**Universal Hash Family**We're considering the following scenario: the universe \( U \) of all possible key values is extremely large compared to \( n \), the number of actual keys we have.If you choose a hash function from all possible functions from \( U \) to \(\{0, 1, ..., t-1\}\), what goes wrong?

Hash table is used to search the elements in constant time. Each hash table contain some keys and…

Universal Hash Family. Here, we're considering the following scenario. The universe U of all possible key values is super large, compared to n, the number of actual keys we have. If you choose a hash function from all possible functions from U to {0, 1,.., t- 1}, what goes wrong?

Universal Hash Family. Here, we're considering the following scenario. The universe U of all possible key values is super large, compared to n, the number of actual keys we have. If you choose a hash function from all possible functions from U to {0, 1,.., t- 1}, what goes wrong?

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Consider a open bucket hash table and closed bucket hash table in which the keys are student identifiers (strings of 6 digits). Assume the following number of buckets and hash function: m = 100; hash(id) = first two digits of id #. Starting with an empty hash table, show the effect of successively adding the following student identifiers: 000014, 990021, 990019, 970036, 000015, 970012, 970023. Use following function as second hash function hash(id) = last digit of id # Show the effect of deleting 000014 from the hash table. (Java Programming DSA)
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Does a hash table of size m always have the same number of linked lists? No matter how long I try, I just can't put my finger on the function of a hash function. Give a case study to demonstrate your point.
Does a hash table of size m always have the same number of linked lists? I just can't get my head around the function of a hash function. Provide an instance to demonstrate your point.
In Python, implement a hashing table data structure of size 20, which can cope with collisions. Thus, you must implement the closed collision system in its three modes: Linear scanQuadratic scanDouble hashing scan. For halt use halt(k)= (k mod (m - 1)) + 1, where m is the size of the table. The data to be managed are students from a certain university who have a 5-digit registration number and it is unique. In addition, the student has a name, age and address. Create a .txt file with 10 students (one for each line) and have them loaded into your program. Build a menu of options where you can: - Find a student and print his data if found and position in the table. Otherwise print a "Not Found" message.
In Python, implement a hashing table data structure of size 20, which can cope with collisions. Thus, you must implement the closed collision system in its three modes: Linear scanQuadratic scanDouble hashing scan. For halt use halt(k)= (k mod (m - 1)) + 1, where m is the size of the table. The data to be managed are students from a certain university who have a 5-digit registration number and it is unique. In addition, the student has a name, age and address. Create a .txt file with 10 students (one for each line) and have them loaded into your program. Build a menu of options where you can: - Enter a new student. For this option, that only allows entering the 5-digit license plate and that random data is automatically generated for the name, age and address. Return the position where it was entered and if it is not possible to enter it, print a message that explains why the entry was not obtained.
A hash map of size 15 has been constructed with Double-Hashing by applying h; (k;) = [ h(k;)+jd(k;)] mod 15. The primary hashing function is h(k;) = k; mod 15. The secondary hashing function is d;(k;) = k; mod 7. In one sentence, describe why the secondary hashing function is poorly chosen.
The Java hash function for Strings computes a hash code based on a fixed maximum number of characters of the string. Given that Strings have no meaningful upper bound in length, describe how an effective, constant-time hashing algorithm can be constructed. (Hint: If you were to pick, say, eight characters to represent a string of length l, which would you choose?)
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Let S be the set consisting of all 10 digits and all upper- and lowercase letters. Let U be the set of all (ordered) strings consisting of exactly four elements from S. Construct an injective hash function h: U -> N with the smallest possible value of maxUh(u). Modify your hash function to give a simply uniform map h' to the set {0, ..., 30}. In other words, if U is a probability space with P(u) = P(u') for all u, u' in U, then the random variable h': U -> {0, ..., 30} should have a uniform distribution..
Generate 11 unique random elements and perform the extendible hashing. The bucket size is three (3) and the Hash function is the following: H(X) => Suppose the global depth is X. Then the Hash Functi9on returns X LSBs. Make sure that in every hashing step there is an equivalent explanation.
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…