Tree and Venn Diagrams

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Tree and Venn Diagrams Tree and Venn Diagrams By: OpenStaxCollege Sometimes, when the probability problems are complex, it can be helpful to graph the situation Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities Tree Diagrams A tree diagram is a special type of graph used to determine the outcomes of an experiment It consists of "branches" that are labeled with either frequencies or probabilities Tree diagrams can make some probability problems easier to visualize and solve The following example illustrates how to use a tree diagram In an urn, there are 11 balls Three balls are red (R) and eight balls are blue (B) Draw two balls, one at a time, with replacement "With replacement" means that you put the first ball back in the urn before you select the second ball The tree diagram using frequencies that show all the possible outcomes follows Total = 64 + 24 + 24 + = 121 The first set of branches represents the first draw The second set of branches represents the second draw Each of the outcomes is distinct In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8 Then the nine RR outcomes can be written as: • • • • • • • • • R1R1 R1R2 R1R3 R2R1 R2R2 R2R3 R3R1 R3R2 R3R3 1/18 Tree and Venn Diagrams The other outcomes are similar There are a total of 11 balls in the urn Draw two balls, one at a time, with replacement There are 11(11) = 121 outcomes, the size of the sample space a List the 24 BR outcomes: B1R1, B1R2, B1R3, a • • • • • • • • • • • • • • • • • • • • • • • • B1R1 B1R2 B1R3 B2R1 B2R2 B2R3 B3R1 B3R2 B3R3 B4R1 B4R2 B4R3 B5R1 B5R2 B5R3 B6R1 B6R2 B6R3 B7R1 B7R2 B7R3 B8R1 B8R2 B8R3 b Using the tree diagram, calculate P(RR) b P(RR) = ( 113 )( 113 ) = 1219 c Using the tree diagram, calculate P(RB OR BR) c P(RB OR BR) = 48 ( 113 )( 118 ) + ( 118 )( 113 ) = 121 2/18 Tree and Venn Diagrams d Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw) d P(R on 1st draw AND B on 2nd draw) = P(RB) = 24 ( 113 )( 118 ) = 121 e Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw) e P(R on 2nd draw GIVEN B on 1st draw) = P(R on 2nd|B on 1st) = 24 88 = 11 This problem is a conditional one The sample space has been reduced to those outcomes that already have a blue on the first draw There are 24 + 64 = 88 possible outcomes (24 24 BR and 64 BB) Twenty-four of the 88 possible outcomes are BR 88 = 11 f Using the tree diagram, calculate P(BB) f P(BB) = 64 121 g Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw) g P(B on 2nd draw|R on 1st draw) = 11 There are + 24 outcomes that have R on the first draw (9 RR and 24 RB) The sample space is then + 24 = 33 24 of the 33 outcomes have B on the second draw The 24 probability is then 33 Try It In a standard deck, there are 52 cards 12 cards are face cards (event F) and 40 cards are not face cards (event N) Draw two cards, one at a time, with replacement All possible outcomes are shown in the tree diagram as frequencies Using the tree diagram, calculate P(FF) 3/18 Tree and Venn Diagrams Total number of outcomes is 144 + 480 + 480 + 1600 = 2,704 P(FF) = 144 144 + 480 + 480 + 1,600 = 144 2, 704 = 169 An urn has three red marbles and eight blue marbles in it Draw two marbles, one at a time, this time without replacement, from the urn "Without replacement" means that you not put the first ball back before you select the second marble Following is a tree diagram for this situation The branches are labeled with probabilities instead of frequencies The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, 11 10 = 110 ( )( ) Total = 56 + 24 + 24 + 110 = 110 110 =1 NOTE If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw You not put back or replace the first marble after you have drawn it You draw without replacement, so that on the second draw there are ten marbles left in the urn Calculate the following probabilities using the tree diagram a P(RR) = a P(RR) = ( 113 )( 102 ) = 1106 4/18 Tree and Venn Diagrams b Fill in the blanks: P(RB OR BR) = ( 113 )( 108 ) + ( _)( _) b P(RB OR BR) = = 48 110 48 ( 113 )( 108 ) + ( 118 )( 103 ) = 110 c P(R on 2nd|B on 1st) = c P(R on 2nd|B on 1st) = 10 d Fill in the blanks P(R on 1st AND B on 2nd) = P(RB) = ( _)( _) = d P(R on 1st AND B on 2nd) = P(RB) = 24 100 24 ( 113 )( 108 ) = 100 e Find P(BB) e P(BB) = ( 118 )( 107 ) f Find P(B on 2nd|R on 1st) f Using the tree diagram, P(B on 2nd|R on 1st) = P(R|B) = 10 If we are using probabilities, we can label the tree in the following general way • • • • Try It P(R|R) here means P(R on 2nd|R on 1st) P(B|R) here means P(B on 2nd|R on 1st) P(R|B) here means P(R on 2nd|B on 1st) P(B|B) here means P(B on 2nd|B on 1st) 5/18 Tree and Venn Diagrams In a standard deck, there are 52 cards Twelve ...Venn Diagrams with Few Vertices Bette Bultena and Frank Ruskey abultena@csr.csc.uvic.ca , fruskey@csr.csc.uvic.ca Department of Computer Science University of Victoria Victoria, B.C. V8W 3P6, Canada Submitted: September 15, 1998, Accepted: October 1, 1998 Abstract An n-Venn diagram is a collection of n finitely-intersecting simple closed curves in the plane, such that each of the 2 n sets X 1 ∩ X 2 ∩···∩X n ,whereeach X i is the open interior or exterior of the i-th curve, is a non-empty connected region. The weight of a region is the number of curves that contain it. A region of weight k is a k-region. A monotone Venn diagram with n curves has the property that every k-region, where 0 <k<n, is adjacent to at least one (k − 1)-region and at least one (k + 1)-region. Monotone diagrams are precisely those that can be drawn with all curves convex. An n-Venn diagram can be interpreted as a planar graph in which the intersection points of the curves are the vertices. For general Venn diagrams, the number of vertices is at least  2 n −2 n−1 . Examples are given that demonstrate that this bound can be attained for 1 <n≤ 7. We show that each monotone Venn diagram has at least  n n/2  vertices, and that this lower bound can be attained for all n>1. Keywords: Venn diagram, dual graph, convex curve, Catalan number. AMS Classification (primary, secondary): 05C10, 52C99. 1 Introduction There has been a renewed interest in Venn diagrams in the past couple of years. Recent surveys have been written by Ruskey [10] and Hamburger [8]. In this paper we tackle a natural problem that has not received any attention: What is the least number of vertices in a Venn diagram of n curves? Figure 1(a) shows the classic Venn diagram of 3 curves, which contains 6 vertices. The Venn diagram of Figure 1(b) is also constructed with 3 curves, but has only 3 vertices. This second diagram has the minimum number of vertices among all Venn diagrams of 3 curves (a complete listing may be found in Chilakamarri, Hamburger, and Pippert [3]). We show that this is the minimum value in Theorem 2.1 in the following section. 1 the electronic journal of combinatorics 5 (1998), #R44 2 (a) Venn Diagram with 3 curves and 6 vertices (b) Venn Diagram with 3 curves and 3 vertices Figure 1: Example of a simple and a non-simple 3-Venn diagram. We give the relevant graph theoretic definitions in the remainder of this section. Section 2 provides a proof of the lower bound for the number of vertices of general Venn diagrams and provides examples of Venn diagrams that have this minimum number if 1 <n≤ 7. Finding a minimum vertex Venn diagram for n>7remains an open problem. In Section 3, we demonstrate that the upper bound of  n n/2  for the minimum number of vertices of a monotone Venn diagrams is attainable for all n>1. This is demonstrated, using a specific and recursively constructed sequence of diagrams. The proof that the number of vertices is as stated involves the Catalan numbers. 1.1 Venn Diagrams and Graphs Let us review Gr¨unbaum’s definition of a Venn diagram [7]. An n-Venn diagram in the plane is a collection of simple closed Jordan curves C = C 1 ,C 2 , ,C n , such that each of the 2 n sets X 1 ∩X 2 ∩ ∩ X n is a nonempty and connected region. Each X i is either the bounded interior or the unbounded exterior of C i , and this intersection can be uniquely identified by a subset of {1, 2, ,n}, indicating the subset of the indices of the curves whose interiors are included in the intersection. To this definition we add the condition that pairs of curves can intersect only at a finite number of points. We say that two Venn diagrams are isomorphic if, by continuous transformation of the plane, one of them can be changed into the other or its mirror image [10]. When analyzing a Venn diagram, we often think of it as a plane graph V ,whose vertices (called Venn vertices) are the intersection points of the curves. The labelled edges of V are of the form C(v, w), Venn Diagrams and Symmetric Chain Decompositions in the Boolean Lattice Jerrold Griggs ∗ Department of Mathematics University of South Carolina Columbia, SC 29208 griggs@math.sc.edu Charles E. Killian Department of Computer Science, Box 8206 North Carolina State University Raleigh, NC 27695 chip killian@acm.org Carla D. Savage † Department of Computer Science, Box 8206 N. C. State University Raleigh, NC 27695 savage@csc.ncsu.edu Submitted: Aug 30, 2002; Accepted: Dec 11, 2003; Published: Jan 2, 2004 MR Subject Classifications: 03E99, 05610, 06A07, 06E10 Abstract We show that symmetric Venn diagrams for n sets exist for every prime n, settling an open question. Until this time, n = 11 was the largest prime for which the existence of such diagrams had been proven, a result of Peter Hamburger. We show that the problem can be reduced to finding a symmetric chain decomposition, satisfying a certain cover property, in a subposet of the Boolean lattice B n , and prove that such decompositions exist for all prime n. A consequence of the approach is a constructive proof that the quotient poset of B n , under the relation “equivalence under rotation”, has a symmetric chain decomposition whenever n is prime. We also show how symmetric chain decompositions can be used to construct, for all n, monotone Venn diagrams with the minimum number of vertices, giving a simpler existence proof. ∗ Research supported in part by NSF grant DMS–0072187. † Research supported by NSA grant MDA 904-01-0-0083 the electronic journal of combinat orics 11 (2004), #R2 1 1 Introduction 1.1 Venn Diagrams Following Gr¨unbaum [Gr¨u75], an n-Venn diagram is a collection of n simple closed curves in the plane, {Θ 1 , Θ 2 , ,Θ n }, with the property that for each S ⊆{1, 2, ,n} the region  i∈S int(Θ i ) ∩  i∈S ext(Θ i ) is nonempty and connected, where int(Θ i )andext(Θ i ) denote the interior and exterior, respectively, of Θ i . For this paper, we require that any two of the curves Θ i intersect in only a finite number of points. Figure 1 shows four 3-Venn diagrams ((b), (c) and (d) are from [CHP96]). A region of the Venn diagram is a maximal connected subset of R 2 −∪ n i=1 Θ i ,whereR 2 denotes the set of all points in the plane. Thus, a Venn diagram partitions R 2 −∪ n i=1 Θ i into exactly 2 n regions, one for each subset of {1, 2, ,n}. A Venn diagram is called simple if no three curves have a common point of intersection. In Figure 1, the Venn diagram (a) is simple, but the others are not. It is known that n-Venn diagrams exist for all n ≥ 1 and constructions of Venn [Ven80] and Edwards [Edw89] are illustrated in [Rus97]. Most of the results, conjectures, and problems we mention in this paper can be found in [Rus97], an excellent survey and expository article on Venn diagrams by Frank Ruskey. 1.2 Symmetric Venn Diagrams A symmetric Venn diagram is one with rotational symmetry. That is, there is a point p in the plane such the each of the n rotations of Θ 1 about p by an angle of 2πi/n, 0 ≤ i ≤ n − 1, coincides with one of the curves Θ 1 , Θ 2 , ,Θ n . For example, in Figure 1, the Venn diagrams (a) and (c) are symmetric, but (b) and (d) are not. Symmetric Venn diagrams have been considered by several researchers including Hen- derson [Hen63], Gr¨unbaum [Gr¨u92, Gr¨u99], Ruskey [Rus97], Edwards [Edw98], and Ham- burger [Ham02]. Henderson [Hen63] proved that symmetric Venn diagrams are not possi- ble when n is not prime, but symmetric Venn diagrams are known for the primes n =3, 5, 7 and, most recently, for n = 11 [Ham02]. It has been an open question whether a sym- metric n-Venn diagram exists for every prime number n [Gr¨u75]. The main result of this paper is a constructive proof to show that the answer is yes - symmetric n-Venn diagrams exist for every prime n. For a survey of results on symmetric Venn diagrams, as well as Hamburger’s construc- tion of the first known symmetric Venn diagram for n = 11 sets see [Ham02]. 1.3 Monotone Venn Diagrams Two distinct Half-Simple Symmetric Venn Diagrams Charles E. Killian Department of Computer Science, Duke University, Durham, NC ckillian@cs.duke.edu Frank Ruskey ∗ Department of Computer Science, University of Victoria, Victoria, BC Carla D. Savage † Department of Computer Science, North Carolina State University, Raleigh, NC savage@csc.ncsu.edu Mark Weston Department of Computer Science, University of Victoria, Victoria, BC mweston@cs.uvic.ca Submitted: Sep 9, 2004; Accepted: Oct 13, 2004; Published: Nov 30, 2004 Mathematics Subject Classifications: 05A10, 05A16, 06A07, 06E10 Abstract A Venn diagram is simple if at most two curves intersect at any given point. A recent paper of Griggs, Killian, and Savage [Elec. J. Combinatorics, Research Pa- per 2, 2004] shows how to construct rotationally symmetric Venn diagrams for any prime number of curves. However, the resulting diagrams contain only  n n/2  inter- section points, whereas a simple Venn diagram contains 2 n − 2 intersection points. We show how to modify their construction to give symmetric Venn diagrams with asymptotically at least 2 n−1 intersection points, whence the name “half-simple.” 1 Introduction Following Gr¨unbaum [5], a Venn diagram for n sets is a collection of n simple closed curves in the plane, {Θ 1 , Θ 2 , ,Θ n }, with the property that for each S ⊆{1, 2, ,n} the region  i∈S int(Θ i ) ∩  i∈S ext(Θ i ) is nonempty and connected. Here int(Θ i )andext(Θ i ) denote the open interior and open exterior, respectively, of Θ i . A Venn diagram is simple if no 3 curves have a common point of intersection. In a Venn diagram the curves are assumed to be finitely intersecting. A Venn diagram is rotationally symmetric if there is a point p such that each of the rotations ∗ Research supported in part by NSERC grant 0-GP-000337999 † Research supported in part by NSF grant DMS-0300034 and NSA grant MDA 904-01-0-0083 the electronic journal of combinatorics 11 (2004), #R86 1 of Θ 1 about p by an angle of 2πi/n,0≤ i ≤ n − 1, coincides with one of the curves Θ 1 , Θ 2 , ,Θ n . A Venn diagram is monotone if every region enclosing k curves, with 0 <k<n, is adjacent to a region enclosing k − 1 curves and a region enclosing k +1 curves. Monotone Venn diagrams are precisely those that can be drawn with all curves convex, as shown in [1]. Venn diagrams for n sets with rotational symmetry cannot exist unless n is prime (Henderson [8]) and it is shown in [4] that they do exist for all prime n,byageneralcon- struction. The symmetric diagrams in [4] contain  n n/2  intersection points, with exactly n points of intersection through which all curves pass. Such diagrams were introduced by Ruskey and Chow [9], who provided examples of them for n =5andn =7. In[7] Hamburger constructed a symmetric Venn diagram for n = 11 with this property. It follows from Euler’s formula (V − E + F = 2) that a simple Venn diagram has 2 n − 2 vertices. The diagrams constructed in [4], both symmetric and non-symmetric, are monotone, and among monotone Venn diagrams they contain the least number of vertices, namely:  n  n 2   ∼ 2 n √ n . (1) This was shown in [2] to be minimum for monotone Venn diagrams. Simple monotone non-symmetric diagrams exist for all n, but simple symmetric Venn diagrams are known only for n =3, 5 and 7: see [9] for some examples. So, we are now motivated to ask if we can we find simple symmetric Venn diagrams for all prime n,orat least ones that are “more simple”, where as a measure of simplicity, we use the number of vertices in the Venn diagram. In this paper we show that for n prime, we can in fact add vertices to the diagrams in [4] to produce symmetric diagrams in which the number of vertices is asymptotically at least 2 n−1 . The technique used in [4] makes use of some novel observations about the Greene-Kleitman symmetric chain decomposition of the Boolean lattice [3]. The paper [4] also includes a construction, for any n, of monotone (non-symmetric) Venn diagrams with the Maximum Multiplicity of a Root of the Matching Polynomial of a Tree and Minimum Path Cover Cheng Yeaw Ku Department of Mathematics, National University of Singapore, Singapore 117543 matkcy@nus.edu.sg K.B. Wong Institute of Mathematical Sciences, University of Malaya, 50603 Kuala Lump ur, Malaysia kbwong@um.edu.my Submitted: Nov 18, 2008; Accepted: Jun 26, 2009; Published: Jul 2, 2009 Mathematics Subject Classification: 05C31, 05C70 Abstract We give a necessary and sufficient condition for the maximum multiplicity of a root of the matching polynomial of a tree to be equal to the minimum number of vertex disjoint paths needed to cover it. 1 Introduction All the graphs in this paper are simple. The vertex set and the edge set of a graph G are denoted by V (G) and E(G) respectively. A matching of a graph G is a set of pairwise disjoint edges of G. Recall that for a graph G on n vertices, the matching polynomial µ(G, x) of G is given by µ(G, x) =  k≥0 (−1) k p(G, k)x n−2k , where p(G, k) is the number of matchings with k edges in G. Let mult(θ, G) denote the multiplicity of θ as a root of µ(G, x). The following results are well known. The proofs can b e fo und in [2, Theorem 4.5 on p. 102]. Theorem 1.1. The maximum multiplicity of a root of the matching polynomial µ(G, x) is at most the minimum number of vertex disjoint paths needed to cover the vertex set of G. the electronic journal of combinatorics 16 (2009), #R81 1 Consequently, Theorem 1.2. If G has a Hamiltonian path, then all roots of its matching polynomial are simple. The above is the source of motivation for our work. It is natural to ask when does equality holds in Theorem 1.1. In this note, we give a necessary and sufficient condition for the maximum multiplicity of a root of t he matching polynomial of a tree to be equal to the minimum number of vertex disjoint paths needed to cover it. Before stating the main result, we require some terminology and basic properties of matching polynomials. It is well known that the roots of the matching polynomial are real. If u ∈ V (G), then G \ u is the graph obtained from G by deleting the vertex u and the edges of G incident to u. It is known that the roots of G \u interlace those of G, that is, the multiplicity o f a root changes by at most one upon deleting a vertex from G. We refer the reader to [2] for an introduction to matching polynomials. Lemma 1.3. Suppose θ is a root of µ(G, x) and u is a vertex of G. Then mult(θ, G) − 1 ≤ mult(θ, G \u) ≤ mult(θ, G) + 1. As a consequence of Lemma 1.3, we can classify the vertices in a graph by assigning a ‘sign’ to each vertex (see [3]). Definition 1.4. Let θ be a root of µ(G, x). Fo r any vertex u ∈ V (G) , • u is θ-essential if mult(θ, G \ u) = mult(θ, G) − 1, • u is θ-neutral if mult(θ, G \ u) = mult(θ, G), • u is θ-positive if mult(θ, G \ u) = mult(θ, G) + 1. Clearly, if mult(θ, G) = 0 t hen there are no θ-essential vertices since the multiplicity of a root cannot be negative. Nevertheless, it still makes sense to talk about θ-neutral and θ-positive vertices when mult(θ, G) = 0. The converse is also true, i.e. any graph G with mult(θ, G) > 0 must have a t least one θ-essential vertex. This was proved in [3, Lemma 3.1]. A further classification of vertices plays an important role in establishing some struc- tural properties o f a graph: Definition 1.5. Let θ be a root of µ(G, x). For any vertex u ∈ V (G), u is θ-special if it is not θ -essential but has a neighbor that is θ-essential. If G is connected and not all of its vertices are θ-essential, then G must contain a θ-special vertex. It turns out that a θ-special vertex must be θ-positive (see [3, Corollary 4.3]). We now introduce the following definition which is crucial in describing our main result. the electronic journal of combinatorics 16 (2009), #R81 2 Definition 1.6. Let G be a graph and Q = {Q 1 , . . . , Q m } be a set of vertex disjoint paths that cover G. Then Q is said to be (θ, G)-extremal if it satisfies the following: (a) ... (R), four yellow (Y) and five blue (B) beads For the coin, P(H) = and P(T) = where H is heads and T is tails 12/18 Tree and Venn Diagrams Find P(tossing a Head on the coin AND a Red bead) 15 36... P(>64) = P(>64 and F) + P(>64 and M) = 0.1356 No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 and F) = 0.0661 15/18 Tree and Venn Diagrams. .. 2013) 11/18 Tree and Venn Diagrams Chapter Review A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize A Venn diagram