Tabulation hashing

Numerical 3-dimensional matching is an NP-complete decision problem. It is given by three multisets of integers $X$, $Y$ and $Z$, each containing $k$ elements, and a bound $b$. The goal is to select a subset $M$ of $X\times Y\times Z$ such that every integer in $X$, $Y$ and $Z$ occurs exactly once and that for every triple $(x,y,z)$ in the subset $x+y+z=b$ holds. This problem is labeled as [SP16] in.

Example

Take $X=\{3,4,4\}$, $Y=\{1,4,6\}$ and $Z=\{1,2,5\}$, and $b=10$. This instance has a solution, namely $\{(3,6,1),(4,4,2),(4,1,5)\}$. Note that each triple sums to $b=10$. The set $\{(3,6,1),(3,4,2),(4,1,5)\}$ is not a solution for several reasons: not every number is used (a $4\in X$ is missing), a number is used too often (the $3\in X$) and not every triple sums to $b$ (since $3+4+2=9\ne b=10$). However, there is at least one solution to this problem, which is the property we are interested in with decision problems. If we would take $b=11$ for the same $X$, $Y$ and $Z$, this problem would have no solution (all numbers sum to $30$, which is not equal to $k\cdot b=33$ in this case).

Related problems

Every instance of the Numerical 3-dimensional matching problem is an instance of both the 3-partition problem, and the 3-dimensional matching problem.

Proof of NP-completeness

NP-completeness of the 3-partition problem is stated by Garey and Johnson in "Computers and Intractability; A Guide to the Theory of NP-Completeness". It is done by a reduction from 3-dimensional matching via 4-partition. To prove NP-completeness of the numerical 3-dimensional matching, the proof is similar, but a reduction from 3-dimensional matching via the numerical 4-dimensional matching problem should be used.

References

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