# 5. GRAPHS ON SURFACES

Save this PDF as:

Size: px
Start display at page:

Download "5. GRAPHS ON SURFACES"

## Transcription

1 . GRPHS ON SURCS.. Graphs graph, by itself, is a combinatorial object rather than a topological one. But when we relate a graph to a surface through the process of embedding we move into the realm of topology. graph is a set, X, of elements together with a relation on X. We often draw pictures of graphs, where the elements of X are represented by dots and the relation by a set of arrows connecting certain dots to others. The elements of X are called vertices and the arrows joining vertices related by the relation are called edges. xample : The following is a graph on the set {,,, }: {(, ), (, ), (, ), (, ), (, ), (, ), (, )}. These are the (directed) edges and they can be represented by the following diagram. There are vertices and edges. However we ll only consider undirected graphs with no loops. loop is an edge (x, x) from a vertex to itself. n undirected graph is one where the relation is symmetric, that is, if x is related to y then y is related to x. So if there s an arrow in one direction there s always one in the opposite direction. In an undirected graph there s no need to use arrows since we know that the relation goes in both directions, so we simply use edges without arrows. xample : The following is an undirected graph with no loops. rom now on, when we use the word graph, we ll mean that it s undirected and has no loops. graph is a combinatorial structure where the only consideration is which vertices are adjacent to which. When we draw a graph the positions of the points representing the vertices are arbitrary. So are the routes of the edges. The edges needn t be straight, they re allowed to cross over other edges, and they could even wind around in more complicated ways. However we usually draw a graph in such a way that it gives as simple a picture as possible. xample : The graph in example could be redrawn as in the diagram on the left, but would look much better when drawn as the one on the right.

2 Notice that in the above example it s possible to draw the graph without any of the edges crossing. This isn t always possible, and we ll be very much concerned with the problem of when it is possible and when it isn t. In a graph we say that two vertices are adjacent if there s an edge between them. xample : In example, vertices and are adjacent but and are not. Two graphs X and Y are equivalent if there s a - and onto map f: X Y such that x and x are adjacent in X if and only if f(x ) and f(x ) are adjacent in Y. xample : The two graphs in example are equivalent. nd both are equivalent to the following graph. (It s essentially the same graph but with the vertices labelled differently.) B C D xample : The following list contains all the graphs with vertices or less. very graph with up to vertices is equivalent to exactly one of these. They ve been systematically classified according to the number of vertices, V and edges,. V = = 0 V = = 0 V = = V = = 0 V = = V = = V = = V = = 0 V = = V = = V = =

3 . V = = V = = V = = The complete graph on n vertices, denoted by K n, is the graph where every vertex is n adjacent to every other vertex. The number of edges in K n is clearly the binomial coefficient xample : The following are the complete graphs on vertices or less. K K K K K K nother important family of graphs consists of the graphs K m,n for various values of m and n (they don t have a name, just a symbol). The graph K m,n has m + n vertices divided into two subsets, one of size m and the other of size n. very vertex in one subset is adjacent to every vertex in the other, but there are no edges connecting two vertices within the same subset. xample : The following is K, : This graph was once featured in an ir New Zealand advertisement, where the vertices consisted of the cities Brisbane, Sydney, Melbourne, uckland, Wellington and Christchurch. The edges represented the trans-tasman routes... mbedding a Graph in a Surface graph has vertices and edges. So has a map, so what s the difference? Maps have faces, while graphs don t. It s the existence of faces on a map that give it its topological significance. Consider the following famous puzzle, called the Utilities Puzzle. You have three houses and three utilities. The utilities are a gasworks, a power station and a water reservoir. They have to pump gas, electricity and water to each of the three houses. But they have to do this so that the pipes and cables don t intersect. You see this is a -dimensional problem. In real life (-dimensional) there s no problem at all. Gas pipes can be routed over electrical cables or under water pipes. But we have to solve the puzzle in dimensions. The problem is to draw the three houses and three utilities on a sheet of paper, and draw lines to represent the pipes and cables, in such a way that they only meet one another at endpoints. You might like to have a go at this problem. 9

4 xample 9: The following is a near solution. Clearly we can t put in the remaining water pipe without crossing the lines we ve drawn already. But that doesn t prove that the problem has no solution. Perhaps you can think of better places to put the six vertices or perhaps you can route the first eight edges in a really clever way so that the last one can be drawn without crossing the others. Don t spend too long on this puzzle, because it s impossible! If you ve tried to solve it for a few minutes you ll come to this conclusion, though you won t have a proof. But beware! Haven t you ever attempted puzzles where, after trying in vain for many minutes, you become quite convinced that it s impossible, only to have someone come along and show you a really clever solution? Not in this case, though. We re going to prove that this puzzle is impossible! In the language of graph theory the network of pipes and cables is the graph K,. We want to draw this graph in the plane so that edges meet only at vertices. Or, to use a new concept that we re just about to define, the problem is to embed K, on a disk. graph can be embedded (is embeddable) on a surface if it can be drawn on that surface so that edges meet only at vertices. So we ll be proving that K, is not embeddable on a disk... Planarity graph is defined to be planar if it can be embedded in a disk. So K, isn t planar. But K, is: So is K : Since a disk can be cut out of a sphere, any planar graph can be embedded in a sphere. On the other hand, if we can embed a graph in a sphere, we can cut out a small hole in the middle of one of the faces and we have an embedding of the graph in a disk. (Remember that a disk is homeomorphic to a sphere with one hole.) 0

5 and = So a graph is planar if and only if it can be embedded in a sphere. This is useful because often a sphere is more convenient to work with. To prove that a graph is planar we can simply draw it, with edges not crossing. But how do we show that a graph, such as, K, or K is not planar? The technique discussed here is to work out the average number of edges per face and compare this to the smallest number of edges per face. But wait a minute. Graphs don t have faces! That s true, but a graph embedded in a surface becomes a map, and maps have faces. So we suppose that the graph is planar, that is, it can be embedded in a sphere. But how can we count the number of faces if we re only supposing that the graph can be embedded? The answer is to use uler s formula: V + = χ or planarity we use χ =, the uler characteristic of the sphere. Why not χ= for the disk? The answer is that we ll be assuming that there s a face on both sides of each edge of the map. If we have boundaries this will not be so. xample 0: K is not planar. Proof: or K we have V = and = 0. Suppose that K is planar. Then embedding it in a sphere we can deduce that the number of faces must be: = + V =. 0 The average number of edges per face must therefore be =. Why not just? The reason is that every edge is associated with two faces one on each side. So if you were to split each edge lengthwise, so that each half edge was associated with only one face, you d have half edges to share among the faces. Now we wanted to prove that K can t be embedded in a sphere and we started out by supposing that it can be. We re clearly looking for a contradiction. So what s contradictory about the average number of edges per face being? What s wrong is that it s less than. very face must be surrounded by at least edges ( face bounded by edges would require that the two edges connect the same two vertices, and a face bounded by just edge would mean that the graph has a loop.) Now the average of a collection of numbers can t be less than the smallest of them. So here we have our contradiction! xample : K, is not planar.

6 < = Proof: Here V = and = 9. Suppose that K, can be embedded in a sphere. The resulting map would have to have faces where + 9 =, that is, it must have faces. The average number of edges per face would therefore be. This isn t less than, so where s the contradiction? The contradiction is that it s less than. You see, in this graph there are no cycles of length. ach edge takes you from one set of vertices to the other. Going along another edge must take you back to a different vertex in the first set. The smallest cycles in this graph therefore have length. The boundary of a face must be a cycle in the graph. So the smallest number of edges for each face is. But the average of these numbers is less than. This can t be, and so we have our contradiction. The girth of a graph is the length of the shortest cycle. The girth of K is but the girth of K, is. We get a contradiction if the average number of edges per face is less than the girth... Proving that a Graph is not mbeddable in a Surface. Suppose a graph has edges and V vertices and the girth is g. If the graph can be embedded in a surface with uler characteristic χ then it will produce a map with faces where = + V. If < g then we have a contradiction. That graph can t possibly be embedded in that surface. But beware. The g is a one-way test. If, on the contrary, we find that g we d be wrong to conclude that the graph is embeddable. It may be embeddable, or on the other hand it may not be embeddable but this test isn t powerful enough to show it. This is one of the most common errors in graph embedding. Remember that essentially the only way to prove that a graph IS embeddable is to actually draw an embedding. 0BTest for Non-mbeddablility only applies to surfaces with no boundaries. B = χ + V < g graph with edges, V vertices and girth g is NOT g The graph may be embeddable. On the other hand it might not be. In this case the TST ILS. embeddable in any surface with uler characteristic χ If the surface has boundaries, remove them and use the corresponding surface with no boundaries. xample : The graph K is embeddable in the torus. b a b b K a torus

7 = =. Non-Proof: or K, V = and = 0. Suppose that K can be embedded in a torus. The resulting map will have faces where: + 0 = 0 (remember that χ = 0 for the torus). So =. Now. The girth of K is. So > g. What does this prove? bsolutely nothing! This whole calculation achieves nothing beyond confirming that the claimed result might be true. Proof: To prove that K is embeddable in a torus we simply take a torus and draw K on it (in such a way that edges don t cross). Of course it s easier to use a polygon with identified edges rather than the surface of a real -dimensional torus. a b b a Note that there are alternative routes we could have followed for the - connection. We should only include one of them. (The apparent double-up of the - connection is an illusion as the - edge at the top and the bottom of the square are, in reality, the same edge.) The duplication of the labels of the vertices comes about because of the identification of the edges. Check that every vertex, from to, is adjacent to every other. Theorem : The largest value of n for which K n is embeddable in the torus is. Solution: or K n, V = n and = ½ n(n ). Suppose that K n can be embedded in a torus. The resulting map will have faces where: n n n + ½ n(n ) = 0. So = n(n ) n(n ) n. The girth of K n n is. So n, in which case n 9n n n. n Hence n n 0. Since n > 0 we may divide this inequality by n to obtain n. We now show that K can be embedded in a torus.

8 < = = < = < < Suppose you have a graph G and a surface S an you want to determine whether or not G is embeddable in S. The first thing you might try is the g test. But before you do, there are a couple of things you should do first to both the graph and the surface: () Clearly holes are irrelevant to the embedding problem. graph is embeddable in a surface with holes if and only if it is embeddable in the corresponding surface with no holes. But the < g test assumes that the surface has no holes. So the very first thing you must do is to: replace the surface by the corresponding surface with no holes. If you re asked whether a graph is embeddable in a disk, or a cylinder, you replace that surface by a sphere. In testing embeddability in a Möbius Band you replace it by a projective plane. () If the graph has a vertex of degree you can clearly remove that vertex, and the edge joined to it, without affecting embedability. urthermore, you can remove a vertex of degree and join the edges on either side into a single edge, without affecting embedability. However removing vertices of degree may reduce the girth, which will weaken the test. So before you apply the g test you should: remove vertices of degrees and remove any vertices of degree if so doing does not change the girth. Having modified the graph and surface appropriately you can now apply the < g test. If it holds, the graph is not embeddable in the surface. If g then the test fails, that is, it is inconclusive. The next thing to do would be to attempt to draw an embedding. xample : The following graph is known as the Petersen Graph: V = 0 = If embedded in a sphere = + 0 = 0 f = g = So the Petersen graph can t be embedded in a sphere. In other words it isn t planar. or a projective plane, χ = and so if this graph can be embedded in a PP, = + 0 = 0 and f = = g. The g test fails. So it might be possible to embed the Petersen graph in a projective plane. On the other hand it might be impossible, which isn t much help! But in fact we can embed the Petersen graph in a projective plane by actually displaying such an embedding:

9 is and.. b 9 9 a 0 a 0 9 b Whenever you display such an embedding it s important to do what we ve done here label the vertices of the original graph and the embedded graph so that it s easy to check that it s indeed the same graph. (Vertex is adjacent to in both graphs, it s not adjacent to vertex in either graph, and so on.) + m + Theorem : If K n is embeddable in a sum of m toruses then n Proof: Suppose n. Then for K n, V = n, = ½ n(n ) and the girth is g =. The uler characteristic of the sum of m toruses is m. So = + χ V = ½ [n n m + ]. n(n ) Since K n can be embedded, = n g =, that is, n n (m ) 0. n m + The zeros of n ± m + ± m + n (m ) are so n + m + The smallest value of, occurring when m = 0. The sum of zero toruses is a sphere and since K and K can be embedded in a sphere the cases n = and n = automatically hold... Printed Circuit Boards. printed circuit board has electronic components laid out on both sides of a board with the connecting tracks printed on the board. They can be considered as graphs where the vertices occur on both sides of the surface and the edges on each side form a planar graph. We insist that each edge lies entirely on one side or the other. graph is -embeddable in a -sided (orientable) surface S if it is the union of two subgraphs, each of which is embeddable in S.

10 and m Theorem : If K n is -embeddable in a sum of m toruses then n Proof: Suppose that K n is -embeddable in mt. Let the numbers of vertices in the subgraphs, the graphs representing the two sides, be V and V repectively and let the number of edges be and. Clearly each V i n and the uler characteristic of the sum of m toruses is m. Suppose that the numbers of faces of the corresponding maps are and. Since each i subgraph must be planar we must have, for i =,. Since every vertex must be connected to every other, on one side or the other, we have + = ½ n(n ). Hence, for each i, i i = ( i + m V i ) ( i + m n) So i n + m. We therefore have ½ n(n ) = + n + m. It follows that n n (m ) 0. The zeros of n ± 9 + 9(m ) n (m ) are i + + 9m so n xample : The largest value of n for which K n can be -embedded in a plane is, 9 or 0. If K n is -embeddable in a plane then from Theorem, n 0 (taking m = 0). The following is a -embedding of K in a plane. Is it possible to -embed K 9 or even K 0?

11 XRCISS OR CHPTR xercise : Show that the following graph may be embedded in a projective plane, but not in a disk. xercise : Determine (with reasons) which of the following graphs are planar, which can be embedded in a Möbius Band and which can be embedded in a torus. (a) (b) xercise : (a) What is the girth of the following graph? (b) Prove that this graph is not planar. (c) Can this graph be embedded in a projective plane? (d) Can this graph be embedded in a torus? (e) Can this graph be embedded in a Möbius Band? Give reasons for your answers to (c) to (e).

12 . xercise : or each of the following graphs determine whether or not it can be embedded in a cylinder and whether or not it can be embedded in a torus with holes. Graph Graph B xercise : (i) Draw a diagram for K,, as a graph on vertices. Draw a square with identified edges that represents the Klein bottle. (ii) mbed K, in the Klein bottle (so that edges don t cross). xercise : or each of the following graphs determine whether or not it can be embedded in (i) a cylinder; (ii) a torus. graph graph B xercise : (a) Suppose the following graph was embedded in a sphere. Calculate the average number of edges per face in the resulting map. xplain how this leads to a contradiction. (b) Can the graph in (b) be embedded in a cylinder (Give reasons.) xercise : (a) Show that if K n can be embedded in a surface with no boundaries and uler characteristic χ then + 9 χ n (b) Hence find the largest value of n for which K n can be embedded in the connected sum of a torus, a projective plane and a disk.

13 = xercise 9: The graph K mn (where m n ) consists of two sets S, T of sizes m, n respectively where every vertex in S is joined to every vertex in T but where vertices within the same set are not joined to each other. (a) Draw pictures of K mn for all m, n with m + n =. (b) Prove that if K mn is embeddable in a surface of weight W then, unless m = n =, W ½(m )(n ). (c) ind all values of m, n for which K mn is planar. (Illustrate these cases with suitable picture.) (d) ind all values of m, n for which K mn is not planar but can be embedded in a torus. (Illustrate each of these with a suitable picture, using a polygon with identified edges to represent the torus.) SOLUTIONS OR CHPTR xercise : Deleting the vertices of degree and we obtain the following graph: But this decreases the girth from to. By adding an extra vertex of degree on the edge from to 0 we increase the girth back to. This graph has V = vertices and = 9 edges. Moreover this graph can be embedded in a sphere if and only 0 9 if the original graph can be. But if it this new graph could be embedded in a sphere it would result in a map with = + V = 0 faces. The average number of edges per face would therefore be 0 =. which is less than the girth, a contradiction. Hence the new graph is not embeddable in a sphere and so the original graph is not embeddable in a sphere. But since a disk is a sphere with one hole, and holes do not affect embedability, the original graph can not be embedded in a disk. The following gives an embedding of the graph in a projective plane. 9 0 xercise : (a) By inspection the girth =. The number of edges,, is and the number of vertices, V, is 0. graph is planar if and only if it can be embedded in a sphere. Suppose it is planar. By uler s formula V + = and so =. 9

14 = < The average number of edges per face is = > so it might be planar. Let s try, 0 9 is equivalent to 9 0 and hence the graph is indeed planar. Hence it is embeddable in a Möbius Band and a torus. (b) If it could be embedded in a Möbius Band plane it could be embedded in a Projective Plane, with χ =. g =, V =, = 9, so if embedded in a Projective Plane, = + V =.. The average number of edges per face would be less than the minimum number of edges per face, a contradiction. Hence the second graph is not planar. (c) If embeddable in the Projective Plane = + = 9. Since f = /9 < we get a contradiction, so the graph is not embeddable in a projective plane. (d) If embeddable in the torus = + 0 =. Since f = / = the graph may be embeddable in the torus. To prove that it is requires a specific embedding, which we show below. (e) Since the Möbius Band is D + P, a graph is embeddable in the Möbius Band if and only if it is embeddable in the Projective Plane. So by (c) it is not embeddable in the Möbius Band. 90

15 = xercise : (a) girth = (b) V =, = (rather than count them all, simply notice that each vertex has degree ( edges to each vertex). Multiplying by the number of vertices, and dividing by (since each edge connects two vertices) we get. If the graph is planar it can be embedded in a sphere, and the number of faces that would result would be = + V = 0. 0 <, so we get a contradiction. This graph is not planar. (c) If it was embedded in a projective plane, = 9 and = 9 <. So it cannot be embedded in a projective plane. (d) or the torus, = and =. The test fails. Perhaps it can be embedded in a torus. If it can t it will need another argument to prove this. So we attempt to draw it on a torus. It s a good idea to locate one point at the corners of the diagram (all identified) and some others on the edges. We don t have to use the edges on the diagram made up of edges in the graph but it makes life simpler if we do. So let s put at the four corners and look for a couple of cycles of edges that intersect only in that vertex. There s a cycle ---- and So we plot these points along the edges in appropriate places. There s still vertex to locate. It will go somewhere in the middle of the square. We ll ignore it for the moment. Now we put in the remaining edges, except for those having as an endpoint. Because vertices have two locations we have choices as to which locations we use. We make choices, with an eye to allowing to be able to access the four vertices it will need to be joined to. 9

16 = Clearly, if we locate in the largest face in this diagram it will be able to reach,, and. xercise : Graph : girth = V =, =. Suppose graph can be embedded in a sphere. Let the number of faces be. Then = + =. Hence < girth. So graph cannot be embedded in a sphere. B C D G H can be embedded in a projective plane as follows: B C a G b D H D b G C B a 9

17 Graph B: The graph can be redrawn as: so it is planar. Hence it can be embedded in any surface, in particular the sphere and the Möbius Band. xercise : Graph can be drawn as: so it is embeddable in the disk and hence in all surfaces, including the cylinder and the torus. or graph B, V =, = and if embedded in the sphere the number of faces would be = + =. So f = = =. < = girth. Hence graph B cannot be embedded in the sphere and hence not in the cylinder. But we can embed B in a torus: can be redrawn as There s just one crossing left. But we can join to by going up and to the right to the middle of the vertical and reappear at the corresponding point on the left (since the rectangle represents a torus). 9

18 = <. xercise : (a) V =, =. If embedded in a sphere the resulting map would have = + = faces and so the average number of edges per face would be f = / = ½. Since this is less than the girth (which is ), we get a contradiction. Hence this graph cannot be embedded in a sphere. (b) Since the cylinder is a sphere with two holes and since holes do not affect embeddability, a surface is embeddable in a cylinder if and only if it is embeddable in a sphere. Hence the above graph is not embeddable in a cylinder. However it is embeddable in a torus. xercise : (a) The number of vertices of K n is V = n and the number of edges is = ½ n(n ). Its girth is g =. Suppose K n can be embedded in S. The number of faces would be = V χ = ½ (n n + χ). n n Then n. n + χ n n Hence n and so n n + χ 0. n + χ + 9 χ Hence n (b) χ 0 (T + P) = + = so χ(t + P) = = so by part (a), if K n can be embedded in T + P + then n. So K n cannot be embedded in T + P if n. But K can be embedded in a torus and hence can be embedded in T + P so the largest value of n for which K n can be embedded in T + P is n =. xercise 9: (a) K K K 9

19 (b) Suppose K mn is embeddable in a surface (with no holes) of weight W. The uler characteristic of this surface is χ = W. Case I: n. Here g =, V = m + n, = mn. Thus = ( W) + mn m n and so mn f =. Thus mn m n + W mn and so mn m n + W W mn m n + = (m )(n ) and so W ½(m )(n ). Case II: n =, m >. Then ½(m )(n ) 0 W. (c) If n =, K mn is embeddable in a plane (and hence a torus). Suppose n and that K mn is embeddable in a torus (with weight W = ). rom (b) (m )(n ). The only solutions of this inequality for positive integers m, n with m n are: n =, m = any value (These are planar so can be ignored); n =, m = ; n =, m = ; n =, m = ; n =, m = ; n =, m =. The last five all include K, which is not planar so none of these is planar. It remains to prove that they are indeed embeddable in the torus. K can be embedded in a torus and hence so can K and K and K. B C That just leaves K. D B C 9

20 9

### Graph Theory Problems and Solutions

raph Theory Problems and Solutions Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles November, 005 Problems. Prove that the sum of the degrees of the vertices of any finite graph is

More information

### Mathematical Induction. Mary Barnes Sue Gordon

Mathematics Learning Centre Mathematical Induction Mary Barnes Sue Gordon c 1987 University of Sydney Contents 1 Mathematical Induction 1 1.1 Why do we need proof by induction?.... 1 1. What is proof by

More information

### Sum of Degrees of Vertices Theorem

Sum of Degrees of Vertices Theorem Theorem (Sum of Degrees of Vertices Theorem) Suppose a graph has n vertices with degrees d 1, d 2, d 3,...,d n. Add together all degrees to get a new number d 1 + d 2

More information

### Graph Theory Lecture 3: Sum of Degrees Formulas, Planar Graphs, and Euler s Theorem Spring 2014 Morgan Schreffler Office: POT 902

Graph Theory Lecture 3: Sum of Degrees Formulas, Planar Graphs, and Euler s Theorem Spring 2014 Morgan Schreffler Office: POT 902 http://www.ms.uky.edu/~mschreffler Different Graphs, Similar Properties

More information

### Induction Problems. Tom Davis November 7, 2005

Induction Problems Tom Davis tomrdavis@earthlin.net http://www.geometer.org/mathcircles November 7, 2005 All of the following problems should be proved by mathematical induction. The problems are not necessarily

More information

### 3. Mathematical Induction

3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)

More information

### 136 CHAPTER 4. INDUCTION, GRAPHS AND TREES

136 TER 4. INDUCTION, GRHS ND TREES 4.3 Graphs In this chapter we introduce a fundamental structural idea of discrete mathematics, that of a graph. Many situations in the applications of discrete mathematics

More information

### Introduction to Graph Theory

Introduction to Graph Theory Allen Dickson October 2006 1 The Königsberg Bridge Problem The city of Königsberg was located on the Pregel river in Prussia. The river divided the city into four separate

More information

### Planar Graphs. Complement to Chapter 2, The Villas of the Bellevue

Planar Graphs Complement to Chapter 2, The Villas of the Bellevue In the chapter The Villas of the Bellevue, Manori gives Courtel the following definition. Definition A graph is planar if it can be drawn

More information

### Pick s Theorem. Tom Davis Oct 27, 2003

Part I Examples Pick s Theorem Tom Davis tomrdavis@earthlink.net Oct 27, 2003 Pick s Theorem provides a method to calculate the area of simple polygons whose vertices lie on lattice points points with

More information

### WRITING PROOFS. Christopher Heil Georgia Institute of Technology

WRITING PROOFS Christopher Heil Georgia Institute of Technology A theorem is just a statement of fact A proof of the theorem is a logical explanation of why the theorem is true Many theorems have this

More information

### If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C?

Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question

More information

### ALGEBRA. sequence, term, nth term, consecutive, rule, relationship, generate, predict, continue increase, decrease finite, infinite

ALGEBRA Pupils should be taught to: Generate and describe sequences As outcomes, Year 7 pupils should, for example: Use, read and write, spelling correctly: sequence, term, nth term, consecutive, rule,

More information

### Solutions to Homework 10

Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x

More information

### Pythagorean Triples. Chapter 2. a 2 + b 2 = c 2

Chapter Pythagorean Triples The Pythagorean Theorem, that beloved formula of all high school geometry students, says that the sum of the squares of the sides of a right triangle equals the square of the

More information

### Comments on Quotient Spaces and Quotient Maps

22M:132 Fall 07 J. Simon Comments on Quotient Spaces and Quotient Maps There are many situations in topology where we build a topological space by starting with some (often simpler) space[s] and doing

More information

### ON SELF-INTERSECTIONS OF IMMERSED SURFACES

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 126, Number 12, December 1998, Pages 3721 3726 S 0002-9939(98)04456-6 ON SELF-INTERSECTIONS OF IMMERSED SURFACES GUI-SONG LI (Communicated by Ronald

More information

### The Graphical Method: An Example

The Graphical Method: An Example Consider the following linear program: Maximize 4x 1 +3x 2 Subject to: 2x 1 +3x 2 6 (1) 3x 1 +2x 2 3 (2) 2x 2 5 (3) 2x 1 +x 2 4 (4) x 1, x 2 0, where, for ease of reference,

More information

### 5.4 The Quadratic Formula

Section 5.4 The Quadratic Formula 481 5.4 The Quadratic Formula Consider the general quadratic function f(x) = ax + bx + c. In the previous section, we learned that we can find the zeros of this function

More information

### Induction. Margaret M. Fleck. 10 October These notes cover mathematical induction and recursive definition

Induction Margaret M. Fleck 10 October 011 These notes cover mathematical induction and recursive definition 1 Introduction to induction At the start of the term, we saw the following formula for computing

More information

### Euclidean Geometry. We start with the idea of an axiomatic system. An axiomatic system has four parts:

Euclidean Geometry Students are often so challenged by the details of Euclidean geometry that they miss the rich structure of the subject. We give an overview of a piece of this structure below. We start

More information

### Pigeonhole Principle Solutions

Pigeonhole Principle Solutions 1. Show that if we take n + 1 numbers from the set {1, 2,..., 2n}, then some pair of numbers will have no factors in common. Solution: Note that consecutive numbers (such

More information

### Math 2443, Section 16.3

Math 44, Section 6. Review These notes will supplement not replace) the lectures based on Section 6. Section 6. i) ouble integrals over general regions: We defined double integrals over rectangles in the

More information

### A couple of Max Min Examples. 3.6 # 2) Find the maximum possible area of a rectangle of perimeter 200m.

A couple of Max Min Examples 3.6 # 2) Find the maximum possible area of a rectangle of perimeter 200m. Solution : As is the case with all of these problems, we need to find a function to minimize or maximize

More information

### Homework Exam 1, Geometric Algorithms, 2016

Homework Exam 1, Geometric Algorithms, 2016 1. (3 points) Let P be a convex polyhedron in 3-dimensional space. The boundary of P is represented as a DCEL, storing the incidence relationships between the

More information

### WHERE DOES THE 10% CONDITION COME FROM?

1 WHERE DOES THE 10% CONDITION COME FROM? The text has mentioned The 10% Condition (at least) twice so far: p. 407 Bernoulli trials must be independent. If that assumption is violated, it is still okay

More information

### Unit 3: Triangle Bisectors and Quadrilaterals

Unit 3: Triangle Bisectors and Quadrilaterals Unit Objectives Identify triangle bisectors Compare measurements of a triangle Utilize the triangle inequality theorem Classify Polygons Apply the properties

More information

### 6.3 Conditional Probability and Independence

222 CHAPTER 6. PROBABILITY 6.3 Conditional Probability and Independence Conditional Probability Two cubical dice each have a triangle painted on one side, a circle painted on two sides and a square painted

More information

### CHAPTER 3 Numbers and Numeral Systems

CHAPTER 3 Numbers and Numeral Systems Numbers play an important role in almost all areas of mathematics, not least in calculus. Virtually all calculus books contain a thorough description of the natural,

More information

### POWER SETS AND RELATIONS

POWER SETS AND RELATIONS L. MARIZZA A. BAILEY 1. The Power Set Now that we have defined sets as best we can, we can consider a sets of sets. If we were to assume nothing, except the existence of the empty

More information

### Shapes Puzzle 1. Shapes Puzzle 2. Shapes Puzzle 3

Shapes Puzzle The twelve pentominoes are shown on the left. On the right, they have been placed together in pairs. Can you show which two pentominoes have been used to make each shape? (Each pentomino

More information

### Section 1.1. Introduction to R n

The Calculus of Functions of Several Variables Section. Introduction to R n Calculus is the study of functional relationships and how related quantities change with each other. In your first exposure to

More information

### Midterm Practice Problems

6.042/8.062J Mathematics for Computer Science October 2, 200 Tom Leighton, Marten van Dijk, and Brooke Cowan Midterm Practice Problems Problem. [0 points] In problem set you showed that the nand operator

More information

### Topological Treatment of Platonic, Archimedean, and Related Polyhedra

Forum Geometricorum Volume 15 (015) 43 51. FORUM GEOM ISSN 1534-1178 Topological Treatment of Platonic, Archimedean, and Related Polyhedra Tom M. Apostol and Mamikon A. Mnatsakanian Abstract. Platonic

More information

### *1. Derive formulas for the area of right triangles and parallelograms by comparing with the area of rectangles.

Students: 1. Students understand and compute volumes and areas of simple objects. *1. Derive formulas for the area of right triangles and parallelograms by comparing with the area of rectangles. Review

More information

### Arrangements And Duality

Arrangements And Duality 3.1 Introduction 3 Point configurations are tbe most basic structure we study in computational geometry. But what about configurations of more complicated shapes? For example,

More information

### Permutation Groups. Rubik s Cube

Permutation Groups and Rubik s Cube Tom Davis tomrdavis@earthlink.net May 6, 2000 Abstract In this paper we ll discuss permutations (rearrangements of objects), how to combine them, and how to construct

More information

### Taylor Polynomials and Taylor Series Math 126

Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will

More information

### 1 if 1 x 0 1 if 0 x 1

Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

More information

### Lecture 17 : Equivalence and Order Relations DRAFT

CS/Math 240: Introduction to Discrete Mathematics 3/31/2011 Lecture 17 : Equivalence and Order Relations Instructor: Dieter van Melkebeek Scribe: Dalibor Zelený DRAFT Last lecture we introduced the notion

More information

### When is a graph planar?

When is a graph planar? Theorem(Euler, 1758) If a plane multigraph G with k components has n vertices, e edges, and f faces, then n e + f = 1 + k. Corollary If G is a simple, planar graph with n(g) 3,

More information

### Polynomial and Rational Functions

Polynomial and Rational Functions Quadratic Functions Overview of Objectives, students should be able to: 1. Recognize the characteristics of parabolas. 2. Find the intercepts a. x intercepts by solving

More information

### We call this set an n-dimensional parallelogram (with one vertex 0). We also refer to the vectors x 1,..., x n as the edges of P.

Volumes of parallelograms 1 Chapter 8 Volumes of parallelograms In the present short chapter we are going to discuss the elementary geometrical objects which we call parallelograms. These are going to

More information

### Smooth functions statistics

Smooth functions statistics V. I. rnold To describe the topological structure of a real smooth function one associates to it the graph, formed by the topological variety, whose points are the connected

More information

### 1. Sorting (assuming sorting into ascending order) a) BUBBLE SORT

DECISION 1 Revision Notes 1. Sorting (assuming sorting into ascending order) a) BUBBLE SORT Make sure you show comparisons clearly and label each pass First Pass 8 4 3 6 1 4 8 3 6 1 4 3 8 6 1 4 3 6 8 1

More information

### Continuous Functions, Smooth Functions and the Derivative

UCSC AMS/ECON 11A Supplemental Notes # 4 Continuous Functions, Smooth Functions and the Derivative c 2004 Yonatan Katznelson 1. Continuous functions One of the things that economists like to do with mathematical

More information

### Chapter 3. Distribution Problems. 3.1 The idea of a distribution. 3.1.1 The twenty-fold way

Chapter 3 Distribution Problems 3.1 The idea of a distribution Many of the problems we solved in Chapter 1 may be thought of as problems of distributing objects (such as pieces of fruit or ping-pong balls)

More information

### 1. LINEAR EQUATIONS. A linear equation in n unknowns x 1, x 2,, x n is an equation of the form

1. LINEAR EQUATIONS A linear equation in n unknowns x 1, x 2,, x n is an equation of the form a 1 x 1 + a 2 x 2 + + a n x n = b, where a 1, a 2,..., a n, b are given real numbers. For example, with x and

More information

### SHARP BOUNDS FOR THE SUM OF THE SQUARES OF THE DEGREES OF A GRAPH

31 Kragujevac J. Math. 25 (2003) 31 49. SHARP BOUNDS FOR THE SUM OF THE SQUARES OF THE DEGREES OF A GRAPH Kinkar Ch. Das Department of Mathematics, Indian Institute of Technology, Kharagpur 721302, W.B.,

More information

### Labeling outerplanar graphs with maximum degree three

Labeling outerplanar graphs with maximum degree three Xiangwen Li 1 and Sanming Zhou 2 1 Department of Mathematics Huazhong Normal University, Wuhan 430079, China 2 Department of Mathematics and Statistics

More information

### ALGEBRA. Find the nth term, justifying its form by referring to the context in which it was generated

ALGEBRA Pupils should be taught to: Find the nth term, justifying its form by referring to the context in which it was generated As outcomes, Year 7 pupils should, for example: Generate sequences from

More information

### The Geometer s Sketchpad: Non-Euclidean Geometry & The Poincaré Disk

The Geometer s Sketchpad: Non-Euclidean Geometry & The Poincaré Disk Nicholas Jackiw njackiw@kcptech.com KCP Technologies, Inc. ICTMT11 2013 Bari Overview. The study of hyperbolic geometry and non-euclidean

More information

### Lines, Segments, Rays, and Angles

Line and Angle Review Thursday, July 11, 2013 10:22 PM Lines, Segments, Rays, and Angles Slide Notes Title Lines, Segment, Ray A line goes on forever, so we use an arrow on each side to indicate that.

More information

### Series Convergence Tests Math 122 Calculus III D Joyce, Fall 2012

Some series converge, some diverge. Series Convergence Tests Math 22 Calculus III D Joyce, Fall 202 Geometric series. We ve already looked at these. We know when a geometric series converges and what it

More information

### Permutation Groups. Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles April 2, 2003

Permutation Groups Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles April 2, 2003 Abstract This paper describes permutations (rearrangements of objects): how to combine them, and how

More information

### COMBINATORIAL PROPERTIES OF THE HIGMAN-SIMS GRAPH. 1. Introduction

COMBINATORIAL PROPERTIES OF THE HIGMAN-SIMS GRAPH ZACHARY ABEL 1. Introduction In this survey we discuss properties of the Higman-Sims graph, which has 100 vertices, 1100 edges, and is 22 regular. In fact

More information

### CS311H. Prof: Peter Stone. Department of Computer Science The University of Texas at Austin

CS311H Prof: Department of Computer Science The University of Texas at Austin Good Morning, Colleagues Good Morning, Colleagues Are there any questions? Logistics Class survey Logistics Class survey Homework

More information

### Chapter 19. General Matrices. An n m matrix is an array. a 11 a 12 a 1m a 21 a 22 a 2m A = a n1 a n2 a nm. The matrix A has n row vectors

Chapter 9. General Matrices An n m matrix is an array a a a m a a a m... = [a ij]. a n a n a nm The matrix A has n row vectors and m column vectors row i (A) = [a i, a i,..., a im ] R m a j a j a nj col

More information

### MATH 131 SOLUTION SET, WEEK 12

MATH 131 SOLUTION SET, WEEK 12 ARPON RAKSIT AND ALEKSANDAR MAKELOV 1. Normalisers We first claim H N G (H). Let h H. Since H is a subgroup, for all k H we have hkh 1 H and h 1 kh H. Since h(h 1 kh)h 1

More information

### 2015 Junior Certificate Higher Level Official Sample Paper 1

2015 Junior Certificate Higher Level Official Sample Paper 1 Question 1 (Suggested maximum time: 5 minutes) The sets U, P, Q, and R are shown in the Venn diagram below. (a) Use the Venn diagram to list

More information

### MATH STUDENT BOOK. 8th Grade Unit 6

MATH STUDENT BOOK 8th Grade Unit 6 Unit 6 Measurement Math 806 Measurement Introduction 3 1. Angle Measures and Circles 5 Classify and Measure Angles 5 Perpendicular and Parallel Lines, Part 1 12 Perpendicular

More information

### Session 6 Number Theory

Key Terms in This Session Session 6 Number Theory Previously Introduced counting numbers factor factor tree prime number New in This Session composite number greatest common factor least common multiple

More information

### Copyrighted Material. Chapter 1 DEGREE OF A CURVE

Chapter 1 DEGREE OF A CURVE Road Map The idea of degree is a fundamental concept, which will take us several chapters to explore in depth. We begin by explaining what an algebraic curve is, and offer two

More information

### Formal Languages and Automata Theory - Regular Expressions and Finite Automata -

Formal Languages and Automata Theory - Regular Expressions and Finite Automata - Samarjit Chakraborty Computer Engineering and Networks Laboratory Swiss Federal Institute of Technology (ETH) Zürich March

More information

### Dedekind s forgotten axiom and why we should teach it (and why we shouldn t teach mathematical induction in our calculus classes)

Dedekind s forgotten axiom and why we should teach it (and why we shouldn t teach mathematical induction in our calculus classes) by Jim Propp (UMass Lowell) March 14, 2010 1 / 29 Completeness Three common

More information

### In mathematics, there are four attainment targets: using and applying mathematics; number and algebra; shape, space and measures, and handling data.

MATHEMATICS: THE LEVEL DESCRIPTIONS In mathematics, there are four attainment targets: using and applying mathematics; number and algebra; shape, space and measures, and handling data. Attainment target

More information

### Lesson 26: Reflection & Mirror Diagrams

Lesson 26: Reflection & Mirror Diagrams The Law of Reflection There is nothing really mysterious about reflection, but some people try to make it more difficult than it really is. All EMR will reflect

More information

### 2. THE x-y PLANE 7 C7

2. THE x-y PLANE 2.1. The Real Line When we plot quantities on a graph we can plot not only integer values like 1, 2 and 3 but also fractions, like 3½ or 4¾. In fact we can, in principle, plot any real

More information

### 7.7 Solving Rational Equations

Section 7.7 Solving Rational Equations 7 7.7 Solving Rational Equations When simplifying comple fractions in the previous section, we saw that multiplying both numerator and denominator by the appropriate

More information

### This puzzle is based on the following anecdote concerning a Hungarian sociologist and his observations of circles of friends among children.

0.1 Friend Trends This puzzle is based on the following anecdote concerning a Hungarian sociologist and his observations of circles of friends among children. In the 1950s, a Hungarian sociologist S. Szalai

More information

### M-Degrees of Quadrangle-Free Planar Graphs

M-Degrees of Quadrangle-Free Planar Graphs Oleg V. Borodin, 1 Alexandr V. Kostochka, 1,2 Naeem N. Sheikh, 2 and Gexin Yu 3 1 SOBOLEV INSTITUTE OF MATHEMATICS NOVOSIBIRSK 630090, RUSSIA E-mail: brdnoleg@math.nsc.ru

More information

### a) x 2 8x = 25 x 2 8x + 16 = (x 4) 2 = 41 x = 4 ± 41 x + 1 = ± 6 e) x 2 = 5 c) 2x 2 + 2x 7 = 0 2x 2 + 2x = 7 x 2 + x = 7 2

Solving Quadratic Equations By Square Root Method Solving Quadratic Equations By Completing The Square Consider the equation x = a, which we now solve: x = a x a = 0 (x a)(x + a) = 0 x a = 0 x + a = 0

More information

### Spring 2007 Math 510 Hints for practice problems

Spring 2007 Math 510 Hints for practice problems Section 1 Imagine a prison consisting of 4 cells arranged like the squares of an -chessboard There are doors between all adjacent cells A prisoner in one

More information

### Normal distribution. ) 2 /2σ. 2π σ

Normal distribution The normal distribution is the most widely known and used of all distributions. Because the normal distribution approximates many natural phenomena so well, it has developed into a

More information

### Pythagorean Theorem Differentiated Instruction for Use in an Inclusion Classroom

Pythagorean Theorem Differentiated Instruction for Use in an Inclusion Classroom Grade Level: Seven Time Span: Four Days Tools: Calculators, The Proofs of Pythagoras, GSP, Internet Colleen Parker Objectives

More information

### Divisor graphs have arbitrary order and size

Divisor graphs have arbitrary order and size arxiv:math/0606483v1 [math.co] 20 Jun 2006 Le Anh Vinh School of Mathematics University of New South Wales Sydney 2052 Australia Abstract A divisor graph G

More information

### Just the Factors, Ma am

1 Introduction Just the Factors, Ma am The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive

More information

### 7 Relations and Functions

7 Relations and Functions In this section, we introduce the concept of relations and functions. Relations A relation R from a set A to a set B is a set of ordered pairs (a, b), where a is a member of A,

More information

### Geometry Chapter 1. 1.1 Point (pt) 1.1 Coplanar (1.1) 1.1 Space (1.1) 1.2 Line Segment (seg) 1.2 Measure of a Segment

Geometry Chapter 1 Section Term 1.1 Point (pt) Definition A location. It is drawn as a dot, and named with a capital letter. It has no shape or size. undefined term 1.1 Line A line is made up of points

More information

### INCIDENCE-BETWEENNESS GEOMETRY

INCIDENCE-BETWEENNESS GEOMETRY MATH 410, CSUSM. SPRING 2008. PROFESSOR AITKEN This document covers the geometry that can be developed with just the axioms related to incidence and betweenness. The full

More information

### Theory of Computation Prof. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras

Theory of Computation Prof. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras Lecture No. # 31 Recursive Sets, Recursively Innumerable Sets, Encoding

More information

### Lecture 16 : Relations and Functions DRAFT

CS/Math 240: Introduction to Discrete Mathematics 3/29/2011 Lecture 16 : Relations and Functions Instructor: Dieter van Melkebeek Scribe: Dalibor Zelený DRAFT In Lecture 3, we described a correspondence

More information

### SHAPE. Work Journal. Student Name:. QLD191SHD03A Certificate I in Core Skills for Employment & Training - Numeracy. Teacher:

SHAPE Work Journal QLD191SHD03A Certificate I in Core Skills for Employment & Training - Numeracy Student Name:. Start Date: End Date: Teacher: 1 ASSESSMENT PROFILE COVER SHEET Student s Name: Assessment

More information

### 55 questions (multiple choice, check all that apply, and fill in the blank) The exam is worth 220 points.

Geometry Core Semester 1 Semester Exam Preparation Look back at the unit quizzes and diagnostics. Use the unit quizzes and diagnostics to determine which topics you need to review most carefully. The unit

More information

### a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.

Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given

More information

### The Taxman Game. Robert K. Moniot September 5, 2003

The Taxman Game Robert K. Moniot September 5, 2003 1 Introduction Want to know how to beat the taxman? Legally, that is? Read on, and we will explore this cute little mathematical game. The taxman game

More information

### WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT?

WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT? introduction Many students seem to have trouble with the notion of a mathematical proof. People that come to a course like Math 216, who certainly

More information

### Vieta s Formulas and the Identity Theorem

Vieta s Formulas and the Identity Theorem This worksheet will work through the material from our class on 3/21/2013 with some examples that should help you with the homework The topic of our discussion

More information

### 3.4 Complex Zeros and the Fundamental Theorem of Algebra

86 Polynomial Functions.4 Complex Zeros and the Fundamental Theorem of Algebra In Section., we were focused on finding the real zeros of a polynomial function. In this section, we expand our horizons and

More information

### CMPSCI611: Approximating MAX-CUT Lecture 20

CMPSCI611: Approximating MAX-CUT Lecture 20 For the next two lectures we ll be seeing examples of approximation algorithms for interesting NP-hard problems. Today we consider MAX-CUT, which we proved to

More information

### Determinants, Areas and Volumes

Determinants, Areas and Volumes Theodore Voronov Part 2 Areas and Volumes The area of a two-dimensional object such as a region of the plane and the volume of a three-dimensional object such as a solid

More information

### 9.4. The Scalar Product. Introduction. Prerequisites. Learning Style. Learning Outcomes

The Scalar Product 9.4 Introduction There are two kinds of multiplication involving vectors. The first is known as the scalar product or dot product. This is so-called because when the scalar product of

More information

### Open and Closed Sets

Open and Closed Sets Definition: A subset S of a metric space (X, d) is open if it contains an open ball about each of its points i.e., if x S : ɛ > 0 : B(x, ɛ) S. (1) Theorem: (O1) and X are open sets.

More information

### The GED math test gives you a page of math formulas that

Math Smart 643 The GED Math Formulas The GED math test gives you a page of math formulas that you can use on the test, but just seeing the formulas doesn t do you any good. The important thing is understanding

More information

### Linear Programming. Solving LP Models Using MS Excel, 18

SUPPLEMENT TO CHAPTER SIX Linear Programming SUPPLEMENT OUTLINE Introduction, 2 Linear Programming Models, 2 Model Formulation, 4 Graphical Linear Programming, 5 Outline of Graphical Procedure, 5 Plotting

More information

### Situation 23: Simultaneous Equations Prepared at the University of Georgia EMAT 6500 class Date last revised: July 22 nd, 2013 Nicolina Scarpelli

Situation 23: Simultaneous Equations Prepared at the University of Georgia EMAT 6500 class Date last revised: July 22 nd, 2013 Nicolina Scarpelli Prompt: A mentor teacher and student teacher are discussing

More information

### . 0 1 10 2 100 11 1000 3 20 1 2 3 4 5 6 7 8 9

Introduction The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive integer We say d is a

More information

### Handout #Ch7 San Skulrattanakulchai Gustavus Adolphus College Dec 6, 2010. Chapter 7: Digraphs

MCS-236: Graph Theory Handout #Ch7 San Skulrattanakulchai Gustavus Adolphus College Dec 6, 2010 Chapter 7: Digraphs Strong Digraphs Definitions. A digraph is an ordered pair (V, E), where V is the set

More information

### Solution to Homework 2

Solution to Homework 2 Olena Bormashenko September 23, 2011 Section 1.4: 1(a)(b)(i)(k), 4, 5, 14; Section 1.5: 1(a)(b)(c)(d)(e)(n), 2(a)(c), 13, 16, 17, 18, 27 Section 1.4 1. Compute the following, if

More information

### E XPLORING QUADRILATERALS

E XPLORING QUADRILATERALS E 1 Geometry State Goal 9: Use geometric methods to analyze, categorize and draw conclusions about points, lines, planes and space. Statement of Purpose: The activities in this

More information