Save this PDF as:

Size: px
Start display at page:

## Transcription

1 MAP6 Combinatorics Answers Guidance on notation: graphs may have multiple edges, but may not have loops. A graph is simple if it has no multiple edges.. (a) Show that if two graphs have the same degree sequence then they have the same number of vertices and the same number of edges. Find two nonisomorphic graphs with the degree sequence (,,,, ). Suppose G and G are graphs with degree sequence (d,..., d n ). Each entry records the degree of one vertex, so both G and G have n vertices. By the Handshaking Theorem, d d n is equal to twice the number of edges in either graph. So G and G have the same number of edges. Two non-isomorphic graphs with degree sequence (,,,, ) are and. (There is also a non-simple graph with this degree sequence.) (b) Find all simple graphs on vertices, up to isomorphism. There are simple graphs on vertices (up to isomorphism). Any such graph has between 0 and 6 edges; this can be used to organise the hunt. (With more vertices, it might also be useful to first work out the possible degree seqences.) The table below show the number of graphs for edge possible number of edges. edges 0 6 number of graphs You might like to work out the reason for the symmetry in the table.

2 . For each of the sequences below, decide whether or not it is the degree sequence of a graph. (If it is, give an explicit example, if not, say why not.) (i) (9,,,, ), (ii) (,,, ), (iii) (0,,, ), (iv) (,,, ). Do any of your answers change if the graph has to be simple? (i) No: by the Handshaking Theorem, the sum of the degrees of a graph is even. (ii) Yes, for example. (iii) No: the vertex of greatest degree can have at most + + edges coming into it, so its degree is at most. (iv) Yes, for example. Suppose G is a simple graph with this degree sequence. Then G has vertices x, y, z of degree, and one further vertex, w, of degree. Each of x, y, z is connected to all the other, so in particular to w. But this says w has degree, a contradiction.. Let G and G be graphs. Suppose that f : V (G) V (G ) is an isomorphism. Let x, y V (G). Use induction to show that that the distance between x and y in G is equal to the distance between f(x) and f(y) in G. Note: I should have added the assumption that G is connected without this there might be no path between x and y, and the distance between them would be undefined. (It follows from the proof below that if G is connected then so is G, so we only have to make this assumption for one of the graphs.) Write d(x, y) for the distance between x and y. If d(x, y) = 0 then x = y, so f(x) = f(y) and so d(f(x), f(y)) = 0. This gives the base case in the induction. Now suppose that d(x, y) = m. Let x,..., x r be the neighbours of x in G. Pick a path of shortest length from x to y. It must go through one of x,..., x r ; we may choose the labelling so that it passes through x. Thus d(x, y) = m. By induction, d(f(x ), f(y)) = m. Since f(x) and f(x ) are adjacent in G, this shows that d(f(x), f(y)) m. Suppose for some i we have d(f(x i ), f(y)) < m. By another application of the inductive hypothesis, d(x i, y) < m, so d(x, y) < m, a contradiction. Hence it takes a least m steps to go from any neighbour of f(x) to f(y), and so at least m steps to go from f(x) to f(y). This shows that d(f(x), f(y)) = m. The inductive step is now complete.

3 . (a) Show that the graphs below all have the same degree sequence.,,. Inspection shows that all vertices have degree. (b) Show that the first two graphs are isomorphic. Hint: start by finding a closed path of length 6 in the first. Following the hint, we start by taking the closed path,,,,, 6 in the first graph, and mapping this to the closed path forming the outside of the second graph. We then check that we can consistently complete the labelling , 0 9, 6 (c) Show that neither is isomorphic to the third. There are many possible arguments. For example, using Question, we could argue that the third graph has two vertices a distance apart, but the first does not.

4 . Let G be a connected graph with k vertices of odd degree, where k > 0. Show that the minimum number of trails with mutually distinct edges needed to cover every edge of G is k/. It follows from the Handshaking Theorem that k is even: say k = r. Say z,..., z r have odd degree. Introduce r further edges {z, z },..., {z r, z r }. This gives a new graph in which every vertex has even degree. By Theorem., this new graph has a closed Eulerian trail. If one cuts out from this graph the r edges not present in G, then we are left with r = k/ trails covering G. We must also show that no fewer trails suffice. If we remove from G all the edges involved in a single trail, then we can change the parity of the degree at most vertices, namely the vertices at the start and end of the trail. So as G has r vertices of odd parity, we need at least r trails. How many continuous pen-strokes are needed to draw the diagram below if it is forbidden to draw any line segment twice? As there are vertices of odd degree, trails are the minimum. You can either use the algorithm above to find suitable trails, or just do it by inspection. 6. (a) Show that the complete graph on n vertices has n(n )/ edges. There are n vertices, each connected to n other vertices. This gives n(n ) edges, but double counting each one (as in the Handshaking Theorem). Or argue that the edges are in bijection with -subsets of n, of which there are ( n ) = n(n )/. (b) Show that if G is a simple graph on 6 vertices which is not connected then G has at most 0 edges. Can equality occur? Suppose G has a connected component with m vertices. This component has at most m(m )/ edges, by the previous part, None of the remaining 6 m vertices are connected to these m, so they can contribute at most (6 m)(6 m )/ edges. So the total number of edges is at most m(m ) + (6 m)(6 m ) Now put in m =,,,, and check that in all cases the number of edges is at most 0. Equality occurs if we have a copy of the complete graph on vertices, with the 6th vertex not connected to any others.

5 Turn over MAP6 Combinatorics Answers Guidance on notation: graphs may have multiple edges, but may not have loops. A graph is simple if it has no multiple edges.. Which of the graphs shown below have (i) a closed Eulerian trail? (ii) a closed Hamiltonian path? d c e c d a b, a b, The first has (a, b, c, d, a) as a closed Hamiltonian path. It doesn t have a closed Eulerian trail as not all vertices have degree (see Lemma. from lectures). It s fairly obvious that any walk in the second graph visiting all the vertices must pass through vertex c twice. Hence it doesn t have a closed Hamiltonian path. It has the closed Eulerian trail (c, e, c, d, c, b, a, c). The third has a closed Hamiltonian path: (,,,,, ) and a closed Eulerian trail: (,,,,,,,,,, ). Further exercise: count the number of such paths and trails.. Show that the graph below is isomorphic to the complete bipartite graph K,. For which n N 0 does K, have (i) a closed path of length n; (ii) a closed trail of length n? The isomorphism is indicated by the colouring: white vertices are connected to each of the black vertices. We don t allow closed paths (or walks or trails) of length 0. As there are no multiple edges, there are no closed paths or closed trails of length. (Any such path looks like (x, y, x) for two vertices x and y.) Clearly there are closed paths of lengths and 6. There are only vertices, so no path can have length >.

6 For trails it it useful to use the alternative representation of K, shown below. a b c d x y z The paths already found give closed trails of lengths and 6. A closed trail of length is (a, x, b, y, c, z, d, y, a). A closed trail of length would be an Eulerian trail (there are = edges), but K, has vertices of degree, so we know no such trail can exist. It remains to show that there is no closed trail of length 0. Any such trail visits 9 vertices, alternately visiting vertices from the sets {x, y, z} and {a, b, c, d}. It must therefore visit vertices in the set {x, y, z} at least times. Without loss of generality we may assume that x is visited twice. This requires edges attached to x, but x only has degree (contradiction).. Use Euler s formula to give an alternative proof that K, is not planar. Hint: adapt the proof given in lectures for K, noting that each face in a planar drawing of K, must be bounded by at least edges. Suppose for a contradiction that K, may be drawn in the plane, with f faces. This graph has 6 vertices and = 9 edges, so by Euler s formula we have f =. Hence f =. Suppose that the faces have e, e,..., e edges respectively. Counting up edges by walking round each face in turn gives e e edges; this counts every edge twice as each edge bounds exactly faces. Thus e + e + e + e + e = 9 =. On the other hand, a bipartite graph can t have any triangles (see Q6(a) for a more general result), so each face is bounded by at least edges. Hence e e = 0 which contradicts the previous equation.

7 Turn over. Professors Beeblebrox, Catalan, Descartes, Euler, Frobenius, Gauss and Hamilton go punting. Gauss considers it beneath his dignity to be in the same punt as anyone except Euler or Frobenius. Euler will tolerate Gauss, but dislikes Descartes. No-one except Hamilton is willing to share a punt with Beeblebrox. (a) Explain the relevance of the graph shown below. E D B G H F C With the given labelling, person X is connected to person Y if and only if they cannot be put in the same punt. Thus if the graph is coloured so that no two adjacent vertices have the same colour, then mathematicians with the same colour may safely be put in the same punt. (You could perhaps make use of the graph in some other way, but I think this is the only way that makes a genuine connection with the theory of graph colourings.) (b) By colouring this graph, or otherwise, determine the minimum number of punts required. How many possible seating plans are there using this number of punts? E D B G H F C It s quite easy to find the -colouring shown. Apart from permuting the colours, the only possible change which does not increase the number of colours is to make Frobenius white instead of grey. So there are just different seating plans using punts: { {E, F, G}, {C, D}, {B, H} } { {E, G}, {C, D, F }, {B, H} }.

8 . Let G be a planar graph. (a) Show that it follows from Euler s formula that G has a vertex of degree or less. Hint: suppose G has n vertices, e edges, and f faces, and that all vertices of G have degree 6 or more. Show that e n and f e/, and then use Euler s formula to get a contradiction. Each face is bounded by at least edges, so by the usual argument (counting up edges by going round faces), f e. On the other hand, the counting argument used in the Hand-shaking theorem shows that 6n e. (Each vertex has at least 6 edges coming out of it, so counting up edges by vertices gives at least 6n edges. This double-counts each edge, so we have 6n e.) Substituting in n e + f gives which contradicts Euler s formula. n e + f e/ e + e/ = 0 (b) Show by induction on the number of vertices of G that G may be coloured using 6 colours. Hint: in the inductive step delete from G the vertex given by (a), along with all the edges to which it belongs. Let x be a vertex of G of degree or less. By induction we can colour the graph obtained from G by deleting x with 6-colours. As x has at most neighbours in G, there is at least one colour which isn t used to colour any neighbour of x. If this colour is used to colour x then we get a 6-colouring of G. Remark: the same ideas can be pushed a bit harder to show that any planar graph is -colourable: see Wilson, Theorem..

9 6. (a) Show that if G is a bipartite graph then every closed walk in G has even length. Let G have bipartition {A, B}. Suppose we have a closed walk in G starting in A. As there are no edges between vertices in A, the next vertex visited in the walk must lie in B. Similarly, the vertex after that must lie in A. So the walk has the form (a, b, a, b,..., a m, b m, a ) for some a i A, b i B. In particular, it has even length m. The argument is similar if the walk starts in B. (b) Now suppose that G is a connected graph and that every closed walk in G has even length. Fix x G. Set A = {y V (G) : there is a walk from x to y of even length}, B = {y V (G) : there is a walk from x to y of odd length}. Show that A B = V (G), A B = and that if {u, v} is an edge of G then either u A and v B or u B and v A. Deduce that G is bipartite. Let y V (G). Since G is connected, there is a walk from x to y. Either this walk has even length, so y A, or it has odd length, so y B. Hence A B = V (G). Suppose y A B. Then there is a walk from x to y of even length, say (z 0, z,..., z r ) where z 0 = x,z r = y, and another, inevitably different, walk from x to y of odd length, say (w 0, w,..., w s ) where w 0 = x, w s = y. Now (z 0,..., z r = w s, w s,..., w 0 ) is a closed walk starting and ending at x of odd length. This contradicts our assumption that all closed walks in G have even length. Hence A B =. Now suppose that (a, a ) is an edge between two vertices in A. We can walk from x to a in an even number of steps. By taking this walk, and then taking one final step to a, we can walk from x to a in an odd number of steps. Therefore a A B =, another contradiction. Similarly there are no edges between any two vertices in B. Hence G is bipartite with bipartition {A, B}.

10 Turn over MAP6 Combinatorics Answers Guidance on notation: graphs may have multiple edges, but may not have loops. A graph is simple if it has no multiple edges. A tree is a connected simple graph with no closed paths. ( ) indicates an optional question, included for interest only.. Find all spanning trees in the following graph:. You should find that there are different spanning trees. Clearly any spanning tree will have the edges {, }, {, }, {, } and {, }. A tree on vertices has edges, so we must take edges from {, 6}, {, }, {, }, {6, }, {6, }. As ( ) = 0, this gives 0 candidates: two of them have a closed path of length so aren t trees. One of the rejected subgraphs is shown below. The other is its mirror image Find the Prüfer codes of the spanning trees in K shown below. 6 6 Which spanning trees in K have Prüfer codes (,,,, ) and (,,,, 6)? Which spanning trees in K have Prüfer codes of the form (i, j, k, l, m) where i, j, k, l, m are mutually distinct? The first has Prüfer code (,,,, 6). The second has Prüfer code (,,,, ).

11 The star with edges {, }, {, },..., {, } has Prüfer code (,,,, ). To find a tree with Prüfer code (,,,, 6) we use the algorithm given in lectures. The initial set of candidate leaves is {,,,,, 6, }. The smallest number not appearing in the code is, so the first edge is {, }. This uses up the leaf, so we remove from the set of candidate leaves, leaving {,,,, 6, }, and then carry on with the remaining part of the code, (,,, 6). The table below shows all the steps. edge added candidate leaves for next step {, } {,,,, 6, } {, } {,,, 6, } {, } {,, 6, } {, } {, 6, } {, 6} {6, } The last edge {6, }. It is now probably wise to draw the tree and check directly that it has the Prüfer code we want. 6. The purpose of this question is to give an alternative proof of Theorem., that the number of edges in a tree on n vertices is n. (a) Let T be a tree. Show that if (x 0, x,..., x m, x m ) is a path in T of greatest possible length then x 0 and x m are leaves. Suppose, for a contradiction, that x 0 is not a leaf. Then there must be some vertex y, other than x, which is joined to x 0. We can t have y {x,..., x m } as if {x k, y} is an edge then (x 0, x,..., x k, y, x 0 ) is a closed path in T. Hence y {x,..., x m }, and so (y, x 0, x,..., x m ) is a path of length m +. We had assumed that (x 0, x,..., x m ) had the greatest possible length of any path in T, so this is a contradiction. (b) Deduce that every non-empty tree has a leaf. Take a path of greatest possible length in the tree. By (a) either of its end vertices is a leaf. (c) Use the previous part and induction to prove that a tree on n vertices has exactly n edges. If our tree has just vertex then it has no edges, so the results holds in this case.

12 Turn over Now suppose our tree has n + vertices. Choose any leaf and remove it and its associated edge. The remaining graph is still a tree (clearly it still has no closed paths, and removing a leaf can t disconnect a graph). By induction it has n edges. We removed one edge to get it, so the original tree has n edges, as required.. Show that a connected simple graph is a tree if and only if there is a unique path between any two of its vertices. if : Let G be a connected graph satisfying the hypothesis that there is a unique path between any two of its vertices. If there is a closed path starting and finishing at the vertex x 0, say (x 0, x,..., x m, x 0 ) then (x 0, x ) and (x 0, x m,..., x ) are two different paths from x 0 to x. So G has no closed paths. By assumption G is connected. Hence G is a tree. only if : Left to you.. Let (s,..., s n ) be the Prüfer code of a tree on {,,..., n}. Show that if vertex j has degree d then j appears exactly d times in the sequence (s,..., s n ). Deduce that the tree has leaves {,,..., n}\{s,..., s n }. We shall prove the following claim by induction on n. (As in lectures, it is necessary to prove something very slightly more general, in order for the induction to go through.) Claim: Let T be a tree on a set X N 0 of size n with Prüfer code (s,..., s n ). If vertex j X has degree d then j appears exactly d times in the sequence (s,..., s n ). Proof: If n = then both vertices of T are leaves and the Prüfer code is (). Hence the claim holds in this case. Suppose we know the claim holds for all trees with n vertices. Let the smallest numbered leaf of T be b. By definition of the Prüfer code, {s, b } is an edge of T. Let T be the tree obtained from T by removing b and the edge {s, b }. Now T is a tree on X\{b } with Prüfer code (s,..., s n ). Let j X and suppose that j has degree d as a vertex of T. If j s, b then j also has degree d as a vertex of T. By induction we know that j appears exactly d times in (s,..., s n ). Hence j appears exactly d times in (s,..., s n ), as required. If j = s then j has degree d as a vertex of T. By induction, j appears exactly d times in (s,..., s n ), so j appears exactly d times in (s,..., s n ). Finally, if j = b then j doesn t appear at all in (s,..., s n ), and correspondingly d =, as j is a leaf. The claim therefore holds for the tree T. This completes the inductive step.

13 6. The diagram below shows a network with source s and sink t. Plain numbers give the capacities, bold numbers one possible flow. u x 0 0 s w y 0 t v z (a) Check that the bold numbers give a valid flow from s to t. Check that in each edge the flow is at most the capacity, and that flow is conserved at all vertices (except the source and the target). (b) Apply one iteration of the Ford Fulkerson algorithm to this flow. We start by setting S = {s}. () Add u as f(s, u) < c(s, u); now S = {s, u}. () Add w as f(u, w) < c(u, w); now S = {s, u, w}. () Add v as f(v, w) > 0; now S = {s, u, v, w} () Add y as f(v, y) < c(v, y); now S = {s, u, v, w, y} () Add x as f(x, y) > 0; now S = {s, u, v, w, y, x} (6) Add t as f(x, t) < c(x, t); now S = {s, u, v, w, y, x, t} We can now stop, as we ve reached t. The route to t is (s, u, w, v, y, x, t). We increase the flow in edges (s, u), (u, w), (v, y) amd (x, t), and decrease the flow in edges (v, w), (x, y) to get the new flow below. u x s 0 w y 0 t v z

14 (c) Find, with proof, a maximal flow in this network. In fact the new flow is maximal. Applying one more iteration of the Ford Fulkerson algorithm gives the cut (S, T ) = ({s, u}, {v, w, x, y, z, t}) which has capacity. No flow can have value greater than the capacity of this cut, and the new flow has value, so it must be maximal.. Let f be a flow in a network with source s, target t and edge list E. By summing f(x, y) over all edges (x, y) E and using conservation of flow show that f(s, y) = f(z, t). y:(s,y) E The conservation of flow equation is f(x, y) = y:(x,y) E z:(z,t) E z:(z,x) E f(z, x) for all vertices x s, t. Sum this over all allowed vertices to get f(z, x). x =s,t y:(x,y) E f(x, y) = x =s,t z:(z,x) E Now notice that if (u, v) is an edge with u s and v t then f(u, v) appears on the left-hand-side (when x = u, y = v) and on the right-hand-side (when x = v, z = u). After cancelling all these terms, we are left with f(x, t) = f(s, x) x:(x,t) E x:(s,x) E which is the required result (up to labelling of the dummy variables).

15 MAP6 Combinatorics Answers Guidance on notation: A partition of a natural number n is a sequence of natural numbers (λ,..., λ k ) such that λ λ... λ k and λ λ k = n. ( ) indicates an optional question, included for interest only.. Let G be the Petersen graph (shown below) Let H be the subgroup of Sym{,,..., 0} consisting of the permutations h such that {i, j} is an edge of G {h(i), h(j)} is an edge of G. (a) Find an element of H of order. The rotational symmetry of the graph gives one such element: h = ( )( ). (b) Find an element of H of order 6. ( Hint: The labelling above is consistent with Example 6. from lectures.) Another way to draw G (used in Example 6.) is From this drawing it is easy to see that k = ( 6)( 9 ) H.

16 (c) ( ) Find an element of H of order. Hence, or otherwise, draw the Petersen graph in a way that exhibits -fold symmetry. The product hk has order : hk = ( )( 0 6)( 9). (And since H is a group, hk H.) This suggests that we might try putting vertices,,, 9 and,, 0, 6 at the corners of two squares. 6 0 Putting in the edges of G which meet these vertices gives a drawing of a subgraph of G with rotation symmetry of order. 6 0 We now add vertices and 9 (and their incident edges) above and below the plane of the octagon. This is rather fiddly to draw, but looks something like: The resulting -dimensional figure has a symmetry of order, given by rotating by 90 and then reflecting in the plane of the octagon.

17 Remark: The graphs shown below are isomorphic, being different ways to draw the Petersen graph. (Definition. defines graph isomorphisms. For other examples of isomorphic graphs see Example.6(a) from lectures and Question (b) on Sheet.). A -bead necklace is made using red and blue beads. Two necklaces should be regarded as the same if one is a rotation of the other. Show that different necklaces can be made. How does this answer change if reflections are also considered? This is a simpler version of the next question. You should find that in this case it doesn t make any difference to allow reflections, i.e. there are necklaces in both cases.. An -bead necklace is made using c different colours of beads. Two necklaces should be regarded as the same if one is a rotation of the other. (Do not consider reflections.) (a) Find the number of different necklaces as a polynomial in c. Let X be the set of all c ways to colour the vertices of an octagon with c different colours. Let h be the permutation of X given by rotating the octagon by. Thus h sends to and so on. Let H be the subgroup of Sym(X) generated by h. The number of different necklaces is equal to the number of orbits of H on X. By the Orbit-Counting Theorem, the number of orbits is given by Fix k = ( Fix e + Fix h + Fix h Fix h ). k H

18 The table below shows the number of fixed points of each element of H. We use the relation Fix k = Fix k to reduce the number of cases. element e c h, h c h, h 6 c h, h c h c # fixed points For example, h has two orbits on the vertices of the octagon: {,,, } and {,, 6, } in the diagram below. 6 If a colouring is fixed by h, vertices,,, must all be the same colour, and so must vertices,, 6,. This gives independent choices, hence there are c colourings fixed by h. The number of different necklaces is therefore ( c + c + c + c ). (b) When c =, how many necklaces have exactly beads of each colour? Let Y be the set of vertex colourings of the octagon in which vertices are coloured white and vertices are black. As ( ) = 0 there are 0 colourings in Y. The orbits of H on Y correspond to necklaces with exactly beads of each colour. element e 0 h, h 0 h, h 6 h, h 0 h 6 # fixed points in Y The hardest case is h. If a vertex colouring in Y is fixed by h then opposite vertices must have the same colour. There are ( ) = 6 ways to choose pairs of opposite vertices to be coloured white. The remaining vertices

19 must then be coloured black. Hence the number of fixed point of h in its action on Y is 6. The number of necklaces with beads of each colour is therefore ( ) = 0. (c) (Extra). Suppose that we want to consider two necklaces of length on c colours to be the same if one is a reflection of the other. There are reflections of the octagon: reflections through pairs of opposite vertices, and reflections through opposite edge midpoints. Element reflection through opposite vertices c reflection through opposite edge midpoints c # fixed points Hence the number of different necklaces up to reflection is ( c + c + c + c + c ). 6 For example, if c =, there are 6 necklaces if reflections are not considered, and 0 if they are.

20 . The conjugate of a partition is obtained by reflecting its Young diagram in its major diagonal. For example (,,, ) has conjugate (,,,, ) since reflects to It is usual to write λ for the conjugate of λ. (a) Show that λ has exactly k parts if and only if k is the the largest part of λ. The partition λ has exactly k parts if and only if its Young diagram has exactly k rows. This holds if and only if the reflected Young diagram has exactly k columns, i.e. if and only if λ has largest part of size k. (b) Show that the number of self-conjugate partitions of n is equal to the number of partitions of n into distinct odd parts. [Hint: There is a bijective proof based on straightening hooks :. 9.] This is just a matter of convincing yourself that this map is bijective. (c) Find a closed form for the generating function of self-conjugate partitions. The generating function for partitions into odd distinct parts is ( + x)( + x )( + x )( + x )( + x 9 ).... (For instance, the partition (9,,, ) corresponds to expanding the product by choosing the x 9, x, x and x terms.) By (b) this is also the generating function for self-conjugate partitions.

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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)

### 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

### GRAPH THEORY LECTURE 4: TREES

GRAPH THEORY LECTURE 4: TREES Abstract. 3.1 presents some standard characterizations and properties of trees. 3.2 presents several different types of trees. 3.7 develops a counting method based on a bijection

### Discrete Mathematics & Mathematical Reasoning Chapter 10: Graphs

Discrete Mathematics & Mathematical Reasoning Chapter 10: Graphs Kousha Etessami U. of Edinburgh, UK Kousha Etessami (U. of Edinburgh, UK) Discrete Mathematics (Chapter 6) 1 / 13 Overview Graphs and Graph

### 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

### Outline 2.1 Graph Isomorphism 2.2 Automorphisms and Symmetry 2.3 Subgraphs, part 1

GRAPH THEORY LECTURE STRUCTURE AND REPRESENTATION PART A Abstract. Chapter focuses on the question of when two graphs are to be regarded as the same, on symmetries, and on subgraphs.. discusses the concept

### 3. Equivalence Relations. Discussion

3. EQUIVALENCE RELATIONS 33 3. Equivalence Relations 3.1. Definition of an Equivalence Relations. Definition 3.1.1. A relation R on a set A is an equivalence relation if and only if R is reflexive, symmetric,

### = 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that

Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without

### 1. Determine all real numbers a, b, c, d that satisfy the following system of equations.

altic Way 1999 Reykjavík, November 6, 1999 Problems 1. etermine all real numbers a, b, c, d that satisfy the following system of equations. abc + ab + bc + ca + a + b + c = 1 bcd + bc + cd + db + b + c

### 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

### Lecture 1: Course overview, circuits, and formulas

Lecture 1: Course overview, circuits, and formulas Topics in Complexity Theory and Pseudorandomness (Spring 2013) Rutgers University Swastik Kopparty Scribes: John Kim, Ben Lund 1 Course Information Swastik

### Discrete Mathematics Problems

Discrete Mathematics Problems William F. Klostermeyer School of Computing University of North Florida Jacksonville, FL 32224 E-mail: wkloster@unf.edu Contents 0 Preface 3 1 Logic 5 1.1 Basics...............................

### Euler Paths and Euler Circuits

Euler Paths and Euler Circuits An Euler path is a path that uses every edge of a graph exactly once. An Euler circuit is a circuit that uses every edge of a graph exactly once. An Euler path starts and

### 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

### 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

### 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)

### Lecture 15 An Arithmetic Circuit Lowerbound and Flows in Graphs

CSE599s: Extremal Combinatorics November 21, 2011 Lecture 15 An Arithmetic Circuit Lowerbound and Flows in Graphs Lecturer: Anup Rao 1 An Arithmetic Circuit Lower Bound An arithmetic circuit is just like

### 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

### 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

### 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

### Every tree contains a large induced subgraph with all degrees odd

Every tree contains a large induced subgraph with all degrees odd A.J. Radcliffe Carnegie Mellon University, Pittsburgh, PA A.D. Scott Department of Pure Mathematics and Mathematical Statistics University

### Social Media Mining. Graph Essentials

Graph Essentials Graph Basics Measures Graph and Essentials Metrics 2 2 Nodes and Edges A network is a graph nodes, actors, or vertices (plural of vertex) Connections, edges or ties Edge Node Measures

### 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,

### 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

### 3. Eulerian and Hamiltonian Graphs

3. Eulerian and Hamiltonian Graphs There are many games and puzzles which can be analysed by graph theoretic concepts. In fact, the two early discoveries which led to the existence of graphs arose from

### 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

### Session 7 Bivariate Data and Analysis

Session 7 Bivariate Data and Analysis Key Terms for This Session Previously Introduced mean standard deviation New in This Session association bivariate analysis contingency table co-variation least squares

### 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

### 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

### Chapter 6: Graph Theory

Chapter 6: Graph Theory Graph theory deals with routing and network problems and if it is possible to find a best route, whether that means the least expensive, least amount of time or the least distance.

### 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

### Connectivity and cuts

Math 104, Graph Theory February 19, 2013 Measure of connectivity How connected are each of these graphs? > increasing connectivity > I G 1 is a tree, so it is a connected graph w/minimum # of edges. Every

### IE 680 Special Topics in Production Systems: Networks, Routing and Logistics*

IE 680 Special Topics in Production Systems: Networks, Routing and Logistics* Rakesh Nagi Department of Industrial Engineering University at Buffalo (SUNY) *Lecture notes from Network Flows by Ahuja, Magnanti

### Chapter 11 Number Theory

Chapter 11 Number Theory Number theory is one of the oldest branches of mathematics. For many years people who studied number theory delighted in its pure nature because there were few practical applications

### 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

### S on n elements. A good way to think about permutations is the following. Consider the A = 1,2,3, 4 whose elements we permute with the P =

Section 6. 1 Section 6. Groups of Permutations: : The Symmetric Group Purpose of Section: To introduce the idea of a permutation and show how the set of all permutations of a set of n elements, equipped

### Decision Mathematics D1. Advanced/Advanced Subsidiary. Friday 12 January 2007 Morning Time: 1 hour 30 minutes. D1 answer book

Paper Reference(s) 6689/01 Edexcel GCE Decision Mathematics D1 Advanced/Advanced Subsidiary Friday 12 January 2007 Morning Time: 1 hour 30 minutes Materials required for examination Nil Items included

### 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

### 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

V. Adamchik 1 Graph Theory Victor Adamchik Fall of 2005 Plan 1. Basic Vocabulary 2. Regular graph 3. Connectivity 4. Representing Graphs Introduction A.Aho and J.Ulman acknowledge that Fundamentally, computer

### TOURING PROBLEMS A MATHEMATICAL APPROACH

TOURING PROBLEMS A MATHEMATICAL APPROACH VIPUL NAIK Abstract. The article surveys touring, a problem of fun math, and employs serious techniques to attack it. The text is suitable for high school students

### 8. Matchings and Factors

8. Matchings and Factors Consider the formation of an executive council by the parliament committee. Each committee needs to designate one of its members as an official representative to sit on the council,

### 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

### 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

### 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

### Lecture 3: Linear Programming Relaxations and Rounding

Lecture 3: Linear Programming Relaxations and Rounding 1 Approximation Algorithms and Linear Relaxations For the time being, suppose we have a minimization problem. Many times, the problem at hand can

### Chapter 3. Cartesian Products and Relations. 3.1 Cartesian Products

Chapter 3 Cartesian Products and Relations The material in this chapter is the first real encounter with abstraction. Relations are very general thing they are a special type of subset. After introducing

### Simple Graphs Degrees, Isomorphism, Paths

Mathematics for Computer Science MIT 6.042J/18.062J Simple Graphs Degrees, Isomorphism, Types of Graphs Simple Graph this week Multi-Graph Directed Graph next week Albert R Meyer, March 10, 2010 lec 6W.1

### 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,

### 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

### Mathematics for Algorithm and System Analysis

Mathematics for Algorithm and System Analysis for students of computer and computational science Edward A. Bender S. Gill Williamson c Edward A. Bender & S. Gill Williamson 2005. All rights reserved. Preface

### Elementary triangle geometry

Elementary triangle geometry Dennis Westra March 26, 2010 bstract In this short note we discuss some fundamental properties of triangles up to the construction of the Euler line. ontents ngle bisectors

### 2. Methods of Proof Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try.

2. METHODS OF PROOF 69 2. Methods of Proof 2.1. Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try. Trivial Proof: If we know q is true then

### The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 72.

ADVANCED SUBSIDIARY GCE UNIT 4736/01 MATHEMATICS Decision Mathematics 1 THURSDAY 14 JUNE 2007 Afternoon Additional Materials: Answer Booklet (8 pages) List of Formulae (MF1) Time: 1 hour 30 minutes INSTRUCTIONS

### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### 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

### MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Systems of Equations and Matrices Representation of a linear system The general system of m equations in n unknowns can be written a x + a 2 x 2 + + a n x n b a

### 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

### 1 Introduction to Counting

1 Introduction to Counting 1.1 Introduction In this chapter you will learn the fundamentals of enumerative combinatorics, the branch of mathematics concerned with counting. While enumeration problems can

### 3 Some Integer Functions

3 Some Integer Functions A Pair of Fundamental Integer Functions The integer function that is the heart of this section is the modulo function. However, before getting to it, let us look at some very simple

### 5.3 The Cross Product in R 3

53 The Cross Product in R 3 Definition 531 Let u = [u 1, u 2, u 3 ] and v = [v 1, v 2, v 3 ] Then the vector given by [u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ] is called the cross product (or

### Kenken For Teachers. Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles June 27, 2010. Abstract

Kenken For Teachers Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles June 7, 00 Abstract Kenken is a puzzle whose solution requires a combination of logic and simple arithmetic skills.

### 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

### 6.2 Permutations continued

6.2 Permutations continued Theorem A permutation on a finite set A is either a cycle or can be expressed as a product (composition of disjoint cycles. Proof is by (strong induction on the number, r, of

### The Triangle and its Properties

THE TRINGLE ND ITS PROPERTIES 113 The Triangle and its Properties Chapter 6 6.1 INTRODUCTION triangle, you have seen, is a simple closed curve made of three line segments. It has three vertices, three

### MATH10040 Chapter 2: Prime and relatively prime numbers

MATH10040 Chapter 2: Prime and relatively prime numbers Recall the basic definition: 1. Prime numbers Definition 1.1. Recall that a positive integer is said to be prime if it has precisely two positive

### 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

### Geometric Transformations

Geometric Transformations Definitions Def: f is a mapping (function) of a set A into a set B if for every element a of A there exists a unique element b of B that is paired with a; this pairing is denoted

### The chromatic spectrum of mixed hypergraphs

The chromatic spectrum of mixed hypergraphs Tao Jiang, Dhruv Mubayi, Zsolt Tuza, Vitaly Voloshin, Douglas B. West March 30, 2003 Abstract A mixed hypergraph is a triple H = (X, C, D), where X is the vertex

### 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

### Math 475, Problem Set #13: Answers. A. A baton is divided into five cylindrical bands of equal length, as shown (crudely) below.

Math 475, Problem Set #13: Answers A. A baton is divided into five cylindrical bands of equal length, as shown (crudely) below. () ) ) ) ) ) In how many different ways can the five bands be colored if

### . 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

### Network Flow I. Lecture 16. 16.1 Overview. 16.2 The Network Flow Problem

Lecture 6 Network Flow I 6. Overview In these next two lectures we are going to talk about an important algorithmic problem called the Network Flow Problem. Network flow is important because it can be

### 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

### 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

### Grade 7/8 Math Circles November 3/4, 2015. M.C. Escher and Tessellations

Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Tiling the Plane Grade 7/8 Math Circles November 3/4, 2015 M.C. Escher and Tessellations Do the following

### Discrete Mathematics: Solutions to Homework (12%) For each of the following sets, determine whether {2} is an element of that set.

Discrete Mathematics: Solutions to Homework 2 1. (12%) For each of the following sets, determine whether {2} is an element of that set. (a) {x R x is an integer greater than 1} (b) {x R x is the square

### 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

### 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.,

### 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

### Sudoku puzzles and how to solve them

Sudoku puzzles and how to solve them Andries E. Brouwer 2006-05-31 1 Sudoku Figure 1: Two puzzles the second one is difficult A Sudoku puzzle (of classical type ) consists of a 9-by-9 matrix partitioned

### arxiv: v2 [math.co] 30 Nov 2015

PLANAR GRAPH IS ON FIRE PRZEMYSŁAW GORDINOWICZ arxiv:1311.1158v [math.co] 30 Nov 015 Abstract. Let G be any connected graph on n vertices, n. Let k be any positive integer. Suppose that a fire breaks out

### Problem Set 7 Solutions

8 8 Introduction to Algorithms May 7, 2004 Massachusetts Institute of Technology 6.046J/18.410J Professors Erik Demaine and Shafi Goldwasser Handout 25 Problem Set 7 Solutions This problem set is due in

### 8.1 Min Degree Spanning Tree

CS880: Approximations Algorithms Scribe: Siddharth Barman Lecturer: Shuchi Chawla Topic: Min Degree Spanning Tree Date: 02/15/07 In this lecture we give a local search based algorithm for the Min Degree

### Unit 1 Number Sense. In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions.

Unit 1 Number Sense In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions. BLM Three Types of Percent Problems (p L-34) is a summary BLM for the material

### 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

### 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

### 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