Artificial Intelligence: A Modern Approach
3rd Edition
ISBN: 9780136042594
Author: Stuart Russell, Peter Norvig
Publisher: Prentice Hall
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Expert Solution & Answer
Chapter 5, Problem 14E
Explanation of Solution
Alpha beta pruning
- The exact statement is that the
algorithms examines bm/2 + bm/2 − 1 nodes at level m...
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Students have asked these similar questions
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.
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
The following solution designed from a problem-solving strategy has been proposed for finding a minimum spanning tree (MST) in a connected weighted graph G:
Randomly divide the vertices in the graph into two subsets to form two connected weighted subgraphs with equal number of vertices or differing by at most Each subgraph contains all the edges whose vertices both belong to the subgraph’s vertex set.
Find a MST for each subgraph using Kruskal’s
Connect the two MSTs by choosing an edge with minimum wight amongst those edges connecting
Is the final minimum spanning tree found a MST for G? Justify your answer.
Chapter 5 Solutions
Artificial Intelligence: A Modern Approach
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- You are given a tree T with n vertices, rooted at vertex 1. Each vertex i has an associated value ai , which may be negative. You wish to colour each vertex either red or black. However, you must ensure that for each pair of red vertices, the path between them in T consists only of red vertices.Design an algorithm which runs in O(n) time and finds the maximum possible sum of values of red vertices, satisfying the constraint above.arrow_forwardConsider 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.arrow_forwardGiven an undirected weighted graph G with n nodes and m edges, and we have used Prim’s algorithm to construct a minimum spanning tree T. Suppose the weight of one of the tree edge ((u, v) ∈ T) is changed from w to w′, design an algorithm to verify whether T is still a minimum spanning tree. Your algorithm should run in O(m) time, and explain why your algorithm is correct. You can assume all the weights are distinct. (Hint: When an edge is removed, nodes of T will break into two groups. Which edge should we choose in the cut of these two groups?)arrow_forward
- make 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 varrow_forwardThe Floyd-Warshall algorithm is a dynamic algorithm for searching the shortest path in a graph. Each vertex pair has its assigned weight. You are asked to draw the initial directed graph and show the tables for each vertex from Mo to Ms by finding all the shortest paths. Below is the algorithm as a guide. Algorithm 1: Pseudocode of Floyd-Warshall Algorithm Data: A directed weighted graph G(V, E) Result: Shortest path between each pair of vertices in G for each de V do | distance|d][d] «= 0; end for each edge (s, p) € E do | distance[s][p] + weight(s, p); end n = cardinality(V); for k = 1 to n do for i = 1 to n do for j = 1 to n do if distancefi][j] > distance/i][k] + distance/k][j] then | distance i]lj] + distancefi|[k] + distance/k|[j]; end end end end Consider the relation R = {(1,4) =4, (2,1)=3, (2,5)=-3, (3,4)=2, (4,2)=1, (4,3)=1, (5,4)=2 } on A = (1,2,3,4,5) solve the Floyd-Warshall Algorithm.arrow_forwardImplement The dynamic programming algorithm for the leveled graph problem. pre-cond: G is a weighted directed layered graph, and s and t are nodes. post-cond: optSol is a path with minimum total weight from s tot, and optCost is its weight, and optNum is the number of possible optimal solutions.arrow_forward
- Algebraic Preis’ AlgorithmAlgorithm due to Preis provides a different way to solve the maximal weightedmatching problem in a weighted graph. The algorithm consists of the followingsteps.1. Input: A weighted graph G = (V, E, w)2. Output: A maximal weighted matching M of G3. M ← Ø4. E ← E5. V ← V6. while E = Ø7. select at random any v ∈ V8. let e ∈ E be the heaviest edge incident to v9. M ← M ∪ e10. V ← V {v}11. E ← E \ {e and all adjacent edges to e}show two ways of implementing this algorithm in Pythonarrow_forwardAlgebraic Preis’ AlgorithmAlgorithm due to Preis provides a different way to solve the maximal weightedmatching problem in a weighted graph. The algorithm consists of the followingsteps.1. Input: A weighted graph G = (V, E, w)2. Output: A maximal weighted matching M of G3. M ← Ø4. E ← E5. V ← V6. while E = Ø7. select at random any v ∈ V8. let e ∈ E be the heaviest edge incident to v9. M ← M ∪ e10. V ← V {v}11. E ← E \ {e and all adjacent edges to e} show two ways of implementing this algorithm in Pythonarrow_forwardEstablish 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).arrow_forward
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