ALGORITHM BruteForceClosestPair(P)
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A: Question : Give Some Examples Greedy Algorithms?
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A: Answer in step2
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A: 4. Here (n+1) ( n-1) has greater function of n as n². Thus it cannot be O(logn) Thus it is false.
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A: Introduction :
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A: Answer to the above question is in step2.
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Q: 1. Graph Algorithms Consider the following graph. Unless otherwise indicated, always visit adjacent…
A:
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A: SUMMARY: -Hence, we discussed all the points
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Q: Breadth first search and best-first search.
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A: I give the code in JAVA with code,output screenshot and also give the time complexity
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A: Lets discuss the solution in the next steps
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A: Given : Operation Number = 1, 2 , 3 , 4 …. , n Cost = 1 , 2 , 1 , 4 , ….
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- In computer science and mathematics, the Josephus Problem (or Josephus permutation) is a theoretical problem. Following is the problem statement: There are n people standing in a circle waiting to be executed. The counting out begins at some point (rear) in the circle and proceeds around the circle in a fixed direction. In each step, a certain number (k) of people are skipped and the next person is executed. The elimination proceeds around the circle (which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom. Given the total number of persons n and a number k which indicates that k-1 persons are skipped and kth person is killed in circle. The task is to choose the place in the initial circle so that you are the last one remaining and so survive. For example, if n = 5 and k = 2, then the safe position is 3. Firstly, the person at position 2 is killed, then person at position 4 is killed, then person at position 1…Create an ABM function that takes the following parameters: n := number of paths to be simulated m := number of discretization points per path S0 := initial starting point dS=μdt+σdW Program the function by using two nested "for loops" def ABM(n,m,S0,mu,sigma,dt): np.random.seed(999) arr = # create 2D zeros array with the correct dimensions arr[,] = #initialize column 0 # fill in array entries for i in : for j in : arr[i,j] = return arrAlgorithm of Preis in AlgebraThe following steps make up the algorithm.A weighted graph G = as the first input (V, E, w)A maximum weighted matching M of G as the output.While E =, choose any v V at random, let e E be the largest edge incident to v, and then perform the following calculations: 3. M e; 4. E e; 5. V V; 6.11. Two different Python implementations of this algorithm are shown by E, E, and all edges that are adjacent to e.
- A wrestling tournament has 256 players. Each match includes 2 players. The winner each match will play another winner in the next round. The tournament is single elimination, so no one will wrestle after they lose. The 2 players that are undefeated play in the final game, and the winner of this match wins the entire tournament. How would you determine the winner? Here is one algorithm to answer this question. Compute 256/2 = 128 to get the number of pairs (matches) in the first round, which results in 128 winners to go on to the second round. Compute 128/2 = 64, which results in 64 matches in the second round and 64 winners, to go on to the third round. For the third round compute 64/2 = 32, so the third round has 64 matches, and so on. The total number of matches is 128 + 64 + 32+ .... Finish this process to find the total number of matches.A certain recursive algorithm takes an input list of n elements. Divides the list into Vn sub-lists, each with yn elements. Recursively solves each of these yn smaller sub- instances. Then spends an additional 0(n) time to combine the solutions of these sub- instances to obtain the solution of the main instance. As a base case, if the size of the input list is at most a specified positive constant, then the algorithm solves such a small instance directly in 0(1) time. a) Express the recurrence relation that governs T(n), the time complexity of this algorithm. b) Derive the solution to this recurrence relation: T(n) = 0(?). Mention which methods you used to derive your solution.Given an array A and a positive integer k, the selection problem amounts to finding the largestelement x ∈ A such that at most k elements of A are less than or equal to x, or nil if no suchelement exists. A simple way to implement it is as follows:SimpleSelection(A, k)1 if k > A.length2 return nil3 else sort A in ascending order4 return A[k]Example A = (29, 28, 35, 20, 9, 33, 8, 9, 11, 6, 21, 28, 18, 36, 1) k = 6
- IN JAVA, USING RECURSION PLEASE Create a method int[][] generateMatrix(int row, int col, int boundary1, int boundary2, int iteration) that generates a random matrix with random numbers between [min(boundary1, boundary2), max(boundary1, boundary2)). The sum of the diagonal and the sub-diagonal should be the same. If not, regenerate it again, until a matrix that satisfies the condition is generated (return that matrix). If you try iteration times and none of the matrixes satisfy the condition, return null.Write an algorithm for a divide-and-conquer algorithm for finding values of both the largest and smallest elements in an array of n numbers. Implement the above-designed algorithm using Java Programming Language. How does this algorithm compare with the brute-force algorithm for this problem?3. R(L) is the worst-case running time of foo when called on an array of length L. static int fo0(int z) { int x = z.length; for (int i=0; i= i) { break; } } } }
- PLEASE USE PYTHONGiven a jungle matrix NxM:jungle = [ [1, 0, 0, 0], [1, 1, 0, 1], [0, 1, 0, 0], [1, 1, 1, 1,]]Where 0 means the block is dead end and 1 means the block can be used in the path fromsource to destination.Task:Starting at position (0, 0), the goal is to reach position (N-1, M-1).Your program needs to build and output the solution matrix – a 4x4 matrix with 1’s inpositions used to get from the starting position (0,0) to the ending position (N-1,M-1)with the following constraints:You can only move one space at a timeYou can only in two directions: forward and down.You can only pass thru spaces on the jungle matrix marked ‘1’If you cannot reach the ending position – print a message that you’re trapped in thejungleAlgorithm:If destination is reachedprint the solution matrixElseMark current cell in the solution matrixMove forward horizontally and recursively check if this leads to a solution If there is no solution, move down and recursively check if this leads to a solution If…A magic square is a square array of non-negative integers where the sums of the numbers on each row,each column, and both main diagonals are the same. An N × N occult square is a magic square with Nrows and N columns with additional constraints:• The integers in the square are between 0 and N, inclusive.• For all 1 ≤ i ≤ N, the number i appears at most i times in the square.• There are at least two distinct positive integers in the square.For example, the following is a 5 × 5 occult square, where the sums of the numbers on each row, eachcolumn, and both main diagonals are 7:0 0 0 3 42 4 0 0 10 0 3 4 05 0 0 0 20 3 4 0 0For a given prime number P, you are asked to construct a P × P occult square, or determine whether nosuch occult square exists.InputInput contains a prime number: P (2 ≤ P ≤ 1000) representing the number of rows and columns in theoccult square.OutputIf there is no P × P occult square, simply output -1 in a line. Otherwise, output P lines, each contains Pintegers…Java Only Solution: Let's consider a rectangular R table containing N rows and M columns. Rows are numbered 1 to N from top to bottom. Columns are calculated from 1 to M from left to right. Each element of R is a non-negative number. R is pronounced steadily if the total number of elements in the ith line is not less and then the total number of elements in the i-line (1) th for each i lapho is 2 Kanye N and the total element in the Nth line is less than or equal to M. Your job is to find the number of fixed tables of different sizes N x M modulo 1 000 000 000 000 Using the Java Programming language. Input: 1 11 Output: 2