a.
Search tree:
Search tree is that tree which consists of the part of the state graph. It is used to solve the eight-puzzle. The start state is the root node of the search tree and children of nodes are those possible states which are reachable from parent node for one production.
Breadth first search:
A breadth first search is an
b.
Explanation of Solution
The state found in the backward search is also a binary tree and the rate of progress of both the search is same.
The total number of nodes considered when the search is conducted can be reduced by conducting two searches at a time as:
- The tree should be binary tree when a search begins from the goal so that the leaf node can be eliminated to reduce the number of nodes form a binary tree when the tree traversal occurs from backward...
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Computer Science: An Overview (12th Edition)
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- Question 1arrow_forwardA heuristic is a way of trying(a) To discover something or an idea embedded in a program (b) To search and measure how far a node in a search tree seems to be from a goal (c) To compare two nodes in a search tree to see if one is better than the other (d) Only (a), (b) and (c).arrow_forwardThe BST remove algorithm traverses the tree from the root to find the node to remove. When the node being removed has 2 children, the node's successor is found and a recursive call is made. One node is visited per level, and in the worst-case scenario, the tree is traversed twice from the root to a leaf. A BST with N nodes has at least log2N levels and at most N levels. Therefore, the runtime complexity of removal is best case O(logN) and worst case O(N). Two pointers are used to traverse the tree during removal. When the node being removed has 2 children, a third pointer and a copy of one node's data are also used, and one recursive call is made. Thus, the space complexity of removal is always O(1)." I have to explain this clearly! and the advantages of the BST algorithimarrow_forward
- Recall that the set of full binary trees is defined as follows:arrow_forwardGiven the following tree, implement DFS(Depth-first Search), IDDFS(iterative Deepening Breath First Search), and BFS(Breadth-First Search) to write out the order in which nodes are visited for each. Each algorithm should be written as a separate function taking in the parameters of the graph (T) and the node being searched for. Please use python and comment your code so I can better understand. Your code should be structured as follows: T={'a':['b','c'],'b':['d','e'],'c':['E','g'],'d':['h','i'1, 'e':['j,'k'],'f':[],'g':['l'],'h':[],'i':['m',in'],'j':[], 'k': [],'1':[],'m':[],'n':[]} def DFS function. def IDDFS function. def BFS function. x=call to DES y=call to IDDFS z=call to BFS print ("DFS:" + x's results) print ("IDDFS:" + y's results) print ("BFS:" + z's results) B mon Narrow_forward4. Write a recursive algorithm in pseudocode that finds the lowest common ancestor (LCA) of two given nodes in a binary tree T. The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself). If either p or q is null, the LCA is null. For this problem, Nodes have left, right, and parent references as well as a field called level which stores the level of the node in the tree. In the sample tree below, node 5 is on level 0, while nodes 4 and 6 are on level 1. Write your solution here. 5 <- level 0 4 6 < level 1 function lowest_common_ancestor (Node P, Node q)arrow_forward
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