What is the best case, worst case time complexity for determiningif a given directed edge exists in a graph. Assume the outgoingedges from each Vertex are stored in an adjacency list for thatvertex. V is the total number of vertices in the graph, and E is thetotal number of outgoing edges in the graph, and you can find thevertex itself in constant time from a HashTable or HashMap ?Best Case: [ Select ]Worst Case: [ Select] /[ Select ]O(1)O(V)O(E)O(V+E)

Suppose, a vertex u∈V exists in a graph, G = (V, E) with E as edge between two nodes. In a graph, G…

Answered: What is the best case, worst case time…

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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Suppose we have a hash table with N slots containing n keys. Suppose that instead of a linked list in open hashing, each slot is implemented as a binary search tree. Give the worst and the best time complexity of adding an entry to this hash table. Explain your answers. Worst Case Best Case
If your hash map is flawed, random integers will be lumped together (in the same LinkedList) even if they have no obvious connection. Explaining how this works will allow you to defeat hash maps.
a. Consider each 3 consecutive digits in your ID as a key value. Using linear propping, insert items with those keys into an empty hash table and show your steps. Example ID: 201710349. You must use your own ID. Key values: 201, 710, 340 tableSize: 3 hash(x) = x mod tableSize b. Calculate the number of edges in a complete directed graph with N vertices. Where N is equal to the 3rd and 4th digits in your ID. Show your steps. Example ID: 201710340. You must use your own ID. N = 17 c. Below an adjacency list representation of an undirected graph where there are no weights assigned to the edges. Draw 1. the graph and 2. the adjacency matrix with this adjacency list representation. 1 2
Develop an API and implementation that uses Dijkstra’s algorithm to solve the multisource shortest-paths problem on edge-weighted digraphs with positive edge weights: given a set of sources, find a shortest-paths forest that enables implementation of a method that returns to clients the shortest path from any source to each vertex. Hint : Add a dummy vertex with a zero-weight edge to each source, or initialize the priority queue with all sources, with their distTo[] entries set to 0.
We have a badly constructed hash map where all values are kept in one bucket (LinkedList). Explain how this defeats the hash map's purpose.
As all the values are now lumped together, we have a hash map that was not constructed with great care (that is, they are all in the same LinkedList). Defend your position by detailing how this negates the need for hash maps.
For now, we're making do with a rudimentary hash map in which all the values are stored in a single bucket (that is, they are all in the same LinkedList). Once a hash map's goal is met, it is considered successful.
a. Consider each 3 consecutive digits in your ID as a key value. Using Open Hashing, insert items with those keys into an empty hash table and show your steps. Example ID: 201710349. You must use your own ID. Key values: 201, 710, 340 tableSize: 2 hash(x) = x mod tableSize b. Calculate the number of edges in a complete undirected graph with N vertices. Where N is equal to the 3rd and 4th digits in your ID. Show your steps. Example ID: 201710340. You must use your own ID. N = 17 c. Below an adjacency matrix representation of a directed graph where there are no weights assigned to the edges. Draw 1. The graph and 2. The adjacency list with this adjacency matrix representation. 2 6
Complete this method so that it computes the degree of a node if the edges or an undirected graph are stored in an Adjacency Matrix called matrix. public int computeDegree (Graphnode node) { int index = node.getMatrixIndex(); // maps this node's ke y to correct index in matrix int degree [ Select ] for (int x=0; x < [ Select ] ; x++) { if ( [ Select ] [: [ Select] return degree; }
A hash-map has been constructed with quadratic-hashing. The hashing function is h(k;) = (3 * k; + 7) mod 17 and the table length is N = 17. How many cells will be probed by the call of insert(2) 1 4 5 6 7 8 10 11 12 13 14 15 16 26 89 11 35 70 59 Note: First row contains the indices and the second row contains the items.
We're using a basic hash map with all values in one bucket (LinkedList). Successful hash maps achieve their goals.
We have a hash map that was not constructed correctly since all of the values were placed into the same bucket (that is, they are all in the same LinkedList). Give an explanation of how this negates the objective of utilizing a hash map in the first place.

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