Explanation of Solution
a.
State space diagram:
The state space of start state number 1 and the successor function for state ‘k’ returns two states, numbers 2k and 2k+1.
So, if the start state is 1, then it will have two states 2(1)=2 and 2(1)+1=3.
For each state, till 15 the corresponding states are obtained from the functions 2k and 2k+1...
Explanation of Solution
b.
Breadth first search:
The breadth first search begins from the root node, then examines the neighbouring nodes and travels to the next level neighbours.
Until the solution is found, the breadth first search creates one tree at a time.
Using the FIFO queue data form, the breadth first search
It expands the shallowest nodes first if the goal state is 11, the order in which the nodes will be visited for breadth first search is
1 →2→3→4→5→6→7→8→9→10→11
Depth limit search:
To prevent the infinite loop in depth first search, it is conducted with a fixed depth limit in depth dependent search technique.
The depth limit search checks for the solution up to a specified depth...
Explanation of Solution
c.
bidirectional search for the problem:
The idea of a bidirectional search is to reduce the search time by simultaneously searching forward from the beginning and back from the goal.
For the given problem, the bidirectional search would work because, the only successor for “n” in the reverse direction floor of (n/2).
Branching factor:
Consider that the breadth first search is done in both forward direction and backward direction, the search node will be as follows,
Consider the visited node as 1, then the forward fringe will be node {2,3}...
Explanation of Solution
“yes”, the reformulation is the branching factor for moving forward is 2. And moving backward is 2...
Explanation of Solution
Explanation of algorithm for the problem and its solution:
The solution for the target number can be read off the binary number.
Write the binary target number.
Since the user can enter only positive integer numbers, these binary expansion being have a 1...
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Check out a sample textbook solutionChapter 3 Solutions
Artificial Intelligence: A Modern Approach
- Consider the following map where S is the start state and G is the goal state. Process the nodes in alphabetical order. State B 4 A 2 B 1 3 G 12 A. The path returned by applying A* search using h2 is: o S->A->D->G o S->A->C->G O S->G o None of the answers B. Consider the heuristics for this problem shown in the table above. o hl is admissible o hl isn't admissible o Information is not enough.arrow_forwardConsider a state space where the start state is number 1 and the successor function for state n returns two states, numbers 2n and 2n + 1. a) Draw the portion of the state space for states 1 to 31. b) Suppose the goal state is 12. List the order in which nodes will be visited for depth first search if we start with the left side of the tree, depth first search if we start with the right side of the tree, and breadth first search.arrow_forwardConsider a state space where the start state is number 1 and the successor function for state n returns two states, numbers 2n and 2n + 1. a) Draw the portion of the state space for states 1 to 31. b) Suppose the goal state is 12. List the order in which nodes will be visited for depth first search if we start with the left side of the tree, depth first search if we start with the right side of the tree, and breadth first search. c) Calculate the total time in terms of number of nodes each algorithm will take to find the goal state 12.arrow_forward
- a) Given a depth-first search tree T, the set of edges in T are referred to as "tree edges" while those not in T are referred to as "back edges". Modify the implementation of the Depth-First Search algorithm to print out the set of tree edges and the set of back edges for the following graph. 1(0 1 1 0 0 1 o) 210 10 0 0 31 10 1 0 0 1 40 0 10 0 0 0 50 0 0 0 0 1 1 6 10 001 70 0 10 1 1 0 0 1arrow_forward3) The graph k-coloring problem is stated as follows: Given an undirected graph G= (V,E) with N vertices and M edges and an integer k. Assign to each vertex v in V a color c(v) such that 1arrow_forward3) The graph k-coloring problem is stated as follows: Given an undirected graph G = (V,E) with N vertices and M edges and an integer k. Assign to each vertex v in Va color c(v) such that 1< c(v)arrow_forward1. Prove by induction that a graph with n vertices has at most n(n-1)/2 edges. 2 10 12 5 2 1 A 6. 8. 3 E 3. 3 4 11 10 2. For the undirected weighted graphs shown above: a. Represent the graphs using the adjacency matrix representation b. Represent the graphs using the adjacency list representation. If a pointer requires four bytes, a vertex label requires two bytes, and an edge weight requires two С. bytes, which representation requires more space for each graph? B.arrow_forward2. Draw the state space that would be generated by the Breadth-First Search algorithm for theinitial state given in Table 2. You can stop when the first goal state is reached. Perform themoves strictly in the following sequence: Up; Down; Left; Right. Do not create more thanone copy of any particular state, and identify the goal state.arrow_forward2. An undirected graph G can be partitioned into connected components, where two nodes are in the same connected component if and only if there is a path connecting them. Design and analyze an efficient algorithm that computes the connected components of a graph G given in adjacency list format. Be sure to give a correctness argument and detailed time analysis. You can use algorithms from class as a sub-procedure, but be sure to use the claims proven about them carefully. A good algorithm has time approximately 0(n + m) where the graph has n nodes and m edges.arrow_forwardQuestion 1: In graph theory, a graph X is a "complement" of a graph F if which of the following is true? Select one: a. If X is isomorph to F, then X is a complement of F. b. If X has half of the vertices of F (or if F has half of the vertices of X) then X is a complement of F. c. If X has the same vertex set as F, and as its edges ONLY all possible edges NOT contained in F, then X is a complement of F. d. If X is NOT isomorph to F, then X is a complement of F. Question 2: Which statement is NOT true about Merge Sort Algorithm: Select one: a. Merge Sort time complexity for worst case scenarios is: O(n log n) b. Merge Sort is a quadratic sorting algorithm c. Merge Sort key disadvantage is space overhead as compared to Bubble Sort, Selection Sort and Insertion Sort. d. Merge Sort adopts recursive approacharrow_forwardReview the 8-puzzle problem. Consider that the initial state is 1 3 48 27 6 5and the goal state is1 2 38 47 6 5Apply the breath-first-search method to obtain the path from the initial state to the goal state. Youneed to show the corresponding search tree. As soon as you find the goal state, you can stop the searchprocess.arrow_forward1. Input: G = (V, E)2. Output: MIS I of G3. I ← ∅4. V ← V5. while V = ∅6. assign a random number r(v) to each vertex v ∈ V7. for all v ∈ V in parallel8. if r(v) is minimum amongst all neighbors9. I ← I ∪ {v}10. V ← V \ {v ∪ N(v)}This algorithm terminates in O(log n)rounds with good probability. The stepsof the algorithm between lines 7–10 can be performed in parallel which will providea speedup.Make Python Implementation this algorithm in sequential formarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_iosRecommended textbooks for you
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