Consider the following undirected binary tree T with 17 vertices. a C d 30 P Starting with the root vertex a, we can use Breadth-First Search (BFS) or Depth-First Search (DFS) to pass through all of the vertices in this tree. Whenever we have more than one option, we always pick the vertex that appears earlier in the alphabet. For example, from vertex a, we go to b instead of c. Clearly explain the difference between Breadth-First Search and Depth-First Search, and determine the order in which the 17 vertices are reached using each algorithm. K

Consider the following undirected binary tree T with 17 vertices. a C d 30 P Starting with the root vertex a, we can use Breadth-First Search (BFS) or Depth-First Search (DFS) to pass through all of the vertices in this tree. Whenever we have more than one option, we always pick the vertex that appears earlier in the alphabet. For example, from vertex a, we go to b instead of c. Clearly explain the difference between Breadth-First Search and Depth-First Search, and determine the order in which the 17 vertices are reached using each algorithm. K

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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4. Consider the graph G, shown below, which is simply a copy of K5. 02 V3 5 V1 24 V5 How many distinct spanning trees does G have? (Hint: Break up your search by the isomorphism type of the tree, as discovered on the previous page. So for example, start by counting the paths of length 5 in G. Then proceed to the next type of tree with 5 vertices. The total number of trees is 125, but please use this answer only to check that your solution is complete!)
depthFirstSearch(digraph G=(V,E)) step 1: time← 0. Step 2. Using the placeholder v, iterate through all vertices of G and execute: If v is still white, do: Run dfs(v). dfs (vertex v) Step 1.: Time ← Time + 1, v.d = Time. coloring v grey. Step 2. [Touch all subsequent descendants of node v] For any unclassified (directed) edge e with Starting node v execute: Let u be the end node of this edge. If the color of u is white then do: Give e the treeEdge classification. Run dfs(u). Other classify e as otherEdge Step 3. [End the visit] ime ← time + 1. Color with the paint black. v.f = time. its not a marked excercise but a self test and i cant really figure it out.
Consider the following graph. Start from vertex x, use the Depth First Search (DFS) algorithm (described in the textbook) to compute the DFS tree(s). Show intermediate results similar to the textbook example and show the final DFS tree(s). It is assumed that in the adjacency list representation, all vertices are stored according to their alphabetical order. In other words, a vertex’s neighbors should be explored by their relative alphabetical order.
Provide a class to assess DAGs for mathematical expressions. Use a vertex-indexed array to store values for every vertex. Suppose that values for leaves have already been determined. Explain a family of mathematical expressions where the expression tree's size is exponentially bigger than the matching DAG's size (so the running time of your programme for the DAG is proportional to the logarithm of the running time for the tree).
You are given a connected, undirected graph G. Devise an algorithm that produces a path that traverses each edge in G exactly once in each direction. A vertex may occur multiple times on the path. Provide a short justification about why your algorithm is correct, and analyze its efficiency.
Use depth-first search starting at vertex SS to construct a spanning tree of the graph below. If we tie-break alphabetically, what is the order of edges we construct our spanning tree? a. {S,T}, {T,X}, {W,X}, {W,Z}, {V,Z}, {X,Y}, {U,Y} b. {S,T}, {S,V}, {V,Z}, {W,Z}, {W,X}, {T,X}, {X,Y}, {U,Y} c. {S,T}, {T,X}, {W,X}, {S,V}, {V,Z}, {Y,Z}, {U,Y} d. {S,T}, {T,X}, {W,X}, {W,Z}, {V,Z}, {Z,Y}, {U,Y}
Question 18 Consider the graph below. Use Prim's algorithm to find a minimal spanning tree of the graph rooted in vertex A. Note: enter your answer as a set of edges {E1, E2, ...} and write each edge as a pair of nodes between parentheses separate by a comma and one blank space e.g. (A, B) 6. D 3 4 1 F 4 C B
Consider a complete undirected graph with vertex set (o, 1, 2, 3, 4). Entry Wij in the matrix W below is the weight of the edge (1.J). What is the minimum possible weight of a spanning tree Tin this graph? 8. 1 4 12 4 12 3 2 4 D. 9 d 10 2. 4. 8.
Consider eight points on the Cartesian two-dimensional x-y plane. a g C For each pair of vertices u and v, the weight of edge uv is the Euclidean (Pythagorean) distance between those two points. For example, dist(a, h) = V4? + 1² = v17 and dist(a, b) = v22 + 0² = 2. Using the algorithm of your choice, determine one possible minimum-weight spanning tree and compute its total distance, rounding your answer to one decimal place. Clearly show your steps.
Using the same graph as in the previous question, list the vertices in the order thatthey will be visited by a breadth-first traversal, starting withe vertex A. Again, when there are multiple adjacent vertexes, save them alphabetically, A to G.Enter the letters with no spaces between them, in the order they would be visited.
I A E H F В G Suppose you run a topological sort on the shown graph starting at vertex B. Use the rule that when DFS has a choice of nodes to visit, it visits them in alphabetical order. What's the first vertex in the resulting topological order? What is the second? What is the third? What is the fourth?
Correct answer will be upvoted else downvoted. You are given an established tree. Every vertex contains simulated intelligence huge loads of gold, which costs ci per one ton. At first, the tree comprises just a root numbered 0 with a0 huge loads of gold and cost c0 per ton. There are q inquiries. Each inquiry has one of two sorts: Add vertex I (where I is a list of inquiry) as a child to some vertex pi; vertex I will have simulated intelligence huge loads of gold with ci per ton. It's surefire that ci>cpi. For a given vertex vi think about the basic way from vi to the root. We need to buy wi huge loads of gold from vertices on this way, going through the base measure of cash. In case there isn't sufficient gold on the way, we purchase everything we can. On the off chance that we purchase x huge loads of gold in some vertex v the leftover measure of gold in it diminishes by x (obviously, we can't accepting more gold that vertex has right now). For each question of the…