2. Honest Goblins are Easy to Find At Hogwarts a new shipment of n goblins has arrived. To be of any use, a goblin must be completely truthful (never lies). Unfortunately, not all of the ngoblins in the shipment are truth tellers. Only some are truth-tellers, and some are deceivers. It is your task to design an algorithm to separate the truth-teller r goblins from the deceiver goblins. To do this, you have one 1 and that you test, the following can be concluded from the goblin responses Geblin A says Goblin B says Conclusion Bis a truth-teller A is a truth-teller both are truth-tellers or both are deceivers B is a truth-teller A is a deceiver at least one is a deceiver Bisa deceiver A is a truth-teller at least one is a deceiver Bis a deceiver A is a deceiver at least one is a deceiver Comparing 2 goblins with each other cests 1"goblin comparison". Assume that more than half of the given n goblins are truth tellers. Design an O(n) goblin comparison algorithm that will identify 1 truth teller.

2. Honest Goblins are Easy to Find At Hogwarts a new shipment of n goblins has arrived. To be of any use, a goblin must be completely truthful (never lies). Unfortunately, not all of the ngoblins in the shipment are truth tellers. Only some are truth-tellers, and some are deceivers. It is your task to design an algorithm to separate the truth-teller r goblins from the deceiver goblins. To do this, you have one 1 and that you test, the following can be concluded from the goblin responses Geblin A says Goblin B says Conclusion Bis a truth-teller A is a truth-teller both are truth-tellers or both are deceivers B is a truth-teller A is a deceiver at least one is a deceiver Bisa deceiver A is a truth-teller at least one is a deceiver Bis a deceiver A is a deceiver at least one is a deceiver Comparing 2 goblins with each other cests 1"goblin comparison". Assume that more than half of the given n goblins are truth tellers. Design an O(n) goblin comparison algorithm that will identify 1 truth teller.

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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Concept explainers

Intelligent Machines

Machines that use artificial intelligence to learn from the environment and past experiences to perform better in the presence of variability and uncertainty are called intelligent machines. Intelligence develops according to environment and experience. It is human-like behavior. When the computer system or any programmed machine learns from the inputs and corresponding outputs, it can be called an intelligent machine.

Question

2. Honest Goblins are Easy to Find
At Hogwarts a new shipment of n goblins has arrived. To be of any use, a goblin must be completely truthful (never lies). Unfortunately, not all of the n goblins in the shipment are truth tellers. Only some are truth-tellers, and some are deceivers. It is your task to design an algorithm to separate the
truth-teller goblins from the deceiver goblins. To do this, you have one tool available: You may combine any two goblins and have them state whether the other goblin is a truth-teller or a deceiver. A truth-teller will always say correctly what the other goblin is, but a deceiver may lie (but also may
sometimes tell the truth to REALLY confuse you). For any two goblins that you test, the following can be concluded from the goblin responses:
Goblin A says
Goblin B says
Conclusion
B is a truth-teller
A is a truth-teller
both are truth-tellers or both are deceivers
B is a truth-teller
A is a deceiver
at least one is a deceiver
B is a deceiver
A is a truth-teller
at least one is a deceiver
B is a deceiver
A is a deceiver
at least one is a deceiver
Comparing 2 goblins with each other costs 1 "goblin comparison". Assume that more than half of the given n goblins are truth tellers. Design an O(n) goblin comparison algorithm that will identify 1 truth teller.
3. Card Flipper: You walk into a room, and see a row of n cards. Each one has a number x; written on it, where i ranges from 1 to n. However, initially all the cards are face down. Your goal is to find a local minimum: that is, a card i whose number is less than or equal to those of its neighbors, x-1= X;
<= Xj+1. The first and last cards can also be local minima, and they only have one neighbor to compare to. There can be many local minima, but you are only responsible for finding one of them.
Obviously you can solve this problem by turning over all n cards, and scanning through them. However, show that you can find such a minimum by turning over only O(log n) cards.
4. Even Faster Multiplication?
Show how to multiply 2 n-digit numbers by breaking the problem down into 5 subproblems of roughly one third the size. Turn this into a recursive multiplication algorithm and analyze the run time.

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You are given a N*N maze with a rat placed at maze[0][0]. Find whether any path exist that rat can follow to reach its destination i.e. maze[N-1][N-1]. Rat can move in any direction ( left, right, up and down).Value of every cell in the maze can either be 0 or 1. Cells with value 0 are blocked means rat cannot enter into those cells and those with value 1 are open.Input FormatLine 1: Integer NNext N Lines: Each line will contain ith row elements (separated by space)Output Format :The output line contains true if any path exists for the rat to reach its destination otherwise print false.Sample Input 1 :31 0 11 0 11 1 1Sample Output 1 :trueSample Input 2 :31 0 11 0 10 1 1Sample Output 2 : false Solution: //// public class Solution { public static boolean ratInAMaze(int maze[][]){ int n = maze.length; int path[][] = new int[n][n]; return solveMaze(maze, 0, 0, path); } public static boolean solveMaze(int[][] maze, int i, int j, int[][] path) {//…
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