# Category Archives: Appendix- Sets and Partitions

## Kth Largest m-Tuple

I think I’m going to move the posts to Wednesday this semester since I teach two 2-hour classes on Tuesday/Thursday.

SP19 is Minimum Sum of Squares

SP20 is Kth Largest Subset and is very similar to our next problem.

The problem: Kth Largest m-Tuple.  This is problem SP21 in the appendix.

The description: Given sets X1 through Xm that all contain positive integers, and integers K and B, are there at least K m-tuples (x1, .., xm) from X1 x X2 x … x Xm for which the sum of the elements in each tuple is at least B?

Example: Let X1 = {1,2,3,4}, X2 = {5,10}, X3 = {1,2}. Notice that the X’s can be different sizes, and can repeat elements.

If B = 15, then the only tuples that sum to at least 15 are {3.10,2), (4,10,1), and  (4,10,2).

Reduction: The paper by Johnson and Mizoguchi presents the reduction pretty densely, but here is what I think they are saying:  We’re going to use a Turing Reduction to solve the Partition problem, which means we’re given an instance of Partition, and assume we can call a subroutine to solve the Kth Largest m-tuple problem multiple times to solve the Partition instance.  Recall that we can use Binary Search on multiple calls of the subroutine to determine (for example) how many tuples sum to some B or more.  (We need multiple calls because the subroutine is a boolean one, just saying yes or no to a given instance).

Updated reduction:

This idea came from Said D. in the comments, and I wanted to make sure it got posted here because it’s so simple and elegant.  If we are given a partition instance S = {s1..sn} and need to know if a subset sums to B (= half the sum of all of the elements in S), then the sets we create for the Kth Largest m-tuple instance are:

{s1+1, 1}, {s2+1, 1}, …(sn+1, 1}

And we number the K-th Largest m-tuple is looking for is B+n.  Each set will contribute 1 to the tuple sum no matter which element is chosen.  It may also contribute si as well, corresponding to whether we “take” that element in the partition instance or not.

That’s way better than what I was trying to do.  Thanks, Said!

The rest of my not very good reduction:

I’ll leave the rest up here for posterity, but this is way worse than Said’s idea above, and glosses over the fact that you can’t just use any elements you want as the sets- you need to set them up in a way that makes sure you don’t repeat elements when you choose the tuples.

So, in a manner similar to that used in the Kth Largest Subset reduction, we can find the number of tuples that sum to any B we want (including half the sum of all of the elements in the partition instance).  The only tricky part is that our “subsets” are tuples of different sizes.  So we need to run this algorithm multiple times:

• Are there are 1-tuples that solve the partition problem?  (There are O(n)  1-tuples, so a binary search requires O(log n) calls)
• Are there any 2 tuples that solve the partition problem?  (There are O(N^2) 2-tuples, so a binary search requires O(2* log n) calls)
• Are there any m-tuples that solve the partition problem? (There are O(N!) n-tuples, so a binary search requires O(n * log n) calls)

Thus, we can do it in a polynomial number of calls to the K-th Largest m-tuple solver.

Difficulty: This is a little harder than the Kth largest subset reduction, which I gave a 5, so I’ll make this a 6.

## Expected Component Sum

Sometimes you need a nudge to see the right way to do a reduction.  The reduction to this problem is based on a reduction for a similar problem, which encouraged me to look at the problem in a way that I probably should have noticed myself.

The problem: Expected Component Sum.  This is problem SP18 in the appendix.

The description: Given a collection V of m-dimensional vectors, where each entry in each vector is a non-negative integer.  We’re also given positive integers K and B.  Can we partition V into K disjoint sets V1 through VK such that:

• For each Vi, we look at each position in each vector (from 1 to m), and we sum up the elements in that position in that Vi
• For each Vi,  we find the position with the largest sum
• We sum together the largest position sums of each Vi.  Is that sum at least B?

Example: Suppose we have 4 vectors, each with 5 elements.

• v1 = (1,2,3,4,5)
• v2 = (9,2,4,6,8)
• v3 = (3,7,1,1,4)
• v4 = (2,9,10,3,11)

If K=2, then we can create:

• V1 = v1 and v2  Column 5 has the highest sum (15)
• V2 = v3 and v4.  Column 2 has the highest sum (16)

The total of the sums from each element of the partition is 15+16 = 31.

Reduction: G&J say to use X3C, and also mention two important facts:

1. The problem is still NP-Complete even if the elements in all vectors are 0 and 1.  This implies to me that the vectors should be Boolean representations of participation in the sets.
2. The problem is no longer NP-Complete if we fix K.  This implies that the K value we choose in the reduction needs to be based on the X3C input somehow, and can’t be a simple number like 2 or 3.

My first pass of working on the reduction had me create a vector for each set in C, with positions in the vector corresponding to elements in X.  (Thus, each vector would have ones in exactly 3 positions and zeroes everywhere else.)  My natural inclination was to set K to 2 (one set for the cover, one set for “everything else”- the sets in C that were not in the cover).  But that ran afoul of the prohibition of a fixed value for K.

I toyed with the idea of inverting the sets but didn’t get very far.  Then in some web-searching for inspiration, I found a paper by Roy, Lakshmanan, and Liu that worked on a similar problem.  They call their version the “Perfect Expected Component Sum” problem and works similarly except they fix B to be equal to |C|, and want the final sum to be exactly equal to B, instead of at least B.

The key idea is to have one vector for each element in X, and to have the dimensionality of the vector to be the size of the sets in C.  So each vector vi has a 1 in position j if set Cj contains element i.  Now each position in the vector (from 1 to |C|)  has exactly three vectors with a 1 in that position (the three elements that make up the set corresponding to that position).  We set K=q and B=3q.

If a C’ exists that is an exact cover of X, then C’ consists of {C’1 .. C’q} – exactly q sets from C that contain each element in X exactly once.  Then we can partition V into sets of 3 vectors that correspond to the three elements of each C’i.  So the first partition has the 3 element vectors that correspond to the 3 elements in C’1, the second partition has the 3 element vectors that correspond to the 3 elements in C’2, and so on down.  Each partition will have one column that has a 1 in all 3 elements, and so the maximum sum of all columns will be 3.  Since we have q different sets of vectors in our partition of V, and each contributes 3 to the sum, our total sum is 3q=B.

If we have a partition of vectors that sums to at least V, notice that no set in C has more than 3 elements, therefore no column sum of any set of vectors in our partition will sum to more than 3.  Thus, the only way to reach a sum of at least B is to have a sum of 3 in each of the sets in the partition.  Since there are 3q vectors in V, this can only be accomplished by having exactly 3 vectors in each set in the partition.  These 3 sets must have at least one column that has 1’s in all entries in that column.  This column tells us which set to choose for the cover of X.

Difficulty: 5.  I’m pretty embarrassed I didn’t come up with the idea of using elements as vectors and sets as boolean entries in the vector.  It’s very similar to the graph theory reductions from 3SAT where we have vertices for edges and clauses, and an edge between a literal vertex and a clause vertex if the literal is in that clause.  There’s a similar property there that each clause has degree 3, that you can exploit.

## 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.