** Set partition problem:** Set partition problem partitions an array of numbers into two subsets such that the sum of each of these two subsets is the same. Let

**S**be a set of numbers and A is a subset of numbers with sum

**S1**, then there exists another subset containing the remainder of the elements

**(S – A)**with sum

**S2**, and

**S1 is equaled to S2**.

** Problem Statement:** Given a set

**S**of

**N**numbers, the task is to determine if the set contains two partitions of

**S**, with both of them having exactly the same sum.

__Explanation__:

An instance of the problem is an input specified to the problem. An instance of the Set Partition problem is a set **S**, and the task is to check whether there exist any two non-overlapping partitions of **S** having a sum of elements as **sum**. Since an NP-Complete problem is a problem which is both in **NP** and **NP-hard**, the proof for the statement that a problem is NP-Complete consists of two parts:

- The problem itself is in NP class.
- All other problems in NP class can be polynomial-time reducible to that. (B is polynomial-time reducible to C is denoted as B ≤ P
^{C})

If the **2nd condition** is only satisfied then the problem is called **NP-Hard**.

But it is not possible to reduce every NP problem into another NP problem to show its NP-Completeness all the time. Therefore, to show a problem is NP-Complete then prove that the problem is in NP and any NP-Complete problem is reducible to that i.e., if B is NP-Complete and B ≤ P^{C }For C in NP, then C is NP-Complete. Thus, it can be concluded that the **Set partition problem** is NP-Complete using the following two propositions:

**Set Partition Problem is in NP:**

If any problem is in NP, then, given a ‘certificate’, which is a solution to the problem and an instance of the problem (a set **S** and two partitions **A** and **A’**, in this case), it can be proved that the certificate in polynomial time. This can be done in the following way:

- For every element
**x**in**A**and**x’**in**A’**, verify that all the elements belonging to**S**are covered. - Let
**S1**is 0,**S2**is 0 - For every element
**x**in**A**add that value to**S1**. - For every element
**x’**in**A’**add that value to**S2**. - Verify that
**S1**is the same as**S2**.

The algorithm takes linear time in the size of the set of numbers.

**Set Partition Problem is NP-Hard:**

In order to prove that the Independent Set problem is NP-Hard, perform a reduction from a known NP-Hard problem to this problem. Carry out a reduction from which the Subset Sum Problem can be reduced to the Set Partition problem. The Subset Problem provides the input as a set of numbers **S** and a target sum **t**, the problem aims at finding the subset **T** belonging to **S** with a sum same as the **t**. Let s be the sum of members of **S**. Now, feed **S’ = S ∪ {s − 2t}** into the Set Partition problem.

Now prove that the problem of computing the set partition indeed boils down to the computation of the subset-sum. The reduction can be proved by the following two propositions:

Now, let us consider a set of the numbers **T** with summation equivalent to **t**(Subset Sum), then the remainder of the elements in **S**(assuming **O**) will have the sum **o = s – t**. Let us assume the original set is equal to **T’ = T ∪ (s – 2t)** which has a sum equal to **t’**.

Now the following observations hold:

o = s – t

o – t = s – 2t, Difference in sum between O and T.

t’ = t + (s – 2t)

= s – t

= o, the sum of T’ and O are equal.

Hence, the original set can be partitioned into two subsets of sum **(s – t)** each. Therefore, the set partition problem is satisfied.

Now suppose an equal-sum partitioning **(A, A’)** of **S’** = **S ∪ {s − 2t}** exists. The sum of each partition is given by:

Consider the partition containing the element **{s – 2t}** to be **A’**. Let **A = A’- {s – 2t}**. The sum of elements in **A** is given by:

A = s – t – {s – 2t}

= t

Also, **S’ – S = {s – 2t}**. So **A** is a subset of **S** with a sum equal to **t**.

Therefore, the subset sum problem is satisfied.

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