One of the earlier applications of crypto-graphic hash functions was the storage of passwords to authenticate usersin computer systems. With this method, a password is hashed after itsinput and is compared to the stored (hashed) reference password.

(a) Assume that you are a hacker and you got access to the hashedpassword list. You would like to impersonate some of the users. Discuss which of the following three attacks below allow this. Exactlydescribe the consequences of each of the attacks:

•Attack A: You break the one-way property ofh.
•Attack B: You can find second preimages forh.
•Attack C: You can find collisions forh.

(b) Why is this technique of storing passwords often extended by theuse of a so-calledsalt? (A salt is a random value appended to thepassword before hashing. Together with the hash, the value of saltis stored in the list of hashed passwords.) Are the attacks aboveaffected by this technique?

(c) Is a hash function with an output length of 80 bit sufficient for this application?

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

true never assume something it makes an a** out of u and me.

answer: internal fragmentation is the difference between the (fixed) size of a memory hole and the data within in that memory. for example, if a 1,024 byte block of memory holds 800 bytes of data, then the 224 bytes difference is considered lost to internal fragmentation.

external fragmentation occurs when the sum of memory fragments is large enough to satisfy a request, but the holes are non-contiguous and the request cannot be satisfied. this can occur when memory is dynamically allocated and freed.

explanation: