Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 11.2, Problem 5E
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To show the worst-case time complexity for hashing with chaining is
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Suppose we have a hash table with N slots containing n keys. Suppose that instead of a linked list in open
hashing, each slot is implemented as a binary search tree. Give the worst and the best time complexity of
adding an entry to this hash table. Explain your answers.
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Suppose we use a hash function h to hash n distinct keys into an array T oflength m. Assuming simple uniform hashing, what is the expected number ofcollisions? More precisely, what is the expected cardinality of ffk; lg W k ¤ l andh.k/ D h.l/g?
Suppose we hash elements of a set U of keys into m slots. Show that if |U| (n − 1)m, then there are at least n keys that all hash to the same slot, so that the worst-case searching time for hashing with linked list to resolve collisions is Θ(n).
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- We have n elements {x1, ..., xn} we want to hash into a table T of size s = 2n. Let us consider the following method of hashing these n elements into T: For each i = 1,..., n, do the following: 1. Pick a permutation of [1,..., s] uniformly at random. Call this permutation T; : [s] → [s], which maps each index to the element which ends up in that index. 2. Set j : : 1. 3. While T[T;(j)] has an element in it, increment j. 4. Place x; in T[T;(j)]. 1.1 Show that while inserting any ;, the probability that there are at least t iterations of the while loop in Step 3 is at most 2-t.arrow_forwardGiven the input set A = {1, 2, 3, 4, 5} and output set B = {1, 2, 3, 4}, is it possible to find a hash function H from A to B that has no hash collision, i.e., LaTeX: \forall ∀ a1, a2 LaTeX: \in ∈A, H(a1) LaTeX: \ne ≠H(a2) If yes, please give the function. Otherwise, please explain. (Hint: pigeonhole principle)arrow_forwardA hash-map has been constructed with quadratic-hashing. The hashing function is h(k;) = (3 * k; + 7) mod 17 and the table length is N = 17. How many cells will be probed by the call of insert(2) 1 4 5 6 7 8 10 11 12 13 14 15 16 26 89 11 35 70 59 Note: First row contains the indices and the second row contains the items.arrow_forward
- A hash map of size 12 has been constructed with Quadratic-Hashing by applying h(k) = (4ks +1 )mod 12. Perform Find(12) and mark in the hash-map below the cells which will be probed. Index 0 12 3 4 5 6 789 10 11 12 Value 3 44 36 11| 43arrow_forwardLinear Probing uses an auxilliary hash function that searches an array for the next empty space to insert a key if the original hash function mapped to an index that already had a key. True False Quadratic probing is meant to resolve a problem that linear probing suffers from, that problem is: O Linear probing tends to create long clusters of full cells making subsequent linear probes take longer and longer. Linear probing tends to overwrite key values after a second probe. Linear probing requires a linked list. O Linear probìng calculations require O(n*lg{n}} time. Collisions are a design problem of the hash table data structure that causes the hash insertion and deletion functions to run O(n^2) time. O True O Falsearrow_forwardA hash-map has been constructed with double-hashing by applying h;(k;) = [h(k;) + jd(k;)] mod N. The primary hashing function is given as h(k;) hashing function as d;(k;) = k;divN where div is integer division. The maximum number of cells probed is N = 13. How many cells will be probed by the call of findElement(38) = k;modN and the secondary Note: The first row contains the indices and the second row contains the elements. 1 2 4 5 6. 7 8 10 11 12 27 14 available 6. 35 25 51arrow_forward
- What is the worst-case performance of a lookup operation in a hashmap and why? Group of answer choices A- O(1), hashmap always has a constant time lookup, and that is why we like using this associative data structure. B- O(lg(n)) hashmap has a log(n) lookup because we are able to perform a binary search on the keys because our hashmap always maintains a sorted order of entries added. C- O(n) because we can have a bad hash function that puts all of our items in the same bucket, thus we would have to iterate through all n items.arrow_forward11.2-5 Suppose that we are storing a set of n keys into a hash table of size m. Show that if the keys are drawn from a universe U with |U| > nm, then U has a subset of size n consisting of keys that all hash to the same slot, so that the worst-case searching time for hashing with chaining is O(n).arrow_forwardL is a list of data items that is linear. Implement the list as: I an ordered list; (ii) a linear open addressed hash table using a suitable hash function of your choosing.Use binary search or hashing to find a collection of keys on representations I and ii), respectively. Compare how well the two approaches performed over list L.arrow_forward
- Hey, Given is a hash table with an initial size of 1000 and a hash function that ensures ahashing, where the keys are chosen randomly under the uniformity assumption.me are chosen. After how many insertions do you have to expect a collision probability of more than 80%?of more than 80%?To keep the number of collisions low during hashing, one can reduce the size of the has-hash table after a certain number of insertions. After which number n of inserted elements must the table be increased for the first time, if no collision occurred in the previous n - 1 elements and the probability of a collision in the n-th insertion should be less than 20%?arrow_forwardSuppose that keys are t-bit integers. For a modular hash function with prime M, prove that each key bit has the property that there exist two keys differing only in that bit that have different hash values.arrow_forwardProve the theorem Suppose that a hash function h is chosen randomly from a universal collection of hash functions and has been used to hash n keys into a table T of size m, using chaining to resolve collisions. If key k is not in the table, then the expected length E Œnh.k/ of the list that key k hashes to is at most the load factor ˛ D n=m. If key k is in the table, then the expected length E Œnh.k/ of the list containing key k is at most 1 C ˛.arrow_forward
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