Paperclip graphs. Steve Butler Erik D. Demaine Martin L. Demaine Ron Graham Adam Hesterberg Jason Ku Jayson Lynch Tadashi Tokieda

Transcription

1 Paperclip graphs Steve Butler Erik D. Demaine Martin L. Demaine Ron Graham Adam Hesterberg Jason Ku Jayson Lynch Tadashi Tokieda Abstract By taking a strip of paper and forming an S curve with paperclips holding the paper together at two key locations it is possible to link the paperclips by pulling the paper taut. We explore what other configurations of linked paperclips might be possible by introducing paperclip graphs and give several constructions as well as impossible configurations. 1 An experiment with paperclips To start, go get a strip of paper and two paperclips. We will wait a minute for you to come back... Take your strip of paper and shape the strip into an S curve and fix it with two paperclips as shown in Figure 1. Figure 1: Putting two paperclips on a strip of paper shaped as an S curve. What happens when we pull the strip taut? As we begin to pull, the paperclips come closer and closer together. If we keep pulling, presumably they will collide and jam... We suggest that, before reading on, the readers try this experiment themselves with their strip of paper and paperclips. Result: At the instant the strip becomes taught, the paperclips pop out (see Figure 2). And there is more: pick up the paperclips that landed. You will see that they are now linked. 1

2 Figure 2: Launching the paperclips. This trick seems to go back to Seattle Magician Bill Bowman who used a dollar bill for the strip of paper. The trick was circulated among the magic community in 1954; Martin Gardner [2] later popularized it in his book on magic which includes variations involving multiple paperclips, string, and a rubber band. It is natural to generalize the trick by making the paper more sinuous and adding more paperclips. For example, with three consecutive turns and three paperclips placed as shown in Figure 3, we are tempted to guess that pulling the strip straight will result in a chain of three Figure 3: Putting three paperclips on a curve. linked paperclips: clip-clip-clip. That is what should happen in the idealization of very elastic paperclips and very strong paper. In practice, however, paperclips are somewhat plastic and the paper of limited strength; a sharp concentration of stress attacks the middle paperclip, which ends up twisting it out of shape and often tearing the paper. This physical effect is interesting, but it is a subject for another time. For now, we want to investigate the idealized paperclip linking, where the focus up front is on the topology of the resulting configuration, while the mechanics of deformable media, which matters if you want to (per)form these experiments, fades into the background. 2 Paperclips as arcs We can represent our strip of paper by a (non-crossing) curve in the plane where the paperclips correspond to points where the curve passes through the same point. We will assume that the two ends of the curve are in the unbounded region of the plane (so that we can pull the strip taut). Finally, at no point does the curve pass through the same point three or more times (in other words paperclips only connect two distinct parts of the strip). 2

3 Let us start with an example. Suppose we have taken a strip of paper and connected it in five locations as shown in Figure 4(a) (the paperclips can be assigned two different colors; this should be done by following the strip along one side and anytime that we would have to break the paperclip to continue the walk we color it red; otherwise color blue). Now treat both the strip and paperclips as elastic so we can start to pull them apart as shown in Figure 4(b). Continuing this until the paper is made taut we will get the diagram shown in Figure 4(c) which consists of a straight line (the strip of paper) and then a set of (red) non-intersecting arcs placed on one side and a set of (blue) non-intersecting arcs placed on the other. The arcs connect two points on the strip of paper which correspond to where the paperclips had connected the strip; arcs of the same color cannot intersect as that would lead to a situation where the paper would have to cross itself. (a) Strip with paperclips (b) Starting to pull apart (c) Making the strip straight Figure 4: From a strip with paperclips to sets of non-intersecting arcs Alternatively, given any collection of non-intersecting arcs on one side of a straight strip and non-intersecting arcs on the other we can find a corresponding way to have a strip of paper with paperclips connecting points on the strip. The key is to treat the paper as elastic and then use the arcs to indicate how to stretch the paper to make the connection. An example of this is shown in Figure 5. (a) Configuration of arcs (b) Stretching the paper Figure 5: Going from a set of arcs to a strip with paperclips We can now glue the two ends of the paper together to form a circle with red and blue arcs (we can place all arcs in the interior with the rule that arcs of the same color cannot intersect). Doing this for Figure 4(c) and Figure 5(a) we get Figure 6. Given a configuration of arcs connecting points on a line there is a unique circle for which it corresponds. On the other hand given a circle there are potentially multiple configurations of arcs connecting points to which it corresponds, e.g., by choosing to cut the circle at different points. 3

4 (a) Based on Figure 4(c) (b) Based on Figure 5(a) Figure 6: Circular representation for arc configurations The last piece we need is to know when two paperclips (represented by two arcs) will intersect. When represented as a sequence of arcs connecting points on a straight line this will be when the endpoints of the pair of arcs alternate. When represented as arcs connecting points on a circle this will be when the corresponding arcs intersect. (Recall that arcs of the same color cannot intersect.) Definition. Given a configuration of red and blue arcs connecting distinct points on a circle so that no pair of red red arcs intersect and no pair of blue arcs intersect, form a graph by letting each arc correspond to a vertex and two vertices are connected if and only if the endpoints of the corresponding arcs alternate. Any graph which can be constructed in this manner is said to be a paperclip graph. As an example, the paperclip graph corresponding to Figure 6(a) is the path graph on five vertices; the paperclip graph corresponding to Figure 6(b) is the cycle graph on six vertices with one additional edge connecting a pair of opposite vertices. Understanding what configurations of paperclips are possible is equivalent to understanding what graphs are paperclip graphs. Since we can naturally split the arcs into the sets of red arcs and blue arcs, we immediately have the following. Observation 1. All paperclip graphs are bipartite. 3 Constructions We start with a few simple propositions. Proposition 1. If G is a paperclip graph and v is any vertex, then the graph H formed by taking G and adding a new vertex v which is adjacent to only v is also a paperclip graph. Proof. We can represent G as a configuration of arcs connecting points on the line. The vertex v corresponds to one of these arcs, and suppose that one of the endpoints of the arc is located at x. We now introduce a new arc of the opposite color with endpoints x ε and x + ε where ε > 0 is chosen sufficiently small so that the only endpoint in the interval is the endpoint for x. In terms of the paperclip graph this has the effect of adding a new vertex which is only adjacent to v, as claimed. Corollary 1. All trees are paperclip graphs. 4

5 Proposition 2. If G is a paperclip graph and v is any vertex, then the graph H formed by taking G and adding a new vertex v which is adjacent to the neighbors of v (i.e., duplicating v) is also a paperclip graph. Proof. We can represent G as a configuration of arcs connecting points on the line. The vertex v corresponds to one of these arcs, and suppose that the endpoints of the arc are located at x and y. We now introduce a new arc of the same color with endpoints x + ε and y ε where ε > 0 is chosen sufficiently small so that there are no endpoints of arcs in the interval between x and x + ε and no endpoints of arcs in the interval between y and y ε. This new arc will then connect to all of the same arcs that the original arc connected. In terms of the paperclip graph this has the effect of adding a new vertex which is adjacent to all neighbors of v, as claimed. Corollary 2. All complete bipartite graphs are paperclip graphs. Proposition 3. If G is a paperclip graph, then any induced subgraph of G is also a paperclip graph. Proof. By removing an arc corresponding to a paperclip we remove the corresponding vertex and all incident edges. No non-incident edges are effected as we have not created or destroyed any other intersections. So to get the induced subgraph, remove the arcs corresponding to the missing vertices. The requirement that this is induced is important as we have already seen that complete bipartite graphs are paperclip graphs and in the next section we will see that there are (bipartite) graphs which are not paperclip graphs. Theorem 1. All bipartite outerplanar graphs are paperclip graphs. Proof. The requirement that the graph is bipartite follows from Observation 1. From Proposition 3 it suffices to show that the graph is an induced subgraph of some paperclip graph. Since any bipartite outerplanar graph is an induced subgraph of a bipartite graph consisting of an even cycle possibly with chords (i.e., keep adding vertices and edges incident to the new vertices as needed until such a structure is formed). (a) The graph C 8 (b) Place to add a chord (c) Reversing one side Figure 7: Forming cycles and adding chords Even cycles are paperclip graphs since we can alternate red and blue arcs on the circle (e.g., the 8-cycle is shown in Figure 7(a)). So it remains to show that we can add our chords. 5

6 For this suppose that we want to add a chord between vertex u and vertex v on the cycle. Because the graph will be bipartite they must correspond to different colored arcs. Our situation is as shown in Figure 7(b) where we have the two arcs marked while in the shaded regions we have some unknown configuration of arcs (the important thing is that the only two arcs going between the two sides are the ones shown). We now take the left side and flip it over by which we mean reverse the location of all the endpoints on one side. Any crossing which exists on either side will still exist, and we have introduced exactly one new crossing, namely between the two arcs which cross between the two sides, as desired. Doing this at any given location has no effect on adding chords at other locations, so we repeat and add as many chords as needed. The preceding result is far from exhaustive as there exist planar paperclip graphs which are not outerplanar as well as many nonplanar paperclip graphs. For example, the complete bipartite graphs K 2,3 and K 3,3 are such examples. Proposition 4. If G and H are paperclip graphs, then so also is G H. Moreover if v V (G) and u V (H) the graph formed by taking G H and adding a path of any length (including 0) between u and v is also paperclip. Proof. Starting with a circular configuration for G we can cut the circle at a point next to one end of the arc corresponding to v so that as a configuration of arcs connecting points on a line the last point corresponds to v. Similarly we can find a configuration of arcs connecting points on a line for H where the first point corresponds to u. To get the result for G H we can concatenate these two configurations. If we want to combine them together by a path we first look at the parity of the path. If the path has even parity we make sure that the arc colors for u and v agree. If the path has odd parity we make sure that they are distinct (this can be accomplished by switching red and blue only for H). Then after concatenation we add in sufficiently many arcs connecting the two together. In the case when the path is a single edge we push the last point for u just past the first point for v. In the case when the path is length 0 (so that u and v become the same vertex), we concatenate and then we remove the tail end of the arc corresponding vertex u and the front end of the arc corresponding to vertex v and then connect the two freed endpoints to create a new arc. (In other words we combine the arcs for u and v together.) In particular if we want to know whether a graph is paperclip it suffices to look at the maximally two-connected components. 4 Impossible configurations In this section, we first examine some examples of non-paperclip graphs. We then discuss two closure properties of non-paperclip graphs, both of which generate infinite families of graphs. 6

7 Examples. From the preceding section, if we want to search for graphs which are not paperclip graphs, then we should look for two-connected graphs which do not have any leaves, duplicated vertices, and are not outerplanar. The first such candidate graph is on seven vertices, shown in Figure 8(a), and turns out to not be a paperclip graph. By Proposition 3 any graph which contains this as an induced subgraph is not a paperclip graph. Moving to eight vertices that leaves us with five graphs to consider, two of which are not paperclip graphs and are shown in Figure 8(b) and (c). b c 1 2 a a b c a b c d d (a) G 7 (b) G 8 (c) G + 8 Figure 8: Non-paperclip graphs on seven and eight vertices Proposition 5. The graphs G 7, G 8 and G + 8 are not paperclip graphs. Proof. If G 7 were a paperclip graph, then it would be possible to find a configuration on the circle in which vertex 1 corresponded to a blue arc and vertices a, b, and c, corresponded to three red arcs each which intersected the blue arc. This leads to the situation shown in Figure 9 (by symmetry it doesn t matter which red arc corresponds to which vertex). Figure 9: Intermediate step in the proof of Proposition 5 We now observe that because of vertices 2, 3, and 4 it must be the case that for any two of the red arcs there is a blue arc that will intersect with exactly those two red arcs. But there is no way to have a blue arc intersect the left and right red arcs without intersecting the middle arc as well. So this is impossible. If G 8 were a paperclip graph then we would again have that the vertices 1, a, b, and c lead to the situation shown in Figure 9. At this point we must place a red arc corresponding to vertex d in such a way that it does not intersect any arc currently drawn, and such that for every existing red arc, one endpoint of the red arc is next to the new arc. But this is impossible, and so the graph is not a paperclip graph. 7

8 The proof for G + 8 is the same as the proof for G 8 with the added constraint that when we add in the next arc it must intersect the blue arc. Again the situation is still impossible, and so the graph is not a paperclip graph. As n increases there are even more forbidden induced subgraphs. As an example, it is easy to take the proof for G 7 and adapt it to show that the graph formed by taking an even cycle on 6 vertices and connecting every other vertex to a central vertex is not a paperclip graph; moreover these are all minimal in the sense that any induced proper subgraphs of these graphs are outerplanar and hence are paperclip graphs. For completeness, the remaining three graphs on eight vertices which have not been shown to be a paperclip graph by the construction tools we have, and have not been ruled out by Proposition 5, are paperclip graphs by using the configurations shown in Figure 10. Figure 10: The remaining paperclip graphs on eight vertices Double Subdivision. A double subdivison of a graph selects a single edge of the graph and replaces that edge with a length three chain. Alternatively we can subdivide a single edge and then immediately subdivide one of the new resulting edges. We now show that non-paperclip graphs are closed under double subdivision. Theorem 2. If G is a non-paperclip graph, then any double subdivision of an edge in G results in a non-paperclip graph. Proof. Without loss of generality, we may assume that G is connected. We prove the converse. Let H be the graph that results from doing the double subdivision of an edge e in the original graph producing edges e 1, e 2, and e 3 and introducing two new 8

9 vertices v 1 and v 2 and suppose that the graph H is paperclip. Then the structure of the graph is as shown in Figure 11(a). In particular because the arcs corresponding to v 1 and v 2 α e 2 β β e 1 v 1 v 2 e 3 e α (a) A realization for H (b) A realization for G Figure 11: Double edge subdivision only have degree two and they already have two intersections, then the arcs which are not shown must be in one of the shaded regions. To construct an arc realization for G we note that we can now combine three arcs together (two of them coming from v 1 and v 2 ) to form a single arc. The only remaining thing to note is that we can transplant the arrangement of arcs in one of the shaded regions (α in Figure 11(b)) so that all other intersections of arcs are preserved. In particular we now have a configuration of arcs corresponding to G, so that G is also a paperclip graph. The result of this theorem cannot be strengthened to be an if and only if statement. For example, G 8 (Figure 8(a)) is a non-paperclip graph and is the double subdivision of the paperclip graph given in Figure 6(b). 5 There are exponentially many paperclip graphs By Theorem 1 we know that the number of paperclip graphs is at least the number of bipartite outerplanar graphs. And this is known to grow exponentially. Theorem 3 (Bodirsky et al. [1]). If g n is the number of bipartite outerplanar graphs on n vertices, then c( )n g n. n 5/2 To get an upper bound for the number of paperclip graphs on n vertices, we use the interpretation of using non-intersecting arcs above and below a strip. So start with n paperclips, let i of them correspond to arcs above the strip and n i below; then there are ( 2n 2i) ways to choose which paperclip endpoints are above the strip (leaving the rest below). Then, given 2i endpoints above, they must be connected in the same way as( balanced parentheses, the number of ways to do which is the ith Catalan number C i = 1 2i ) i+1 i ; similarly, the number of ways to connect the arcs on the bottom is C n i (for more on the Catalan numbers see 9

10 Stanley [3]). So the total number of possible configurations is i=n i=0 ( ) 2n C i C n i 2i i=n i=0 2 2n n C i C n i = 22n n C n+1 24n n 2. There are situations in which exponentially many configurations which were produced in the above argument yield the same graph. For example there are at least 2 n 2 ways to produce the path on n vertices. We see the result is true for P 1 and P 2 where in P 2 we will consider the arc on the right the active arc. Now we continue to extend by adding a leaf (see Proposition 1) to the active arc and note that at each step we have one of two possible places to attach, and then the new arc becomes active. Given the final arrangement we can reconstruct our choices made giving 2 n 2 configurations. 6 Open problems We still do not have a complete characterization of paperclip graphs. For small graphs it is not hard to exhaustively check all possibilities of arc configurations. But there do not yet seem to be any clear patterns about what is possible and what cannot be done. As a warmup exercise in this direction two of the three graphs shown in Figure 12 are paperclip graphs and one of them is not a paperclip graph. Try to determine which one is not a paperclip graph. Figure 12: Two of these three graphs are paperclip graphs; one of them is not We have also established some exponential upper and lower bounds for the number of paperclip graphs. There is still a significant gap between these bounds; can this be improved? Going back to the circle interpretation we had a set of disjoint red arcs and a set of disjoint blue arcs and then then put in edges for vertices where the corresponding pairs of arcs that intersected. One could generalize to having k sets of disjoint arcs of different colors and then again putting in edges for vertices where the corresponding pairs of arcs intersected. Such graphs would necessarily have to be k-partite. What graphs are possible now? Acknowledgment The authors thank Zachary Abel for discussions related to this paper. References [1] Manuel Bodirsky, Éric Fusy, Mihyun Kang, and Stefan Vigerske, Enumeration and asymptotic properties of unlabeled outerplanar graphs, Electronic Journal of Combinatorics 17 (2007), #R66, 24 pp. 10

11 [2] Martin Gardner, Encyclopedia of Impromptu Magic, Magic Inc., Chicago, Ill (1978), [3] Richard Stanley, Catalan Numbers, Cambridge University Press,