Question 1. Let H : {0, 1}* →→ {0,1}" be a hash function that achieves the one-way and collision resistance security properties. (a). Let LSB(x) denote the least significant bit of string x and || denote the concatenation of two strings. We define the following function H': {0, 1}* {0,1}+¹ as H'(x) = H (x)||LSB(x). Show that the new function H' achieves collision resistance.
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- We have a crude hash map where all values are concatenated into a single bucket. A hash map's goal has been accomplished.item h1(item) h2(item) Hash table 1 b 5 5 1 3 2 d 4 4 e 3 4 a f 6. 5 b 5 1 Suppose you are building a hash table of the items a through g. You have two hash functions, h1 and h2, with values as given. You have already inserted a, b, and c and are attempting to insert d, when you have a collision. Where would chaining store item d? Oin a new cell 00 03 06 0.00 points out of 1.00 Where would linear probing store item d? Oin a new cell 02 03 06 0.00 points out of 1.00 Where would double hashing store item d? Oin a new cell 02 03x 0.00 points out of 1.00 3. 4. boConstruct a Linear Hashing index for the following data entries: 1, 3, 5, 6, 7, 10, 14, 18, 21, 26, 27, 28, 30, 31, 32, 40, 46, 62, 63, 66, assuming there are four initial primary bucket pages and each bucket can hold up to four data entries. Show the snapshot of the index whenever an overflow page is created causing a bucket split.
- Experiment with hashCode() in Java on a large dataset (of size M) to show that the hash functions distribute keys fairly evenly between 0 to M-1.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…QUESTION 7 Given a Count-Min Sketch of size 10 x 2 (w=10, d=2) that operates on two hash functions: h1(x) = (x*7+4) % w h2 (x) = (X*13+9) % w After counting a stream of data, our sketch has the following contents: 6 7 8 9 Column Index 0 1 2 3 4 Row 1 Value 0 25 14 22 00 55 0 18 0 Row 2 Value 94 0 70 0 64 79 0 0140 We would like to add two elements of value "42", and "51", respectively, into our sketch. Please specify which counters will be changed, and to what values after the operation. You can use the Column Index, and Row number to reference the cell being changed. Please clearly explain how you come up with the
- Insert the following sequence of keys in the hash table.Keys = {4, 2, 1, 3, 5, 6, 8}Use linear probing technique for collision resolution: h(k, i) = [h(k) + i] mod mHash function, h(k) = 3k + 1Table size, M = 10 Draw the hash table, compute the index where the hashvalue will be inserted for each key, and show how you would use the technique when a collisionoccurs.Consider the following hash table, and a hash function of key % 5. What would bucket 3's list be after the following operations? hashTable: + 25 95 50 1 2 3 68 + 33 + 48 84 54 HashRemove(hashTable, 25) HashRemove(hashTable, 48) Hashlnsert(hashTable, item 53) О а. 33, 53 O b.68, 33, 53 О с. 68, 33, 48 O d. 68, 33, 48, 53Use the hash function h(x) =x mod 11 to load the following values 25, 14, 36, 47 using separate chaining.
- Given the hash function H (key) = (3 * key) mod 11, use square probe to solve the conflict. Please construct a hash table for the keyword sequence {6,8,10,17,20,23,33,41,59,60} in the hash table whose length is 11.Given input {4371, 1323, 6173, 4199, 4344, 9679, 1989} and a hashfunction h(x) = x mod 10, show the resulting:a. Separate chaining hash table.b. Hash table using linear probing.c. Hash table using quadratic probing.d. Hash table with second hash function h2(x) = 7 − (x mod 7).Consider a Hash Table with 6 bins. Show the content of the Hash Table (that uses chaining to handle collisions) after the following numbers are inserted into the table in order: 0 1 6 7