cessed according to the increasing order of timestamps (discuss why it is impossible for the CS to b
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Show that in the Ricart-Agrawala's
increasing order of timestamps (discuss why it is impossible for the CS to be granted out of
timestamp order).
Step by step
Solved in 2 steps
- Implement Algorithm to illustrates the search procedure on a skip list S for a key K. The algorithm employs functions next(p) and below(p) to undertake the search.I need to show the worst-case and average-case time complexities for the number of assignments of records performed by the Exchange Sort algorithm, given byThe 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 ∈ ?? if j ≠ m + 1 extracted[j] = i let L be the smallest value greater than j for which KL exists KL = KL U Kj, destroying ????return 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…
- 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, destoying Kjreturn extracted 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.…Question 11mah.Full explain this question and text typing work onlyProvide an algorithm with the same time complexity as Bellman-Ford so that: it sets dist [w] to −∞for allvertices w for which there is a vertex that belongs to a negative-weighted cycle on some path from the sourcevertex to w, and outputs False if there is no such vertex w.
- Computing the union of two sets using Boolean values is simple and efficient. Since the union of two sets is a combination of the members of bothsets, we can build a new union set by Oring the corresponding elements ofthe two BitArrays. In other words, a member is added to the new set if thevalue in the corresponding position of either BitArray is True.Computing the intersection of two sets is similar to computing the union;only for this operation we use the And operator instead of the Or operator.Similarly, the difference of two sets is found by executing the And operatorwith a member from the first set and the negation of the corresponding member of the second set. We can determine if one set is a subset of another set byusing the same formula we used for finding the difference. For example, if:setA(index) && !(setB(index))evaluates to False then setA is not a subset of setB.The BitArray Set Implementation Write The code for a CSet class based on a BitArray.We have seen that the cost of removing a value from the PriorityVector takes linear time. If elements were stored in descending order, thiscould be reduced to constant time. Compare the ascending and descendingimplementations, discussing the circumstances that suggest the use of one implementation over the other.We have seen that the cost of removing a value from the PriorityVector takes linear time. If elements were stored in descending order, thiscould be reduced to constant time. Compare the ascending and descendingimplementations, discussing the circumstances that suggest the use of one implementation over the other.Perform Dijkstra’s algorithm (while (Q is not empty){...}, where Q isa min priority queue) on the following graph for source node “s”, writing outevery execution of the RELAX function, the update to the priority queue and theupdates of distance from the source node and parent for the currently visited node.
- Problem 3. Recollect that the standard implementation of the Dijkstra's algorithm uses a priority queue that supports the Extract-min() and Decrease-Key() operations. Describe a method to implement Dijkstra's algorithm in O((m + n) log n) time without the use of Decrease-Key operation, i.e., your algorithm can use the Extract-Min() operation but not the Decreased-Key() operation. Solution:Suppose that in applying the minimization algorithm in Section 2.6, weestablish some fixed order in which to process the pairs, and we follow thesame order on each pass a. What is the maximum number of passes that might be required?Describe an FA, and an ordering of the pairs, that would require thisnumber.b. Is there always a fixed order (depending on M) that would guaranteethat no pairs are marked after the first pass, so that the algorithmterminates after two passes?For the storing files on tape problem (without frequencies), why does sorting by smallest file size produce the best lookup cost in expectation? Because, for any two files with equal file size, it does not matter what order they are in Because, sorting them is greedy and greedy algorithms usually work Because, for any two files with the larger file preceding the smaller, swapping the files reduces the cost in expectation Because, for any three files that are out-of-order a random permutation of those files reduces the cost in expectation
