The hash coding method described in the text links all the entries having the same hash code together on a linked list. An alternative method is to have only a single n- slot table, with each table slot having room for one key and its value (or pointers to them). If the hashing algorithm generates a slot that is already full, a second hashing algorithm is used to try again. If that one is also full, another is used, and so on, until an empty is found. If the fraction of the slots that are full is R, how many probes will be needed, on the average, to enter a new symbol?

The hash coding method described in the text links all the entries having the same hash code together on a linked list. An alternative method is to have only a single n- slot table, with each table slot having room for one key and its value (or pointers to them). If the hashing algorithm generates a slot that is already full, a second hashing algorithm is used to try again. If that one is also full, another is used, and so on, until an empty is found. If the fraction of the slots that are full is R, how many probes will be needed, on the average, to enter a new symbol?

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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I am trying to implement a hash table of key-value pairs, where the key determines the location of the pair in the hash table, the key is what the hash function is used on. The hash table that I am trying to implement will use separate chaining (with unordered linked lists) for collision resolution. I need help changing the code where: 1. The "search" method of the code returns the node that was found or "None" 2. The "Node" implementation has a "key" and "value" member This is what I have so far class Node: def __init__(self,initdata): self.data = initdata self.next = None def getData(self): return self.data def getNext(self): return self.next def setData(self,newdata): self.data = newdata def setNext(self,newnext): self.next = newnext class UnorderedList: def __init__(self): self.head = None def isEmpty(self): return self.head == None def add(self,item): temp = Node(item) temp.setNext(self.head)…
However, the hashmap we're using is somewhat rudimentary, since it stores every value in one big bucket (that is, they are all in the same LinkedList). The hash map's intended function has been completed effectively.
write a simple hashing function that uses addition and multiplication and then find a pair of strings that will return the same hash value for different strings (i.e you will cause a Hash Collision). The algorithm uses multiplication based on the position of a letter in the hash to avoid a hash collision when two letters are transposed like in 'ABCDE' and 'ABDCE'. Your strings need to be at least three characters long and no more than 10 characters long. Here is the code that computes your hash: while True: txt = input("Enter a string: ") if len(txt) < 1 : break hv = 0 pos = 0 for let in txt: pos = ( pos % 2 ) + 1 hv = (hv + (pos * ord(let))) % 1000000 print(let, pos, ord(let), hv) print(hv, txt) For simplicity we will only use upper and lower case ASCII letters in our text strings. Enter a string: ABCDE A 1 65 65 B 2 66 197 C 1 67 264 D 2 68 400 E 1 69 469 469 ABCDE Enter a string: BACDE B 1 66 66 A 2 65 196 C 1 67 263 D 2 68 399 E 1 69 468 468 BACDE
3. Double hashing is one of the methods to resolve collision. Write a function to implement this method. The equations used in this method are given below. Note: implement everything within the double hash function. P = (P + INCREMENT(Key)) mod TABLE_SIZE INCREMENT(Key) = 1 + (Key mod INCR)
Our hash map data are all saved in a single LinkedList because of the shoddy way the map was built. The hash map's utility is reduced as a consequence.
Task on Hashing Given an array containing Strings, you need to write a code to store them in a hashtable. Assume that the Strings contain a combination of capital letters and numbers, and the String array will contain no more than 9 values.Use the hash function to be the (total number of consonants*24 + summation of the digits) %9. In case of a collision, use linear probing. For a String "ST1E89B8A32", it's hash function will produce the value=(3*24+(1+8+9+8+3+2))%9=4, hence it will be stored in index 4 of the hash table.
Task on Hashing Given an array containing Strings, you need to write a code to store them in a hashtable. Assume that the Strings contain a combination of capital letters and numbers, and the String array will contain no more than 9 values.Use the hash function to be the (total number of consonants*24 + summation of the digits) %9. In case of a collision, use linear probing.For a String “ST1E89B8A32”, it’s hash function will produce the value=(3*24+(1+8+9+8+3+2))%9=4, hence it will be stored in index 4 of the hash table. Instructions Complete the following problem using concepts of Key index searching,sorting and hashing You may use JAVA or PYTHON to complete the tasks. If you are using JAVA, you must include the main method as well which should test your other methods and print the outputs according to the tasks. If you are using PYTHON, make sure your code has the methods invoked and proper printing statements according to the tasks.
The hash map we're using is rather terrible, since it stores every value in the same place (that is, they are all in the same LinkedList). In this case, a hash map has successfully completed its intended task.
2-A linear probing hash table of length 10 uses the hash function h(x) = x mod 10 + 1. mod is the reminder opertatro. For example the h(42) = 42 mod 10 + 1 = 2 + 1 = 3 so we insert 42 in t he position 3 in the array. After inserting 6 integer keys into an initially empty hash table, the array of keys is 1 2 3 4 5 7 8 9. 42 23 34 52 46 33 a- insert the key 35 b- insert the key 12 c- insert the key 10
A 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.
We use Linked Lists to store all of our hash map data due to the sloppy manner in which the map was originally constructed. This substantially reduces the efficacy of a hash map.
A hash table is an efficient data structure to store and access data via keys. In this problem, you need to find the longest subarray of distinct elements. For example, given an array A = <milk, water, 6724, water, soda, beer, apple, grape, wine>, the longest subarray is <milk, water, 6724, soda, beer, apple, grape, wine> Hint: you need to keep track and update the positions of the elements. a) Explain how many unique subarrays are there from a given array b) Describe your ideas on how to solve the problem, especially how hash tables can be used to keep track of the important information c) Write the most efficient algorithm (pseudo code) to solve this problem with complexity analysis. You can assume that a good hash function is given.