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Assuming that the graph G = (V, E) is represented in Adjacency List format, justify in detail the fact Greedy independent sets can be implemented in O(n 2 + m) worst-case running time, where n = |V |, m = |E|. This will require you to take real care in how the adjustment of Adj is done in line 10. The key is to only update/delete what is really necessary for the
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- Note: Your solution should have O(n) time complexity, where n is the number of elements in l, and O(1) additional space complexity, since this is what you would be asked to accomplish in an interview. Given a linked list l, reverse its nodes k at a time and return the modified list. k is a positive integer that is less than or equal to the length of l. If the number of nodes in the linked list is not a multiple of k, then the nodes that are left out at the end should remain as-is. You may not alter the values in the nodes - only the nodes themselves can be changed.An Unfair AlgorithmFinding the MWM of a graph is as easy as selecting edges from a list in order, including the edge in matching, and removing all neighbouring edges from the list. This is done by sorting edges according to their weights in a non-increasing sequence. 1. Input: A weighted graph G = (V, E, w)2. Output: A maximal weighted matching M of G 3. M ← Ø 4. sort E and store the list in Q 5. while Q = Ø6. remove first element e from Q7. M ← M ∪ e8. Q ← Q\ { all adjacent edges to e }Let's attempt to use the preceding methods to create this algorithm in Python.We discussed how Prim’s algorithm runs on a graph G = (V,E)in O(|V|2). However, we also noted that an alternative implementation exists using heaps that runs in O(|E| log |V|). You will need to execute Prim’s algorithm on a large, complete graph. Which of the two implementations should you choose based solely on the Big-Oh complexity analyses listed above? You must justify your answer using Big-Oh notation in terms of |E|and |V|.
- For any given AVL tree T of n elements, we need to design an algorithm that finds elements x in T such that a <= x <= b, where a and b are two input values. Assume that T has m elements x satisfying a <= x <= b. What is the best time complexity for such an algorithm? A. O(m) B. O(m log n) C.O(m + log n) D.O(log m + log n)A Greedy AlgorithmA simple way to find MWM of a graph is to sort edges with respect to their weightsin non-increasing order and to select edges from this list in order, include the edgein matching and remove all adjacent edges from the list as shown below.1. Input: A weighted graph G = (V, E, w)2. Output: A maximal weighted matching M of G3. M ← Ø4. sort E and store the list in Q5. while Q = Ø6. remove first element e from Q7. M ← M ∪ e8. Q ← Q\ { all adjacent edges to e }Let us try to implement this algorithm in Python with the above steps.Consider a sort of items, according to their keys, that inserts all the items one at a time into an initially empty regular binary search tree and then applies an in-order traversal to complete the sort. Assume that all items have distinct keys. Using big-Theta notation, what is the worst-case complexity of the sort? What is the average-case complexity of the sort? Now answer the same two questions if an AVL tree is used instead of a regular binary search tree.
- An unweighted graph G = (V, E) does not have weights associated with its edges asnoted. A simple strategy to find the maximal matching in a greedy way is to randomlyselect an edge in the graph, include it in the matching and remove all of its adjacentedges from the graph as in the steps below.1. Input: An unweighted graph G = (V, E)2. Output: A maximal matching M of G3. M ← Ø4. E ← E5. while E = Ø6. randomly select e ∈ E7. M ← M ∪ {e}8. E ← E \ {e∪ all adjacent edges to e}implement this algorithm in PythonThis assignment is an exercise in finding the average-case complexity of an algorithm.Rather than looking at how long an algorithm can run in the worst case as in worst-case analysis, we are looking at how long an algorithm runs on average. This is doneby computing the average number of comparisons and operations executed until thealgorithm ends.Bogosort is a sorting algorithm that orders a list in increasing order by taking thelist, checking to see if the list is ordered increasingly, if the list is not ordered increasinglythen the list is randomly shuffled, and then repeating this process until the list is orderedincreasingly.Expressed in pseudocode:Algorithm 1 BogosortRequire: list: a1, a2, . . . , an of real numbersEnsure: list is sorted in increasing order1: procedure bogo(list)2: while not sorted(list) do ▷ Checks to see if list is sorted3: shuffle(list) ▷ Shuffle the current list if not sorted4: end while5: end procedure Problem: Consider the Bernoulli trial where a success is…This assignment is an exercise in finding the average-case complexity of an algorithm.Rather than looking at how long an algorithm can run in the worst case as in worst-case analysis, we are looking at how long an algorithm runs on average. This is doneby computing the average number of comparisons and operations executed until thealgorithm ends.Bogosort is a sorting algorithm that orders a list in increasing order by taking thelist, checking to see if the list is ordered increasingly, if the list is not ordered increasinglythen the list is randomly shuffled, and then repeating this process until the list is orderedincreasingly.Expressed in pseudocode:Algorithm 1 BogosortRequire: list: a1, a2, . . . , an of real numbersEnsure: list is sorted in increasing order1: procedure bogo(list)2: while not sorted(list) do ▷ Checks to see if list is sorted3: shuffle(list) ▷ Shuffle the current list if not sorted4: end while5: end procedure We will now find the average-case time complexity for…
- The Gale-Shapley algorithm is upper bounded by ≤ n^2 for n men and n women, since each man can only make ≤n proposals.However, we haven’t shown a lower bound on number of iterations. To show the lower bound is also inthe order of n^2, please give a way to construct the preference lists for n men and n women such that theGale-Shapley algorithm will run for Θ(n^2) iterations. (For simplicity, you can assume that the algorithmalways chooses the unmatched man with the smallest index at each iteration).Considering the search problem, we have a list of ?n integers ?=⟨?1,?2,⋯??⟩A=⟨v1,v2,⋯vn⟩. We want to design an algorithm to check whether an item ?v exists or not such that it should return either the index, ?i, if it was found or −1−1 otherwise, when not found.The off-line minimum problem maintains a dynamic set T of elements from the domain {1, 2,...,n}under the operations INSERT and EXTRACT-MIN. A sequence S of n INSERT and m EXTRACT-MIN calls are given, where each key in {1, 2,...,n} is inserted exactly once. Let a sequence S berepresented by I1 , E, I2, E, ... , E, Im+1 , where each Ij stands for a subsequence (possibly empty) ofINSERT and each E stands for a single EXTRACT-MIN. Let Kj be the set of keys initially obtainedfrom insertions in Ij. The algorithm to build an array extracted[1..m], where for i = 1, 2, ..., m,extracted[i] is the key returned by the ith EXTRACT-MIN call is given below: Off-Line-Minimum(m, n)for i = 1 to n determine j such that i ∈ Kj if j ≠ m + 1 extracted[j] = i let L be the smallest value greater than j for which KL exists KL = KL U Kj, Kjreturn extracted (1) Given the operation sequence 9, 4, E, 6, 2, E, E, 5, 8, E, 1, 7, E, E, 3; where eachnumber stands for its insertion. Draw a…