Consistent Heuristic: An admissible heuristic h is consistent (or monotone) if for each node N andeach child N' of N, the following inequality holds:h(N) < c(N, N') + h(N'),where c(X,Y) is the cost of the edge between X and Y.Prove that whenever A* chooses to expand a goal node, the path to this node is optimal.

Given: h(N) < c(N, N) + N(N!), where c(X,Y) is the cost of the edge between X and Y Calculate:…

Answered: Consistent Heuristic: An admissible…

Operations Research : Applications and Algorithms

4th Edition

ISBN:9780534380588

Author:Wayne L. Winston

Publisher:Wayne L. Winston

Chapter9: Integer Programming

Section9.7: Implicit Enumeration

Problem 4P

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True or False: - Best-first search is optimal in the case where we have a perfect heuristic (i.e., h(?) = h∗(?), the true cost to the closest goal state). - Suppose there is a unique optimal solution. Then, A* search with a perfect heuristic will never expand nodes that are not in the path of the optimal solution.- A* search with a heuristic which is admissible but not consistent is complete.
Be G = (V. E) a connected graph and u, vEV.The distance Come in u and v, denoted by du, v), is the length of the shortest path between u and v, Meanwhile he width from G, denoted as A(G) is the greatest distance between two of its vertices. Dice k EN such that k>0, consider the following decision problem: k-WIDTH: • I«WIDTH = {G| G is a graph} - L«WIDTH = {G | Gis a connected graph such that A(G) > k} Show that k-WIDTH EP. Hint:Study algorithms that find the shortest path between two vertices of a graph.
he heuristic path algorithm is a best-first search in which the objective function is f(n) = (4 −w)g(n) + wh(n). For what values of w is this algorithm guaranteed to be optimal? What kind of search does this perform when w = 0? When w = 1? When w = 4?
Please answer the following question in full detail. Please be specifix about everything: You have learned before that A∗ using graph search is optimal if h(n) is consistent. Does this optimality still hold if h(n) is admissible but inconsistent? Using the graph in Figure 1, let us now show that A∗ using graph search returns the non-optimal solution path (S,B,G) from start node S to goal node G with an admissible but inconsistent h(n). We assume that h(G) = 0. Give nonnegative integer values for h(A) and h(B) such that A∗ using graph search returns the non-optimal solution path (S,B,G) from S to G with an admissible but inconsistent h(n), and tie-breaking is not needed in A∗.
Establish the Shortest-Paths Optimality Conditions in the Proposition P. Then, for any v reachable from s, the value of distTo[v] is the length of some path from s to v, with distTo[v] equal to infinity for all v not reachable from s. Let G be an edge-weighted digraph. s is a source vertex in G. distTo[] is a vertex-indexed array of path lengths in G. These numbers represent the lengths of the shortest pathways if and only if each edge e from v to w satisfies the condition that distTo[w] = distTo[v] + e.weight() (i.e., no edge is eligible).
We know that when we have a graph with negative edge costs, Dijkstra’s algorithm is not guaranteed to work. (a) Does Dijkstra’s algorithm ever work when some of the edge costs are negative? Explain why or why not. (b) Find an algorithm that will always find a shortest path between two nodes, under the assumption that at most one edge in the input has a negative weight. Your algorithm should run in time O(m log n), where m is the number of edges and n is the number of nodes. That is, the runnning time should be at most a constant factor slower than Dijkstra’s algorithm. To be clear, your algorithm takes as input (i) a directed graph, G, given in adjacency list form. (ii) a weight function f, which, given two adjacent nodes, v,w, returns the weight of the edge between them. For non-adjacent nodes v,w, you may assume f(v,w) returns +1. (iii) a pair of nodes, s, t. If the input contains a negative cycle, you should find one and output it. Otherwise, if the graph contains at least one…
Question 8 Greedy best-fırst search is equivalent to A* search with all step costs set to 0. O True O False Question 9 If you had implemented Uniform Cost Search (the graph search version) in Programming Assignment 1, it would have found an optimal solution. (You may assume that the path costs are kept with the nodes on the frontier and explored lists and checked when comparing newly generated states to what has been seen before.) O True O False Question 10 A* search with an admissible heuristic always expands fewer nodes than depth-first search. O True O False
Algorithm A* for a monotonic evaluation function.A-Star(G, s, r)in: graph G = (V , E); start vertex s; goal vertex rout: mapping π : V → Vlocal: open list S; cost function g(u v); heuristic lower bound estimate h(u v)
Suppose we have a heuristic h that over-estimates h* by at most epsilon (i.e., for all n, 0<= h(n) <= h*(n)+epsilon). Show that A* search using h will get a goal whose cost is guaranteed to be at most epsilon more than that of the optimal goal.
Consider the following algorithm for the maximum cut problem, based on the technique of local search. Given a partition of V into sets, the basic step of the algorithm, called flip, is that of moving a vertex from one side of the partition to the other. The following algorithm finds a locally optimal solution under the flip operation, i.e., a solution which cannot be improved by a single flip. The algorithm starts with an arbitrary partition of V. While there is a vertex such that flipping it increases the size of the cut, the algorithm flips such a vertex. (Observe that a vertex qualifies for a flip if it has more neighbors in its own partition than in the other side.) The algorithm terminates when no vertex qualifies for a flip. Show that this algorithm terminates in polynomial time, and achieves an approximation guarantee of 1/2.
Algorithm for Alpha-beta pruning using minimax.Minimax-Alpha-Beta(v, α, β)in: node v; alpha value α; beta value βout: utility value of node v
group of n people are lying on the beach. The beach is represented by the real line R and the location of the i-th person is some integer x; e Z. Your task is to prevent people from getting sunburned by covering them with umbrellas. Each umbrella corresponds to a closed interval I = [a, a+ L] of length Le N, and the i-th person is covered by that umbrella if r; e I. Design a greedy algorithm for covering all people with the minimum number of umbrellas. The input consists of the integers x1,..., Xn, and L. The output of your algorithm should be the positions of umbrellas. A For example, if the input is x1 = 1, x2 = 3, x3 = 5, and L = 2, then an optimum solution is the set of two umbrellas placed at positions 2 and 5, covering intervals [1,3] and [4, 6]. %3D %3! %3D 3 4 5 6 1 2 The running time of your algorithm should be polynomial in n.

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