# Open Addressing Collision Handling technique in Hashing

**Open Addressing:**

Like separate chaining, open addressing is a method for handling collisions. In Open Addressing, all elements are stored in the **hash table** itself. So at any point, the size of the table must be greater than or equal to the total number of keys (Note that we can increase table size by copying old data if needed). This approach is also known as closed hashing. This entire procedure is based upon probing. We will understand the types of probing ahead:

Insert(k):Keep probing until an empty slot is found. Once an empty slot is found, insert k.Search(k):Keep probing until the slot’s key doesn’t become equal to k or an empty slot is reached.Delete(k):. If we simply delete a key, then the search may fail. So slots of deleted keys are marked specially as “deleted”.Delete operation is interesting

The insert can insert an item in a deleted slot, but the search doesn’t stop at a deleted slot.

## Different ways of Open Addressing:

**1. Linear Probing:**

In linear probing, the hash table is searched sequentially that starts from the original location of the hash. If in case the location that we get is already occupied, then we check for the next location.

The function used for rehashing is as follows: rehash(key) = (n+1)%table-size.

**For example,** The typical gap between two probes is 1 as seen in the example below:

Let

hash(x)be the slot index computed using a hash function andSbe the table sizeIf slot hash(x) % S is full, then we try (hash(x) + 1) % S

If (hash(x) + 1) % S is also full, then we try (hash(x) + 2) % S

If (hash(x) + 2) % S is also full, then we try (hash(x) + 3) % S

…………………………………………..

…………………………………………..Let us consider a simple hash function as “key mod 7” and a sequence of keys as 50, 700, 76, 85, 92, 73, 101.

**Challenges in Linear Probing :**

**Primary Clustering:**One of the problems with linear probing is Primary clustering, many consecutive elements form groups and it starts taking time to find a free slot or to search for an element.**Secondary Clustering**Secondary clustering is less severe, two records only have the same collision chain (Probe Sequence) if their initial position is the same.**:**

**Example:** Let us consider a simple hash function as “key mod 5” and a sequence of keys that are to be inserted are 50, 70, 76, 93.

**Step1:**First draw the empty hash table which will have a possible range of hash values from 0 to 4 according to the hash function provided.

**Step 2:**Now insert all the keys in the hash table one by one. The first key is 50. It will map to slot number 0 because 50%5=0. So insert it into slot number 0.

**Step 3:**The next key is 70. It will map to slot number 0 because 70%5=0 but 50 is already at slot number 0 so, search for the next empty slot and insert it.

**Step 4:**The next key is 76. It will map to slot number 1 because 76%5=1 but 70 is already at slot number 1 so, search for the next empty slot and insert it.

**Step 5:**The next key is 93 It will map to slot number 3 because 93%5=3, So insert it into slot number 3.

**2. Quadratic Probing**

If you observe carefully, then you will understand that the interval between probes will increase proportionally to the hash value. Quadratic probing is a method with the help of which we can solve the problem of clustering that was discussed above. This method is also known as the **mid-square** method. In this method, we look for the **i ^{2}‘th** slot in the

**i**iteration. We always start from the original hash location. If only the location is occupied then we check the other slots.

^{th}let hash(x) be the slot index computed using hash function.

If slot hash(x) % S is full, then we try (hash(x) + 1*1) % S

If (hash(x) + 1*1) % S is also full, then we try (hash(x) + 2*2) % S

If (hash(x) + 2*2) % S is also full, then we try (hash(x) + 3*3) % S

…………………………………………..

…………………………………………..

**Example:** Let us consider table Size = 7, hash function as Hash(x) = x % 7 and collision resolution strategy to be f(i) = i^{2 }. Insert = 22, 30, and 50.

**Step 1:**Create a table of size 7.

**Step 2**– Insert 22 and 30- Hash(22) = 22 % 7 = 1, Since the cell at index 1 is empty, we can easily insert 22 at slot 1.
- Hash(30) = 30 % 7 = 2, Since the cell at index 2 is empty, we can easily insert 30 at slot 2.

**Step 3:**Inserting 50- Hash(50) = 50 % 7 = 1
- In our hash table slot 1 is already occupied. So, we will search for slot 1+1
^{2}, i.e. 1+1 = 2, - Again slot 2 is found occupied, so we will search for cell 1+2
^{2}, i.e.1+4 = 5, - Now, cell 5 is not occupied so we will place 50 in slot 5.

**3.**** Double Hashing**** **

The intervals that lie between probes are computed by another hash function. Double hashing is a technique that reduces clustering in an optimized way. In this technique, the increments for the probing sequence are computed by using another hash function. We use another hash function hash2(x) and look for the i*hash2(x) slot in the **i ^{th}** rotation.

let hash(x) be the slot index computed using hash function.

If slot hash(x) % S is full, then we try (hash(x) + 1*hash2(x)) % S

If (hash(x) + 1*hash2(x)) % S is also full, then we try (hash(x) + 2*hash2(x)) % S

If (hash(x) + 2*hash2(x)) % S is also full, then we try (hash(x) + 3*hash2(x)) % S

…………………………………………..

…………………………………………..

**Example: **Insert the keys 27, 43, 692, 72 into the Hash Table of size 7. where first hash-function is** h1(k) = k mod 7** and second hash-function is **h2(k) = 1 + (k mod 5)**

**Step 1:**Insert 27- 27 % 7 = 6, location 6 is empty so insert 27 into 6 slot.

**Step 2:**Insert 43- 43 % 7 = 1, location 1 is empty so insert 43 into 1 slot.

**Step 3:**Insert 692- 692 % 7 = 6, but location 6 is already being occupied and this is a collision
- So we need to resolve this collision using double hashing.

h

_{new}= [h1(692) + i * (h2(692)] % 7

= [6 + 1 * (1 + 692 % 5)] % 7

= 9% 7

= 2Now, as 2 is an empty slot,

so we can insert 692 into 2nd slot.

**Step 4:**Insert 72- 72 % 7 = 2, but location 2 is already being occupied and this is a collision.
- So we need to resolve this collision using double hashing.

h

_{new}= [h1(72) + i * (h2(72)] % 7

= [2 + 1 * (1 + 72 % 5)] % 7

= 5 % 7

= 5,Now, as 5 is an empty slot,

so we can insert 72 into 5th slot.

See this for step-by-step diagrams:

**Comparison of the above three:**

- Linear probing has the best cache performance but suffers from clustering. One more advantage of Linear probing is easy to compute.
- Quadratic probing lies between the two in terms of cache performance and clustering.
- Double hashing has poor cache performance but no clustering. Double hashing requires more computation time as two hash functions need to be computed.

S.No. | Separate Chaining | Open Addressing |
---|---|---|

1. | Chaining is Simpler to implement. | Open Addressing requires more computation. |

2. | In chaining, Hash table never fills up, we can always add more elements to chain. | In open addressing, table may become full. |

3. | Chaining is Less sensitive to the hash function or load factors. | Open addressing requires extra care to avoid clustering and load factor. |

4. | Chaining is mostly used when it is unknown how many and how frequently keys may be inserted or deleted. | Open addressing is used when the frequency and number of keys is known. |

5. | Cache performance of chaining is not good as keys are stored using linked list. | Open addressing provides better cache performance as everything is stored in the same table. |

6. | Wastage of Space (Some Parts of hash table in chaining are never used). | In Open addressing, a slot can be used even if an input doesn’t map to it. |

7. | Chaining uses extra space for links. | No links in Open addressing |

**Note:** Cache performance of chaining is not good because when we traverse a Linked List, we are basically jumping from one node to another, all across the computer’s memory. For this reason, the CPU cannot cache the nodes which aren’t visited yet, this doesn’t help us. But with Open Addressing, data isn’t spread, so if the CPU detects that a segment of memory is constantly being accessed, it gets cached for quick access.

**Performance of Open Addressing:**

Like Chaining, the performance of hashing can be evaluated under the assumption that each key is equally likely to be hashed to any slot of the table (simple uniform hashing)

m = Number of slots in the hash table

n = Number of keys to be inserted in the hash table

Load factor α = n/m ( < 1 )

Expected time to search/insert/delete < 1/(1 – α)

So Search, Insert and Delete take (1/(1 – α)) time

**Related Articles:** Hashing | Set 1 (Introduction), Hashing | Set 2 (Separate Chaining)

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