3-partition problem

1. The set ${\displaystyle S=\{20,23,25,30,49,45,27,30,30,40,22,19\}}$ can be partitioned into the four sets ${\displaystyle \{20,25,45\},\{23,27,40\},\{49,22,19\},\{30,30,30\}}$, each of which sums to T = 90.
2. The set ${\displaystyle S=\{1,2,5,6,7,9\}}$ can be partitioned into the two sets ${\displaystyle \{1,5,9\},\{2,6,7\}}$ each of which sum to T = 15.
3. (every integer in S is strictly between T/4 and T/2): ${\displaystyle S=\{4,5,5,5,5,6\}}$, thus m=2, and T=15. There is feasible 3-partition ${\displaystyle \{4,5,6\},\{5,5,5\}}$.
4. (every integer in S is strictly between T/4 and T/2): ${\displaystyle S=\{4,4,4,6,6,6\}}$, thus m=2, and T=15. There is no feasible solution.

The 3-partition problem remains NP-complete even when the integers in S are bounded above by a polynomial in n. In other words, the problem remains NP-complete even when representing the numbers in the input instance in unary. i.e., 3-partition is NP-complete in the strong sense or strongly NP-complete. This property, and 3-partition in general, is useful in many reductions where numbers are naturally represented in unary.

The 3-partition problem is similar to the partition problem, in which the goal is to partition S into two subsets with equal sum, and the multiway number partitioning, in which the goal is to partition S into k subsets with equal sum, where k is a fixed parameter. In 3-Partition the goal is to partition S into m = n/3 subsets, not just a fixed number of subsets, with equal sum. Partition is "easier" than 3-Partition: while 3-Partition is strongly NP-hard, Partition is only weakly NP-hard - it is hard only when the numbers are encoded in non-unary system, and have value exponential in n. When the values are polynomial in n, Partition can be solved in polynomial time using the pseudopolynomial time number partitioning algorithm.

In the unrestricted-input variant, the inputs can be arbitrary integers; in the restricted-input variant, the inputs must be in (T/4, T/2). The restricted version is as hard as the unrestricted version: given an instance S_u of the unrestricted variant, construct a new instance of the restricted version S_r ≔ {s + 2 T  | s ∈ S_u}. Every solution of S_u corresponds to a solution of S_r but with a sum of 7 T  instead of T, and every element of S_r is in [2 T , 3 T ] which is contained in ( T /4, 7 T /2).

In the distinct-input variant, the inputs must be in (T/4, T/2), and in addition, they must all be distinct integers. It, too, is as hard as the unrestricted version.[1]

In the unrestricted-output variant, the m output subsets can be of arbitrary size - not necessarily 3 (but they still need to have the same sum T). The restricted-output variant can be reduced to the unrestricted-variant: given an instance S_u of the restricted variant, construct a new instance of the unrestricted variant S_r ≔ {s + 2{{hsp|T}} | s ∈ S_u}, with target sum 7 T . Every solution of S_u naturally corresponds to a solution of S_r. In every solution of S_r, since the target sum is 7 T  and each element is in ( T /4, 7 T /2), there must be exactly 3 elements per set, so it corresponds to a solution of S_u.

The 4-partition problem is a variant in which S contains n = 4 m  integers, the sum of all integers is ${\displaystyle mT}$, and the goal is to partition it into m quadruples, all with a sum of T. It can be assumed that each integer is strictly between T/5 and T/3.

The ABC-partition problem is a variant in which, instead of a set S with 3 m  integers, there are three sets A, B, C with m integers in each. The sum of numbers in all sets is ${\displaystyle mT}$. The goal is to construct m triplets, each of which contains one element from A, one from B and one from C, such that the sum of each triplet is T. [2] This problem can be reduced to 3-partition as follows. Construct a set S containing the numbers 1000a+100 for each ${\displaystyle a\in A}$; 1000b+10 for each ${\displaystyle b\in B}$; and 1000c+1 for each ${\displaystyle c\in C}$. Every solution of the ABC-partition instance induces a solution of the 3-partition instance with sum ${\displaystyle 1000(a+b+c)+111=1000T+111}$. Conversely, in every solution of the 3-partition instance, all triplet-sums must have the same hundreds, tens and units digits, which means that they must have exactly 1 in each of these digits. Therefore, each triplet must have exactly one number of the form 1000a+100, one 1000b+10 and one 1000c+1. Hence, it induces a solution to the ABC-partition instance.

• The ABC-partition problem is also called numerical 3-d matching, as it can also be reduced to the 3-dimensional matching problem: given an instance of ABC-partition, construct a tripartite hypergraph with sides ${\displaystyle A,B,C}$, where there is an hyperedge ${\displaystyle (a,b,c)}$ for every three vertices in ${\displaystyle A,B,C}$ such that ${\displaystyle a+b+c=T}$. A matching in this hypergraph corresponds to a solution to ABC-partition.

Garey and Johnson (1975) originally proved 3-Partition to be NP-complete, by a reduction from 3-dimensional matching.[3] The classic reference by Garey and Johnson (1979) describes an NP-completeness proof, reducing from 3-dimensional matching to 4-partition to 3-partition.[4]

The NP-hardness of 3-partition was used to prove the NP-hardness rectangle packing, as well as of Tetris[5][6] and some other puzzles,[7] and some job scheduling problems.[8]