Consider a hash table of size M using separate chaining with ordered lists and the hash function hash(k) = k mod M.   a) Assume N items have already been inserted in the table. What is, in average, the cost (number of operations) needed for searching an item in the table? Justify your answer. b) Assume N items have already been inserted in the table. What is, in average, the cost (number of operations) needed for inserting the next (N+1)th item? Justify your answer.

The hash function h(k)= [(m(kA mod 1)] for A = 2-³ is used to insert the keys 61, 62, 63, 64, and 65 into a hash table of size m = 1000. The order of the hash codes in the table are given by a. 625, 125, 0, 750, 875 b. 625, 750, 625, 125, 0 c. 625,125, 0, 875, 750 d. 625, 750, 875, 0, 125 e. 625, 750, 125, 875, 0

Given a sequence of 5 keys < 53, 66, 32, 41, 88 > for searching task: a) Given the hash function H(k) = (3.k) mod 7, insert the keys above according to its original order (from left to right) into a hash table of 7 slots. Indicate the cases of collision. Clearly show the steps and calculation with a table. b) In case of any collisions found above in a) part, determine the new slot for each collided case using Linear Probing to solve the collision problem. Clearly show your answer for each identified case. No step of calculation required. o (Answer with “NO collision found”, in case there is no collisions found above) c) In case of any collisions found above in a) part, determine the new slot ONLY for the first collided case using Double-Hashing (with functions below) to solve the collision problem. Clearly show the steps of your calculation. d1 = H(k) = (3.k) mod 7 di = (di−1 + ((5·k) mod 6) + 1) mod 7 , i ≥ 2 o (Answer with “NO collision found”, in case there is no collisions…

Assume that linear probing is used for hash-tables. To improve the time complexity of the operations performed on the table, a special AVAILABLE object is used to mark a location when an item is removed from the location. Assuming that all keys are positive integers, the following two techniques were suggested instead of marking the location as AVAILABLE: i) When an entry is removed, instead of marking its location in the table as AVAILABLE, indicate the key in the location as the negative value of the removed key (e.g., if the removed key was 16, indicate the key as -16). Searching for an entry with the removed key would then terminate once a negative value of the key is found (instead of continuing to search if AVAILABLE is used). ii) Instead of using AVAILABLE, find a key in the table that should have been placed in the location of the removed entry, then place that key (the entire entry of course) in that location (instead of setting the location as AVAILABLE). The motive is to…