What Exactly is Hash Collision

Question

Hash Collision or Hashing Collision in HashMap is not a new topic and I\'ve come across several blogs and discussion boards explaining how to produce Hash Collision or how t

Answer 1

Actually I think the hash collision is Normal. Let talk about a case to think. We have 1000000 big numbers(the set S of x), say x is in 2^64. And now we want to do a map for this number set. lets map this number set S to [0,1000000] .

But how? use hash!!

Define a hash function f(x) = x mod 1000000. And now the x in S will be converted into [0,1000000), OK, But you will find that many numbers in S will convert into one number. for example. the number k * 1000000 + y will all be located in y which because (k * 1000000 + y ) % x = y. So this is a hash collision.

And how to deal with collision? In this case we talked above, it is very difficult to delimiter the collision because the math computing has some posibillity. We can find a more complex, more good hash function, but can not definitely say we eliminate the collision. We should do our effort to find a more good hash function to decrease the hash collision. Because the hash collision increase the time cost we use hash to find something.

Simplely there are two ways to deal with hash collision. the linked list is a more direct way, for example: if two numbers above get same value after the hash_function, we create a linkedlist from this value bucket, and all the same value is put the value's linkedlist. And another way is that just find a new position for the later number. for example, if number 1000005 has took the position in 5 and when 2000005 get value 5, it can not be located at position 5, it then go ahead and find a empty position to took.

For the last question : Does Java generate or at least try to generate unique hashCode per class during object initiation?

the hashcode of Object is typically implemented by converting the internal address of the object into an integer. So you can think different objects has different hashcode, if you use the Object's hashcode().