Mega Menger Graphs Allan Bickle Calvin College February

Mega. Menger Graphs Allan Bickle Calvin College February 26, 2015 Mega. Menger Graphs 1

Fractals • A fractal is a type of mathematical object. • Many well-known fractals are self-similar. That is, part of the object is isomorphic to the whole. • Often, a fractal can be formed by iterating some operation infinitely many times on a set. 11/6/2020 [Project Name] 2

The Cantor Set • For example, consider the Cantor set. Begin with the interval [0, 1] and remove the middle third. This leaves the set which contains two intervals of length 1/3. Next, remove the middle thirds of both of these subintervals. • At each step, remove the middle third of each subinterval remaining. Iterating this operation infinitely many times produces the Cantor set. 11/6/2020 [Project Name] 3

The Cantor Set • Note that the portion of the Cantor set in is isomorphic to the entire set. • Here we see levels 0 -5 of the construction of the Cantor Set. 11/6/2020 [Project Name]

The Sierpinski Carpet • The Cantor set is a subset of a line, which is onedimensional. We can generalize it to higher dimensional fractals. • One way of doing this is to start with a square, divide it into nine smaller equal-sized squares, and remove the middle square. Then repeat the operation on each smaller square ad infinitum. • This produces a fractal called the Sierpinski carpet. This can be seen to contain many copies of the Cantor set. 11/6/2020 [Project Name]

Level 0 -5 Sierpinski Carpets 11/6/2020 [Project Name] 6

The Menger Sponge • One way of generalizing this idea to a three dimensional object is to start with a cube, and divide it into 27 smaller equal-sized cubes. Then remove the middle cube from each face and the center cube, leaving 20 smaller cubes. Then repeat the operation on each smaller cube ad infinitum. • This produces a fractal called the Menger sponge. This can be seen to contain many copies of the Sierpinski carpet. • It was first described in 1926 by Karl Menger, who is famous in graph theory for Menger's Theorem on connectivity. 11/6/2020 [Project Name]

Level 0 -3 Menger Sponges 11/6/2020 [Project Name] 8

Discrete Models of Fractals • If we want to build real world models of these fractals, we cannot perform infinitely many iterations of an operation. Even performing finitely many operations eventually becomes impossible if it requires us to remove smaller and smaller portions of a real world object. • An alternative way to construct real world models of fractals is to start with a given unit and build larger and larger models out of more and more copies of this unit. We will refer to these as levels of the fractal model. 11/6/2020 [Project Name]

Discrete Models of Fractals • Thus level 0 for the Sierpinski carpet is a square, and level 1 is built by arranging eight of these squares into a ring. Then level 2 is built by arranging eight copies of level 1 into a ring, and so on. • For the Menger sponge, level 0 is a cube, and level 1 is built out of twenty of these cubes. Level two is built out of twenty copies of level 1, and so on. 11/6/2020 [Project Name]

Discrete Models of Fractals • We would like a way to describe the locations of the level 0 units in these constructions. To describe the location of a subinterval in the Cantor set, label it 0, 1, or 2 depending which third of the interval it is contained in. Then append a 0, 1, or 2 for which third of this subinterval it is contained in. • Continuing in this fashion, we obtain a string of 0's, 1's, and 2's that describes the location of the subinterval. But if the subinterval is actually in the Cantor set, 1 cannot appear. • We can think of such a string of length i as a ternary number between 0 and 3 i -1 describing the location of a subinterval in level i of the model of the Cantor set. • For example, 0220 represents 270+92+32+10=90. 11/6/2020 [Project Name]

Discrete Models of Fractals • Since the Sierpinski carpet is a subset of the plane, we assign two coordinates to indicate the location of a square in it. • Each coordinate is described by a ternary string constructed analogously to that for the Cantor set. • A square is contained in the Sierpinski carpet if and only if its coordinates do not both contain a 1 in the same position. 11/6/2020 [Project Name]

Discrete Models of Fractals • For the Menger sponge, we assign three coordinates to indicate the location of a cube in it. Each coordinate is described by a ternary string constructed analogously to that for the Cantor set. • A cube is contained in the Menger sponge if and only if there is at most one 1 in any position of its three coordinates. • For example, (10, 01, 02) is in the Menger sponge, but (10, 01, 12) is not. 11/6/2020 [Project Name]

The Mega. Menger Project • The goal of the Mega. Menger project was to build twenty level 3 models of the Menger sponge out of business cards at sites around the world. • Each level one cube was constructed out of six cards, and has side length two inches with two flaps on each face that were used for linking it with other cubes. • Once all the cubes were linked together, another card was attached to each face to hide the exposed flaps, add aesthetic appeal, and provide additional structural support. • The level 3 model has side length 4. 5 feet and weighs approximately 170 pounds. 11/6/2020 [Project Name]

The Mega. Menger Project • The Mega. Menger project was organized by Queen Mary University of London. I participated in Calvin College's project, which was organized by Randy Pruim and Gerard Venema and took place during October 2014. • More information is available at http: //www. megamenger. com/ and http: //www. calvin. edu/academics/departmentsprograms/mathematics-statistics/mega-menger/. • Following are some photos that illustrate the process. 11/6/2020 [Project Name]

Build a Level 0 Box 11/6/2020 [Project Name] 16

Build a Lot of Them 11/6/2020 [Project Name] 17

Connect Them in a Ring 11/6/2020 [Project Name] 18

Build a Level 1 Cube 11/6/2020 [Project Name] 19

Connect Some Level 1 Cubes 11/6/2020 [Project Name] 20

A Level 2 Cube 11/6/2020 [Project Name] 21

A Ring of Level 2 Cubes 11/6/2020 [Project Name] 22

A Solid Foundation 11/6/2020 [Project Name] 23

Tight Spaces 11/6/2020 [Project Name] 24

Lifting the Top Ring 11/6/2020 [Project Name] 25

Finishing Touches 11/6/2020 [Project Name] 26

Complete Level 3 Menger Sponge 11/6/2020 [Project Name] 27

The Final Day’s Volunteers 11/6/2020 [Project Name] 28

Graph Theory Models • This project inspired me to ask and answer a number of questions about models of the Sierpinski carpet and Menger sponge. • We will use graph theory to model the real world models of the Sierpinski carpet and Menger sponge. • A graph is composed of a set of vertices and a set of edges, which is a subset of the set of 2 -element subsets of the vertex set. • In a Sierpinski carpet graph, each vertex represents a square of the carpet. Vertices are adjacent if their squares share a common edge. 11/6/2020 [Project Name]

Constructing the Graph SC 2 11/6/2020 [Project Name]

Graph Theory Models • In a Menger sponge graph, each vertex represents a cube of the sponge. Vertices are adjacent if their cubes share a common face. • We will denote the level i Sierpinski carpet graph as SCi, and the level i Menger sponge graph as MSi. • We will use graph theory to find information about the structure of the Sierpinski carpets and Menger sponges. We begin by calculating some parameters of these graphs. 11/6/2020 [Project Name]

Orders of the Graphs • The order of a graph, n(G), is the number of vertices in graph G. • Each iteration in the construction of the Cantor set can be formed from two copies of the previous one. This gives us the recurrence relation for the number of intervals ci = 2 ci-1, with initial condition c 0 = 1. • This immediately gives the exponential solution ci = 2 i. 11/6/2020 [Project Name]

Orders of the Graphs • The Sierpinski carpet is formed from eight copies of the previous level, so the order n(SCi) = 8·n(SCi-1) and n(SC 0) = 1. Thus n(SCi) = 8 i. • The Menger sponge is formed from twenty copies of the previous level, so n(MSi) = 20·n(MSi-1) and n(MS 0) = 1. Thus n(MSi) = 20 i. 11/6/2020 [Project Name]

Sizes of the Sierpinski Carpets • The size m(G) of a graph is the number of edges it contains. • To determine the sizes of the Sierpinski carpet graphs, we must develop a recurrence relation. • Obviously, m(SC 0) = 0 and m(SC 1) = 8. • Each level is formed from eight copies of the previous level with some added edges in between. 11/6/2020 [Project Name]

The Sierpinski Carpet Graph SC 3 11/6/2020 [Project Name] 35

Sizes of the Sierpinski Carpets • There are 3 i-1 edges between each copy of SCi. Since there are eight edges in SC 1, there are eight sets of 3 i-1 edges added. Thus we find the recurrence relation m(SCi) = 8·m(SCi-1)+8· 3 i-1. • This is a nonhomogeneous linear recurrence relation. The homogeneous part is exponential, so it has a solution of the form A 8 i. • A theorem on recurrence relations tells us that the nonhomogeneous part has a solution of the form B 3 i. 11/6/2020 [Project Name]

Sizes of the Sierpinski Carpets • Thus the solution should have the form m(SCi) = A 8 i + B 3 i. • Plugging in the initial conditions, we find 0 = A + B and 8 = 8 A + 3 B. • Solving this system of linear equations, we find A = 8/5 and B = -8/5. Thus m(SCi) = (8/5)8 i – (8/5)3 i. • The first few terms of this sequence are 0, 8, 88, 776, 6424, . . . • This also implies that the average degree of the Sierpinski carpet graphs approaches 16/5 = 3. 2 as i grows large. 11/6/2020 [Project Name]

Sizes of Menger Sponge Graphs • To determine the sizes of the Menger sponge graphs, we use a similar method. • We see m(MS 0) = 0 and m(MS 1) = 24. • Each level is formed from 20 copies of the previous level with some added edges in between. There are 8 i-1 edges between each copy of MSi, since each face of the Menger sponge is a level i-1 Sierpinski carpet. Since there are 24 edges in MS 1, there are 24 sets of 8 i-1 edges added. • Thus we find the recurrence relation m(MSi) = 20·m(MSi-1)+24· 8 i-1. 11/6/2020 [Project Name]

Sizes of Menger Sponge Graphs • The homogeneous part is exponential, so it has a solution of the form A 20 i. The nonhomogeneous part has a solution of the form B 8 i. Thus the solution should have the form m(MSi) = A 20 i+B 8 i. • Plugging in the initial conditions, we find 0 = A + B and 24 = 20 A + 8 B. Solving this system of linear equations, we find A = 2 and B = -2. Thus m(MSi) = 2· 20 i-2· 8 i. • The first few terms of this sequence are 0, 24, 672, 14976, 311808, . . . • This also implies that the average degree of the Menger sponge graphs approaches 4 as i grows large. 11/6/2020 [Project Name]

Surface Area of Menger Sponges • In constructing a model of the Menger sponge, we added extra cards on the outside faces to improve its aesthetics and stability. • To determine how many cards were needed, we want to find the surface area of the Menger sponge. • When constructing a level from the twenty copies, we see that some outside faces will be covered up. • Thus it is convenient to split the surface area into the inside surface area and outside surface area. 11/6/2020 [Project Name]

Surface Area of Menger Sponges • For the inside surface area, we note that IS (MS 1) = 24, since each of the twelve cubes of degree two has two interior faces. • Now level i is composed of twenty copies of level i-1. In addition, there are 24 inside faces that are level i-1 Sierpinski carpets. • Thus we find the recurrence relation IS(MSi) = 20·IS(MSi-1)+24· 8 i-1. 11/6/2020 [Project Name]

Surface Area of Menger Sponges • This is the same recurrence relation as for the size of the Menger sponge! • It also has the same initial condition, so it has the same solution, IS(MSi) = 2· 20 i-2· 8 i. • One way of illustrating the relationship between the size and inside surface area is to note that in the level one cube, each vertex of degree two is incident with two edges (and this counts all the edges exactly once) and the corresponding cube has two inside faces. 11/6/2020 [Project Name]

Surface Area of Menger Sponges • For the outside surface area, we note that there are six outside faces. Each face is a level i Sierpinski carpet, which has order 8 i. Thus the outside surface area is 6· 8 i. • Adding the inside and outside surface areas, we see that the total surface area of MSi is 2· 20 i+4· 8 i. • The first few terms of this sequence are 6, 72, 1056, 18048, 336384, . . . 11/6/2020 [Project Name]

Partite Sets of Menger Sponges • Calculating the chromatic number of the Sierpinski carpet and Menger sponge graphs is not an interesting problem since these graphs are easily seen to be bipartite. • A more interesting problem is determining the cardinalities of the two partite sets. • First we note that all of the corners of any Menger sponge graph are in the same partite set since any two corners are an even distance apart. • We denote the cardinality of this partite set as n’i, and the cardinality of the other partite set as n"i. • We see that n’ 0 = 1, n"0 = 0, n’ 1 = 8, and n"1 = 12. 11/6/2020 [Project Name]

Partite Sets of Menger Sponges • When we construct level i, set n’i consists of the sets n’i-1 for the corner level i-1 cubes of level i, and also contains the sets n"i-1 of the non-corner level i-1 cubes of level i. • Thus some of the sets 'switch places', so the recurrences for each partite set depend on both. • Specifically, we find n’i = 8 n’i-1+12 n"i-1 and n"i = 12 n’i-1+8 n"i-1. We can write this in matrix form. 11/6/2020 [Project Name]

Partite Sets of Menger Sponges • To solve this discrete dynamical system, we find the eigenvalues of the coefficient matrix A. • We see det(A-λI) = (8 -λ)2 -122 = λ 2 -16λ-80 = (λ-20)(λ+4), so the eigenvalues are λ 1 = 20 and λ 2 = -4. • For λ 1, we see A-λ 1 I = , so the corresponding eigenvector is v 1 = (1, 1). • For λ 2, we see A-λ 2 I = , so the corresponding eigenvector is v 2 = (1, -1). 11/6/2020 [Project Name]

Partite Sets of Menger Sponges • It is known in linear algebra that (n’i, n"i) = A(λ 1)iv 1+B(λ 2)iv 2. We plug in the initial conditions, finding • It is easy to see that A = 1/2 , B = 1/2 is the solution. Thus we find that n’i = (1/2)20 i + (1/2)(-4)i and n"i = (1/2)20 i –(1/2)(-4)i. Notably, which set is larger alternates between successive levels. • The first few terms of the sequence for n’ are 1, 8, 208, 3968, 80128, . . . and the first few terms of the sequence for n" are 0, 12, 192, 4032, 79872, . . . 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • The degree of a vertex is the number of edges incident with it. • Examining the vertices of the Sierpinski carpet graphs, we find vertices of degree two, three, and four. How many are there of each? • We denote by d 2 k(i) the number of vertices of degree k in SCi. • We note that SC 1 has eight vertices of degree two and none of degree three or four. • Each level i Sierpinski carpet is formed from eight copies of level i-1 with eight sets of 3 i-1 edges added in between. 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • We refer to the set of vertices for which one of the dimensions is as large (or small) as possible as a boundary of the carpet. • Each set of edges added between level i-1 carpets will increase the degrees of the vertices in two boundary sets. Thus we must add in or subtract out the vertices in these 16 boundary sets to count the vertices of a particular degree correctly. • In addition, there are four vertices contained in two boundary sets, so we must also make sure that these vertices are counted correctly. 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • We need to know the degrees of the vertices in the boundary sets of a level i carpet. • For level 1, we have three vertices of degree two. To construct the next level, we take three copies of this side, which makes four vertices of degree three. • Let d 12(i) be the number of vertices of degree two in the boundary of the level i Sierpinski carpet and d 13(i) be the number of vertices of degree three. 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • Then we see that d 13(i) = 3 d 13(i-1)+4 and d 12(i) = 3 d 12(i-1)-4. • Thus d 12(i) = A 3 i+B, d 12(1) = 3, and d 12(2) = 5. • Thus 3 = 3 A+B and 5 = 9 A+B. • Solving, we find A = 1/3 and B = 2, so d 12(i) = (1/3)3 i+2 = 3 i-1+2. • Also, d 13(i) = A 3 i+B, d 13(1) = 0, and d 13(2) = 4. • Thus 0 = 3 A+B and 4 = 9 A + B. • Solving, we find A = 2/3 and B = -2, so d 13(i) = (2/3)3 i-2 = 2· 3 i-1 -2. 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • To find the number of vertices of degree two in level i of the Sierpinski carpet, we see that 16 sets of d 12(i-1) vertices have their degrees increased from two to three and must be removed. • Note however that this leads to four vertices whose degrees increase from two to four. They are subtracted out twice and must be added back in once. Thus the recurrence relation is d 22(i) = 8 d 22(i-1)-16(3 i-2+2)+4 = 8 d 22(i-1)-16· 3 i-2 -28. 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • We would suspect that this recurrence relation has a solution of the form A 8 i + B 3 i + C. • To solve it, I used a spreadsheet to calculate many values of the sequence. Then I calculated the ratio between these and 8 i. • These ratios appear to converge to a limit. I guessed the value A that they converged to, and subtracted A 8 i from the sequence. • Then I calculated the ratios of this sequence to 3 i. These ratios appear to converge to a limit B. Subtracting B 3 i from the previous sequence, we find a sequence that appears to converge to a constant C. (See Excel example) 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • For the number of vertices of degree two, this leads to the solution d 22(i) = (1/10)8 i+(16/15)3 i + 4. • To verify that this is correct, we check that d 22(1) = 8 and plug the formula into the recurrence relation. We see d 22(i+1) = 8((1/10)8 i+(16/15)3 i+4)-16· 3 i-1 -28 = (1/10)8 i+1+(16/15)3 i+1+4, verifying the formula. • The first few terms of this sequence are 8, 20, 84, 500, 3540, . . . 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • For the number of vertices of degree three, we see that 16 sets of d 12(i-1) vertices have their degrees increased from two to three, and so must be added in. • Meanwhile, 16 sets of d 13(i-1) vertices have their degrees increased from three to four, and so must be removed. • There also four vertices of degree two which have their degrees increased to four. They were added in twice, and so must be subtracted out twice. 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • Thus the recurrence relation is d 23(i) = 8 d 23(i-1)+16(3 i-2+2)-16(2· 3 i-2 -2)-8 = 8 d 23(i-1)-16· 3 i-2+56. • Using a similar method to that for the previous problem, we find the solution d 23(i) = (3/5)8 i+(16/15)3 i-8. • We check that d 23(1) = 0 and d 23(i+1) = 8((3/5)8 i+(16/15)3 i-8)-16· 3 i-1+56 = (3/5)8 i+1+(16/15)3 i+1 -8, verifying the formula. • The first few terms of this sequence are 0, 40, 328, 2536, 19912, . . . 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • For the number of vertices of degree four, we see that 16 sets of d 13(i-1) vertices have their degrees increased from three to four, and so must be added in. • There also four vertices of degree two which have their degrees increased to four, and so must be added in. • Thus the recurrence relation is d 24(i) = 8 d 24(i-1)+16(2· 3 i-2 -2)+4 = 8 d 24(i-1)+32· 3 i-2 -28. 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • Using the spreadsheet method, we find the solution d 24(i) = (3/10)8 i-(32/15)3 i+4. • We check that d 23(1) = 0 and d 24(i+1) = 8((3/10)8 i-(32/15)3 i+4)+32· 3 i-1 -28 = (3/10)8 i+1(32/15)3 i+1+4, verifying the formula. • The first few terms of this sequence are 0, 4, 100, 1060, 9316, . . . 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • Summarizing, we have found that for i≥ 1, 11/6/2020 [Project Name]

Vertex Degrees in Sierpinski Carpets • How can we be certain that we have not made a mistake in deriving these formulas? These formulas should count each vertex exactly once between them. Thus we see that d 22(i)+d 23(i)+d 24(i) = 8 i = n(SCi). • Further, we see that 2 d 22(i)+3 d 23(i)+4 d 24(i) = (16/5)8 i-(16/5)3 i = 2 m(SCi), so the First Theorem of Graph Theory is satisfied. • These formulas also imply that as i grows large, approximately 1/10 of the vertices will have degree 2, 3/5 of the vertices will have degree 3, and 3/10 of the vertices will have degree 4. 11/6/2020 [Project Name]

Degrees in Faces of Menger Sponges • We would like to determine similar formulas for the Menger sponge graphs. Before we can do this, we need to know how many vertices of each degree are in the faces of Menger sponge graphs. • This is related to the formulas for the Sierpinski carpet graphs, but not identical since some of the vertices in the face of a Menger sponge will be adjacent to a vertex in its interior. • We denote by f 2 k(i) the number of vertices of degree k in a face of MSi. 11/6/2020 [Project Name]

Degrees in Faces of Menger Sponges 11/6/2020 [Project Name]

Degrees in Faces of Menger Sponges • We find that all of the vertices of degree four in the Sierpinski carpet become vertices of degree five in the face of a Menger sponge, so f 25(i) = d 24(i) = (3/10)8 i-(32/15)3 i+4. • All of the vertices of degree two in the Sierpinski carpet remain such in the face of a Menger sponge except for the four corners. Thus f 22(i) = d 22(i)-4 = (1/10)8 i+(16/15)3 i. • For vertices of degree three and four, we use recurrence relations. I will omit the details of these cases. 11/6/2020 [Project Name]

Degrees in Faces of Menger Sponges • Summarizing, we have found that for i≥ 1, • As a check, we note f 22(i)+f 23(i)+f 24(i)+f 25(i) = 8 i = n(SCi). 11/6/2020 [Project Name]

Vertex Degrees of Menger Sponges • Now we are ready to determine how many vertices of degree two through six are in the Menger sponge graphs. • We denote by d 3 k(i) the number of vertices of degree k in MSi. • The level i Menger sponge graph is formed from twenty copies of level i-1. There are 24 edges in MS 1, and so there are 48 copies of faces of level i-1 that are joined in building level i. • There also 8· 3 = 24 boundaries of level i-1 Sierpinski carpets that are joined to two level i-1 cubes. There are 8· 3 = 24 corners that are joined to two level i-1 cubes, and eight vertices that are joined to three level i-1 cubes. 11/6/2020 [Project Name]

Vertex Degrees of Menger Sponges • To determine the number of vertices of degree two, we must subtract out f 22(i-1) vertices of degree two on the faces of the Menger sponge. This will subtract out 3 i-2 vertices on the boundaries of these faces twice, so we must add them back in. • Thus the recurrence relation is d 32(i) = 20 d 32(i -1)-48((1/10)8 i-1+(16/15)3 i-1)+24· 3 i-2 = 20 d 32(i 1)-(3/5)8 i-(7/25)3 i. 11/6/2020 [Project Name]

Vertex Degrees of Menger Sponges • Using the spreadsheet method, we find the solution d 32(i) = (1/17)20 i+(2/5)8 i+(216/85)3 i. • We check that d 32(1) = 12 and d 32(i+1) = 20((1/17)20 i+(2/5)8 i+(216/85)3 i)-(3/5)8 i+1 -(72/5)3 i+1 = (1/17)20 i+1+(2/5)8 i+1+(216/85)3 i+1, verifying the formula. • The first few terms of this sequence are 12, 744, 11256, 201960, . . . • Note that the decimal form of the fraction 1/17 is. 0588235294117647 repeating, which was difficult to recognize using the spreadsheet method. 11/6/2020 [Project Name]

Vertex Degrees of Menger Sponges • We use similar techniques to determine 11/6/2020 [Project Name]

Vertex Degrees of Menger Sponges • As a check, we note that d 32(i)+d 33(i)+d 34(i)+d 35(i)+d 36(i) = 20 i = n(MSi), so the order is correct. • Further, we see that 2 d 32(i)+3 d 33(i)+4 d 34(i)+5 d 35(i)+6 d 36(i) = 4· 20 i-4· 8 i = 2 m(MSi), so the First Theorem of Graph Theory is satisfied. • These formulas also imply that as i grows large, approximately 1/17 of the vertices will have degree 2, 24/85 of the vertices will have degree 3, 32/85 of the vertices will have degree 4, 14/85 of the vertices will have degree 5, and 2/17 of the vertices will have degree 6. 11/6/2020 [Project Name]

Degeneracy of Menger Sponges • Suppose we successively delete a vertex of minimum degree of a graph until none remain. • The maximum degree of any of these vertices when deleted is called the degeneracy D(G) of a graph G. A graph is k-degenerate if its degeneracy is at most k. • By reversing the sequence of vertices formed by deleting a vertex of minimum degree, we can construct the graph so that each vertex, when added, is adjacent to at most D(G) vertices. 11/6/2020 [Project Name]

Degeneracy of Menger Sponges • This is relevant to constructing a physical model of the Menger sponge, since if it constructed by adding one level 0 cube at a time, it is easier to add a cube adjacent to as few other previously added cubes as possible. • Note that in practice, we constructed the level 3 Menger sponge by constructing smaller chunks (mainly level 1 and 2 cubes) and then combining them together, missing some possible connections. 11/6/2020 [Project Name]

Degeneracy of Menger Sponges • It is not hard to see that any Menger sponge graph is 3 -degenerate. Note that MSi is a subgraph of the Cartesian product of three paths, P 3 i×P 3 i. • Any subgraph of Pa×Pb×Pc has a vertex of degree at most three, which can be found by maximizing (or minimizing) the first coordinate, then maximizing (or minimizing) the second coordinate, then maximizing (or minimizing) the third coordinate. (Aside from some degenerate cases, any such graph will have at least eight such vertices. ) • Thus any Menger sponge graph is 3 -degenerate. 11/6/2020 [Project Name]

Degeneracy of Menger Sponges • We see MS 1 is 2 -degenerate, but what about larger Menger sponge graphs? • The k-core of a graph is the maximal subgraph of minimum degree at least k of the graph (if it exists). • Building a level 2 cube, we see that any face of a level 1 cube has a ring of eight vertices. Some of these faces will be face-to-face with other faces of level 1 cubes. In the Menger sponge graphs, this creates the subgraph C 8×K 2, which is 3 -regular. • Thus for i≥ 2, the Menger sponge graph MSi has a 3 -core, so D(MSi) = 3. 11/6/2020 [Project Name]

3 -Cores of Menger Sponges • How many vertices are in the 3 -core of a Menger sponge graph? • As seen above, any vertex in a face of a level 1 cube adjacent to a vertex in another level 1 cube is in the 3 -core. • Vertices for which this is not the case must have degree 2 or 3 in the entire graph. Vertices with degree 2 are certainly not in the 3 -core. • A vertex of degree 3 is the corner of a level 1 cube. Since it is not adjacent to vertices in any of the three faces it is contained it, its level 1 cube must be in the corner of its level 2 cube. 11/6/2020 [Project Name]

3 -Cores of Menger Sponges • Proceeding similarly, we see that this vertex must be in the corner of the entire cube. Then its three neighbors all have degree two, so it is not in the 3 core. • Thus the only degree three vertices not in the 3 core are the eight corners of the cube. • Thus the order of the 3 -core n(C 3(MSi)) = n(MSi)d 32(i)-8 = 20 i-((1/17)20 i+(2/5)8 i+(216/85)3 i-8 = (16/17)20 i-(2/5)8 i-(216/85)3 i-8. • The first few terms of this sequence are 0, 320, 7248, 148736, 2998032, . . . 11/6/2020 [Project Name]

3 -Cores of Menger Sponges • What can we say about the structure of the 3 -core of a Menger sponge graph? We note that any Menger sponge has three planes of symmetry that are parallel to its faces. • In any nontrivial sponge, the vertices on these planes all have degree 2. This is easily verified by induction. • This implies that the 3 -core is disconnected, and its number of components is divisible by 8. For MS 2, there are exactly eight components, each of which is composed of three partially overlapping copies of C 8×K 2. 11/6/2020 [Project Name]

3 -Cores of Menger Sponges • We note that each of these components contains vertices in three of the faces of the level 2 cube. • When level 2 cubes are combined to form level three cubes, vertices that are in the three faces of a corner cube that are face-toface with other faces, or vertices in those faces, will be contained in a single component of the 3 -core. 11/6/2020 [Project Name]

3 -Cores of Menger Sponges • Seven of the eight components of the 3 -core of the corner cube will be part of this larger component, along with four components of the 3 -cores of each of the three cubes adjacent to it. • This larger component also contains vertices in three of the faces of the level 3 cube. • Proceeding similarly, we find that for i≥ 1, the 3 -core of MSi has 8(i-1) components, which can be divided into i-1 classes of eight isomorphic components. 11/6/2020 [Project Name]

Thank You! • Thanks to Gerard Venema and Allen Schwenk for their comments. • See also: • http: //www. megamenger. com/ • http: //www. calvin. edu/academics/department s-programs/mathematics-statistics/megamenger/ 11/6/2020 [Project Name]