...3 I J Consider the graph above (not drawn to scale). Given the heuristic values for the distance to city F: h(A) = 5 h(B) = 3 h(C) = 6 h(D) = 3 h(E) = 2 h(F) = 0 h(G) = 7 h(H) = 8 h(I) = 29 h(J) = 9 h(K) = 8 Draw the search trees resulting from i) BFS can be performed on the graph ii) DFS iii) Uniform Cost iv) A* search on the graph with start node A. Q2) Consider the problem of finding a path in the grid shown below from the position s to the position g. A piece can move on the grid horizontally and vertically, one square at a time. No step may be made into a forbidden shaded area. 1. On the grid, number the nodes in the order in which they are removed from the frontier in a depth-first search from s to g, given that the order of the operators you will test is: up, left, right, then down. Assume there is a cycle check. 2. Number the nodes in order in which they are taken off the frontier for an A* search for the same graph. Manhattan distance should be used as the heuristic function. The Manhattan distance between two points is the distance in the x-direction plus the distance in the y-direction. It corresponds to the distance traveled along city streets arranged in a grid. For example, the Manhattan distance between g and s is 4. What is the path that is found? 3. Based on this experience, discuss which of the two algorithms is best suited for this problem. 4. Suppose that the graph extended infinitely in all directions...
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...Here is a worked example and process for constructing a network diagram: Using this example: Activity Duration (days) Predecessor(s) A 5 --- B 2 --- C 4 A D 6 B E 3 C, D F 1 D Follow this sequence: 1. Construct the Network diagram. 2. Check the network diagram. 3. Add durations to the activities. 4. Identify all paths through the network. 5. Identify the critical path (CP) and scheduled duration. 6. Calculate slack times (float) for each activity. 1. Construct the Network diagram. Using the AIB method: Draw a box on the left labeled "Start" Add boxes to the right of this box for activities with no predecessors, in the example A, and B. Connect these boxes to the box labeled start, but not to each other! I always start with box A at the top, and others in sequence underneath, vertically. Add arrows to the end of the lines where they connect to the activity (A and B) boxes. (This is important for step 4). Draw a box to the right of activity A for any activities which list A as a predecessor. (C in our example). Connect this box to activity A, don't forget the arrow! Draw a box to the right of activity B for any activities which list B as a predecessor. (D in our example). Connect this box to activity B, don't...
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...discrete geometry involve finite or discrete sets of basic geometric objects, such as points, lines, planes, circles, spheres, polygons, and so forth. The subject focuses on the combinatorial properties of these objects, such as how they intersect one another, or how they may be arranged to cover a larger object. Discrete geometry has large overlap with convex geometry and computational geometry, and is closely related to subjects such as finite geometry, combinatorial optimization, digital geometry, discrete differential geometry, geometric graph theory,toric geometry, and combinatorial topology. Although polyhedra and tessellations have been studied for many years by people such as Kepler and Cauchy, modern discrete geometry has its origins in the late 19th century. Early topics studied were: the density of circle packings by Thue,projective configurations by Reye and Steinitz, the geometry of numbers by Minkowski, and map colourings by Tait, Heawood, and Hadwiger. László Fejes Tóth, H.S.M. Coxeter and Paul Erdős, laid the foundations of discrete geometry. "This is an introduction to the field of discrete geometry understood as the investigation of combinatorial properties of configurations of (usually finitely many) geometric objects … . The book is written in a lively and stimulating but very precise style and contains many figures. It gives a good impression of the richness and the relevance of the field. Statement of the Problem Discrete geometry is not in a scope...
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...Lecture 12: Mean Shift and Normalized Cuts CAP 5415 Fall 2006 Each Pixel Data Vector Example Once we have vectors… • Group the vectors into clusters • Algorithms that we talked about last time: – K-means – EM (Expectation Maximization) • Today: (From Comanciu and Meer) – Mean-Shift – Normalized Cuts Mean-Shift • Like EM, this algorithm is built on probabilistic intuitions. • To understand EM we had to understand mixture models • To understand mean-shift, we need to understand kernel density estimation (Take Pattern Recognition!) Basics of Kernel Density Estimation • Let’s say you have a bunch of points drawn from some distribution • What’s the distribution that generated these points? Using a Parametric Model • Could fit a parametric model (like a Gaussian) • Why: – Can express distribution with a few number of parameters (like mean and variance) Non-Parametric Methods • We’ll focus on kernel-density estimates • Basic Idea: Use the data to define the distribution • Intuition: – If I were to draw more samples from the same probability distribution, then those points would probably be close to the points that I have already drawn – Build distribution by putting a little mass of probability around each data-point • Why not: – Limited in flexibility Example Formally Kernel • Most Common Kernel: Gaussian or Normal Kernel • Another way to think about it: (From Tappen – Thesis) – Make an image, put 1(or more) wherever you have...
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...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 circuit = closed Euler path Theorem: A graph that has an Euler circuit must have all vertices of even degree Graph: If 2 vertices of odd degree or ==> Euler path 0 vertices of odd degree...
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...Length: 3 Parts: See Assignment Details Points Possible: 75 Graphs and Trees Task Background: Graphs and trees provide you with ways to visualize data sets, and the opportunity to do analysis on the data (e.g., shortest path). Knowing the structure of a database enables you to choose a proper algorithm for searching for data within a database. Primary Task Response: Within the Discussion Board area, write up to 3 paragraphs that respond to the following questions with your thoughts, ideas, and comments. This will be the foundation for future discussions by your classmates. Be substantive and clear, and use examples to reinforce your ideas. Part I (25 points – distributed as follows) Trees are somewhat less complicated than graphs, which makes things like data searching easier, when a data has the structure of a tree. However, not all data can be represented by a tree. Give an example of a data set that cannot be represented by a tree, but that can be represented by a more general graph. 1) Create, show, and describe your data set. (5 points) V = {Bill, John, Kim, James, Chris, Destiny, Noah, Paul} E = {(Bill, John), (Kim, James), (Chris, Destiny), (Noah, Paul), (Bill, Kim), (John, Chris), (Destiny, Noah)} These are people that are employees at a store. Some work on the same shift together and associate with each other. 2) Then, show by building a graph, how your data is represented by a graph. (5 points) Bill Bill John John Chris Chris Kim Kim ...
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...Phase 3 DB Graphs and Trees Elie De Jesus MATH203-1302A-01 – Discrete Mathematics Professor Andrew Halverson April 24, 2013 Part I Graphs and trees are a little more complicated to understand than what I thought. Based on the information that I found they give you a way to visualize your sets and use the data that you have to find the shortest path. So because of this it shows that Trees cannot contain a cycle, so a set would be Y=COS(X); which can be a general graph but not a tree. The one example that I understood was the one about “the mileage on a bike”. Now I don’t quite understand the example but it shows that the graph would have a decrease in mileage where as it would increase in time. That is not how a tree is explained because there is no sequence to be shown for the data. This is the examples graph: So based on that example I understand that the tree encoding defines a root node or one path between two nodes that represent the output of a solution. A tree is still a graph but without multiple paths. So to be a tree it has to start from any node and be able to reach another, there can be no cycles, and you must have more nodes that edges. Part II To first answer this question one must know the meaning of a Breadth-first or a Depth-first. A Breadth-first search is a strategy for searching in a graph when search is limited to essentially two operations: (a) visit and inspect a node of a graph; (b) gain access to visit the nodes that neighbor the currently...
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...数据结构基本英语词汇 I like ITPUB! 数据结构基本英语词汇 数据抽象 data abstraction 数据元素 data element 数据对象 data object 数据项 data item 数据类型 data type 抽象数据类型 abstract data type 逻辑结构 logical structure 物理结构 phyical structure 线性结构 linear structure 非线性结构 nonlinear structure 基本数据类型 atomic data type 固定聚合数据类型 fixed-aggregate data type 可变聚合数据类型 variable-aggregate data type 线性表 linear list 栈 stack 队列 queue 串 string 数组 array 树 tree 图 grabh 查找，线索 searching 更新 updating 排序（分类) sorting 插入 insertion 删除 deletion 前趋 predecessor 后继 successor 直接前趋 immediate predecessor 直接后继 immediate successor 双端列表 deque(double-ended queue) 循环队列 cirular queue 指针 pointer 先进先出表（队列） first-in first-out list 后进先出表（队列） last-in first-out list 栈底 bottom 栈定 top 压入 push 弹出 pop 队头 front 队尾 rear 上溢 overflow 下溢 underflow 数组 array 矩阵 matrix 多维数组 multi-dimentional array 以行为主的顺序分配 row major order 以列为主的顺序分配 column major order 三角矩阵 truangular matrix 对称矩阵 symmetric matrix 稀疏矩阵 sparse matrix 转置矩阵 transposed matrix 链表 linked list 线性链表 linear linked list 单链表 single linked list 多重链表 multilinked list 循环链表 circular linked list 双向链表 doubly linked list 十字链表 orthogonal list 广义表 generalized list 链 link 指针域 pointer field 链域 link field 头结点 head node 头指针 head pointer 尾指针 tail pointer 串 string 空白（空格）串 blank string 空串（零串） null string 子串 substring 树 tree 子树 subtree 森林 forest 根 root 叶子 leaf 结点 node 深度 depth 层次 level 双亲 parents 孩子 children 兄弟 brother 祖先 ancestor 子孙 descentdant ...
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...Q1. The above represents a random looking undirected graph where the circles represent nodes and the boundary circumscribes all the nodes of the graph. The length of the segment joining two nodes represents the length of the shortest path from one node to another. Let’s assume nodes A and B are the farthest pair of nodes according to the definition in the question. Thus A and B have to be on the circumscribing boundary. Thus AB represents the diameter of the graph. Let C be any random node in the graph. Let D be a point which is farthest away from C. So if a BFS is launched on node C, the depth(in terms of the summation of the edges) of the BFS tree rooted at C will be equal to CD. As a result CD >= CA and CD>=CB. Now let AB = d. So from triangle inequality CA+CB>=d. Thus d/2 <= max(CA,CB)<=d. Now since CD is greater than or equals to both CA and CB, it has to be greater than or equals to d/2 as well. CD has to be less than d as well because if not then CD would have been the diameter. But that is not the case. Hence d/2<=CD<=d thus if we can find the length of CD we are done with approximating the value of the diameter within a factor of 2 and that we can do by creating BFS tree from C and calculating depth of the tree which is nothing but...
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...Task Background: Graphs and trees are useful in visualizing data and the relations within and between data sets. Conversely, it is also important to be able to represent graphs as databases or arrays, so that programs for processing the data can be written. Part I: Adjacency Matrix and Shortest Path Construct a graph based on the adjacency matrix that appears below. Label all nodes with indices consistent with the placement of numbers within the matrix. ⌈0 | 6 | 0 | 5 | 0⌉ | | 6 | 0 | 1 | 0 | 3 | | | 0 | 1 | 0 | 4 | 8 | | | 5 | 0 | 4 | 0 | 0 | | ⌊0 | 3 | 8 | 0 | 0⌋ | | | | | | * Describe the graph and why it is consistent with the matrix. The Graph above is an undirected graph. As the lines are not directed towards a particular node, the lines or edges go both ways. It is consistent with the matrix because the matrix is defining the edges. * How many simple paths are there from vertex 1 to vertex 5? Explain. There are 3 paths. 1-2-5, 1-2-3-5, 1-4-3-5. * Which is the shortest of those paths? That would be the 1-2 path. As the sum equals 9 and is the shortest path. 1-2-3-5=15 and 1-4-3-5=17 Part II: Trees * Construct and describe a tree that indicates the following: * A college president has 2 employees who answer directly to him or her, namely a vice president and provost. * The vice president and provost each have an administrative assistant. * Three deans answer to the provost, and the heads of finance and alumni relations...
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...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 b. Draw one spanning tree for this weighted graph and determine its weight. Show your work. (1 point) [Hint: One way to do this is to make a copy of the graph using cut/paste, then show your spanning tree.] Answer: Total weight: Part II. Case Study The Case of the Yip-it-tee DooDahs This week, the cast and crew of the “Patty Madeye Mysteries” will be filming...
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...Assignments Section 9.1 Exercise 2 The graph is not a tree, because there more than one paths from one vertex to another. Exercise 3 The graph is not a tree, because not all of the vertices of the graph are connected. Thus, there is no an existing unique path between some vertices. Exercise 9 The height of the tree in the Exercise 8 is 4 Exercise 10 [pic] The height of this tree is 5 Exercise 12 An example of the hierarchical relationship is a family tree: [pic] Section 9.2 Exercise 5 Siblings of Ares: Hades, Poseidon and Zeus. Exercise 6 [pic] Exercise 8 The parent of c is b, while the parent of j is e. Exercise 9 The children of d are h and i, while the only child of e is j. Exercise 11 The siblings of f are e and g. The sibling of h is i. Exercise 12 The terminal vertices are: j, f, g, h, i. Exercise 18 If two vertices in a rooted tree have same ancestors, they must share same simple path to their parent, which means that they are siblings. Exercise 19 The only vertex in the rooted tree, that does not have ancestors, is a root itself. Exercise 20 By definition, a tree cannot have two vertices sharing a child, because this will create two possible paths to that child and will violate the definition of a tree. Exercise 21 A vertex in a rooted tree that has no decedents is a leaf (a terminal vertex). Exercise 29 According to the Theorem 9.2.3, if a graph is a tree, then it is acyclic and has n – 1 edges. The graph in this problem contains a cycle (v1...
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...Vijay V. Vazirani College of Computing Georgia Institute of Technology Copyright c 2001 Approximation Algorithms Springer Berlin Heidelberg NewYork Barcelona Hong Kong London Milan Paris Singapore Tokyo To my parents Preface Although this may seem a paradox, all exact science is dominated by the idea of approximation. Bertrand Russell (1872–1970) Most natural optimization problems, including those arising in important application areas, are NP-hard. Therefore, under the widely believed conjecture that P = NP, their exact solution is prohibitively time consuming. Charting the landscape of approximability of these problems, via polynomial time algorithms, therefore becomes a compelling subject of scientific inquiry in computer science and mathematics. This book presents the theory of approximation algorithms as it stands today. It is reasonable to expect the picture to change with time. The book is divided into three parts. In Part I we cover a combinatorial algorithms for a number of important problems, using a wide variety of algorithm design techniques. The latter may give Part I a non-cohesive appearance. However, this is to be expected – nature is very rich, and we cannot expect a few tricks to help solve the diverse collection of NP-hard problems. Indeed, in this part, we have purposely refrained from tightly categorizing algorithmic techniques so as not to trivialize matters. Instead, we have attempted to capture, as accurately as possible, the individual character...
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...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. Sec 11.1 /8 Figure 11.10 shows an undirected graph representing a section of a department store. The vertices indicate where cashiers are located; the edges denote unblocked aisles between cashiers. The department store wants to set up a security system where (plainclothes) guards are placed at certain cashier locations so that each cashier either has a guard at his or her location or is only one aisle away from a cashier who has a guard. What is the smallest number of guards needed? a b c Sec 11.1 /11 10. Give an example of a connected graph G where removing any edge of G results in a disconnected graph. 11. Let G be a graph that satisfies the condition in Exercise 10. (a) Must G be loop-free? (b) Could G be a multigraph? (c) If G has n vertices, can we determine how many edges it has? Sec 11.1 /15 15. For the undirected graph in Fig. 11.12, find and solve a recurrence relation for the number of closed v-v walks of length n ≥ 1, if we allow such a walk, in this case, to contain or consist of one or more loops. Sec 11.1 /16 16. Unit-Interval Graphs. For n ≥ 1, we start with n closed intervals of unit length and draw the corresponding unit-interval graph on n vertices, as shown in Fig...
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...8. Figure 11.10 shows an undirected graph representing a section of a department store. The vertices indicate where cashiers are located; the edges denote unblocked aisles between cashiers. The department store wants to set up a security system where (plainclothes) guards are placed at certain cashier locations so that each cashier either has a guard at his or her location or is only one aisle away from a cashier who has a guard. What is the smallest number of guards needed? I think {a}→{a,d,e,b},{i}→{i,h,e,j},{g}→{f,c,k,g}, where {x}→{T} is meant that we put a guard at x that covers all elements in T. Now we have to prove that we cant get what we desire with a smaller number of guards. Therefore we take 2 guards, and if not possible with 2 its not with one, as well. To cover the elements in the set {d,e,h} we requires 2 guards from {d,e,h,I,a}. If it is the 2nd case then no guards from this set can cover for instance g, Therefore we would have once again at least 3 guards. 11. Let G be a graph that satisfies the condition in Exercise 10. (a) Must G be loop-free? (b) Could G be a multigraph? (c) If G has n vertices, can we determine how many edges it has? Eliminating an edge from the loop will still lead to a connected graph, therefore it is a necessity. It cant be since eliminating from a vertex with degree more than 1 will still result in a connected graph. It can be verified that these sorts of graphs are tress, therefore the number of edges...
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