The following hashtable has a capacity of 10, and __ marks empty slots. The index of...
The following hashtable has a capacity of 10, and __ marks empty slots. The index of each slot is shown below for convenience. Using linear probing (probe length = 1), you insert a new key x (where x is your student id) into the hashtable. Next, you insert a second new key y (where y is your student id + 10). In which slot will y be inserted?(JAVA)
[ 60 __ __ 53 23 __ 36 46 27 __ ]
<0> <1> <2> <3> <4> <5> <6> <7> <8> <9>
Homework Answers
Linear probing:
Linear probing is a form of open addressing for resolving
collisions in hash table. Here, gap between two probes is 1.
If the key slot is full then we find the next empty slot and then insert key.
| Key | 60 | __ | __ | 53 | 23 | __ | 36 | 46 | 27 | __ |
|---|---|---|---|---|---|---|---|---|---|---|
| Index | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Hash function : hash(key) = key mod 10
Index : hash(key) % capacity
Inserting x :
Let's assume x = 71
So, hash(x) = 71 % 10 = 1.
So, the index = 1 % 10 = 1
since, index = 1 is empty so, we insert 71 at index 1.
Inserting y:
Given y = x + 10.
So the hash function will give the same value.
So, hash(x) = hash(x+10) = hash(y).
because, x%10 = (x+10)%10
So, y = 71 + 10 = 81 (since x = 71)
hash(y) = 81 % 10 = 1
index of y = 1 % 10 = 1
Since, index = 1 is occupied by x so we probe until we find the
next empty slot.
next empty index is index = 2.
So, we insert y at index 2.
| Key | 60 | x=71 | y=81 | 53 | 23 | __ | 36 | 46 | 27 | __ |
|---|---|---|---|---|---|---|---|---|---|---|
| Index | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
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