I will mark you Brainliest if you could guess how old i am :3

Design a data structure to support the following two operations for a set S of inte- gers, which allows duplicate values: • INSE

umka2103 [35]

Answer and Explanation:

Note that we are free to use any data structure that allows for arbitrary insertion and deletion of data

As an underlying data structure, we’ll use an (unsorted) array. INSERT(S, x) will simply append x to the array, increasing its length. This has a constant runtime, so we’ll say its cost is 1.

DELETE-LARGER-HALF(S) will work as follows: first, use SELECT to find the median. Next, use PARTITION around the median to make sure that the upper half is stored within the last $[|S|/2]$ elements. Finally, we delete these elements, reducing the size of the array.This has a linear running time, so we’ll say its cost is n.

To show that any m operations can run in O(m) time, we define a potential function $\phi(S) = 2|S|$ . The amortized cost of INSERT is thus $1 + \delta \phi = 1 + 1 = 1$ ; the amortized cost of DELETE-LARGER-HALF is $n +\delta\phi\leq n-2(n/2) = 0$ . So the amortized cost of any m operations is O(m).

This answer essentially captures the idea behind the problem. However, there are some technical points to clear up. (Calling the real-time costs 1 and n are not among them; this underestimates the running time by at most a constant. Ignoring constants like that is necessary to make concise arguments about amortized costs.)

First, an array does not support arbitrary insertions. Possible remedies include:

(1) using a dynamic array along the lines of §17.4, or (2) using a different structure like a linked list, and having DELETE-LARGER-HALF convert it to an array and back in linear time so that the SELECT and PARTITION algorithms may be used.

Second, it’s important to know which median to partition around and how to delete the upper half of the elements: a mistake could lead to incorrect behavior when the array has an odd size or repeated elements. We should select the lower median, $[|S|/2]$ , since that’s the number of elements we want in the lower set: as written, the CLRS Partition function will put elements less than or equal to the

pivot in the left set, and strictly larger elements in the right set. (If the partition function is defined differently, the answer should be different as well. You generally should give a brief description of how your partition function works.) After a call to Partition, it is safe simply to keep the first $[|S|/2]$ elements and drop the rest. On the other hand, it is not safe to go around deleting every element with

a sufficiently large value—take an array of zeros as a drastic example. If you wish to take that approach, you’ll have to count the number of elements equal to the median and delete the correct number of them.

Finally, the argument only shows that the amortized cost with respect to $\phi$ is O(m). The conclusion we’re asked for requires a technical condition: the potential $\phi$ never drops below its initial value. This is true for the usual reason: initially, $\phi$ = 0 because S is empty; during execution, $\phi \geq 0$ by definition.

7 0

3 years ago

One of the earlier applications of crypto-graphic hash functions was the storage of passwords to authenticate usersin computer s

vagabundo [1.1K]

Answer: provided in the explanation part.

Explanation:

This is actually quite long but nevertheless i will make it as basic as possible.

Question (a)

Attack A:

One way property of hash means that we can't find the input string if given the hash value. The calculation of hash from input string is possible but it is not possible to calculate the input string when given the hash. If the hash function is properly created to have one-way property then there is no way of finding the exact input string. So this attack won't work as the one-way property of hash function can't be broken if the hash function is properly created.

Attack B:

Suppose h() is the hash function. And h(x) = m where x is the string and m is the hash. Then trying to find another string y such that h(y) = m is called finding out the second pre-image of the hash.

Although we can't know the exact initial string for sure, we can by using brute force method find out a second preimage.

This attack will take a very long time. It has the time complexity of 2n. It requires the attacker to have an idea about the kind of passwords that might be used and then brute force all of them to find the string that has the same hash. Each try will have a chance of 1/2n to succeed.

Rainbow attack using rainbow table is often used for such brute-force attack. This comprises a rainbow table which contains passwords and their pre-hashed values.

Therefore, it is not possible to determine the second preimages of h so easily.

Attack C:

Collisions refer to finding out m and m' without knowing any of them. Finding out collisions is easier than finding preimages. This is because after finding out 2n pairs of input/output. The probability of two of them having the same output or hash becomes very high. The disadvantage is that we can't decide which user's hash to break. However, if I do not care about a particular user but want to get as many passwords as possible, then this method is the most feasible.

It has the time complexity of 2n/2.

Hence, this is the attack which has the most success rate in this scenario.

Question (b)

The brute force way of finding out the password usually involves using a rainbow attack. It comprises a rainbow table with millions of passwords and their hashes already computed. By matching that table against the database, the password can be recovered.

Therefore it is often preferred to salt the password. It means we add some random text to the password before calculating the hash.

The salts are usually long strings. Although users usually do not select long passwords, so a rainbow table with hashes of smaller passwords is feasible. But once salt is used, the rainbow table must accommodate for the salt also. This makes it difficult computationally. Although password might be found in the rainbow table. The salt can be anything and thus, make brute-force a LOT more difficult computationally.

Therefore salt is preferred to be added to passwords before computing their hash value.

Question (c)

A hash output length of 80 means there can be exactly 280 different hash values. This means there is at least one collision if 280+1 random strings are hashed because 280 values are used to accommodate all the possible strings. It is not hard with today's computation power to do match against more than this many strings. And doing so increases the probability of exposing a probable password of a user.

Hence, 80 is not a very secure value for the hash length.

cheers i hope this helps!!!!

6 0

3 years ago

How will advertising and communications change as technology improves?????? HELP PLEASE :0 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

drek231 [11]

Technology changes advertising and communications by increasing the outreach of people that will get the message, and increasing the speed at which they get the message. Which in turn makes the advertising more relevent because they can release more product to a wider audience during a specific time period.

6 0

3 years ago

………………….. is the process of causing a system variable to conform to some desired value. Options Control feedback Design none of

frutty [35]

Control is the process of causing a system variable to conform to some desired value. Options Control feedback Design none of the above

4 0

2 years ago

16.

dimaraw [331]

Answer:

r = 15 cm

Explanation:

formula

V = πr²×h/3

replace

4950 = 22/7×r²×21/3

4950 = 22/7×r²×7

4950 = 22×r²

r² = 4950/22

r² = 225

r = √225

r = √15²

r = 15 cm

3 0

2 years ago