Hashing Techniques in Data Structures Part2

Hashing Techniques in Data Structures Part2

• 1.
  Hash Function Examples Let h(k) = k % 15. Then, if k = 25 129 35 2501 47 36 h(k) = 10 9 5 11 2 6 Storing the keys in the array is straightforward: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 _ _ 47 _ _ 35 36 _ _ 129 25 2501 _ _ _ Thus, delete and find can be done in O(1), and also insert, except…
• 2.
  Hash Function Whathappens when you try to insert: k = 65 ? k = 65 h(k) = 5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 _ _ 47 _ _ 35 36 _ _ 129 25 2501 _ _ _ 65(?) This is called a collision.
• 3.
  Handling Collisions Chaining (Hashing with Chaining) Open Addressing – Linear Probing
• 4.
  Handling Collisions Chaining
• 5.
  Separate Chaining Leteach array element be the head of a chain. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14       47 65 36 129 25 2501  35 Where would you store: 29, 16, 14, 99, 127 ?
• 6.
  Separate Chaining Leteach array element be the head of a chain: Where would you store: 29, 16, 14, 99, 127 ? 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14          16 47 65 36 127 99 25 2501 14    35 129 29 New keys go at the front of the relevant chain.
• 7.
  Separate Chaining: Disadvantages • Parts of the array might never be used. • As chains get longer, search time increases to O(n) in the worst case. • Constructing new chain nodes is relatively expensive . • Is there a way to use the “unused” space in the array instead of using chains to make more space?
• 8.
  Handling Collisions LinearProbing
• 9.
  Linear Probing Letkey k be stored in element h(k)=t of the array 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 47 35 36 129 25 2501 65(?) What do you do in case of a collision? If the hash table is not full, attempt to store key in the next array element (in this case (t+1)%N, (t+2)%N, (t+3)%N …). until you find an empty slot.
• 10.
  Linear Probing Wheredo you store 65 ? [Here N is 15]. 65%15=5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 47 35 36 65 129 25 2501    attempts Where would you store: 29?
• 11.
  Linear Probing Ifthe hash table is not full, attempt to store key in array elements (t+1)%N, (t+2)%N, … 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 47 35 36 65 129 25 2501 29  attempts Where would you store: 16?
• 12.
  Linear Probing Ifthe hash table is not full, attempt to store key in array elements (t+1)%N, (t+2)%N, … [16%15=1] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 16 47 35 36 65 129 25 2501 29  Where would you store: 14? [14%15=14]
• 13.
  Linear Probing Ifthe hash table is not full, attempt to store key in array elements (t+1)%N, (t+2)%N, … [14%15=14] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 14 16 47 35 36 65 129 25 2501 29   attempts Where would you store: 99?
• 14.
  Linear Probing Ifthe hash table is not full, attempt to store key in array elements (t+1)%N, (t+2)%N, … [99%15=9] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 14 16 47 35 36 65 129 25 2501 99 29     attempts Where would you store: 127 ?
• 15.
  Linear Probing Ifthe hash table is not full, attempt to store key in array elements (t+1)%N, (t+2)%N, … [127%15=7] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 16 47 35 36 65 127 129 25 2501 29 99 14   attempts
• 16.
  Linear Probing •Eliminates need for separate data structures (chains), and the cost of constructing nodes. • Leads to problem of clustering. Elements tend to cluster in dense intervals in the array.      • Search efficiency problem remains.