A well designed hash function normally has multiple properties, including collision resistance, which means it is very hard for one to find a pair of inputs x≠y, such that h(x) = h(y). Could we just use a collision resistant function to do encryption? i.e., Enc(k,m)=f(k,m), where f is a collision resistant function.

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.

All the values were thrown into the same hash bucket, therefore the hash map isn't particularly good (that is, they are all in the same LinkedList). If this is the case, please elaborate on how using a hash map in the first place will be rendered ineffective.

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.

Suppose the length of a hash table is 14, and the hash function is h(key)=key%11. There are only four nodes in the table: H(15)=4, H(38)=5, H(61)=6, H(84)=7, and the remaining positions are empty. If the conflict is handled by the linear detection method, what is the node address of 49? 8. 3 9. None of the above

We have a poorly designed hash map in which all values have been placed in the same bucket (that is, they are all in the same LinkedList). Explain how this goes against the purpose of utilizing a hash map in the first place.

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?)

Suppose we were supplied a hash function h(k) = k mod 10. Let our hash table T consist of a modest 10 slots, addressed by index values 0 through 9:  and the remaining part is continued in the image

Rehash the hash table below (separate chaining, length 5) to a new length of 10. Draw the new hash table after rehashing, Assume that hashCode (x) = x for this question. Assume that we always traverse a chain from head to tail and always insert at the head of the chain example 0:15->30 1:6->21->11 2: 3: A 4:-14