is similar to the argument for equilateral triangles that the network length was an even function of wobble. Technically, we have only shown that if there is a perimeter-minimizing shape of fixed area, it is a circle. Butour approximation strategy can also be turned into a proof that the circle does minimize. Left.
A circle.
Middle.
The standard double bubble.
Right.
The standard triple bubble.Figure 33: Tinkertoys giving rise to bubbletoys.Of course, we might say: forget mathematics, and let Nature be our guide. By carrying outsimple experiments, we should be able to see which configurations are predicted by physics. Right?Unfortunately, the same combinatorial explosion that plagued soap bubble computers in §3.3 afflictsplanar bubble configurations. One argument is that every tinkertoy gives rise to a bubble config-uration, simply by adding arcs to the outside as illustrated in Fig. 33. I call these bubbletoys .Incidentally, the two pictured bubbletoys are conjectured to be the minimal planar configurations forfour and five equal-area bubbles. This suggests that solving the planar bubble problem is
NP-hard ,and even finding a bubble configuration which encloses the volumes A , A , . . . , A n is NP-complete ,or possibly
NP-hard as well. Nature will take increasingly long times to converge on her “conjec-tures”, solve the wrong problem, or both. Either way, we cannot get physics to magically solve our More generally, we will have to bend the inner walls to make the sure the pressure difference is balanced bytension. See Exercise 4.11 for more details. I haven’t been able to find this statement in the literature, and would be grateful if anyone could point mein the right direction. But it seems much harder than minimal networks! Unlike tinkertoys, where the problemcould be easier when we stiched together small tinkertoys, here, there are no external nodes so we only have large tinkertoys. And finding the minimal configuration is much, much harder , since we not only have an exponential P-complete problems for us!
Exercise 4.11.
Bubble radii and pressure cocycles. (cid:21)
We show the standard double and triple bubble for bubbles of different radii below.This exercise uses physics to simply relate the bubble radii! (There are also deriva-tions from the ◦ rule, but they are much messier [17].)(a) Let’s start with the double bubble. By considering pressure differences asacross interfaces, explain why R = 1 R + 1 R . (18) Hint.
Use Exercise 4.7(b).(b) We can make this observation more general. Consider moving around a loopon a bubble network. Across each interface, there are pressure differences ∆ P , ∆ P , . . . , ∆ P n . Show that, along the loop, ∆ P + ∆ P + · · · ∆ P n = 0 . This is called the pressure cocycle condition in the mathematics literature.(c) Using the pressure cocycle condition for the triple bubble, calculate that inaddition to (18), we have R = 1 R + 1 R = 1 R + 1 R . Check that the results of executing a loop around the inner junction areconsistent with these relations. number of tinkertoys, but an infinite set of configurations that arise from splitting and empty pockets. No wonderwe know almost nothing about bubbles! Bubbles in three dimensions
So far, we’ve only considered two-dimensional networks, while the real world has three dimensions.Thankfully, removing the plexiglass changes less than you might expect! Let’s start by summarizingwhat we know about bubble networks. The key result from §2 was the ◦ rule. In §4, we learnedthat bubble walls are straight or arcs of circles, so that they have constant radius of curvature(Exercise 4.7). We also discovered from our treatment of the isoperimetric problem that perimeter-minimizing wall do not have “kinks”. We can encode these insights as “laws” for bubble networks: Box 5.1.
Bubble network laws I. No kinks.
Edges are smooth, i.e. no vertices attached to one or two edges.2.
Constant curvature.
Edges have constant radius of curvature.3.
The ◦ rule. Three edges meet at a junction, separated by ◦ .There is another way to motivate the ◦ rule that will prove very useful in three dimensions.The law forbidding kinks means that the fewest edges that can meet at a junction is three. Moreover,meeting at angles of ◦ is the most symmetric way for incoming edges to be separated. Way backin §2.1, we saw that symmetry had an important role to play in minimizing the length of the networkon an equilateral triangle, so perhaps it’s unsurprising that the two are connected here. We call thisthe minsym principle . It lets us reformulate our network laws in a slightly different way: Box 5.2.
Bubble network laws II. No kinks.
Edges are smooth, i.e. no vertices attached to one or two edges.2.
Constant curvature.
Edges have constant radius of curvature. ¯3 . Minsym.
At a junction, the minimal number of edges meet symmetrically.Generalizing to three dimensions is now “easy”!
Viewed through a dimensional lens, a planar bubble network is a set of two-dimensional cellsseparated by one-dimensional bubble walls. But when bubbles can roam around in three dimensions,the cells are three-dimensional volumes separated by two-dimensional walls. Although it seems like awhole differen kettle of fish, three-dimensional bubbles are governed by almost exactly the same lawsas their planar counterparts. The three-dimensional laws are called
Plateau’s laws , after the Belgianphysicist
Joseph Plateau (1801–1883) who guessed them by assiduously observing bubbles [23].“No kinks” seems straightforward: bubble walls are smooth and cannot suddenly terminate. But there are subtleties for the remaining two rules. In a bubble network, edges have constantradius of curvature. What is the analogous statement for surfaces? It turns out in three and moredimensions, the notion of the curvature of a surface is not unique, and different definitions are useful Unless there is something for them to end on, e.g. a bubble blower. We’ll return to this problem below. constant mean curvature . This isa technical notion, and requires a bit more explanation.Figure 34:
Left.
A surface, with principal “snug” circles.
Right.
Straight slices through a point.Suppose we have a two-dimensional surface like the one in Fig. 34 (left). If we take variousstraight slices through the black point (shown in Fig. 34 (right)), each will give rise to a radius ofcurvature, i.e. the radius of a circle which fits “snugly” onto the curve at that point, and which isperpendicular to the surface. As we rotate the red slice in Fig. 34 (right), the radius of curvature R will vary, producing a maximum value R max and minimum value R min . The reciprocals /R max and /R min are called the principal curvatures . Note that a radius curvature can be negative if it isoutside the surface, as in Fig. 34 (left). The mean curvature H is defined as the sum of principal curvatures: H = 1 R max + 1 R min . (19)A constant mean curvature (CMC) surface is one where the mean curvature H is the same every-where on the surface. Notice that, as in Fig. 34 (right), it is always the case that the principlecurvatures R max and R min are measured along orthogonal slices. We call this the orthogonal circletheorem . To generalize the constant curvature rule from planar bubble networks, we take bubblesurfaces to be CMC. You can explore some of the physics behind this in Exercise 5.2.
Exercise 5.1.
Spheres are CMC.
Show that a sphere of radius R has constant mean curvature H = 2 /R . Hint.
Theslice normal to the sphere at any point is a great circle.
Exercise 5.2.
Young-Laplace II. (cid:21)
In Exercise 4.7, we saw the Young-Laplace law (13) for a bubble wall: ∆ P = 2 σR , for ∆ P = P out − P in and R the radius of curvature of the wall. Sometimes, outside and inside aren’t well-defined, so you just make an aribtrary choice, and attach a minus signto any circles which are outside. The sign of curvature depends on this choice. Unfortunately, it would take us too far afield to prove it here. /R max = 0 .(b) Using the orthogonal circle theorem, deduce that H = 1 /R is the meancurvature of the wall. Hence, the Young-Laplace law can be written ∆ P = 2 σH. (20)This turns out to be the correct form for an arbitrary surface!(c) If we dip two identical circular bubble blowers in soap film (red below), thesurface that results is typically like the one below left, rather than a cylinder:Give a qualitative explanation, using (20) and mean curvature.(d) Using Exercise 5.1, what is the smallest spherical bubble that can form in theatmosphere? Atmospheric pressure is P = 10 N/m and the surface tensionof soapy water is σ = 7 × − N/m. Can a spherical bubble form in space?
Finally, we have to generalize the “minsym” principle to three dimensions. The “no kink” requirementmeans that we cannot have two walls meeting at an angle, since that would introduce a kink, andif there is no angle, they may as well be count as part of the same wall. Thus, we must have atleast three walls meet at any junction of walls. According to the minsym principle, precisely threefaces should meet (minimum) separated by ◦ (symmetry), as in Fig. 35 (left). In fact, this isexactly what we need to get the ◦ rule in a planar bubble network, since the walls are secretlytwo-dimensional and vertical oriented between the plexiglass plates (Fig. 35 (middle)).The edge along which three bubble walls meet is called a Plateau border . These borders them-selves can intersect! Minsym requires us to figure out the minimum number to avoid kinks, andthe most symmetric arrangement thereof. Clearly, we need at least three, since otherwise we canarrange a junction of three walls with a kink, as in Fig. 35 (right).Figure 35:
Left.
Three faces meeting at a border.
Middle.
Vertical bubble walls.
Right.
A kink.43an we have exactly three? It’s not hard to see that the three sets of three faces cannot beconnected smoothly, simply because we have an odd number of faces! You can check the details inExercise 5.3. This exercise also shows that it is possible to connect the faces smoothly for four setsof Plateau borders. Thus, the minsym principle suggests that precisely four Plateau borders shouldmeet in the most symmetric arrangement. Symmetry is maximized by shooting out the Plateauborders tetrahedrally , i.e. from the center towards the corners of a regular tetrahedron (Fig. 36). Ifyou like vectors, you can play around with the geometry in Exercise 5.4.Figure 36: Left.
Four Plateau borders meeting tetrahedrally.
Right.
Smoothly connected walls.Having defined constant mean curvature, and worked out the implications of the minsym prin-ciple in three dimensions, we are finally in a position to state the laws Plateau discovered [23]:
Box 5.3.
Plateau’s laws. No kinks.
The faces in a soap film are smooth.2.
Constant curvature.
Any face has constant mean curvature.3.
Minsym I.
Three faces always meet at a Plateau border, separated by ◦ .4. Minsym II.
Plateau borders always meet tetrahedrally at a vertex.These are empirical observations about bubbles. While the constant curvature condition followsfrom the Young-Laplace law (Exercise 5.2), and Minsym I from the ◦ rule, it is not at all obviousthat a tetrahedral arrangement of Plateau borders minimizes area. Minimizing subject to whatconstraints? (Feel free to have a guess now.) Is every configuration satisfying Plateau’s laws alocal minimum, subject to these constraints? And does every such locally minimal solution satisfyPlateau’s laws? (These are harder to figure out without a doctorate in math.) Read on to find out! Exercise 5.3.
Plateau borders.
Let’s check we need four Plateau borders in order to smoothly connect walls. Con-sider some number of Plateau borders meeting at a vertex. Each border has threeassociated bubble walls.(a) Explain why “no kinks” requires each wall to connect smoothly to another.(b) Argue that this is impossible for an odd number of borders meeting at a node.(c) Show explicitly it is possible for four Plateau borders to smoothly connect.
Hint.
Add the two remaining walls in Fig. 36 (right).44 xercise 5.4.
Simplices. (cid:24)
The equilateral triangle and the tetrahedron are part of a family of symmetricshapes called simplices . We can describe them using vector analysis.(a) We can embed the vertices of an equilateral triangle in three dimensions as ∆ = { (1 , , , (0 , , , (0 , , } . Why is this a maximally symmetric arrangement of three points?(b) The center of the triangle is just the average of the vertices. Show that thevectors from center to vertices have length √ and are given by V = (cid:8) ( − , , , (1 , − , , (1 , , − (cid:9) . (c) Using the formula θ = cos − (cid:18) u · v | u || v | (cid:19) , check that the vectors in V make angle ◦ = cos − ( − / with each other.(d) We can embed the tetrahedron in four dimensions as ∆ = { (1 , , , , (0 , , , , (0 , , , , (0 , , , } . Show that the vectors from center to vertex have length √ , and make angles θ = cos − (cid:18) − (cid:19) ≈ . ◦ . The tetrahedron is just a higher-dimensional version of an equilateral triangle! Wecan continue in this fashion, defining the n -simplex ∆ n as a maximally symmetricarrangement of n points in n dimensions: ∆ n = { (1 , , . . . , , (0 , , . . . , , . . . , (0 , . . . , , } . (e) Check that the distance from the center of ∆ n to each vertex is √ n , and thatany two such vectors make an angle θ = cos − (cid:18) − n (cid:19) . As n gets large, confirm these vectors are almost orthogonal.(f) Extrapolate the minsym principle to higher-dimensional foams. In otherwords, if the universe has n dimensions, make a guess at Plateau’s laws.45 .3 Bubbles and wireframes As you might have guessed, Plateau’s laws are related to the three-dimensional version of the planarbubble configuration problem outlined in Box 4.1. Instead of enclosing areas A , A , . . . , A n , we wantto enclose volumes V , V , . . . , V n with a bubble film of minimal surface area. As before, we allowfor empty pockets and split bubbles. We state the optimization problem as follows: Box 5.4.
The Minimal Bubble Problem.
Find the connected bubble film of smallest area enclosing volumes V , V , . . . , V n ,allowing air pockets and split bubbles.Surfaces with bubbles of fixed volume, and which locally minimize area, also satisfy Plateau’s laws,as mathematican Jean Taylor proved in her 1976 tour-de-force [28]. The converse is not true,since we can find bubbles satisfying Plateau’s laws that are not stable (Fig. 38).The planar bubble configuration problem (Box 4.1) is a special case of the three-dimensionalbubble configuration problem, where we put the foam between plates. This implies that localminima satisfy the bubble network laws (Box 5.1), since these are simply Plateau’s laws in thecase where bubble walls are vertical, and because they are vertical, there are no vertices at whichPlateau borders intersect. And since planar bubbles are hard, three-dimensional bubbles are hard!Physically speaking, we expect that foams will take longer and longer to converge to a minimum,or answer the wrong question, if we force them to compute for us.Even when Nature does make conjectures, they can be bewilderingly hard to prove. The simplestexample is a single bubble of volume V . Experience suggests that a lone bubble is always spherical,as in Fig. 37 (left). The corresponding conjecture is that the area-minimizing surface of volume V is a sphere. This is the three-dimensional version of the isoperimetric theorem for circles in §4.4.Figure 37: Left.
A single spherical bubble.
Middle.
The standard double bubble.
Right.
Thestandard triple bubble.The proof is remarkably similar. We split the surface into many small patches of area A , A , . . . , A k , with mean curvature H , H , . . . , H k . If these areas are “pushed out” a distance I want to point out that the ◦ rule was more or less proved as soon the problem was stated. It took over 100years for the teatrahedral rule to go from empirical observation to mathematical proof. It’s much harder! , t , . . . , t k normal to the surface, the volume and area change as ∆ V = A t + A t + · · · + A k t k ∆ A = A t H + A t H + · · · + A k t k H k . If the wobbling preserves volume, then ∆ V = 0 , and if area is locally minimized, then ∆ A = 0 asbefore. We can then repeat our argument word for word to conclude that H = H = · · · = H k .Mean curvature is the same everywhere, and we have a CMC surface! In the plane, there was exactly one way to have a smooth curve with constant radius of curvature.In three dimensions, there are all sorts of exotic CMC surfaces. But it turns out that the sphereis the only
CMC surface that enclose a finite volume, as proved by
Aleksandr Aleksandrov (1912–1999) in 1958 [2]. The same argument we gave in Exercise 4.8 shows that empty pockets andsplitting bubbles will not help. Thus, we have proved the isoperimetric theorem in three dimensions:the sphere is the surface of smallest area enclosing a given volume V .The next simplest problem is two bubbles of equal volume V . Again, Nature seems to preferthe “standard double bubble”, with two spheres fused at a single Plateau border (Fig. 37 (middle))over its non-standard competitors. One of these competitors is the “donut” configuration, where asingle bubble is squeezed into the shape of an apple core by a donut-shaped ring around the outside(Fig. 38). This bubble satisfies Plateau’s laws, but turns out to be unstable, and jiggling the donutwill cause it to collapse into the standard double bubble [27]. It wasn’t until 2002 that the standarddouble bubble was shown to be minimal [20]. Similarly, we often observe the standard triple bubblefor three cells of volume V (Fig. 37 (right)). No one knows if this is truly minimal, so it remainsthe triple bubble conjecture .Figure 38: An unstable double bubble, consisting of an apple core wrapped in a donut.If there is a three-dimensional bubble configuration problem, it stands to reason there should bea three-dimensional minimal network problem. In the minimal network problem, Box 2.1), we hadto find a network of shortest length connecting some set of fixed nodes. A node is a zero-dimensional object—it has no extent at all! If we raise the number of dimensions of the configurable object, going It’s not hard to see that volume changes this way, but area is the tricky one, and I won’t prove it here. This proof actually generalizes to higher dimensions, where the mean curvature is H = 1 /R + 1 /R + · · · + 1 /R n for n mutually orthogonal principal curvatures. Well, almost. If the surface is allowed to intersect itself, there is an odd three-lobed donut called the
Wente torus ,but this is something of an embarassment so we ignore it. fixed curves . The suggests the following task:
Box 5.5.
The Wireframe Problem.
Given some fixed curves C , C , . . . , C n in three-dimensional space, find a soap filmof minimal area that connects them.These fixed curves are called wireframes , since physically speaking, we can implement them withtwisted pieces of wire. Dunking wire into soapy water gives soap bubbles something to hold onto,and as with plexiglass and screws, we have an analogue computer to solve our problem for us. Wegive two very beautiful examples in Fig. 39: the catenoid , a surface forming between two rings,and the tesseract formed when we dip a wireframe cube. The cube creates a second, slightly puffedout inner cube, and then connects corresponding corners with Plateau borders.Figure 39: Left.
The catenoid, a surface with mean curvature zero, which forms between identicalwireframe rings.
Right.
The tesseract formed from a wireframe cube.Since this generalizes the minimal network problem (the screws are a particularly boring typeof wireframe), the wireframe problem is
NP-hard . We can dip some arbitrarily complicated piece ofwire into the soap, but when we pull it out, it may take a very long time—longer than the age ofthe universe in some cases—for the soap bubbles to converge on a stable solution. Or it will solvea different problem altogether. This is yet another physical prediction!But it’s not obvious there is a solution at all. If we dunk some random wireframe into the water,the bubble film that connects them must satisfy Plateau’s laws, except along the wire itself, in thesame way that minimal networks satisfy the ◦ rule at a hub but not at a fixed node. But doPlateau’s laws always allow a solution? Perhaps we can defeat Nature by giving it some wacky curveit cannot connect with soap film. The intuitive physical argument is that we can simply dip ourwireframe in and see what comes out. But as we’ve just argued, for a complicated enough problem,it may take a very, very, very long time to converge. And I find an argument less convincing if Iam guaranteed to die before it successfully terminates!The question of the existence of a solution to the wireframe task is called Plateau’s problem .In the 1930s, mathematicians
Jesse Douglas (1897–1965) and
Tibor Radó (1895–1965) inde-pendently showed these solutions always exist [9, 25]. Even if it takes longer than the lifetime ofuniverse, Nature will eventually get there. To ensure borders meet tetrahedrally. .4 Space-filling foams In this final section, we’ll consider the three-dimensional honeycomb problem , that is, how to opti-mally partition space into equal-volume cells. To minimize surface area per cell, the partition mustsatisfy Plateau’s laws. The ◦ rule had dramatic consequences for large bubble networks inthe plane. We will see that, in three dimensions, the tetrahedral law has similar (if less dramatic)implications for the structure of three-dimensional foams. To explore these, we first need to extendEuler’s formula (6) to include multiple bubbles. In a finite configuration of soap bubbles, let N denote the number of vertices (where Plateau borders join), E the number of Plateau borders, F the number of bubble faces, and C the number of enclosed bubble cells. As before, we will countthe region outside the bubble configuration as a cell as well.Recall from §4.2 that Euler’s formula applies to a polyhedron like the cube, possessing two cells:the inside and the outside. We can divide up the internal cell by adding inner walls. If there are C cells altogether, there are C − internal cells, and C − “extra” internal cells compared to aregular polyhedron. For each extra cell, we can remove an internal face so that two neighbouringcells become one. This leaves something we can flatten into a planar graph, with F (cid:48) = F − ( C − faces, and hence by Euler’s formula N − E + F (cid:48) = 2 . Rearranging gives Euler’s “foamula”: N − E + F = 2 + ( C −
2) = C. (21)Equation (21) is true for any polyhedron with mutiple internal cells, whether or not it satisfiesPlateau’s laws.The ◦ rule, tetrahedral rule, and foamula together show that bubble faces tend to have less than six sides. More precisely, if F avg is the average number of faces per bubble cell, E avg theaverage number of edges around the boundary of a cell, and e avg the average number of edges perface, you can show in Exercise 5.5 that F avg = 13 E avg + 2 = 126 − e avg . (22)Since F avg is positive, it follows that e avg < , so faces tend to be sub-hexagonal. Exercise 5.5.
Sub-hexagonal faces. (cid:21) (a) From Plateau’s fourth law and the handshake lemma (1), argue that E = 2 N .(b) Let F avg denote the average number of faces per cell and E avg the averagenumber of edges per cell. Show that F avg = 2 FC , E avg = 3
EC .
Hint.
You may assume that, like in a bubble network, a face in a bubblefoam always has different cells on either side. Clearly the surface area is infinite, so how can it be minimal? Like I say, our goal is to minimize average surfacearea per cell . This obeys Plateau’s laws because the laws are either local , applying in the vicinty of a point (minsymI), line (minsym II), or face (no kinks). The CMC rule, on the other hand, can be derived at the level of a cell . F avg − E avg = 6 . This is analogous to the result F − E = 6 for bubble networks.(d) Let e avg be the average number of edges per face. Derive the relation F avg = 126 − e avg . Hint.
Write E avg in terms of F avg and e avg . Don’t forget to handshake!Since we are discussing averages , they continue to make sense even if the foam is infinite! Thesimplest infinite foams are those in which each bubble is the same, so the bubbles form a space-fillingtessellation . This is the three-dimensional analogue of the plane tessellations we saw in Fig. 29. Inour quest for infinite foams, we will first rule out simple tessellations which simply extrude theseplane tessellations into three-dimensional prisms. Exercise 5.6.
Prisms. (a) One way to tessellate space is to take a regular tessellation of the plane, thenextend it in the perpendicular direction to form a layer of prisms (as below).We can then stack these layers on top of each other to tessellate space.Explain why no prism-based tessellation satisfies Plateau’s laws.(b) The gyrobifastigium is made from two triangular prisms joined with a twistat their bases. All faces of the solid are regular polygons. Like prisms, youcan arrange gyrobifastigia into layers, and stack layers to fill space:Does this tessellation satisfy Plateau’s laws?50ur next step is to consider the analogue of regular polygons, the
Platonic solids (Fig. 40).These are polyhedra whose faces are identical regular polygons. Could any of these describe aninfinite soap foam? Of these solids, only the cube can tessellate space by itself, but since this is aprism-based tessellation (it is an extruded rectangular tiling), Exercise 5.6 rules it out. We need towork harder!Before we move on, it would be remiss not to mention that the Platonic solids can also beinterpreted as regular tessellations of the sphere . (You can collect all such tessellations in Exercise5.7.) In fact, the solids with three edges meeting at a node—the tetrahedron, cube, dodecahedrondisplayed in Fig. 41—obey the ◦ rule. If we “flatten” them in the same way we did the cube (Fig. 26), we get the bubble networks shown on the bottom row of Fig. 41. It is remarkable thatthe tetrahedral pattern, which so beautifully exhibits the ◦ rule, also shows up as a triple bubbleconfiguration (Fig. 32), in Plateau’s laws, and as a spherical tessellation!Figure 40: From left to right: tetrahedron, cube, octahedron, icosahedron, dodecahedon.Figure 41: The tetrahedron, cube, and dodecahedron as tessellations of a sphere (above) andbubble networks (below). Exercise 5.7.
Platonic solids. (cid:23)
In this exercise, we’ll classify the regular tessellations of the sphere. We’ll useEuler’s formula, N − E + F = 2 .(a) Suppose each face of the tessellation has a edges, and each node joins up with This is the positively curved counterpart to the hyperbolic tiling we saw in Exercise 4.6. Technically, we have drawn a very special flattening called the stereographic projection . This is what the verticeswould look like to an observer positioned on top of the sphere, or if a lantern at the same point cast the shadows ofeach edge onto the plane. edges. Show that E = N a = F b. (b) Using Euler’s formula, deduce that E = 2 ab a + b ) − ab . (23)(c) Since E is a whole number, so is the RHS of (23). We will find all possiblesolutions. To begin with, argue that we can interchange the roles of a and b , so a tessellation with a edges per face and b edges per node also gives atessellation with b edges per face and a edges per node. These tessellationsare said to be dual to each other.(d) If a = 2 , what are the possible values of b ? Draw the corresponding patternson the sphere and their duals.(e) The denominator of (23) must be positive. Show that this implies a < bb − , and hence there are no solutions for b ≥ and a ≥ .(f) The numerator in (23) is even, for a, b whole numbers. Argue that, in orderfor E to be a whole number, at most one of a and b can be odd.(g) Finally, using part (f), conclude that only five combinations of a and b areallowed for ≤ a, b ≤ . Check that each of these gives a Platonic solid.Like Exercise 4.6, the best partition depends on how much honey we want to store in each cell.But for the equal area cells represented by Fig. 40, are the regular tessellations the best way tosplit up equal cells on the surface of the sphere? Or does some irregular tiling do better? The“spherical honeycomb conjecture” is that regular tessellations are best. Although known to be truefor a dodecahedron and tetrahedron [19, 11], it remains a conjecture for the cube. Enough about spheres. Let’s return to the problem of space-filling foams, where, if you recall,we had concluded there was no way to tessellate space with Platonic solids so as to satisfy Plateau’slaws. Platonic solids are maximally symmetric, in the sense that every vertex looks alike, and everyface looks alike. But there are many more possibilities when we relax these contraints! The nextsimplest shapes are the “semi-regular” polyhedra, comprise by the 13
Archimedean solids , whosevertices all look alike but faces differ, and the 13
Catalan solids , whose faces all look alike butvertices differ. Only one from each class can tessellate space:• the rhombic dodecahedron , a Catalan solid with twelve rhombic faces;• the truncated octahedron , an Archimedean solid we get by snipping off an octahedron’s corners.These are shown in Fig. 42, including the “snipping” of a single octahedral corner. And in case you’re wondering, our hexagonality argument from §4.3 does not apply simply because the spherehas finite surface area. There’s not enough space for a network to get large! In fact, these are dual to each other in the sense of Exercise 5.7.
Left.
The rhombic dodecahedron, with 12 rhombic faces.
Middle.
Snipping off thecorner of an octahedron.
Right.
Doing this for all six corners gives the truncated octahedron.There is one more possibility left in our who’s who of space-filling solids. Take each face of therhombic dodecahedron and extrude it to form a pyramid, with each rhombic face replaced by fourtriangles. The result is the stellated rhombic dodecahedron , with “stellated” meaning “star-like”. Itis also called
Escher’s solid , since it features in Escher’s marvellous lithograph
Waterfall (Fig. 43).Remarkably, the extrusions interlock in such a way that Escher’s solid continues to tessellate space,which I like to call a “testellation”. Admittedly, I’ve included Escher’s solid mainly for the sake ofthis pun! We can now use (22) to eliminate all but one candidate on our shortlist.Figure 43: Stellated rhombic dodecahedron. Adapted from
Waterfall (1961), M. C. Escher.
Exercise 5.8.
The Kelvin structure.
We have three remaining candidates for an infinite foam with regular cells:• the rhombic dodecahedron, with two rhombic faces;• Escher’s solid, with 48 triangular faces; and• the truncated octahedron, with eight hexagonal faces and six squares.Show that only the truncated octahedron satisfies (22).53his truncated octahedron tessellation (Fig. 44) is called the
Kelvin structure in honor ofphysicist
William Thomson, 1st Baron Kelvin (1824–1907), who conjectured it was the mostefficient way to separate equal volume cells. Kelvin’s conjecture, often called the
Kelvin problem ,is the three-dimensional version of the honeycomb conjecture. To prove it, we must show thereare no irregular , equal-volume tessellations of space more efficient than the Kelvin structure.Figure 44: The Kelvin structure, made by tiling truncated octahedra.Like the hexagonal lattice, the Kelvin structure is the only regular tessellation which is locallyminimal. But the space of possibilities is much richer in three dimensions than in two. In 1993,
Denis Weaire and
Robert Phelan discovered [30] they could improve on the Kelvin structureby weaving together two funny-shaped cells of equal volume:• an irregular dodecahedron A , with twelve pentagonal faces; and• a -hedron A with two hexagonal and twelve pentagonal faces.The arrangement is called the Weaire-Phelan structure , shown in Fig. 45.Figure 45:
Left.
The -hedron A . Middle.
The irregular dodecahedron A . Right.
A chunk ofthe Weaire-Phelan structure.While we have (cautiously) extolled the virtues of soap bubble computers for solving minimiza-tion problems, Weaire and Phelan took the amusing approach of simulating foams! They were usingthe software of the physical universe, but not the hardware. Experimentally speaking, the problemis that real soap films are finnicky, and it is challenging to arrange equal-volume bubbles [29]. Even Note that we need to bend the edges a little to ensure they meet at θ ≈ . ◦ , in accord with Plateau’s laws. Appropriate to four-dimensional bees. NP Hardness predicts (§4.1).But stubborn though it is, Nature is also wise. It knew about Weaire and Phelan’s oddity longbefore the tool-wielding monkeys it evolved to observe itself! In 1931, chemists noticed that layers oftungsten formed by electrolysis had an unusual chemical structure; a couple of years later, the samestructure was observed in chromium silicide Cr Si , and in 1953, in the superconducting compoundvanadium silicide V Si . This got the physicists excited! Since then, many more superconductingcompounds with the same underlying atomic arrangement have been discovered. Chemists F.C. Frank and
J. S. Kasper began investigating the mathematical properties of these silicidearrangements, and some close cousins, together called tetrahedrally closed-packed (TCP) structures [13, 14]. The original TCP structure is called the
A15 phase , shown in Fig. 46 (left). Here is thepunchlne: this is precisely what you get if you put an atom at the center of each polyhedron inWeaire and Phelan space-filling pattern!Figure 46:
Left.
The A15 phase in superconducting vanadium silicide.
Right.
Deep sea methaneclathrate hydrate, from Wikipedia.Above, we introduced the - and -sided polyhedra A and A . Frank and Kasper built theTCP structures out of four polyhedra, pictured below (Fig. 47). Each has twelve pentagonal faces,and , , or hexagonal faces, with A i referring to the solid with i hexagons. No one has classifiedall the combinations possible with these TCP polyhedra, though it seems there may be an infinite number! Weaire-Phelan is the current TCP record-holder, but whether it is globally optimal in thevasts of TCP space is an open problem. See [27] for further discussion. While exploring this fullspace of possibilities is well beyond us (and indeed, professional mathematics), the “foamula” (21)gives some simple constraints, derived in Exercise 5.9. Exercise 5.9.
TCP structures. (a) Suppose we can build a tessellation out of the TCP polyhedra A , A , A , A in the ratio a : a : a : a . Using (22), what are the constraints on the possible ratios?(b) The Weiare-Phelan structure (A15 phase) interleaves A and A polyhedra. This is the same metal ancient light bulb filaments are made from. This means vanadium silicide exhibits no electrical resistance , at least when cooled below − ◦ C. A , A , A and A appearing in TCP structures.What is their ratio?(c) The Z phase has A , A and A in the ratio a : 2 : 2 . What is a ?(d) The C15 phase uses A and A polyhedra. What is the ratio?(e) Finally, show that we can write any ratio for a TCP structure as a combina-tion of A15, C15 and Z ratios.Although different structures can have the same ratio, this is a useful way tounderstand the space of possibilities.There is a second route to Weaire-Phelan through chemistry. Instead of placing atoms at thecentre of the alternating polyhedra, we can place them at vertices where Plateau borders join. TheWeaire-Phelan structure is then called the Type I clathrate structure , and the class of compoundsthey occur in the clathrate hydrates . Roughly speaking, this means “water cage”, since clathratecompounds are tiny, elaborate cages made from ice. Regular ice doesn’t form cages, since thehydrogen bonds are too strong, collapsing the cage into the usual crystalline arrangement. But ifyou trap a few gas atoms inside—such as methane, carbon dioxide, or neon—it weakens the bondsenough for the cage to persist! It’s a jail that only exists when it has a prisoner. A chunk of deepsea methane hydrate is pictured in Fig. 46 (right).Clathrates are found in all sorts of exotic locales, from the deep ocean floor to the outer solarsystem. Since these ice cages can trap natural gases like methane, they provide a vast but non-renewable energy source [32]. Ironically, clathrates also offer a possible means of capturing carbondioxide and therefore mitigating climate change. So, our journey, which started on the train, hasled via a graph of associated minimization problems to superconductors, trans-Neptunian objectsand climate change. Though long by some measures, I suspect we have followed the shortest pathconnecting these fixed vertices in concept space. “Have you guessed the riddle yet?” the Hatter said, turning to Alice again. “No, I give it up,”Alice replied: “what’s the answer?” “I haven’t the slightest idea,” said the Hatter. “Nor I,” saidthe March Hare. Alice sighed wearily. “I think you might do something better with the time,”she said, “than waste it in asking riddles that have no answers.” The Type I clathrate is built from A and A cages, as we expect. There is also a Type II clathrate structure,built from A and A cages, which can assemble into a C15 phase. “Clathrate” is from the Latin clathratus , meaning “with bars”, while “hydrate” is from the Ancient Greek hydor ( ¨˘ υδω(cid:37) ) for water. Mixing Greek and Latin like this is considered very poor form in some circles. eferences [1] Aaronson, S.
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