Question

Given a set S of integers, we say that S can be partitioned if it can be split into two sets U and V so that considering all u ∈ U and all v ∈ V, Σu = Σ v. Let PARTITION = { <S>| S can be partitioned }.

a. (5) Show that PARTITION ∈ NP by writing either a verifier or an NDTM

b. (15) Show that PARTITION is NP-complete by reduction from SUBSET-SUM.

Answer 1

Given Set s of integerd, we say that I can be Partitione if it can be splet into thro ness U & v So that consis sidering all ие о and reV. Eu, Zv: ter PARTITION <s> is can be partitioned} word To show cay PARTITION & Np by by writing einer a verifier or an NDTM. (by PARTITION IS Np- complete by reduction from SUBSET-Sum Proof PARTITION ENP Here is a polynomial-time verifier for PARTITION : va" on input (S,u,vy: .. زد Check that U U vas & UAV = 0 (where Un union & n - Intersection) sudo • check the sum of elements in equals' w! the sum of elements in v. © ارس . of so accept i otherwise, reject. By the given Condition Eu Evt & HEUS VE V lence, PARTIOTION ENP The partition partition Problem is in the class .

To Shar that A B Af - Complete he red to any problem Show for things : 0 There i3 Gorna a nou-determinista polynomial-time algorithm that solves A ie A ENP ② Any Ap-complete problem B can be redreed to A. ③ The reduction of B to A works in Polynomral time Đ The original has a solusion, problem A has a solution it and only if B We show that SET-PARTIT lort 13 Ap- complete. It is is easy to see that SET-PARTITION lan be venhed in polynomial time ja Given Partitions U, V just som the two and verify that their Sum equals each other, which i obviously a polynomial time verification Beaute Summation is a polynomial operation we are performing at most Isl many Summations). The core of the proof is in reducing SUBSET SUM to PARTITION ; To that end giren sex s & a ralvet (The Subser sum query)

we form new set sasu {xty, where X- x E nes To see that this li a redvehan =) Assume there exists Some PCS such that t=E {x then we would have har x KEP X-ts Σ xé - U{x+ } X-t - x E me s' I Cpu{X-26}) we would have har Pu{X-4} s' (vfx-2) forma partition of s. (E) Suppose that there is a Partition pí, of s' such that x 2 E MER Ex. MEPL Puso Notice that در this indret a natural partition P & P ₂ of s such that INLOG we have that X-at t E E 2 x net, NGP, X-24 Ext f Exp Ex nap neri {x net nop,

> S-24 + 2 x 22 nep, 2 x zt MEPI E Hence, from a Solution ta we can form a partition pia [ atp BU {X-4} . s' \(pu (x-+) and P2 = Convensely form a partition we can form a Solution x S nEp. 1-24 & therefore the mapping f: (s, t) s 13 a reduction (because (s, t) is the language / ser SUBSET SUM o s' =f(s, t) is in the langrage /set PARTITION] And, at is clear to see that the transformann done in polynomial time: & Hence, PARTITION is Np- complete N