# 6. GRAPH AND MAP COLOURING

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 6. GRAPH AND MAP COLOURING 6.1. Graph Colouring Imagine the task of designing a school timetable. If we ignore the complications of having to find rooms and teachers for the classes we could propose the following very simple model. We suppose that we have a certain number of subjects taken by a certain number of students. We ll suppose that all subjects require just one class. In setting up the timetable we must ensure that if two subjects have at least one student in common we must allocate different times. So imagine that the subjects are represented by the vertices of a graph and that we have an edge between subjects that have one or more common students. If we colour the vertices in such a way that adjacent vertices have different colours we can arrange for all subjects that were coloured with the same colour to be held at the same time. We could simply colour each vertex with a different colour. This would correspond to timetabling each subject at a different time. But there aren t enough hours in the week and it s important to try to use as few time-slots, or as few colours, as possible. This is one application that would benefit from a solution to the problem of colouring graphs with as few colours as possible. An n-colouring of a graph is an assignment of colours 1, 2,, n to the vertices of a graph so that adjacent vertices have different colours. The chromatic number, (G), of a graph G, is the smallest n for which an n-colouring exists. Example 1: For this graph G, (G) = colouring 4-colouring 3-colouring No 2-coloring exists. Around the pentagon we d have to alternate colours, but because of the odd number of sides we d end up with two adjacent vertices with the same colour. So 3 is the minimum number of colours. Example 2: The chromatic number of the complete graph K n is n. Proof: Since every vertex is adjacent to every other we need to use n distinct colours. = 3 = 5 95

2 Every graph that contains a subgraph with chromatic number n must itself have chromatic number at least n. So, for example, the chromatic number of any graph that contains a triangle must be at least 3. 2 if n is even Example 3: The chromatic number of an n-sided polygon is: 3 if n is odd = 2 = 3 Example 4: The chromatic number of K m,n is 2. Proof: Colour the m vertices one colour and the n vertices the other colour. (K 4,3 ) = 2 Theorem 1: The chromatic number of a tree (connected graph with no cycles) is 2 (except where there s only one vertex). Proof: Choose a vertex V 0. Each vertex has a unique distance from V 0. Use one colour to colour those vertices whose distance from V 0 is even and the other colour for those vertices whose distance is odd. Clearly this is a valid 2-colouring The Chromatic Polynomial of a Graph The chromatic polynomial of a graph G is the number of ways of colouring G with k colours (a polynomial in k). It is denoted by (G)(k). The chromatic number of G is the smallest value of k such that (G)(k) is positive. For simplicity, unless we need to specify the parameter k, we ll write (G)(k) simply as (G). Theorem 2: (K n ) = k(k 1) (k n + 1). Proof: Order the vertices in some way. There are k colours that can be assigned to the first vertex. Any one of the remaining colours can be assigned to the second vertex, and so on. 96

3 Example 5: (K 3 ) = k(k 1)(k 2). k colours k 1 k 2 colours colours A tree is a connected graph in which there are no cycles. In a tree there s a unique path (sequence of edges) from any vertex to any other and so there s a unique distance (counted in terms of the number of edges) between any two vertices. Any vertex whose distance from a given one is greatest must clearly have degree 1. So trees will always have vertices of degree 1. Theorem 3: If G is a tree with n vertices then (G) = k(k 1) n 1. Proof: We use induction on the number of vertices. It s clearly true for n = 1. Suppose the result is true for trees with n vertices and that we have one with n + 1 vertices. Choose any vertex of degree 1 and remove it, together with the associated edge. By induction the resulting tree can be coloured with k colours in k(k 1) n 1 ways. Reinstating the deleted vertex, it s adjacent to only one coloured vertex and so can be coloured in k 1 ways, giving a total number of colourings of k(k 1) n. k 1 k 1 k 1 k k 1 k 1 k 1 k 1 k 1 k 1 We now show how to compute the chromatic polynomial of a graph. The method is inductive. From a given graph G we produce two simpler graphs G and G = and we use their chromatic polynomials to obtain the chromatic polynomial of G. Let G be a graph. Select any two adjacent vertices A and B. Let G be the same graph with the edge AB deleted. Let G = denote the graph G with vertices A and B identified. (This means they become a single vertex and any edge having either A or B as an endpoint now has, instead, this combined vertex.) A B A B A = B G G G = Example 6: A B A B A = B G G G = 97

4 A A A = B B G G G = B Theorem 4: Let G be a graph. Select any two adjacent vertices A and B and let G and G = be defined as above. Then (G) = (G ) (G = ). Proof: If we k-colour G we re free to colour A and B the same colour. But we don t have to. We could just as validly colour them different colours (provided this was consistent with the colourings at the other vertices). There are two types of colourings of G in terms of what happens to A and B A, B are given the SAME colour A, B are given DIFFERENT colours A B A B G Each of these colourings of G gives a valid colouring of G =. G Each of these colourings of G gives a valid colouring of G. G = G Hence G (k) = G= (k) + G (k). The result now follows algebraically. Example 7: ( ) = ( ) ( ) = k(k 1) 3 k(k 1)(k 2) = k(k 1)[(k 1) 2 (k 2)] = k(k 1)(k 2 3k + 3). Example 8: ( ) = ( ) ( ) Now ( ) = ( ).(k 1) = k(k 1) 2 (k 2) since once the triangle has been coloured the remaining vertex can be coloured any colour, except the colour of the vertex to which it is attached. and ( ) = ( ) ( ) = k(k 1) 4 k(k 1)(k 2 3k + 3) from example 7. Hence ( ) = k(k 1) 4 k(k 1)(k 2 3k + 3) k(k 1) 2 (k 2) = k(k 1)[(k 1) 3 (k 2 3k + 3) (k 1)(k 2)] = k(k 1)[k 3 3k 2 + 3k 1 k 2 +3k 3 k 2 + 3k 2] = k(k 1)(k 3 5k 2 + 9k 6) = k(k 1)(k 2)(k 2 3k + 3). It follows that the chromatic number of is 3. This was pretty obvious without all that calculation, but for a much more complicated graph, where the chromatic number is not so obvious, this could be a useful technique. Also, it could form the basis for a computer program to compute the chromatic number of a graph. 98

5 Theorem 5: If C n is a cycle with n vertices then (C n ) = (k 1) n + ( 1) n (k 1). Proof: Induction on n. (C 3 ) = (K 3 ) = k(k 1)(k 2) so it holds for n = 3. Suppose that it holds for n and let G = C n+1. ( ) = ( ) ( ). (C n+1 ) = k(k 1) n (C n ) = k(k 1) n (k 1) n ( 1) n (k 1) = (k 1) n+1 + ( 1) n+1 (k 1). Hence it holds for n + 1 and so, by induction, for all n Map Colouring Graph colouring is a purely combinatorial exercise that doesn t involve any topology. However it s a useful lead-in to map colouring. As with graph colouring, when colouring maps, we must colour adjacent objects with different colours. But in the case of map colouring the objects are the regions, or faces, in the map and adjacent means that they share a common boundary. Adjacent counties in a map of the world are coloured differently to make the borders clear. Though there s no need to use the minimum number of colours it s an interesting question to ask what is the minimum number of colours we can use?. An n-colouring of a map on a surface is a way of assigning one of n labels (colours) to each face so that adjacent faces have different colours. A map is n-colourable if an n-colouring of it exists. The chromatic number, (M), of a map, M, is the smallest n for which M is n-colourable. Example 9: The chromatic number of the following map is 2: 4-colouring 3-colouring 2-colouring but no 1-colouring exists Example 10: The chromatic number of the following map is 3: 3-colouring but no 2-colouring exists 99

6 Example 11: The chromatic number of the following map is 4: 4-colouring but no 3-coloring exists The chromatic number, (S), of a surface, S, is the largest chromatic number for any map that can be drawn on it. Now there s no a priori reason why this couldn t be infinite. However we ll show that every surface has a finite chromatic number. This means that there s an upper bound to the chromatic numbers of maps on a given surface. Upper bounds are harder to arrive at than lower bounds. To find a lower bound for the chromatic number of a surface we only have to draw a map on the surface, and the chromatic number of the surface will have to be at least as big as the chromatic number of that map. For example, the fact that in example 9 we have a map on a disk with chromatic number 2 means that (disk) 2. Example 12: (disk) 4. Example 11 gives a map, M, on a disk with (M) = 4. Hence (disk) 4. If there existed another map on the disk that needs 5 colours, we would have (disk) 5. But in fact it can be shown that every map on a disk can be coloured validly with 4 colours so in fact (disk) = 4. This is the famous Four Colour Theorem, proved in Recall that the degree of a vertex is the number of edges at that vertex. With a map we define the degree of a face to be the number of edges surrounding that face. Example 13: a vertex of degree 5 a face of degree 6 Clearly holes in a surface don t affect its chromatic number. So the chromatic number of a disk (sphere with one hole) will be the same as that of a sphere, or a cylinder (sphere with two holes). If required to find the chromatic number of a surface therefore, the first step is to remove all holes that is replace the surface by the corresponding surface with no holes. This is exactly what we do for embedding questions. Also, like embedding, we remove vertices of degrees 1 and 2 from a map. When removing a vertex of degree 1 we also remove the corresponding edge. When removing a vertex of degree 2 we combine the two edges into a single edge. In what follows we assume that this has already been done and that the surfaces have no boundaries and the degrees of the vertices are at least 3. In the following theorem we make use of the average degree of the vertices and the average degree of the faces. If v is the average degree of the vertices then v = 2E V where E is the number of edges and V is the number of vertices. Why not E V? The answer is that each edge contributes 100

7 to the degree of each of its two endpoints. If we chopped each edge in two, so that the half edges were associated with only one vertex, then we d have 2E of these half edges. by two faces. Similarly, if f is the average degree of the faces, then f = 2E F. Again, each edge is shared Example 14: Here we ve labelled the vertices and faces with their degrees. Note that the outside has to be considered too, because we re regarding this as a map on a sphere. The average degree of the vertices is thus The average degree of the faces is = Note that E = 13, so this is simply 2E V. = Note that this is simply 2E F. Theorem 6: Suppose M is a map (with all vertices of degree 3) on a surface S (with no holes), with Euler characteristic. Let V, F and E be the numbers of vertices, faces and edges for M and let v be the average degree of the vertices and let f be the average degree of the faces. Then: (f 2)(v 2) = 4 2f V = 4 2v F. Proof: In counting the total number of edges by adding the degrees of the vertices or faces, each edge gets counted twice (it joins two vertices) and each face gets counted twice (it is adjacent to two faces). Hence ff = vv = 2E. Thus f V = v F, establishing the second equality of the theorem. Now v = 2E V = 2 + 2v f 2 V = 2(V + F ) V Thus vf = 2f + 2v 2f V = 2 + 2F V 2 V 2f and so vf 2f 2v + 4 = (f 2)(v 2) = 4 V Polyhedra A polyhedron is a region in 3-dimensional space, homeomorphic to a sphere and bounded by flat sides each of which is a polygon. There s no topological reason why the faces have to be flat, or the edges straight so it s convenient to consider polyhedra as maps on a sphere. Theorem 1 can therefore be applied to polyhedra by taking = 2. Theorem 7: The average number of edges per face for any polyhedron is less than 6. Proof: Since v 3 we have f 2 (f 2)(v 2) = 4 4f V < 4, giving f <

8 A consequence of this theorem is that there s no polyhedron where every face is a hexagon. There are, however, polyhedra where every face is a triangle (f = 3) or a square (f = 4) or even a pentagon (f = 5). But that s as far as we can go. If we extend the idea of a polyhedron to maps on other surfaces we can make the following conclusion about toroidal polyhedra (those which are homeomorphic to a torus). Theorem 8: For a map on a torus we have: (f 2)(v 2) = 4. Proof: This follows from theorem 1 by putting = 0. Example 15: The following is a 3 3 array of cubes, with the centre one removed. It can be considered as a map on a torus. There are eight vertices of degree 3, 16 of degree 4 and eight of degree 5. The average degree of the vertices is therefore v = 4. Since each face is a square the average degree of the faces is f = 4. A consequence of theorem 6 is that for a regular polyhedron on a torus the faces must be hexagons, squares or triangles (since v 2 must be an integer and so f 2 must divide 4) Heawood s Theorem Theorem 9: For a map with F faces on a surface of weight 0 the average face degree is at most Proof: (f 2)(v 2) = 4 2v F. Hence f 2 = 6 + 6( 0 2) F 4 v 2 2 F v v 2 = 4 v 2 2 F v

9 Thus f = v 2 1 F 6 1 F, because v 3. So f 6 + 6( 0 2) F. Corollary: If 0 2 then f Proof: Since f F 1 (a given face cannot border more than all the remaining faces) we have: f 6 + 6( 0 2) f + 1 so f 2 5f + 6(1 0 ) 0. The greater of the two zeros of this quadratic is Theorem 10: (Heawood 1910) The chromatic number of a surface S with no boundaries and weight 0 is at most the integer part of if 0 2. If 0 = 0 or 1 (sphere or projective plane) (S) 6. Proof: Suppose 0 2 and let k = Let n be the integer part of k. We prove by induction on the number of faces that every map on S can be n-coloured. A map with one face can be coloured with a single colour and since n 4 we can easily do it with n colours. Now consider the general case. Suppose we have a certain map with at least two faces and assume that any map with fewer faces can be n-coloured. Since f k 1 there exists a face of degree k 1 and hence of degree n 1. Shrink that face to a point. All edges having one of the vertices of the chosen face as an endpoint will now meet at the same vertex. By induction we can n-colour this new map. Now replace the face in the now coloured map. It remains to show that this face can be validly coloured. Since it borders at most n 1 other faces, at most n 1 colours must be avoided if we are to colour this remaining face. With n colours available this can be done. If 0 = 1 or 2 we can repeat the above argument by taking n = 6. Heawood only gives an upper bound. But surprisingly there are only two surfaces for which this upper bound is bigger than it need be. For all other surfaces Heawood s upper bound can be shown to be the actual value. The exceptions are the Klein bottle and the sphere. Heawood says (KB) 7 but in fact (KB) = 6. And Heawood says (sphere) 6 while in fact (sphere) =

10 If S is a surface with no boundaries and weight 0 then: INT if (S) =. 6 if S is the Klein bottle 4 if S is the sphere In many cases we ll verify the chromatic number by using Heawood to get an upper bound and by constructing a suitable map to get the same value as a lower bound. Example 16: (projective plane) = 6. Proof: Heawood says that INT = INT = 6. The following map on the projective plane requires 6 colours so 6. Hence = 6. a b b a Note that, because of the orientation of the edges, if we cross over the edge a from region 6 we enter region 1, but it appears upside down, relative to what it would appear if we reached region 1 by travelling through region 4. Example 17: (torus) = 7. Proof: Heawood says 7. The following map on the torus requires 7 colours and so = 7. a b b a Each face is adjacent to every other face. 104

11 It s remarkable that we can readily establish the chromatic number of surfaces like the projective plane and the torus while something so simple as the sphere (or equivalently the disk) is much more elusive. We ll soon prove that an upper bound for the sphere is 5. But the proof that in fact 4 is an upper bound (and hence the actual chromatic number) is exceedingly complicated The 5-Colour Theorem Theorem 11: (sphere) 5. In other words, every map on a sphere can be coloured using 5 colours. Proof: Suppose there s a map M on the sphere that isn t 5-colourable. We may suppose that we have a minimal counterexample that is, that every map with fewer faces is 5-colourable. Let f be the average number of edges per face. Since f < 6, M has a face F with at most 5 edges. Remove one of these edges and combine the two faces on either side into a single face. This map has fewer faces than M and so is 5-colourable. Having 5-coloured this slightly smaller map, now replace the edge and leave F uncoloured. If there are less than 5 different colours around F then there s one available with which to colour F. But this contradicts our assumption that M is not 5-colourable. Hence F must be adjacent to exactly 5 faces and these must be coloured using 5 distinct colours. Since there s no colour left that we can use for F we ll need to modify the existing colouring. 105

12 Let the faces surrounding F be F 1, F 2, F 3, F 4 and F 5 (in order around the outside of F, either clockwise or anticlockwise). Let the corresponding colours be C 1, C 2, C 3, C 4 and C 5. Delete all faces except those coloured C 1 or C 3. This may well disconnect M into several components. separate components If F 1 and F 3 are in different components of this new map (in other words if you can t go from F 1 to F 3 by sticking to faces coloured C 1 and C 3 ) we could swap colours C 1 and C 3 in one of these components so that now F 1 and F 3 have the same colour (and, of course, maintaining a valid colouring). But now F would be surrounded by only 4 colours and so it could be validly coloured, to give a 5-colouring of M. Since our assumption is that M doesn t have a 5-colouring we conclude that F 1 and F 3 must be in the same component. That is, we can only get from one to the other by passing through C 1 or C 3 coloured faces. Such a closed path must separate F 2 and F 4. That is, one of F 2 and F 4 is on the inside of this closed path and the other is on the outside. (Of course, being on a sphere we can t say unambiguously which is inside and which is outside, but that doesn t matter.) C 1 C 2 C 3 C 4 C 5 106

13 This means that if we were to delete from M all faces except those coloured C 2 and C 4 the faces F 2 and F 4 must be in separate components. So we could swap colours C 2 and C 4 in one of these components and not in the other so that F 2 and F 4 now have the same colour. As before, this frees up one of the 5 colours with which to colour F and so complete the 5-colouring of M. Again this is a contradiction. So we can t escape a contradiction. It follows that our assumption that there exists a non 5- colourable map on the sphere must be false. Hence every map on the sphere can be 5-coloured. EXERCISES FOR CHAPTER 6 Exercise 1: Find the chromatic number of the following graph: Exercise 2: (i) Find the chromatic polynomial of the following graph and express it in factorised form. (ii) Use this polynomial to find the number of ways of colouring the graph in (i) with 2 colours? How many ways with 3 colours? Exercise 3: (i) Let C n be the graph consisting of the vertices and edges of an n-sided polygon. Prove by induction on n that for n 3 the chromatic number of C n is: (C n ) = (k 1)[(k 1) n 1 ( 1) n ]. (ii) Hence find the number of ways of colouring the vertices of the following graph with 3 colours so that adjacent vertices have different colours. 107

14 Exercise 4: Find the number of ways of colouring the vertices of a square with k colours so that adjacent vertices have different colours. Exercise 5: Find the number of ways of colouring the vertices of the graph: with 6 colours available, so that adjacent vertices have different colours. Exercise 6: (i) Draw a diagram for K 6, the complete graph on 6 vertices. Draw a square with identified edges that represents the Klein bottle. (ii) Embed K 6 in the Klein bottle so that edges don t cross. (iii) Use this to draw, on a separate diagram, a map on the Klein bottle with six regions, each adjacent to the other five. (Finish with a clear picture in which only the edges of the map, and not the original graph, are visible. Label the regions 1 to 6. (iv) Let be chromatic number of the Klein bottle. What does your answer to (iv) tell you about? (v) Heawood s formula gives an upper bound for. Calculate this upper bound. (vi) Combine your answers to (v) and (vi) to determine as closely as possible. (Only use what you have proved. Do not simply quote the known value of.) Exercise 7: Let S be the sum of 3 projective planes and let n be the chromatic number of S. (a) Find n. (b) Construct a polygon with identified edges, P, to represent S. (b) Draw an embedding of K n, the complete graph on n vertices, on S. (c) Hence, or otherwise draw a map on P that requires n colours. Exercise 8: Find the chromatic numbers of the following surfaces: (i) the Möbius Band; (ii) a cylinder with 3 holes; (iii) a sum of 10 projective planes. [You may assume that Heawood s upper bound is the actual value for all surfaces with no boundaries except the sphere and the Klein Bottle.] 108

15 SOLUTIONS FOR CHAPTER 6 Exercise 1: Since the graph contains a cycle of length 5 there is clearly no 2-colouring. The following is a 3-coluring The chromatic number of this graph is therefore 3. Exercise 2: Let G 1 = G 2 = and G 3 = Then (G 1 ) = (G 2 ) (G 3 ). 109

16 Let G 4 = and G 5 = Then (G 2 ) = (G 4 ) (G 5 ). Clearly (G 4 ) = k(k 1) 6 and (G 5 ) = k(k 1) 4 (k 2). Hence (G 2 ) = k(k 1) 6 k(k 1) 4 (k 2) = k(k 1) 4 (k 2 3k + 3). Let G 6 = and G 7 = Then (G 3 ) = (G 6 ) (G 7 ). Clearly (G 6 ) = k(k 1) 4 (k 2) and (G 7 ) = k(k 1) 2 (k 2) 2. Hence (G 3 ) = k(k 1) 4 (k 2) k(k 1) 2 (k 2) 2 = k(k 1) 2 (k 2)(k 2 3k + 3). So (G 1 ) = k(k 1) 4 (k 2 3k + 3) k(k 1) 2 (k 2)(k 2 3k + 3) = k(k 1) 2 (k 2 3k + 3) 2. (ii) There are 2 ways of colouring the graph with 2 colours and 108 ways of colouring it with 3 colours. Exercise 3: (i) (C 3 ) = k(k 1)(k 2) = (k 1)[(k 1) 2 1] so the result is true for n = 3. Suppose it is true for n. (C n+1 ) = (L n+1 ) (C n ) where L n+1 is the graph with n + 1 vertices and n edges. Now (L n+1 ) = k(k 1) n and, by the induction hypothesis, (C n ) = (k 1)[(k 1) n 1 ( 1) n-1 ]. Hence (C n+1 ) = k(k 1) n (k 1)[(k 1) n 1 ( 1) n-1 ] = (k 1)[k(k 1) n 1 (k 1) n 1 ( 1) n ] = (k 1)[(k 1) n ( 1) n ]. Hence the result is true for n + 1. (ii) ( ) = ( ) ( ) ( ) = (k 1)[(k 1) 5 + 1] = = 66 when k = 3. ( ) = ( ) ( ) = k(k 1)(k 2)(k 1) 2 k(k 1)(k 2)(k 1) = k(k 1) 2 (k 2) 2 = 12 when k = 3. Hence ( ) = = 54 when k = 3. Exercise 4: = k(k 1) 3 k(k 1)(k 2) = k(k 1)(k 2 2k + 1 k + 2) = k(k 1)(k 2 3k + 3). 110

17 Exercise 5: (i) a b b a (ii) K 6 embedded in the KB 5 4 (iii) Joining a point in the middle of each face to the midpoint of each of the surrounding edges we get a map with 6 faces, each of which is adjacent to the other 5. This map requires 6 colours (iv) We need 6 colours to colour this map on the KB so (KB) 6. (v) Heawood s formula gives (KB) INT so (KB) 7. (vi) So (KB) = 6 or 7. (In fact it is 6 but we haven t shown that.) Exercise 7: (i) Draw a diagram for K 6, the complete graph on 6 vertices. Draw a square with identified edges that represents the torus. (ii) Embed K 6 in the torus (so that edges don t cross). 111

18 (iii) Use this to draw, on a separate diagram, a map on the torus with six regions, each adjacent to the other five. (Finish with a clear picture in which only the edges of the map, and not the original graph, are visible.) Label the regions 1 to 6. (iv) Let be chromatic number of the torus. What does your answer to (iii) tell you about? (v) Heawood s formula gives an upper bound for. Calculate this upper bound. Exercise 7: (a) The weight of 3P is 3. By Heawood s formula n = INT = 7. (b) (c) To obtain a map with 7 regions, each adjacent to the other 6, we take a point inside each face of the map obtained in (b) and join it to a point along each edge. Removing the original edges we obtain a map on 3P in which each of the 7 regions are adjacent to the other 6. This map needs 7 colours. 112

19 Exercise 8: (i) ( Möbius Band) = (Projective Plane) = INT = 6. (ii) ( cylinder with 3 holes) = (sphere with 5 holes) = (sphere) = 4. (iii) (sum of 10 projective planes) = INT =

20 114

### 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.,

### 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

### 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

### 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

### Class One: Degree Sequences

Class One: Degree Sequences For our purposes a graph is a just a bunch of points, called vertices, together with lines or curves, called edges, joining certain pairs of vertices. Three small examples of

### Cycles in a Graph Whose Lengths Differ by One or Two

Cycles in a Graph Whose Lengths Differ by One or Two J. A. Bondy 1 and A. Vince 2 1 LABORATOIRE DE MATHÉMATIQUES DISCRÉTES UNIVERSITÉ CLAUDE-BERNARD LYON 1 69622 VILLEURBANNE, FRANCE 2 DEPARTMENT OF MATHEMATICS

### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.

### A Turán Type Problem Concerning the Powers of the Degrees of a Graph

A Turán Type Problem Concerning the Powers of the Degrees of a Graph Yair Caro and Raphael Yuster Department of Mathematics University of Haifa-ORANIM, Tivon 36006, Israel. AMS Subject Classification:

### Surface bundles over S 1, the Thurston norm, and the Whitehead link

Surface bundles over S 1, the Thurston norm, and the Whitehead link Michael Landry August 16, 2014 The Thurston norm is a powerful tool for studying the ways a 3-manifold can fiber over the circle. In

### On planar regular graphs degree three without Hamiltonian cycles 1

On planar regular graphs degree three without Hamiltonian cycles 1 E. Grinbergs Computing Centre of Latvian State University Abstract. Necessary condition to have Hamiltonian cycle in planar graph is given.

### Star and convex regular polyhedra by Origami.

Star and convex regular polyhedra by Origami. Build polyhedra by Origami.] Marcel Morales Alice Morales 2009 E D I T I O N M O R A L E S Polyhedron by Origami I) Table of convex regular Polyhedra... 4

### 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

### 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

### A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions

A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved First Draft February 8, 2006 1 Contents 25

### 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

### Answer: (a) Since we cannot repeat men on the committee, and the order we select them in does not matter, ( )

1. (Chapter 1 supplementary, problem 7): There are 12 men at a dance. (a) In how many ways can eight of them be selected to form a cleanup crew? (b) How many ways are there to pair off eight women at the

### Approximation Algorithms

Approximation Algorithms or: How I Learned to Stop Worrying and Deal with NP-Completeness Ong Jit Sheng, Jonathan (A0073924B) March, 2012 Overview Key Results (I) General techniques: Greedy algorithms

### Analysis of Algorithms, I

Analysis of Algorithms, I CSOR W4231.002 Eleni Drinea Computer Science Department Columbia University Thursday, February 26, 2015 Outline 1 Recap 2 Representing graphs 3 Breadth-first search (BFS) 4 Applications

### SPERNER S LEMMA AND BROUWER S FIXED POINT THEOREM

SPERNER S LEMMA AND BROUWER S FIXED POINT THEOREM ALEX WRIGHT 1. Intoduction A fixed point of a function f from a set X into itself is a point x 0 satisfying f(x 0 ) = x 0. Theorems which establish the

### All trees contain a large induced subgraph having all degrees 1 (mod k)

All trees contain a large induced subgraph having all degrees 1 (mod k) David M. Berman, A.J. Radcliffe, A.D. Scott, Hong Wang, and Larry Wargo *Department of Mathematics University of New Orleans New

### (Refer Slide Time: 01.26)

Discrete Mathematical Structures Dr. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras Lecture # 27 Pigeonhole Principle In the next few lectures

### HOLES 5.1. INTRODUCTION

HOLES 5.1. INTRODUCTION One of the major open problems in the field of art gallery theorems is to establish a theorem for polygons with holes. A polygon with holes is a polygon P enclosing several other

### The Union-Find Problem Kruskal s algorithm for finding an MST presented us with a problem in data-structure design. As we looked at each edge,

The Union-Find Problem Kruskal s algorithm for finding an MST presented us with a problem in data-structure design. As we looked at each edge, cheapest first, we had to determine whether its two endpoints

### Catalan Numbers. Thomas A. Dowling, Department of Mathematics, Ohio State Uni- versity.

7 Catalan Numbers Thomas A. Dowling, Department of Mathematics, Ohio State Uni- Author: versity. Prerequisites: The prerequisites for this chapter are recursive definitions, basic counting principles,

### The Clar Structure of Fullerenes

The Clar Structure of Fullerenes Liz Hartung Massachusetts College of Liberal Arts June 12, 2013 Liz Hartung (Massachusetts College of Liberal Arts) The Clar Structure of Fullerenes June 12, 2013 1 / 25

### Vectors 2. The METRIC Project, Imperial College. Imperial College of Science Technology and Medicine, 1996.

Vectors 2 The METRIC Project, Imperial College. Imperial College of Science Technology and Medicine, 1996. Launch Mathematica. Type

### UPPER BOUNDS ON THE L(2, 1)-LABELING NUMBER OF GRAPHS WITH MAXIMUM DEGREE

UPPER BOUNDS ON THE L(2, 1)-LABELING NUMBER OF GRAPHS WITH MAXIMUM DEGREE ANDREW LUM ADVISOR: DAVID GUICHARD ABSTRACT. L(2,1)-labeling was first defined by Jerrold Griggs [Gr, 1992] as a way to use graphs

### ABSTRACT. For example, circle orders are the containment orders of circles (actually disks) in the plane (see [8,9]).

Degrees of Freedom Versus Dimension for Containment Orders Noga Alon 1 Department of Mathematics Tel Aviv University Ramat Aviv 69978, Israel Edward R. Scheinerman 2 Department of Mathematical Sciences

### Odd induced subgraphs in graphs of maximum degree three

Odd induced subgraphs in graphs of maximum degree three David M. Berman, Hong Wang, and Larry Wargo Department of Mathematics University of New Orleans New Orleans, Louisiana, USA 70148 Abstract A long-standing

### Warm up. Connect these nine dots with only four straight lines without lifting your pencil from the paper.

Warm up Connect these nine dots with only four straight lines without lifting your pencil from the paper. Sometimes we need to think outside the box! Warm up Solution Warm up Insert the Numbers 1 8 into

### An Innocent Investigation

An Innocent Investigation D. Joyce, Clark University January 2006 The beginning. Have you ever wondered why every number is either even or odd? I don t mean to ask if you ever wondered whether every number

### Large induced subgraphs with all degrees odd

Large induced subgraphs with all degrees odd A.D. Scott Department of Pure Mathematics and Mathematical Statistics, University of Cambridge, England Abstract: We prove that every connected graph of order

### FIXED POINT SETS OF FIBER-PRESERVING MAPS

FIXED POINT SETS OF FIBER-PRESERVING MAPS Robert F. Brown Department of Mathematics University of California Los Angeles, CA 90095 e-mail: rfb@math.ucla.edu Christina L. Soderlund Department of Mathematics

### On Integer Additive Set-Indexers of Graphs

On Integer Additive Set-Indexers of Graphs arxiv:1312.7672v4 [math.co] 2 Mar 2014 N K Sudev and K A Germina Abstract A set-indexer of a graph G is an injective set-valued function f : V (G) 2 X such that

### Ph.D. Thesis. Judit Nagy-György. Supervisor: Péter Hajnal Associate Professor

Online algorithms for combinatorial problems Ph.D. Thesis by Judit Nagy-György Supervisor: Péter Hajnal Associate Professor Doctoral School in Mathematics and Computer Science University of Szeged Bolyai

### 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

### SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS

SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS Assume f ( x) is a nonconstant polynomial with real coefficients written in standard form. PART A: TECHNIQUES WE HAVE ALREADY SEEN Refer to: Notes 1.31

### 11 Ideals. 11.1 Revisiting Z

11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(

### Finding and counting given length cycles

Finding and counting given length cycles Noga Alon Raphael Yuster Uri Zwick Abstract We present an assortment of methods for finding and counting simple cycles of a given length in directed and undirected

### On the maximum average degree and the incidence chromatic number of a graph

Discrete Mathematics and Theoretical Computer Science DMTCS ol. 7, 2005, 203 216 On the maximum aerage degree and the incidence chromatic number of a graph Mohammad Hosseini Dolama 1 and Eric Sopena 2

### Algebra Unpacked Content For the new Common Core standards that will be effective in all North Carolina schools in the 2012-13 school year.

This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these tools to better serve teachers. Algebra

### Optimal 3D Angular Resolution for Low-Degree Graphs

Journal of Graph Algorithms and Applications http://jgaa.info/ vol. 17, no. 3, pp. 173 200 (2013) DOI: 10.7155/jgaa.00290 Optimal 3D Angular Resolution for Low-Degree Graphs David Eppstein 1 Maarten Löffler

### Some Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.

Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.com This paper contains a collection of 31 theorems, lemmas,

### Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011

Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011 A. Miller 1. Introduction. The definitions of metric space and topological space were developed in the early 1900 s, largely

### A Sublinear Bipartiteness Tester for Bounded Degree Graphs

A Sublinear Bipartiteness Tester for Bounded Degree Graphs Oded Goldreich Dana Ron February 5, 1998 Abstract We present a sublinear-time algorithm for testing whether a bounded degree graph is bipartite

### Mathematical Research Letters 1, 249 255 (1994) MAPPING CLASS GROUPS ARE AUTOMATIC. Lee Mosher

Mathematical Research Letters 1, 249 255 (1994) MAPPING CLASS GROUPS ARE AUTOMATIC Lee Mosher Let S be a compact surface, possibly with the extra structure of an orientation or a finite set of distinguished

### Total colorings of planar graphs with small maximum degree

Total colorings of planar graphs with small maximum degree Bing Wang 1,, Jian-Liang Wu, Si-Feng Tian 1 Department of Mathematics, Zaozhuang University, Shandong, 77160, China School of Mathematics, Shandong

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### 3. Linear Programming and Polyhedral Combinatorics

Massachusetts Institute of Technology Handout 6 18.433: Combinatorial Optimization February 20th, 2009 Michel X. Goemans 3. Linear Programming and Polyhedral Combinatorics Summary of what was seen in the

### Chapter 7: Products and quotients

Chapter 7: Products and quotients Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 7: Products

### Graphs without proper subgraphs of minimum degree 3 and short cycles

Graphs without proper subgraphs of minimum degree 3 and short cycles Lothar Narins, Alexey Pokrovskiy, Tibor Szabó Department of Mathematics, Freie Universität, Berlin, Germany. August 22, 2014 Abstract

### Baltic Way 1995. Västerås (Sweden), November 12, 1995. Problems and solutions

Baltic Way 995 Västerås (Sweden), November, 995 Problems and solutions. Find all triples (x, y, z) of positive integers satisfying the system of equations { x = (y + z) x 6 = y 6 + z 6 + 3(y + z ). Solution.

### Extremal Wiener Index of Trees with All Degrees Odd

MATCH Communications in Mathematical and in Computer Chemistry MATCH Commun. Math. Comput. Chem. 70 (2013) 287-292 ISSN 0340-6253 Extremal Wiener Index of Trees with All Degrees Odd Hong Lin School of

### 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

### H.Calculating Normal Vectors

Appendix H H.Calculating Normal Vectors This appendix describes how to calculate normal vectors for surfaces. You need to define normals to use the OpenGL lighting facility, which is described in Chapter

### ON INDUCED SUBGRAPHS WITH ALL DEGREES ODD. 1. Introduction

ON INDUCED SUBGRAPHS WITH ALL DEGREES ODD A.D. SCOTT Abstract. Gallai proved that the vertex set of any graph can be partitioned into two sets, each inducing a subgraph with all degrees even. We prove

### 4. CLASSES OF RINGS 4.1. Classes of Rings class operator A-closed Example 1: product Example 2:

4. CLASSES OF RINGS 4.1. Classes of Rings Normally we associate, with any property, a set of objects that satisfy that property. But problems can arise when we allow sets to be elements of larger sets

### Rolle s Theorem. q( x) = 1

Lecture 1 :The Mean Value Theorem We know that constant functions have derivative zero. Is it possible for a more complicated function to have derivative zero? In this section we will answer this question

### ON FIBER DIAMETERS OF CONTINUOUS MAPS

ON FIBER DIAMETERS OF CONTINUOUS MAPS PETER S. LANDWEBER, EMANUEL A. LAZAR, AND NEEL PATEL Abstract. We present a surprisingly short proof that for any continuous map f : R n R m, if n > m, then there

### Max Flow, Min Cut, and Matchings (Solution)

Max Flow, Min Cut, and Matchings (Solution) 1. The figure below shows a flow network on which an s-t flow is shown. The capacity of each edge appears as a label next to the edge, and the numbers in boxes

### SUBGROUPS OF CYCLIC GROUPS. 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by

SUBGROUPS OF CYCLIC GROUPS KEITH CONRAD 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by g = {g k : k Z}. If G = g, then G itself is cyclic, with g as a generator. Examples

### FIBRATION SEQUENCES AND PULLBACK SQUARES. Contents. 2. Connectivity and fiber sequences. 3

FIRTION SEQUENES ND PULLK SQURES RY MLKIEWIH bstract. We lay out some foundational facts about fibration sequences and pullback squares of topological spaces. We pay careful attention to connectivity ranges

### ( ) FACTORING. x In this polynomial the only variable in common to all is x.

FACTORING Factoring is similar to breaking up a number into its multiples. For example, 10=5*. The multiples are 5 and. In a polynomial it is the same way, however, the procedure is somewhat more complicated

### Notes on Factoring. MA 206 Kurt Bryan

The General Approach Notes on Factoring MA 26 Kurt Bryan Suppose I hand you n, a 2 digit integer and tell you that n is composite, with smallest prime factor around 5 digits. Finding a nontrivial factor

### Polynomial Degree and Finite Differences

CONDENSED LESSON 7.1 Polynomial Degree and Finite Differences In this lesson you will learn the terminology associated with polynomials use the finite differences method to determine the degree of a polynomial

### Best Monotone Degree Bounds for Various Graph Parameters

Best Monotone Degree Bounds for Various Graph Parameters D. Bauer Department of Mathematical Sciences Stevens Institute of Technology Hoboken, NJ 07030 S. L. Hakimi Department of Electrical and Computer

### 12-1 Representations of Three-Dimensional Figures

Connect the dots on the isometric dot paper to represent the edges of the solid. Shade the tops of 12-1 Representations of Three-Dimensional Figures Use isometric dot paper to sketch each prism. 1. triangular

### On the independence number of graphs with maximum degree 3

On the independence number of graphs with maximum degree 3 Iyad A. Kanj Fenghui Zhang Abstract Let G be an undirected graph with maximum degree at most 3 such that G does not contain any of the three graphs

### it is easy to see that α = a

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore

### GREEN CHICKEN EXAM - NOVEMBER 2012

GREEN CHICKEN EXAM - NOVEMBER 2012 GREEN CHICKEN AND STEVEN J. MILLER Question 1: The Green Chicken is planning a surprise party for his grandfather and grandmother. The sum of the ages of the grandmother

### 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

### TOPOLOGY OF SINGULAR FIBERS OF GENERIC MAPS

TOPOLOGY OF SINGULAR FIBERS OF GENERIC MAPS OSAMU SAEKI Dedicated to Professor Yukio Matsumoto on the occasion of his 60th birthday Abstract. We classify singular fibers of C stable maps of orientable

### Mean Ramsey-Turán numbers

Mean Ramsey-Turán numbers Raphael Yuster Department of Mathematics University of Haifa at Oranim Tivon 36006, Israel Abstract A ρ-mean coloring of a graph is a coloring of the edges such that the average

### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra

U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory

### 13. Write the decimal approximation of 9,000,001 9,000,000, rounded to three significant

æ If 3 + 4 = x, then x = 2 gold bar is a rectangular solid measuring 2 3 4 It is melted down, and three equal cubes are constructed from this gold What is the length of a side of each cube? 3 What is the

### 2010 Solutions. a + b. a + b 1. (a + b)2 + (b a) 2. (b2 + a 2 ) 2 (a 2 b 2 ) 2

00 Problem If a and b are nonzero real numbers such that a b, compute the value of the expression ( ) ( b a + a a + b b b a + b a ) ( + ) a b b a + b a +. b a a b Answer: 8. Solution: Let s simplify the

### 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.

9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role

### December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B. KITCHENS

December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B KITCHENS The equation 1 Lines in two-dimensional space (1) 2x y = 3 describes a line in two-dimensional space The coefficients of x and y in the equation

### CONTINUED FRACTIONS AND FACTORING. Niels Lauritzen

CONTINUED FRACTIONS AND FACTORING Niels Lauritzen ii NIELS LAURITZEN DEPARTMENT OF MATHEMATICAL SCIENCES UNIVERSITY OF AARHUS, DENMARK EMAIL: niels@imf.au.dk URL: http://home.imf.au.dk/niels/ Contents

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.

FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is

### This chapter is all about cardinality of sets. At first this looks like a

CHAPTER Cardinality of Sets This chapter is all about cardinality of sets At first this looks like a very simple concept To find the cardinality of a set, just count its elements If A = { a, b, c, d },

### Solutions to old Exam 1 problems

Solutions to old Exam 1 problems Hi students! I am putting this old version of my review for the first midterm review, place and time to be announced. Check for updates on the web site as to which sections

### SHORT CYCLE COVERS OF GRAPHS WITH MINIMUM DEGREE THREE

SHOT YLE OVES OF PHS WITH MINIMUM DEEE THEE TOMÁŠ KISE, DNIEL KÁL, END LIDIKÝ, PVEL NEJEDLÝ OET ŠÁML, ND bstract. The Shortest ycle over onjecture of lon and Tarsi asserts that the edges of every bridgeless

### FURTHER VECTORS (MEI)

Mathematics Revision Guides Further Vectors (MEI) (column notation) Page of MK HOME TUITION Mathematics Revision Guides Level: AS / A Level - MEI OCR MEI: C FURTHER VECTORS (MEI) Version : Date: -9-7 Mathematics

### Lecture 6 Online and streaming algorithms for clustering

CSE 291: Unsupervised learning Spring 2008 Lecture 6 Online and streaming algorithms for clustering 6.1 On-line k-clustering To the extent that clustering takes place in the brain, it happens in an on-line

### The degree of a polynomial function is equal to the highest exponent found on the independent variables.

DETAILED SOLUTIONS AND CONCEPTS - POLYNOMIAL FUNCTIONS Prepared by Ingrid Stewart, Ph.D., College of Southern Nevada Please Send Questions and Comments to ingrid.stewart@csn.edu. Thank you! PLEASE NOTE

### UNCORRECTED PAGE PROOFS

number and and algebra TopIC 17 Polynomials 17.1 Overview Why learn this? Just as number is learned in stages, so too are graphs. You have been building your knowledge of graphs and functions over time.

### MATH 90 CHAPTER 6 Name:.

MATH 90 CHAPTER 6 Name:. 6.1 GCF and Factoring by Groups Need To Know Definitions How to factor by GCF How to factor by groups The Greatest Common Factor Factoring means to write a number as product. a

### The Goldberg Rao Algorithm for the Maximum Flow Problem

The Goldberg Rao Algorithm for the Maximum Flow Problem COS 528 class notes October 18, 2006 Scribe: Dávid Papp Main idea: use of the blocking flow paradigm to achieve essentially O(min{m 2/3, n 1/2 }

### On the crossing number of K m,n

On the crossing number of K m,n Nagi H. Nahas nnahas@acm.org Submitted: Mar 15, 001; Accepted: Aug 10, 00; Published: Aug 1, 00 MR Subject Classifications: 05C10, 05C5 Abstract The best lower bound known

### The 2010 British Informatics Olympiad

Time allowed: 3 hours The 2010 British Informatics Olympiad Instructions You should write a program for part (a) of each question, and produce written answers to the remaining parts. Programs may be used

### ON DEGREES IN THE HASSE DIAGRAM OF THE STRONG BRUHAT ORDER

Séminaire Lotharingien de Combinatoire 53 (2006), Article B53g ON DEGREES IN THE HASSE DIAGRAM OF THE STRONG BRUHAT ORDER RON M. ADIN AND YUVAL ROICHMAN Abstract. For a permutation π in the symmetric group

### MA651 Topology. Lecture 6. Separation Axioms.

MA651 Topology. Lecture 6. Separation Axioms. This text is based on the following books: Fundamental concepts of topology by Peter O Neil Elements of Mathematics: General Topology by Nicolas Bourbaki Counterexamples

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### Fundamentals of Media Theory

Fundamentals of Media Theory ergei Ovchinnikov Mathematics Department an Francisco tate University an Francisco, CA 94132 sergei@sfsu.edu Abstract Media theory is a new branch of discrete applied mathematics