Slideshow:
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
Initial state:
Insert a search key x that has a hash value 0101
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
Since n=2 (# bits in hash value = ⌈log2(n)⌉ = 1), we insert the search key in bucket 1
Insert the search key in the bucket...
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
Then we check the occupancy:
Occupancy = 4/(2 × 2) = 1.0 > τ (0.85) -- add a physical bucket
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
We add physical bucket 10 (binary):
And we must re-hash the search keys in bucket Suffix-1( 10 (bin) ) = 0 (bin)
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
Re-hashing search keys in bucket 0 (bin):
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
Final state - insertion is complete:
Next: how do we find a search key x that has a hash value 1111 ???
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
Since n=2 (# bits in hash value = ⌈log2(n)⌉ = 2), we will use the "virtual" bucket 11:
Because bucket 11 (bin) is virtual, we will use bucket[ Suffix(11) = 01 (bin)] !!
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
Example continues: insert a search key with a hash value 0001:
Search key is inserted in a full disk block → use overflow block !
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
After insertion, we check the occupancy...
Occupancy = 5/(3 × 2) = 0.8333 ≤ 0.85 -- done ! --- Next: insert 1011...
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
Next: insert 1011: since n=2 (# bits in hash value = ⌈log2(n)⌉ = 2), we will use the "virtual" bucket 11:
Because bucket 11 (bin) is virtual, we will use bucket[ Suffix(11) = 01 (bin)] --- insert in bucket 01 (bin).
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
After inserting 1011:
Check occupancy: 6/(3 × 2) = 1 > τ (0.85) → add a bucket !
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
We add the physical bucket 11 (bin):
And we must re-hash the search keys in the bucket[ Suffix-1 ( 11 (bin) ) = 01 (bin)]
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
Re-hashing the search keys in the bucket[ Suffix-1 ( 11 (bin) ) = 01 (bin)]:
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
After the re-hash operation is completed:
We can optionally compact the bucket[ Suffix-1 ( 11 (bin) ) = 01 (bin)]...
Example Linear Hashing with
# keys/block γ = 2 and
threshold τ = 0.85
If we compact:
- Parameters:
Max # search keys in 1 block (γ) = 2 Threshold avg occupance (τ) = 0.85
- Initial State:
Average occupancy:
r 3 ------- = ------- = 0.75 (< 0.85) n × γ 2 × 2
- Insert search key
K such that
h(K) = 0101
- Insert
0101:
Result:
Average occupancy:
r 4 ------- = ------- = 1 > τ (0.85) n × γ 2 × 2
- Insert
0101:
- We must add
an new bucket:
- Add bucket
2 (= 10 (binary)):
- Re-hash the
search keys from
/-----> Bucket 00 Bucket 0 --- \-----> Bucket 10Graphically:
Result:
Average occupancy:
r 4 ------- = ------- = 0.6666 ≤ τ (0.85) n × γ 3 × 2Notice that:
- n = 3 (changed) !!!
- i = 2
(changed) !!!
- We can find 1111 as follows:
Explanation:
- 1111
will lead to
the bucket
11
(using the last i (= 2) bits)
which is a
"ghost" bucket
- The search algorithm will use the real bucket 01 instead !!!
- 1111
will lead to
the bucket
11
(using the last i (= 2) bits)
which is a
"ghost" bucket
Example continued....
- Add bucket
2 (= 10 (binary)):
- Insert search key
K such that
h(K) = 0001
- Insert
0001:
Result:
So: Linear Hashing uses overflow blocks !!!
Average occupancy:
r 5 ------- = ------- = 0.833333 ≤ τ (0.85) n × γ 3 × 2No need too add another bucket.....
Example contiued....
- Insert
0001:
- Insert search key
K such that
h(K) = 1011
- Insert
1011:
Result:
Average occupancy:
r 6 ------- = ------- = 1 > τ (0.85) n × γ 3 × 2
- Insert
1011:
- We must add
an new bucket:
- Add bucket
3 (= 11 (binary)):
- Re-hash
search keys from:
/----> Bucket 01 Bucket 01 --- \----> Bucket 11Graphically:
Result:
Average occupancy:
r 6 ------- = ------- = 0.75 ≤ τ (0.85) n × γ 4 × 2Notice that:
- n = 4
(changed) !!!
- Every search key
in the Linear Hash table is
now
identified by their
last 2 bits:
- Add bucket
3 (= 11 (binary)):