Make Python Implementation of given algorithm
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1: procedure Bellman_Ford(G,s)
2: for all {v} ∈ V do initialize
3: d[v]←∞
4: p[v] ← Ø
5: end for
6: d[s] ← 0 s has 0 cost
7: for k = 1 to n − 1 do
8: for all (u, v) ∈ E do relaxation
9: if d[v] > d[u] + w[u, v] then
10: d[v] = d[u] + w[u, v]
11: p[v] ← u
12: end if
13: end for
14: end for
15: for all (u, v) ∈ E do
16: if d[v] > d[u] + W[u, v] then
17: return Ø a negative cycle detected
18: end if
19: end for
20: return d, P
21: end procedure
Make Python Implementation of given algorithm
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- Design a divide-and-conquer algorithm for the problem of finding the longest sequence of consecutive negative numbers in a list of negative and positive numbers. Find the time complexity of your algorithm (type recurrence relation first). Example input: [2, 5, 0, -3, -5, 0, -1, -2, -1, 2]; Output: [-1, -2, -1] assum : In this example there are two consecutive sequences of negative numbers. [-3, -5] and [-1, -2, -1]. Since the second one is longer, that is the output.#com# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…Lecture: Analysis of Algorithms >> Time Complexity and Recursion << Please consider the following function and determine the Big-O notation using recurrence relation. int fun(int n){ if (n == 0 || n == 1) return n; if (n%3 != 0) return 0; return fun(n/3); }
- Apply Binary Search Algorithm on: 2, 8, 12, 15, 18, 20, 23, 30, 45, 85, 96, 97 where x=65 Show all necessary steps;#plea# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…A subsequence of a sequence is obtained by deleting zero or more values in the sequence. For example, acdfh is a subsequence of abcdefghi. The order must be preserved. The longest common subsequence problem takes two strings as input and outputs the length of the longest string that is a subsequence of both input strings. Describe an efficient dynamic programming algorithm for the longest common subsequence problem. Start by writing recursive equations using the shop around principle. Concentrate on the last values in the input strings. If they are the same, keep them and look for a longest common subsequence of the two strings with their last characters removed. If the last two characters are different, you can only keep one of them. Try keeping one, and throwing the other one away. Then keep the other one. Take a maximum, since you are looking for a longest common subsequence. Then implement the equations using memoization.
- ## Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…Computer Science Design a divide-and-conquer algorithm for finding the minimum andthe maximum element of n numbers using no more than 3n/2comparisons.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…
- Given an array of integers, find the longest non-decreasing subsequences (the subsequence doesnot need to be consecutive). For example, A = [8,5,2,10,3,6,9,7] contains the longest subsequences[2,3,6,9] and [2,3,6,7].(1) Formulate the recursive relation of the optimal solution (do not miss the base case);(2) Design a bottom-up (iterative) algorithm to calculate the length of the longest non-decreasingsubsequences (pseudo code);(3) Analyze the complexity of your algorithmConsider a divide-and-conquer algorithm that calculates the sum of all elements in a set of n numbers by dividing the set into two sets of n/2 numbers each, finding the sum of each of the two subsets recursively, and then adding the result. What is the recurrence relation for the number of operations required for this algorithm? Answer is f(n) = 2 f(n/2) + 1. Please show why this is the case.Assignment (Math application):Write a program that prompts the user to enter a 3 x 3 matrix of double values and testswhether it is a positive Markov matrix. An n x n matrix is a positive Markov matrix if the following is true:o If each of the elements is positiveo The sum of the elements in each column is 1 Sample Program running Enter a 3 x 3 matrix by row0.15 0.875 0.3750.55 0.005 0.2250.30 0.12 0.4The sum of the columns1.0 1.0 1.0It is a Markov matrix Enter a 3 x 3 matrix by row-0.2 0.875 0.3750.75 0.005 0.2250.45 0.12 0.4The sum of the columns1.0 1.0 1.0It is not a Markov matrix Please note the following requirements: Include a comment before each method explaining what the methods will do All methods called from the main methods There will be two methods which will be called from the main method:public static double [] [] createArray() 1. Creates a 3 by 3 two dimensional array of doubles2. Prompts the user for values as shown in the sample run3. Stores the numbers in the…