To determine the longest probe while performing insertion in hash table.

Given Information: The probability of i^th insertion in hash table by using uniform hashing is atmost 2-k.

Explanation:

In the uniform hashing, where keys are inserted uniformly in available hashes. Suppose there is an open address hash table and keys are inserted one by one then average probe will be (1/(1-a)), where a is equals to load factor. So, while inserting keys into empty hash table, for the first hash slot, unsuccessful expected number of probe will be

  Pr{X≥i}<=αi−1 .

Since n<=m/2,α≤1/2 are given. Putting i=k+1 , in the given equation.

  Pr{X≥k+1}≤(1/2)k+1−1

So, the at most probability will be Pr{X≥k+1}=2−k .

Show that the ith insertion requires more than 2lg n probe and probability Pr{Xi>2lg n}  in the hash table of size m is = O(1/n2) .

Given Information: An open addressing hash table having size m and contains n≤m/2 items.

Explanation: From the above section, where the probability of i^th insertion of an item in the hash table by using uniform hashing is at most 2-k.

Where k is the no. of probes. Now from the given condition, put the value of probe k is equals to 2 logn.

  =2 −k=2 −2lg n= ( 2lg n  ) −2=n −2=1/n2 

So, the complexity will be O(1/n2) .

To show the asymptotic bounds for the probes required is equals to O(1/n) .

Given Information: For a particular event A, probability of probes required Pr{Ai}≤1/n2 .

Explanation: For an event A, required probes X>2lg n, and for the ith insertion, required probes for an event Ai, Xi>2log n . From the above explanation, probability of an event A_i for i^th insertion is less than or equals to 1/n².

An event Ai=A1∪A2∪A3∪A4∪…∪An , where 1≤i≤n

  Pr(A)≤Pr(A1)+Pr(A2)+Pr(A3)+Pr(A4)+……….+Pr(An) ,

by Boole’s inequality

  ≤n.1/n2=1/n

So, the asymptotic bound for the probe required is equals to O(1/n) .

To determine the expected length of the longest probe sequence is O(lg n) .

Given Information: Probability of number of probes required Pr{X=k}, where k is equals to no. of probes.

Explanation:The expected length of the longest probe E[X] can be dependent on the given probability Pr{X=k} . To get the desired outcome, expected length E[X] can be divided in such a way that the total sum would not affected.

  E[X]=∑k.Pr{X=k}, 1≤k<n

Here,break the above equation into two parts by using some breaking point, in this case its 2lg n.

  E[X]=∑k.Pr{X=k} +∑k.Pr{X=k}, 1≤k<n