Numerical Matching With Target Sums

In an effort to make my semesters easier, during breaks I do most of the research on the problems and write quick sketches of the reductions out.  This way when I get to the weekly post, most of the hard math work is done, and I don’t get surprised by a super hard problem.

(I’m doing something similar over our winter break at the present.  I’ve  got sketches up through the middle of April, and I’m currently working on problem SR13- “Sparse Matrix Compression”- which is an “unpublished manuscript”  problem that I’m having a lot of trouble with.  Keep your fingers crossed).

Anyway, I was looking through my notes today and I realized that I’d skipped this problem!  Luckily, I think the reduction is pretty easy.

The problem: Numerical Matching With Target Sums.  This is problem SP17 in the appendix.

The description: Given two sets X and Y, each with the same number (m) of elements, and each with a positive integer size.  We’re also given a “target vector” V, also of M elements, consisting of positive integer entities.  Can we create m sets A1 through Am such that:

  • Each Ai has one element from X and one element from Y
  • Each element in X and Y appears exactly once in some Ai
  • The sum of the sizes of the elements in each Ai is exactly Bi?

Example: I’ll use an example derived from last week’s Numerical 3-Dimensional Matching example because I think it will illustrate how the reduction will work:

  • X = {12,11,7,5}
  • Y = {1,1,4,5}
  • B = {13,12,11,10}

(W from last week was {1,2,3,4}, and B was 14.)

Letting A1 be the first elements of X and Y, A2 being the second elements of X and Y, and so on down, gives us a solution.

Reduction: G&J say to use Numerical 3-Dimensional Matching, and don’t even bother to mark it as “unpublished results”, probably because they think it’s so easy.

Our Numerical 3DM instance is three sets: W, X, and Y, and a bound B.  We need 2 sets and a “bound vector” for the instance of the Numerical Matching problem.  So what we do is:

  • X’ = X
  • Y’ = Y
  • Each bi in the B vector will be set to B-wi.  This is the amount we need the element from X and Y to add up to, so that when we add in the element from W, we get B.

If we have a solution to the Numerical 3-Dimensional Matching solution, then each Ai in that solution consists of 3 elements: wi, xj and yk that sum to B.  Then in our Numerical Matching With target Sums instance, we have a set Ai‘ where xj + yk sum to B-wi.  The same is true in the reverse direction.

Difficulty: 3, which may be too high.  I can see people getting confused by the fact that the sets in the 3DM instance can be taken in any order, but the B vector in the Target Sum matching problem needs to have Ai‘s element sum exactly to bi, and wondering how to force the correct W element to be in that spot?

(The answer is that you define it when you build B.  We set b1 to be “the sum that works when you use wi“, so it (or something with the exact same size, so we can swap the elements) has to go in that position in the vector).

Numerical 3-Dimensional Matching

SP15 is 3-Parttion.

The next problem is one of G&J’s “unpublished results” problems.  I tried figuring out an elegant way to doing it, but couldn’t make it happen.

The problem: Numerical 3-Dimensional Matching.  This is problem SP16 in the appendix.

The description: Given three sets W, X, and Y, each containing the same amount (m) of elements with positive “sizes” and a positive bound B.  Can we create m sets A1 through Am (containing 3 elements each), such that:

  • Each Ai has exactly one element from W, X, and Y
  • The sum of the sizes of the elements in each Ai is exactly B
  • Each element in W, X and Y is in some Ai

Example: Suppose we have the following sets:

  • W has elements with sizes {1,2,3,4}
  • X has elements with sizes {12,11,7,5}
  • Y has elements with sizes {1,1,4,5}  (the ability to allow repeat numbers is why we define the sets as elements with sizes rather than sets of integers)

If B=14, then the partition where A1 is the first element in W, X, and Y, A2 is the second element in W, X, and Y, and so on gives each Ai set a sum of 14.  Obviously, we don’t need to choose corresponding elements from W, X, and Y to form the sets (for example, rearranging the elements in X to be in increasing order doesn’t change whether the problem can be solved, just the exact composition of the Ai sets)

Reduction: I tried doing a reduction using 3-partition, but got stuck (I’ll show it below, in case you want to try to fix it).  G&J refer you to Theorem 4.4 in the book, which is the proof that 3-partition itself is NP-Complete.

We can follow that along and do similar steps to our problem:

  • Theorem 4.3 shows how to turn 3DM into 4-partition (a problem like 3-partition but each set in the solution has 4 elements instead of 3).  Since the sets that are created in the 4-partition solution come from 4 different places (page 98 calls then a “ui, a wi[·], an xj[·], and a yk[·]).  Since these partitioned sets all add to the same total (B) and come from 4 disjoint parent sets, we can see how we could do basically the same reduction and show that the “numerical 4-dimensional matching problem” is NP-Complete.
  • Theorem 4,4 shows how to turn a 4-parition problem into a 3-partition problem.  The idea is to add enough “pairing” and “filler” elements to the 3-partition instance to make any 4-partition set be split into two 3-partition sets, each consisting of 2 elements from the 4-parittion, plus the “pairing” element of one of the 2 elements chosen.  We can do something similar converting numerical 4-dimensional matching to numerical 3-dimentional matching.  (The difference is that we are given specifically which sets the elements are coming from)  So, if we’re given W, X, Y, and Z in our numerical 4DM instance, we construct W’ to be elements from W and Y, X’ to be elements from X and Z, and Y’ to be the pairing elements of pairs from W’ and X’.  We then need to add enough filler elements to our 3 sets in a similar way to the 3-partition proof (again, the difference is that we have to specifically assign them to W’, X’, or Y’.  But that can be determined by how the 3-partition proof allocates the items)

Difficulty: If you have gone over the 3-partition reduction, this is probably a 6.  Lots of tricky math but doable in a (hard) homework.    But keep in mind you’re tacking it on to the difficulty 8 of understanding the 3-partition reduction in the first place.

My reduction I couldn’t get to work: I really want there to be an easier way to do this.  I tried reducing from 3-partition directly because the problems are so similar.  Here is where I got to:

We’re given a 3-parititon instance S, and an integer B.  Our goal is to split S into sets of size 3 that all add up to B.

So, let’s use S to create 3 sets in our numerical 3DM instance:

  • W has all of the elements in S
  • X has all of the elements in S, but the sizes are each increased by 10B.
  • Y has all of the elements in S, but the sizes are each increased by 100B.

This would make B’ be 111B.

S has a 3-parititon, then for each set {si, sj, sk}, we take the three sets {wi, xj, yk}, {wj, xk, yi}, and {wk, xi, yj}  This will solve the numerical 3DM instance.

My problem comes showing this in the other direction.  If we have a numerical 3DM solution, we can only construct the 3-partition instance if the sets in the 3DM solution arrange themselves nicely like they do above.  I need to show that if the 3DM solution has {wi, xj, yk}, then the set in the 3DM solution that contains wj also contains xi (or xk) and yk (or yi).  I think you can get there by using the rules about how the bounds of the elements in the 3-partition instance work, but the work you need to do to show that it’s true makes this way of doing things no longer “easier” than the Theorem 4.4 proof I sketched above, so I gave up on it.

I still wish there was a more elegant way to transform this problem, though.

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