Spanning Trees. John Martinez. November 12, 2012

Transcription

1 Spanning Trees John Martinez November 12, 2012

2 Abstract Following an article of D.R. Shier [S], this paper deals with the formation of spanning trees given the bipartite graph K 2,n. In addition to a more straightforward combinatorial approach, we will show how the solution can be obtained by using in order Recursion, the Principle of Inclusion and Exclusion and finally the geometric correspondence with the n-cube. Relevant basic graph concepts are presented at the beginning of the paper. 1

3 Contents 1 Introduction 3 2 Basic Definitions 3 3 A Counting Problem 4 4 Solutions Counting Solving by Applying a Recursive Formula Applying the Principle of Inclusion and Exclusion Using the Correspondence Between the Bipartite Graph, K 2,n and the Vertices of an n-dimensional Cube References 17 2

4 1 Introduction In a world of technological connectivity, may it be high tech satellite communication, information highways based on fiber optic or more down to earth technologies like irrigation systems, keeping a golf course green, there will be due to cost efficiency and functionality an interest of keeping the system slim thus avoiding unnecessary links in the system. An understanding of graph theory will hereby be of essential interest. Historically, ever since Leonhard Euler solved the puzzle of The Seven Bridges of Königsberg 1736 [M] and laid the foundations of what today is called graph theory, there has been a tremendous development within this field along with the number of ways it is applied. In this work we will look into the specific task of numbering the various ways there are of setting up a graph of nodes and interlinking them so that there will exist exactly one path trough every pair of nodes. We will work with a graph based on two sets of nodes. The first set, containing only two nodes, may be interpreted as satellites with the function of handling communication between the other set of nodes, which can be of any number. The nodes in this latter group may be interpreted as ground stations. For efficiency reasons this bipartite graph should contain no cycles as implied earlier. We will present various mathematical ways of solving the problem based on an article of Douglas R. Shier [S]. The layout of this work is as follows: In Section 2 we will briefly describe some basic definitions and concepts in the field of bipartite graph theory. As we move to Section 3, there will be a precision of the problem dealt within this paper. In Section 4.1 we will give a few straightforward solutions to the problem using counting and combinatorics. A somewhat more elaborate way of solving the problem will be presented in Section 4.2 as we use a recursive formula. In Section 4.3, the principle of inclusion and exclusion will prove to be the most extensive approach here presented, as we will follow through a proof of a proposition which it leads to. Finally, in the last Section 4.4, we display a geometrical approach to the problem using the n-cube. 2 Basic Definitions In order to simplify the discussions in this paper we need some basic definitions. Let us start by defining what we mean by a graph. A graph we 3

5 define to be a nonempty set of interlinked nodes that we call vertices V. The vertices being endpoints of a set, E of undirected edges. We write this as G = (V,E). Another basic structure is the term tree. A tree is an undirected graph in which any two vertices are connected by one simple path. A tree therefore is without cycles. We continue by defining a Spanning tree which is a connected acyclic graph i.e. to say a tree, involving all the vertices of the graph. Formally, let G = (V,E) be a connected graph. A Spanning tree is any tree T = (V,E ), so that E E. Since most of the discussions in the following will be involving the structure of a bipartite graph, it is also to be defined in this section. Therefore, the graph G = (V,E) is bipartite if the vertices can be divided into two single disjoint subsets. Which is to say, G = (V,E) is bipartite if V = V 1 V 2 and V 1 V 2 =, and every edge of G connecting a vertex, a with b is of the form {a,b} with a V 1 and b V 2. If there exists a connecting edge between every vertex in the set V 1 with every vertex in the set V 2 we have a Complete bipartite graph, denoted K m,n. It is now necessary to define what we mean by the degree, d(v) of a vertex, v. The degree of a vertex is defined as the number of times a vertex serves as an endpoint of an edge in a graph. To determine the number of edges in a graph we need to count the total sum of the degrees of every vertex in that graph. For that purpose we will describe the very simple rule of how to determine the relation between the sum of the degrees of a graph s vertices and the number of edges in the graph. The rule is named the Edge-degree lemma. Lemma 2.1. [M][Edge-degree lemma]: Let G be any graph. Let V be the nonempty set of vertices and E the set of edges. Then, d(v) = 2 E. (1) v V 3 A Counting Problem The problem that will be discussed in this paper is about showing how to determine the number of Spanning trees in a bipartite graph. Of didactic interest is that this can be done using several mathematical approaches. We therefore consider the complete bipartite graphk 2,n consisting of two subsets of the vertices, S 1 and S 2. The former subset consists of vertices A and B and the latter subset of n vertices. From each of the 2 vertices in S 1 there is a connecting edge to every vertex of the set S 2, altogether forming a graph involving 2n edges. Imagine now that the vertices in the set S 1 represent two satellites which should be able to communicate with n ground stations. The 4

6 ground stations should also be able to communicate with one another. Furthermore this should be done in an efficient way meaning that transferring data from one vertex to another should not involve unnecessary edges. That is to say cycles should be left out in any connecting path between vertices. In order to form a graph within these restrictions we need to select exactly n+1 edges out of in all 2n edges not isolating any ground station i.e. vertex in the subset S 2. Selecting n+1 edges means that neither of the 2 vertices in S 1 will be isolated since there are no more than n edges belonging to either of the vertices A and B. 4 Solutions 4.1 Counting Before going into somewhat more elaborated ways of solving the problem of forming spanning trees out of the complete bipartite graph K 2,n, let us start by examining the bipartite graph of K 2,1, K 2,2 and K 2,3. In the case of K 2,1 we note that the complete bipartite graph itself forms a spanning tree. As we add a ground station, receiving K 2,2, the graph then consist of 4 edges of which one must be removed in order to form a spanning tree. This leads to 4 possibilities. In the last case, of K 2,3 we have in total 6 edges of which 2 must be removed to form a spanning tree. They must not both be connected to the same ground station as not to leave any vertex disconnected. This gives us ( 6 2) 3 = 12 possible spanning tree altogether. The principle is illustrated in fig.1 and fig.2 below in the case of K 2,5. As we increase the number of ground stations this approach of counting the number of possible ways of constructing a spanning trees gets increasingly impractical. Figure 1: The complete bipartite graph K 2,5 5

7 Figure 2: A spanning tree of K 2,5 If we change approach using the observation that the ground stations can be divided into 2 groups who are connected to either satellite A or satellite B, then given the division we need to choose exactly one ground station of which to be connected to both satellitesaandb. The reason there can be no more than one such ground station connected to both satellites is of course to avoid cycles in the graph. As the latter ground station can be chosen in n ways and the remaining n 1 ones can be connected to either A or B, we count n2 n 1 spanning trees. A modified version of this approach can be set up by observing that there are ( n k) ways of selecting k ground stations out of n to be connected with satellite A. For each selection there need to be one ground station out of the k selected to be connected with satellite B as well. This forming altogether n k=1 k( ) n k ways. This may be written as n k( n k). If we use that in general n n ( k) x k = (1+x) n, then by differentiating we get n k( n k) x k 1 = n(1+x) n 1. Now setting x = 1 we receive again n2 n 1. Finally in this section we show that by excluding edges instead of selecting them we reach yet another approach. Referring to the above there will ben 1 edges to be removed in forming a Spanning tree out of the2n ground stations. Forming a permutation, the selection can be lined up in n 2 j=0 (2n 2j) ways, since by each selection we loose 2 edges from the selected ground stations to choose among in order to still keep it connected. Since there are n 1 removals and n ground stations, this leaves us with exactly one ground station keeping both edges. Now taking into accountant that we are interested of combinations and not orders of selection we will have to divide with (n 1)!. This yields 2n 2(n 1) 2(n 2)...2(n (n 3))2(n (n 2)) = 2(n 1) n! = n2 (n 1) yet again. (n 1)! (n 1)! 4.2 Solving by Applying a Recursive Formula We now turn to using a recursive approach in solving the problem. Let t n be the the number of different ways there is of constructing a Spanning tree in the graph of K 2,n. The addition of one ground station i.e. one vertex to the subset S 2 can be done in two principal ways. The first way is by simply connecting it to one of the two vertices of S 1 i.e., either vertex A or vertex 6

8 B. These possibilities will form twice as many Spanning trees as there were before. The second way is by connecting the additional vertex to both A and B. As there can be only one vertex in the subset of ground stations holding a double connection in order to avoid the existence of cycles in the graph, all the ground stations i.e. vertices except the one that was added will now hold one single edge to the subset of S 1. If we assume there were n 1 vertices in S 2 before the addition of the one single vertex, there will now exist 2 n 1 ways to set up the connections of these n 1 ground stations. Altogether this means that by adding a new vertex to the subset of S 2 of n 1 vertices we get the following recurrence equation considering the initial condition t 0 = 0, t n = 2t n 1 +2 n 1, n 1. (2) Multiplying (2) by x n and summing to infinity we obtain, t n x n = 2 t n 1 x n + n=1 n=1 2 n 1 x n, n 1. n=1 Setting since t 0 = 0, we get T(x) = t n x n (3) T(x) = 2x t n 1 x n n=1 (2x) n, n=1 T(x) = 2xT(x)+ x 1 2x, T(x)(1 2x) = x 1 2x, x T(x) = (1 2x) 2. (4) Applying generating functions in order to solve this equation we consider the geometric series which converges for real x where x < 1, Differentiating we obtain 1 1 x = 1+x+x d 1 dx1 x = 1+2x+3x2 +4x

9 Hence, 1 (1 x) = 2 1+2x+3x2 +4x and therefore after multiplication by x on both sides, x (1 x) = 2 x+2x2 +3x 3 +4x (5) Rewriting (4) and replacing the variable in (5) with 2x instead of x yields 2T(x) = consequently, 2x (1 2x) 2 = 0+1(2x)+2(2x)2 +3(2x) 3 +4(2x) T(x) = x+2 2 x x x using the definition of T(x) in (3) T(x) = t n x n = and thus n2 n 1 x n t n = n2 n 1. (6) 4.3 Applying the Principle of Inclusion and Exclusion Counting the number of spanning trees that leave none of the n ground stations isolated means selecting these from a larger universal set of ( ) 2n n+1 combinations of edges including some in which there are isolated vertices. Using the principle of inclusion and exclusion it is possible to count the number of selections in which no vertex is isolated. The principle is based on gradually excluding sets in which one or more vertex is isolated. In this process it is necessary to repeatedly compensate each exclusion by an inclusion of the intersection of these sets so that no combination of vertices, containing one or more isolated vertex, is excluded more than exactly once in the expression. Denote the sets of combinations that isolate the vertex i by E i (Note that any other vertex may or may not be isolated and therefore the different sets of combinations will intersect). The number we are looking for is t n = Ē1 Ē2 Ēn. According to the principle of inclusion and exclusion this can be written; t n = Ē1 Ē2 Ēn = N E i + E i E j +( 1) n E 1 E 2 E n (7) i i<j 8

10 Isolating k vertices corresponds to the selection of n+1 edges out of 2n 2k possible edges. This can be done in ( ) 2n 2k n+1 ways given the vertices that are isolated. The choice of k vertices among the n vertices can be done in ( ) n k ways. This means that there exist ( )( n 2n 2k ) k n+1 ways to form a combination of n+1 edges out of the edges belonging to n k vertices and at the same time selecting which k vertices are to be isolated. To the original bipartite graph belongs 2n edges which means that selecting n+1 edges will exclude 2n (n+1) = n 1 edges. Meaning that there will be sets decrementally ranging from n 1 to 1 isolated vertices. In the equation 2 of (8) below, N stands for the universal set of ( 2n n+1) which we introduced in the beginning of this section. Thus, t n = N + n 1 2 ( n ( 1) k k k=1 )( ) 2n 2k n+1 Letting the summation start at k = 0 instead of k = 1 will obviously cause the summation to include the term of N shortening the expression. We now have to show that the sum does in fact equal n2 n 1. Let us first write (8) as, n ( )( ) n 2n 2k t n = ( 1) k (9) k n+1 observing that we may for convenience follow trough the summation to the index value of k = n instead of k = n 1 as the terms in the summation equal 2 zero for higher values of k than n 1 2. Proposition 1. For all n 0, n ( )( ) n 2n 2k t n = ( 1) k = n2 n 1. (10) k n+1 The following proof was shown to me by A. Meurman; (8) Proof. First, looking back at (3) recalling T(x) = we write t n x n, xt(x) = f(x) = 9 t n x n+1. (11)

11 Before continuing we also need to note that the following expression may be rewritten as, ( )( ) n 2n 2k n!(2n 2k)! = k n+1 k!(n k)!(n+1)!(n 2k 1)! = 1 n+1 (2n 2k)! k!(n k)!(n 2k 1)!. We then rewrite the left side of the equality in (10) with the help of this new expression as, t n = n ( )( ) n 2n 2k ( 1) k = k n+1 which together gives us, f(x) = n ( 1) k (2n 2k)! (n+1)k!(n k)!(n 2k 1)!, n x n+1 ( 1) k 1 n+1 (2n 2k)! k!(n k)!(n 2k 1)! (n k)! (n k)!. (12) Now differentiating f(x), f (x) = n x n ( 1) k (2n 2k)! (n k)!(n k)! (n k)! k!(n 2k 1)!. (13) Setting j = n k we get n 2k 1 = n 2(n j) 1 = 2j n 1, changing the expression of f (x) as; f (x) = = = n ( ) 2j x n ( 1) n+j j! j (n j)!(2j n 1)! j=0 n ( )( 2j j x n ( 1) n+j j n j j=0 ( 2j j j=0 2j )x j n=j 2j n 2j n ) (2j n) (14) ( ) j ( 1) n j (2j n)x n j (15) n j As showed above we continued by changing the order of the double sum. At equation (13) we differentiated the sum by differentiating each of its terms. Since we are dealing with infinite sums we must make sure that these operations are correctly made. Weierstrass Majorant Theorem states sufficient conditions which will be useful proving specifically the second operation i.e. differentiation. 10

12 Theorem 4.1. [BC]Weierstrass Majorant Theorem. Let u n (x), n = 1,2,..., be a functional series defined on an interval, E. Suppose there exists a sequence of positive constants M n independent of x so that u n (x) M n, when x E, n = 1,2,..., and n=1 M n <, then the functional series n=1 u n(x) converges uniformly on E. We shall apply Weierstrass Majorant Theorem to the functional series that corresponds to the terms of the differentiated sum of (13). Therefore consider, n u n(x) = x n ( 1) k (2n 2k)! (n k)!(n k)! (n k)! k!(n 2k 1)! n 2 ( 2(n k) = x n ( 1) k n k We estimate, if x E = [ 1 / 9, 1 / 9 ], then Thus, u n(x) sup x E )( n k n 2 ( )( 2(n k) n k x n n k k k ) (n 2k). ) (n 2k) (16) ( 1 9 )n 2 2n 2 n n 2 (17) ( 1 9 )n 2 2n 2 n (1+ε) n (18) 0.9 n = M n. M n = 0.9 n = <. In accordance with Weierstrass Majorant Theorem the functional series n=1 u n(x) of (16) converges uniformly on x E. Some comments on the above will follow. At (17) we were using that m ( m n) = (1+1) m = 2 m, and therefore ( m n) 2 m, from which follow that ( 2(n k) ) n k 2 2n and ( ) n k k 2 n. We also estimate (n 2k) n. Considering that the total number of terms in the summation are being less or equal to n, we estimate the value of the sum in (17) to ( 1)n 22n 2n n2. In general 9 n 2 (1+ε) n for n 0. The inequality that follows (18) is implying, that 8 (1+ε) 0.9 which is setting the upper boundary for ε. 9 Since we are dealing with infinite sums we shall need yet another theorem in order to show that the functional series of u n(x) is in fact differentiated 11

13 from the functional series of u n (x), meaning that equation (12) is the primitive function of equation (13). Thus consider the following theorem; Theorem 4.2. [BC](Theorem 7,2 ). If the functions u n (x), n = 1,2,..., are continuous on the compact interval E and the series n=1 u n(x) converges uniformly on E then, E ( n u n (x)dx) = k=1 n k=1 E u n (x). In accordance with theorem 4.2 follows that equation (12) is in fact the primitive function of equation (13). Finally in this section we need to show that interchanging the order of summation done at equation (15) is not affecting the sum. We will use the fact that the properties of a positive sum is totally unaffected of in which order the summation is done [BC] and therefore if a double sum is absolutely convergent then interchanging the order of summation is not affecting the sum [BC]( Theorem 3.2). In accordance hereof we will now show that the double-sum of (14) i.e. n ( )( ) 2j j x n ( 1) n+j (2j n) j n j j=0 is absolutely convergent which will be sufficient to guarantee the correctness of the swap of sums that was made going from equation (14) to (15). Let x E = [ 1 / 9, 1 / 9 ], for some A. x n n j= n 2 x n ( )( ) 2j j (2j n) j n j n 2 2j 2 j n j= n 2 x n (2 3 ) n n 2 A+ x n (2 3 (1+ε)) n < 12

14 Convergence will follow if x 8 (1 + ε) < 1 and x < 1 1 = 1, if 8 (1+ε) 9 ε = 1, since the sum will then converge as a geometric sum. As we have 8 shown that the series is absolutely convergent, it follows that the order of summation is not affecting the value of the sum. Some comments on the above will follow; While replacing the number of terms in the sum with the factor n, we have again been using that in general n 2 (1+ε) n for n 0. It may also be convenient commenting the change of order of the summation and the change of indexes further. In changing the order of summations and the range of indices of the double sum between (14) and (15) we observe that ( j n j) = 0 when j < n j i.e. n > 2j. This means it is sufficient to let the index n end at the value of 2j. It may also be noted that if we imagine the summation being done over pairs of integers in a Cartesian coordinate system with the vertical axis displaying the value of n and the horizontal axis the value of j, the idea of the changes in the summation may be further illustrated. Imagining this Cartesian coordinate system, in (14) the summations are being done in a horizontal order between the vertical n-axis and the line j = n progressively to the right by incrementing the index n followed by the index of j. As in (15) we are instead letting the summation follow a vertical order progressively incrementing the index of j, letting the summation run between the lines n = j and n = 2j, from the left to the right leaving out values were ( j n j) = 0 which are when j < n j i.e. n > 2j as said. Now setting n j = m we get; f (x) = = = = j=0 j=0 ( ) 2j j x j j ( 2j j ( 2j j j=0 ( 2j j j=0 m=0 j 1 )x j ( ) j ( 1) m (j m)x m m ( ) j 1 ( 1) m m m=0 ) j 1 x j j m=0 ) jx j (1 x) j 1 ( ) j 1 ( 1) m m j (j m)xm (j m) x m 13

15 Before continuing we will be using the expression below on the left side which follows, ( ) 1/2 ( 4) j = ( 4) j( 1/2)( 3/2) (1/2 j) j j! = 2 j1 3 (2j 1) = (2j)! ( ) 2j =. j! j!j! j As we further elaborate the expression of f (x) we also extend it with 1 x. 1 x In equation (19) below we will be referring to the following Taylor development as we continue developing the expression of f (x); ( ) a (1+t) a = t k, t ] 1,1[. k Still referring to equation (19), using t = 4x + 4x 2 in this Taylor development and factoring out ( 4x+4x 2 ) from the summation, we rewrite the factor ( 3/2 j=1 j 1) ( 4x+4x 2 ) j 1 as (1 4x+4x 2 ) 3/2. Thus, f (x) = = 1 (1 x) 1 (1 x) = 1/2 (1 x) ( ) 1/2 j( 4x+4x 2 ) j j j=1 ( ) 3/2 ( 1/2) j ( 4x+4x 2 ) j j 1 j ( ) 3/2 ( 4x+4x 2 ) j 1 ( 4x+4x 2 ) (19) j 1 j=1 j=1 = ( 1/2)( 4x+4x2 ) (1 4x+4x 2 ) 3/2 1 x 2x = (1 4x+4x 2 ) 1 4x+4x 2 = 2x 1+1 (1 2x) = 3 (1 2x) 3 (1 2x) 2. Now integrating f (x), f(x) = f (x)dx = 1 4 (1 2x) (1 2x) 1 +C = 1 d 8dx (1 2x) (1 2x) +C (20) 14

16 Referring back to the original equation of (11), as f(0) = 0, we receive C = 1 4 thus, f(x) = 1 8 n2 n x n n=1 2 n x n The geometric sum being differentiated is differentiable for 2x ] 1,1[. We refer to [BC] for proof of that matter. The coefficient t n of x n+1 in this expression is, t n = 1 8 (n+2)2n n+1, n > 0 = (n+2)2 n n 1 = n2 n 1 which confirms the equality of (8) with our earlier result. 4.4 Using the Correspondence Between the Bipartite Graph, K 2,n and the Vertices of an n-dimensional Cube Consider the n-dimensional cube being an analogue of a square, n = 2 and a cube, n = 3. Let each corner of the n-cube hold a vertex. Placing the cube in an n-dimensional Cartesian coordinate system, the vertices of the cube may then be positioned either at the position of 0 in the origin, or in the position of 1 in each axis. The position of 0 for a vertex in the n:th axis will represent that there exist an edge to the vertex A for the n:th ground station and similarly the position of 1 represent an edge to the vertex B for the same n:th ground station. The position of a vertex in the n-cube will then correspond to a binary n-vector of the elements 0 and 1. As a Spanning tree will be formed by connecting n 1 ground stations to either the satellite A or B and letting one ground station be connected to both of these satellites, each possibility there is forming a Spanning tree will correspond to the selection of 2 vertices in the n-cube that are distinguished only by same i:th element which will hold different binary numbers. The set of possible Spanning trees will therefore be in a 1 1 correspondence with every non-diagonal edge in the cube marking the selection of a pair of vertices according to what has just been said. The general idea is diplayed in the figures below. In the case of the 3-cube (fig.3), the line between (0,1,0) and (1,1,0) is made bold as to mark one specific correspondence to a spanning tree in K 2,3. Extending to the 4-dimensional case there will be a Spanning tree corresponding to the 15

17 edge between the vectors, (1,1,0,0) and (1,1,0,1) which is made bold in fig.4. The total number of possibilities is therefore = 32 in this case Figure 3: The 3-cube Figure 4: The 4-cube The number of edges in this hypercube is equivalent with each possible way of forming a Spanning tree in this bipartite case. We can now use the Edge-degree lemma to determine the number of edges in the hypercube and therefore the number of Spanning trees in the bipartite graphg = (V,E). Let V be the nonempty set of vertices and E the set of edges. According to the Edge-degree lemma [M], already declared at (2.1): In any graph G = (V,E); d(v) = 2 E. v V We apply this and use that for each vertex, v in the hypercube, d(v) = n. The number of vertices in the hypercube is now 2 n. This gives us the value of the left side in the above lemma, n2 n = 2 E and therefore the value of E will be n2 n 1 which we recognize from the earlier sections of this paper. 16

18 Acknowledgements I like to acknowledge Prof. A. Meurman who has been the supervisor of this paper for great guidance in helping me write this work and comprehending the theory. I thank my colleague Roger Bengtsson, at Spyken high school in Lund for helpful remarks. I also wish to thank Rickard Björklund who assisted with the illustrative figures. 5 References [B] Lars-Christer Böiers, Diskret Matematik, Studentlitteratur, [BC] Lars-Christer Böiers and Tomas Claesson, Analys för funktioner av flera variabler. Lecture Notes, Dept. of Mathematics, Lund University, [M] H.F Mattson, Jr, Discrete Mathematics with applications, John Wiley & Sons, Inc, 1993 [S] Douglas R. Shier, Spanning Trees: Let Me Count the Ways, Mathematics Magazine, Vol. 73, No. 5 (Dec.,2000), pp