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 be represented by I1 , E, I2, E, ... , E, Im+1 , where each Ij stands for a subsequence (possibly empty) of INSERT and each E stands for a single EXTRACT-MIN. Let Kj be the set of keys initially obtained from 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 Kj return extracted   Given the operation sequence 9, 4, E, 6, 2, E, E, 5, 8, E, 1, 7, E, E, 3; where each number stands for its insertion. Draw a table showing the building process of extracted[1..6]. You have to show all Kj from Ij in the table and how they change (e.g. some have more elements due to merge and some disappeared due to merge).   Use the attached table for the operation sequence 4, 8, E, 3, E, 9, 2, 6, E, E, E, 1, 7, E, 5 as a guide

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 ∈ ??   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…

Interchange Base Points AlgorithmOutput: a base B = [131,132..... 13j-1, 13L.+1, 13j, 13j+2, 13j+3..... 13k] for G; a strong generating set relative to B; procedure interchange(var B: sequence of points; vat S: set of elements; j: 1..k-1);Interchange Base Points AlgorithmInput: a group G; a base [131,132..... 13k] for G and a strong generating set; an integer j between 1 and k-l; Output: a base [131,132..... 13k] for G and a strong generating set.

Problem 2: In this problem we assume that h: U → {0, 1, . . . , m − 1} is a good hashfunction, that is, every key k has the same probability 1/m to map to any place in the tableT of length m.(i) What is the probability that three pairwise distinct elements u1, u2, u3 ∈ U aremapped by the function h to the same place in the table (that is, h(u1) = h(u2) =h(u3))?(ii) We insert three elements into an empty hash table T using the hash function h. Ifcollision is solved by chaining, what is the probability that T[0] and T[1] are empty?

The Dining Philosopher solution using monitors shown in the code snippet below does not suffer from deadlock. However, starvation is still possible. Explain in details why. Each philosopher i invokes the operations pickup() and putdown() in the following sequence: DiningPhilosophers.pickup(i);                    EAT DiningPhilosophers.putdown(i);