Suppose you have n items with non-negative weights w1, ... , wn. For each subset S of the items, define S's weight as w(S) = [_{jES) wj, the sum of the weights of the items in S. In the problem SPLIT-EVEN, the goal is to split the items into sets S and S = {1,..., n} \ S such that max{w(S), w(S)} is minimized. Greedy Split(w1, ..., wn) A=Ø, B=Ø for i=1 to n if w(A) ≤ w(B) A = AU {i} else B = BU {i} return (A, B) Let OPT denote the minimum, over all subsets S of the items, of max{w(S), W(S)} Prove that OPT 2 maxj wj and OPT ≥ 1/2 [_{j=1}^{n} wj

Suppose you have n items with non-negative weights w1, ... , wn. For each subset S of the items, define S's weight as w(S) = [_{jES) wj, the sum of the weights of the items in S. In the problem SPLIT-EVEN, the goal is to split the items into sets S and S = {1,..., n} \ S such that max{w(S), w(S)} is minimized. Greedy Split(w1, ..., wn) A=Ø, B=Ø for i=1 to n if w(A) ≤ w(B) A = AU {i} else B = BU {i} return (A, B) Let OPT denote the minimum, over all subsets S of the items, of max{w(S), W(S)} Prove that OPT 2 maxj wj and OPT ≥ 1/2 [_{j=1}^{n} wj

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

See similar textbooks

Similar questions

09..
This problem is taken from the delightful book "Problems for Mathematicians, Young and Old" by Paul R. Halmos. Suppose that 931 tennis players want to play an elimination tournament. That means: they pair up, at random, for each round; if the number of players before the round begins is odd, one of them, chosen at random, sits out that round. The winners of each round, and the odd one who sat it out (if there was an odd one), play in the next round, till, finally, there is only one winner, the champion. What is the total number of matches to be played altogether, in all the rounds of the tournament? Your answer: Hint: This is much simpler than you think. When you see the answer you will say "of course".
There are n students who studied at a late-night study for final exam. The time has come to order pizzas. Each student has his own list of required toppings (e.g. mushroom, pepperoni, onions, garlic, sausage, etc). Everyone wants to eat at least half a pizza, and the topping of that pizza must be in his reqired list. A pizza may have only one topping. How to compute the minimum number of pizzas to order to make everyone happy?
It's Gary's birthday today and he has ordered his favourite square cake consisting of '0's and '1's . But Gary wants the biggest piece of '1's and no '0's . A piece of cake is defined as a part which consist of only '1's, and all '1's share an edge with each other on the cake. Given the size of cake N and the cake, can you find the count of '1's in the biggest piece of '1's for Gary ?Input Format :The first line of input contains an integer, that denotes the value of N. Each of the following N lines contain N space separated integers.Output Format :Print the count of '1's in the biggest piece of '1's, according to the description in the task.Constraints :1 <= N <= 1000Time Limit: 1 secSample Input 1:21 10 1Sample Output 1:3 Solution: /////////////////public class Solution { static int[][] dir = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } }; public static int dfs(String[] edge, int n) { boolean[][] visited = new boolean[n][n]; int max = 0; for(int…
7. For n 2 1, in how many out of the n! permutations T = (T(1), 7(2),..., 7 (n)) of the numbers {1, 2, ..., n} the value of 7(i) is either i – 1, or i, or i +1 for all 1 < i < n? Example: The permutation (21354) follows the rules while the permutation (21534) does not because 7(3) = 5. Hint: Find the answer for small n by checking all the permutations and then find the recursive formula depending on the possible values for 1(n).
Consider an n by n matrix, where each of the n2 entries is a positive integer. If the entries in this matrix are unsorted, then determining whether a target number t appears in the matrix can only be done by searching through each of the n2 entries. Thus, any search algorithm has a running time of O(n²). However, suppose you know that this n by n matrix satisfies the following properties: • Integers in each row increase from left to right. • Integers in each column increase from top to bottom. An example of such a matrix is presented below, for n=5. 4 7 11 15 2 5 8 12 19 3 6 9 16 22 10 13 14 17 24 1 18 21 23 | 26 | 30 Here is a bold claim: if the n by n matrix satisfies these two properties, then there exists an O(n) algorithm to determine whether a target number t appears in this matrix. Determine whether this statement is TRUE or FALSE. If the statement is TRUE, describe your algorithm and explain why your algorithm runs in O(n) time. If the statement is FALSE, clearly explain why no…
ProblemGiven a value `value`, if we want to make change for `value` cents, and we have infinitesupply of each of coins = {S1, S2, .. , Sm} valued `coins`, how many ways can we make the change?The order of `coins` doesn't matter.For example, for `value` = 4 and `coins` = [1, 2, 3], there are four solutions:[1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3].So output should be 4. For `value` = 10 and `coins` = [2, 5, 3, 6], there are five solutions: [2, 2, 2, 2, 2], [2, 2, 3, 3], [2, 2, 6], [2, 3, 5] and [5, 5].So the output should be 5. Time complexity: O(n * m) where n is the `value` and m is the number of `coins`Space complexity: O(n)""" def count(coins, value): """ Find number of combination of `coins` that adds upp to `value` Keyword arguments: coins -- int[] value -- int """ # initialize dp array and set base case as 1 dp_array = [1] + [0] * value) ++.
There are n people who want to carpool during m days. On day i, some subset ???? of people want to carpool, and the driver di must be selected from si . Each person j has a limited number of days fj they are willing to drive. Give an algorithm to find a driver assignment di ∈ si each day i such that no person j has to drive more than their limit fj. (The algorithm should output “no” if there is no such assignment.) Hint: Use network flow. For example, for the following input with n = 3 and m = 3, the algorithm could assign Tom to Day 1 and Day 2, and Mark to Day 3. Person Day 1 Day 2 Day 3 Driving Limit 1 (Tom) x x x 2 2 (Mark) x x 1 3 (Fred) x x 0
Problem1Given a value `value`, if we want to make change for `value` cents, and we have infinitesupply of each of coins = {S1, S2, .. , Sm} valued `coins`, how many ways can we make the change?The order of `coins` doesn't matter.For example, for `value` = 4 and `coins` = [1, 2, 3], there are four solutions:[1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3].So output should be 4. For `value` = 10 and `coins` = [2, 5, 3, 6], there are five solutions: [2, 2, 2, 2, 2], [2, 2, 3, 3], [2, 2, 6], [2, 3, 5] and [5, 5].So the output should be 5. Time complexity: O(n * m) where n is the `value` and m is the number of `coins`Space complexity: O(n)""" def count(coins, value): """ Find number of combination of `coins` that adds upp to `value` Keyword arguments: coins -- int[] value -- int """ # initialize dp array and set base case as 1 dp_array = [1] + [0] * value.. (+.
You are organizing a conference that has received n submitted papers. Your goal is to get people to review as many of them as possible. To do this, you have enlisted the help of k reviewers. Each reviewer i has a cost sij for writing a review for paper j. The strategy of each reviewer i is to select a subset of papers to write a review for. They can select any subset S; C {1,2, ..., n}, as long as the total cost to write all reviews is less than T (the time before the deadline): 2 Sij 1. (b) Show that for B = 2 this fraction is close to 1/3. [Hint: You can consider an instance with 3n + 1 papers and only n will be reviewed.]
We are given three ropes with lengths n₁, n2, and n3. Our goal is to find the smallest value k such that we can fully cover the three ropes with smaller ropes of lengths 1,2,3,...,k (one rope from each length). For example, as the figure below shows, when n₁ = 5, n₂ 7, and n3 = 9, it is possible to cover all three ropes with smaller ropes of lengths 1, 2, 3, 4, 5, 6, that is, the output should be k = 6. = Devise a dynamic-programming solution that receives the three values of n₁, n2, and n3 and outputs k. It suffices to show Steps 1 and 2 in the DP paradigm in your solution. In Step 1, you must specify the subproblems, and how the value of the optimal solutions for smaller subproblems can be used to describe those of large subproblems. In Step 2, you must write down a recursive formula for the minimum number of operations to reconfigure. Hint: You may assume the value of k is guessed as kg, and solve the decision problem that asks whether ropes of lengths n₁, n2, n3 can be covered by…
Suppose there are a set of n precincts P1 . . . Pn with m voters in each precinct. Each voter supportsone of two possible political parties. (In this country, only 2 political parties exist).We are required to carve the precincts into two electoral districts which each have exactly n/2 precincts.We know exactly how many voters in each precinct support each party. The goal is to strategicallyassign the precincts to the two districts.Describe an algorithm which determines whether it is possible to define the two districts insuch a way that the same political party holds the majority in both districts. Assume n is even.Example: Consider n = 4, with the voters distributed as in the table.In this case. By grouping precincts 1 and 3 into a district and precincts 2 and 4 into a district, theRepulsive party will have a majority in both districts. Though, by total count, the Repulsive party onlyhas a tiny advantage over the Demonic party (205 to 195).