...4). b c d e f 5). b e g f 6). b e f This would = 6 ways. 6). If a, b are distinct vertices in a connected undirected graph G, the distance from a to b is defined to be the length of a shortest path from a to b (when a = b the distance is defined to be 0). For the graph in Fig. 11.9, find the distances from d to (each of) the other vertices in G. d to e = 1 d to f = 1 d to c = 1 d to k = 2 d to g = 2 d to h = 3 d to j = 3 d to l = 3 d to m = 3 d to i = 4 Section 11.2 1). Let G be the undirected graph in Fig. 11.27(a). a) How many connected subgraphs of G have four vertices and include a cycle? 3 b) Describe the subgraph G1 (of G) in part (b) of the figure first, as an induced subgraph and second, in terms of deleting a vertex of G. G1 = (U) where U = {a, b, d, f, g, h, I, j}; G1 = G – {c} c) Describe the subgraphG2 (of G) in part (c) of the figure first, as an induced subgraph and second, in terms of the deletion of vertices of G. G2 = (W), where W = {b, c, d, f, g, i, j}; G2 = G − {a, h} d) Draw the subgraph of...
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...answer right into this document. Part I. Basic Computations 1. (4 points) The plan for a four-room house is shown below. Draw a graph that models the connecting relationships between the areas in the floor plan. [Your graph does not [Your graph does not need to be fancy. You may use any drawing software such as Visio or Creatly.com] Answer: I used viso for graph[pic] 2. a. Identify all the vertices in the above graph with odd degree. Identify the degree of each of these vertices. (2 points) Answer: The odd number of edges is a odd degree vertices are D,E,F from the graph 3,1,3 it has a odd number. So D,E,F, is odd and the rest is even. b. Describe two paths of different lengths that start at vertex A and which end at vertex F. Specify the length of each path. (2 points) Answer: A-B-C-D length 4 A-B-C-C-D-F length 5 c. Describe a circuit of length 3. (2 points) Answer: the circuit path is A-B-C d. Describe two different circuits of length 4 (1 point) Answer: Two circuits with 4 in length...
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...Discrete Applied Mathematics 158 (2010) 1644–1649 Contents lists available at ScienceDirect Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam The k-in-a-tree problem for graphs of girth at least k W. Liu a , N. Trotignon b,∗ a Université Grenoble 1, Joseph Fourier, France b CNRS, LIAFA, Université Paris 7, Paris Diderot, France article info Article history: Received 10 July 2009 Received in revised form 28 May 2010 Accepted 3 June 2010 Available online 1 July 2010 Keywords: Tree Algorithm Three-in-a-tree k-in-a-tree Girth Induced subgraph abstract For all integers k ≥ 3, we give an O(n4 )-time algorithm for the problem whose instance is a graph G of girth at least k together with k vertices and whose question is ‘‘Does G contains an induced subgraph containing the k vertices and isomorphic to a tree?’’. This directly follows for k = 3 from the three-in-a-tree algorithm of Chudnovsky and Seymour and for k = 4 from a result of Derhy, Picouleau and Trotignon. Here we solve the problem for k ≥ 5. Our algorithm relies on a structural description of graphs of girth at least k that do not contain an induced tree covering k given vertices (k ≥ 5). © 2010 Elsevier B.V. All rights reserved. 1. Introduction Many interesting classes of graphs are defined by forbidding induced subgraphs; see [1] for a survey. This is why the detection of several kinds of induced subgraph is interesting; see [5], where many such...
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...Unit 7: Trees - Assignment Part I. Basic Computations 1. Using the following tree, name two vertices that are considered the following. Explain in your own words how you know: (a) parent-child (2 points). Answer: 202 - 401 Explanation: the parent of a vertex is the vertex connected to it on the path to the root, 202 is the parent and 401 is the child (b) sibling nodes (2 points): Answer: 301, 302, 303 Explanation: If two vertices are children of the same parent, then these two vertices are called siblings, 301, 302, 303 have the same parent that is 201 (c) leaf nodes (2 points) Answer: 301, 302, 303, 401 Explanation: the leaves are all terminal vertices 2. Determine if each of the following graphs is considered a tree. Explain why or why not, using what you learned in this unit. a. (2 points) Answer: not a tree Explanation: A tree is a connected graph with no cycles, and there is cycles in this graph b. (2 points) Answer: yes Explanation: A tree is a connected graph with no cycles, and there is no cycles in this graph c. (1 point) Answer: no Explanation: A tree is a connected graph with no cycles, and there a cycle in this graph 3. Determine and sketch two different spanning trees for this graph: a. (1 point) b. (1 point) 4. Consider this graph: a. Determine the total weight for this graph. Show your work. (1 point) Answer: 122 Explanation: 5 + 10 +5 +10 +17 +15 + 5 +4+4+8+5+6+13+7+8 ...
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...Vertex R&D portfolio Decision Joshua Boger, CEO of Vertex has to decide on two out of four R&D portfolios that are to be fully funded by Vertex and to decide on the fate of the other two portfolios i.e. whether to partner or hold them as backups. In order to decide on the R&D portfolio, an objective quantitative analysis might not be suitable considering the high levels of uncertainities and consequently the risks involved in pharmaceutical research projects. It is important to have a qualitative analysis of the situation as a whole that includes Vertex’s own financial position, strategic implications, a quantitative analysis of its Portfolios with realistic estimations and a risk analysis of the portfolios. 1. Vertex finacial analysis As per Vertex’s income statement(exhibit- 2B), it is clear that Vertex R&D expenditures in most of the preceeding years until 2002 has exceeded its revenue to the tune of 120% of its revenue in 2002 . The net income has been negative for all these years and the company is yet to prove itself in the stock market and gain investor’s confidence (exhibit-5). As per exhibit-2a, though Vertex’s cash position is strong, most of it (~50%) is through convertible debt and unless Vertex creates a breakthrough in the market through block buster drugs or substantial revenue it is unlikely that Vertex can attract funds or generate interestes in the market for additional funding for its projects. Also in oder to scale its operations e.g. sales and marketing...
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...determined by the collection set of their vertex-deleted subgraphs. The content of this paper will focus its attention toward finite, simple, undirected graphs and provide a brief introduction of the conjecture and properties of graphs that have been uncovered through the conjecture. Furthermore, this paper will review an approach to the conjecture due to Kelly (1957), and other theorems and corollaries that branched from Kelly’s original lemma. Introduction Paul J. Kelly, in 1957, wrote his doctoral dissertation under the supervision of Polish-American Mathematician Stanislaw M. Ulam. Kelly’s dissertation proved the Reconstruction Conjecture held true for trees. Ulam officially published a statement of the conjecture, which was known to him since 1929, in 1960. It is said that Ulam published the conjecture as a result of collecting various mathematical problems proposed by some of his graduate students. As a result of this, his credit for the proposal of the unsolved problem was questioned. This led to some confusion as to whose name would be associated with the conjecture. None the less, the conjecture has been commonly accepted by the name the Kelly-Ulam conjecture. For the purpose of this paper, it will be generally referred to as the Reconstruction Conjecture. The Reconstruction Conjecture asserts that any finite, simple, undirected graph with at least three vertices can be uniquely determined by its collection of vertex-deleted subgraphs. The conjecture has been...
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...Closed path = 1st Vertex = Last Vertex Simple path = different edges (Vertex can be used twice) Closed simple path: different edges + 1st vertex = last vertex (vertices can be used twice) Cycle = closed simple path (1st vertex = last vertex + different edges) + different vertices Distinct vertices ==> different edges Cycle= closed path e1…en of length at least 3 + distant vertices = path e1…en with n >= 3 + 1st vertex = last vertex + distinct vertices A path has all vertices distinct ==> different edges + no cycles G and H are isomorphic if there exists an isomorphism and γ = V(G) ---> V(H) such that if {u,v} edge in G then {γ(u), γ (v)} edge in H each vertex goes to another vertex degrees are the same shapes are mapped too to prove two graphs are isomorphic check the degree lists….if they match find a mapping between the two graphs Euler path = simple path which goes through each edge exactly once Euler...
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...9-604-101 REV: JUNE 20, 2006 GARY PISANO LEE FLEMING Vertex Pharmaceuticals: R&D Portfolio Management (A) EEEddduuucccaaatttiiiooonnnaaalll mmmaaattteeerrriiiaaalll sssuuuppppppllliiieeeddd bbbyyy TTThhheee CCCaaassseee CCCeeennntttrrreee CCCooopppyyyrrriiiggghhhttt eeennncccooodddeeeddd AAA777666HHHMMM---JJJUUUJJJ999KKK---PPPJJJMMMNNN999III OOOrrrdddeeerrr rrreeefffeeerrreeennnccceee FFF222555222999888333 I’ve never made a bad decision. I’ve just had bad data. — Joshua Boger, CEO and Founder of Vertex Pharmaceuticals Like many New Englanders on this bright October morning in 2003, Josh Boger, CEO of Vertex Pharmaceuticals, had been up until 2:00 a.m. the previous evening watching the Boston Red Sox playoff game. The game, predictably, ended in a heartbreaking loss for the Red Sox, but Boger’s lingering disappointment (and regret over staying up so late) quickly faded as he strode down the halls of the Cambridge, Massachusetts company he had founded 15 years earlier. Vertex had four promising drugs in various stages of clinical development, and Boger was excited by the possibilities: “The portfolio is playing out exactly as we hoped. We’ve got a stream of revenues from our partnered project that will help fund our development costs. There are multiple paths for us to become profitable. We’re in a position to choose.” While the company had revenue from various corporate partnerships and roughly $600...
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...Chapter-1 INTRODUCTION TO GRAPH THEORY A graph G = (V, E) is a set of objects, V = {v1, v2, . . .} called vertices, and a set E = {e1, e2, . . .} called edges, such that an edge ek is determined with an unordered pair (vi, vj) of vertices. The vertices vi, vj that belong to an edge ek are called the end vertices of ek. A graph is generally represented by a diagram, with points representing vertices and line segments joining these vertices representing edges. This diagram is generally referred to as a graph. For example, the figure below depicts a graph. [1] Figure 1: A graph with 5 vertices and 10 edges According to the definition above, an edge is associated with a vertex pair (vi,vi). A self loop is an edge that has the same initial and final vertices. Edge e1 in the above figure is a self-loop. Hence, a given pair of vertices may have more than one edge, for example, edges e4 and e5. These edges are termed parallel edges. A simple graph is a graph that does not have any self loops or parallel edges. There are many applications of graph theory in various fields. In engineering, in physical, social, and biological sciences, in linguistics, and in numerous other areas graph theory helps to solve many problems. Any physical involving discrete objects and their relationships can be shown by using graphs. The concept of Graph theory began in 1736 when Euler considered the Konigsberg bridge problem. [1] 1.1 Konigsberg Bridge Problem: One of the classic and well...
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...Selena Matthews Homework 7 August 11, 2013 6. What is the size of the largest induced Kn in Figure 6.9? A complete graph on n vertices is Kn n vertices v1, v2, . . . , vn with an edge for each pair of distinct vertices a. P3 no pair of vertices b. not complete c. not complete d. K2 or C2 e. C5 not complete Largest induced Kn is K2. 10. What can you say about a five vertex simple graph in which every vertex has degree four? Five vertices *4 degrees = 20 edges V = {1, 2, 3, 4, 5} E = {(1,2), (1,3), (1,4), (1,5), (2,1), (2,3), (2,4), (2,5), 3,1, 3,2, 3,4, 3,5, 4,1, 4,2, 4,3, 4,5, (5,1), (5,2), (5,3), (5,4)} I can say: 1. The simple graph is complete because all pairs of end points are joined by an edge. 2. I can say that it is not a tree because it contains a cycle. 3. That it is connected but undirected. 4. Adding all the degrees 4+4+4+4+4 for odd vertices and even degrees provides an even amount of edges =20. 14. Are there graphs with v vertices and v-1 edges and no cycles that are not trees? No Give a proof to justify your answer. Let G be a graph with v vertices and e edges Let G1, G2, G3…,Gk be G's connected components Let vi be the number of vertices of Gi Let ei be the number of edges of Gi Prove G has e = v-1 and no cycles but is not a tree A tree T has v vertices and v-1 edges T=(V,E) v=V and vi=Vi v-1=E ei=E∩Vi2 Induce that ei=vi-1. A tree has v vertices and...
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...Applications and the TSP Question 1 A Hamiltonian cycle is a closed loop within a graph that visits all its vertices exactly once. An example of the Hamiltonian cycle is the Travelling salesman problem. The solution to the minimum length of a Hamiltonian cycle is an NP complete problem that cannot be computed in polynomial time. This means that the minimum possible path cannot be computed by a deterministic machine. When completing a Hamiltonian cycle, one has to make sure that there are only two edges getting in and out of a vertex. In addition one has to ensure that there are no sub-cycles in the cycle. In a Hamiltonian cycle with n vertices, the number of different cycles that can be completed is (n-1)! /2 in a complete undirected graph and (n-1)! In a complete directed graph (Narasimhan, 2009). Question 2 A Euler cycle is a path that passes through all the edges of a graph exactly once. It usually starts and ends at the same vertex. For one to construct a Euler cycle all the vertices in the graph must have an even degree. Therefore, one can conclude that any graph with all vertices of an even degree connected is a Euler cycle. Unlike the Hamiltonian cycle, the Euler cycle can be computed in polynomial time. The Euler cycle can be constructed using the Fleury’s or Hierholzer’s algorithm. Question 3 The minimum length of a Hamiltonian cycle is the shortest path that can visit all the vertices of a graph exactly once. To evaluate it one needs to compare all possible...
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...Resource Allocation Graphs Roger Henriksson Department of Computer Science Lund University The possibility of deadlock is an undesired property (to say the least) of safety-critical real-time systems. Therefore, a method to verify that a system cannot deadlock, or to detect possible deadlock situations, is of tremendous value. One such method is resource allocation graphs. As stated in Operating System Concepts by Peterson and Silberschatz [PS85], the occurrence of deadlock requires among other things that hold-wait situations exist and a circular chain of such hold-waits must exist. A resource allocation graph attempts to graphically illustrate all the hold-wait situations in a system. In this graph it is possible to search for cases of circular hold-wait. In their book, Peterson and Silberschatz [PS85] (Section 8.2.2) introduce a method for drawing resource allocation graphs. However, in their version the resource allocation graph shows all the hold-wait-states that exist at any one given point in time. This is a good tool for illustrating a transient state in the system, but in order to use their version of the graphs for detecting any possible deadlock situation we would have to draw a graph for each and every possible combination of execution state for all threads in the system. Then, each and every one of these graphs would have to be analysed for cycles indicating circular wait. This is clearly unpractical. Instead, we modify the method of drawing resource...
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...Business Applications typically found in RIT’sAlgebra for Mgmt. Science (1016-225) and Calculus for Mgmt. Science (1016-226) I. Total Revenue, Total Cost, Total Profit a. Total Revenue (R) from the sale of x units = R(x): R(x) = (price per unit)(# of units) R(x) = (price per unit)(x) b. Total Cost (C) of production and sale of x units = C(x): C(x) = (cost per unit)(x) + fixed costs Total cost is made up of 2 parts: 1. fixed costs (e.g. rent, utilities, etc) 2. variable costs (costs directly related to the number of units produced) c. Total Profit (P) from the production and sale of x units = P(x): P(x) = R(x) – C(x) Total profit is the difference between the amount received from sales (revenue) and cost of production. II. Marginal Revenue, Marginal Cost, and Marginal Profit a. Marginal Revenue MR ( ) 1. If R(x) is linear, then MR is the slope of the revenue function. 2. Calculus for Management Science: MR is the derivative of the revenue function b. Marginal Cost MC ( ) ( ) 1. If C(x) is linear, then MC is the slope of the cost function. 2. Calculus for Management Science: MC is the derivative of the cost function III. a. Marginal Profit MP 1. If P(x) is linear, then MP is the slope of the profit function. 2. Calculus for Management Science: MP is the derivative of the profit function IV. Break Even Point a. The point at which the revenue equals the cost: which is another way of saying… Find x such that R(x) = C(x) Find x such that P(x) = 0 ...
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...of G. For example, if G is the free, shown in the figure then PG (k) = K (K-1)2 d) Chromatic Index of G: if G is K colorable (e) but not (K-1) colorable (e) we say that the chromatic index of G is K and write X` (G) = K. For example, the figure shown for which X’ (G) = 4. e) MAP: A map is defined as a representation, usually on a flat surface of a whole or part of an area. The job of a map is to describe spatial relationships of specific features that the map aims to represent. Define A flow of a network: A flow in a network is a function p that assigns to each are a non-negative real number p(a), called the flow in a, in such a way, that, i) For each are - a, p(a) ≤(( (a); ii) The out-degree and in-degree of each vertex, other than v or w, are equal. What is edge-disjoint paths? Edge-disjoint paths: The maximum number of paths from v to w, no two of which have an edge in common such path are called edge-disjoint paths. For this figure, the edge-disjoint paths is E1 = {ps, qs, ty, tz} and E2 = {uw, xw, yw, zw}. 1) Matrix representation of a graph 2) The eight circle problem 3) Six people at a party. Discuss the application domain of Explain that any simple graph with in...
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...lower end of coronal suture SUTURES * Sagittal suture - This lies in between two parietal bone. * Coronal suture - This lies in between the frontal and parietal bone on either side. * Frontal suture - This lies in between two frontal bone. * Lambdoid suture - It lies in between the parietal and occipital bone on either side. * Clinical importance of suture:- * These suture permit gliding movement of one bone over other during moulding of the head in the vertex presentation , as a result the diameter of the head get smaller so passage of head through the birth canal become easier. * Position of fontanelle and sagittal suture can identify attitude and position of vertex. * From the digital palpation of the sagittal suture during labour, degree of internal rotation and degree of moulding of the head can be noticed. * In deep transverse arrest, this sagittal suture lies transversely at the level of the ischial spines. AREA OF SKULL * Vertex - It is the quadrangular area bounded anteriorly by the bregma and coronal sutures behind by the lambda and the lambdoid sutures and laterally by the line passing through the...
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