Q29. The adjacency list takes less memory space than adjacency matrix to store the graph. O True O False
Q: neural networks Program to generate random numbers from 11 to 21 as class 1 and generate…
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A: First lets understand path and cyclePath : is traversing a sequence of vertices where vertices and…
Q: Q29. The adjacency list takes less memory space than adjacency matrix to store the graph. O True O…
A: An adjacency matrix is a square matrix used to represent a finite graph.
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A: The complete code is given below.
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Q: The adjacency matrix that represents the following graph is
A: Answer: Adjacent Matrix: It is basically a labeled graph with some rows and columns in the form of…
Q: Why is it better to use an adjacency matrix to represent a graph rather than an adjacency list?
A: Matrix of Adjacencies: Memory use is O(n2). It doesn't take long to look up and determine whether or…
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A: Each transaction involved in the cycle is said to be deadlocked.
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A: The correct answer is as follows:-
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Q: Path is a walk in which all edges are distinct.
A: TRUE
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![Q29. The adjacency list takes less memory
space than adjacency matrix to store the
graph.
O True
O False](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4f60099-d601-41ef-86bc-e9d112aefad2%2F261136aa-c877-47e9-9bef-2589a18a95c1%2Fzq28va_processed.jpeg&w=3840&q=75)
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- Bishops on a binge def safe_squares_bishops(n, bishops): A generalized n-by-n chessboard has been taken over by some bishops, each represented as a tuple (row, column) of the row and the column of the square the bishop stands on. Same as in the earlier version of this problem with rampaging rooks, the rows and columns are numbered from 0 to n - 1. Unlike a chess rook whose moves are axis-aligned, a chess bishop covers all squares that are on the same diagonal with that bishop arbitrarily far into any of the four diagonal compass directions. Given the board size n and the list of bishops on that board, count the number of safe squares that are not covered by any bishop. To determine whether two squares (r1, c1) and (r2, c2) are reachable from each other in one diagonal move, use abs(r1-r2) == abs(c1-c2) to check whether the horizontal distance between those squares equals their vertical distance, which is both necessary and sufficient for the squares to lie on the same diagonal. This…Vectors are synchronized while ArrayLists are not. Group of answer choices True Falseclass BinaryImage: def __init__(self): pass def compute_histogram(self, image): """Computes the histogram of the input image takes as input: image: a grey scale image returns a histogram as a list""" hist = [0]*256 return hist def find_otsu_threshold(self, hist): """analyses a histogram it to find the otsu's threshold assuming that the input hstogram is bimodal histogram takes as input hist: a bimodal histogram returns: an optimal threshold value (otsu's threshold)""" threshold = 0 return threshold def binarize(self, image): """Comptues the binary image of the the input image based on histogram analysis and thresholding take as input image: an grey scale image returns: a binary image""" bin_img = image.copy() return…
- Rooks on a rampage def safe_squares_rooks(n, rooks): A generalized n-by-n chessboard has been invaded by a parliament of rooks, each rook represented as a two-tuple (row, column) of the row and the column of the square that the rook is in. Since we are again computer programmers instead of chess players and other normal folks, our rows and columns are numbered from 0 to n - 1. A chess rook covers all squares that are in the same row or in the same column. Given the board size n and the list of rooks on that board, count the number of empty squares that are safe, that is, are not covered by any rook. To achieve this in reasonable time and memory, you should count separately how many rows and columns on the board are safe from any rook. Because permuting the rows and columns does not change the answer to this question, you can imagine all these safe rows and columns to have been permuted to form an empty rectangle at the top left corner of the board. The area of that safe rectangle is…Two-dimensional list tictactoe represents a 3x3 tic-tac-toe game board read from input. List tictactoe contains three lists, each representing a row. Each list has three elements representing the three columns on the board. Each element in the tic-tac-toe game board is 'x', 'o', or '-'. If all the elements at column index 0 are 'o', output 'A win at column 0.' Otherwise, output 'No win at column 0.'Fix an error and show it please?
- The function that will a single items to the list is Oextend (...) Oadd(...) Oappend(...) Olengthen(...)Q2: a. Write an algorithm that searches a sorted list of n items by dividing it into three sublists of almost n/3 items. This algorithm finds the sublist that might contain the given item and divides it into three smaller sublists of almost equal size. The algorithm repeats this process until it finds the item or concludes that the item is not in the list. Dry run the above algorithm to find the value 240. A[] = {10,15,20,60,65,110,150,220,240,245,260,290,300,460,470,501}True/False 3. Arrays are usually heterogeneous, but lists are homogeneous.
- Change the following code for Depth First Search Strategy so that the graph (graph values in dictionary) generates randomly. # Using a Python dictionary to act as an adjacency listgraph = { '5' : ['3','7'], '3' : ['2', '4'], '7' : ['8'], '2' : [], '4' : ['8'], '8' : []} visited = set() # Set to keep track of visited nodes of graph. def dfs(visited, graph, node): #function for dfs if node not in visited: print (node) visited.add(node) for neighbour in graph[node]: dfs(visited, graph, neighbour) # Driver Codeprint("Following is the Depth-First Search")dfs(visited, graph, '5')Background: When searching for an item in a list, each item that we examine (compare) is considered to be interrogated. If we search for John, the following names are interrogated: Harry, Larry, John (in that order). If two names tie for the middle position, choose the first of the two names for the middle.If we search this same list for John using the Sequential search we would interrogate all the names from Alice through John. We would start with Alice, move to Bob, move to Carol and so forth until we reached John. Directions: Use the original list of names (Alice - Oliver) to answers questions 1-8. Using a sequential search, what names are interrogated to find Carol? Using a sequential search, what names are interrogated to determine that Sam is not in the list? Using a binary search, what names are interrogated to find Carol? Using a binary search, what names are interrogated to determine that Sam is not in the list? Will a binary search or sequential search find Alice…data structures in java
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