Tag Archives: No G&J reference

Dynamic Storage Allocation

Since Bin Packing was a redo, here is the first real problem in the Storage and Retrieval section.

The problem: Dynamic Storage Allocation.  This is problem SR2 in the appendix.

The description: Given a set A of items.  Each item a in A has size s(a), arrival time r(a) and departure time d(a) (all positive integers).  We’re also given a storage size D.  Can we allocate the items to D “slots” of storage such that:

  • Each item is stored in consecutive slots.  So an element a has to be contained in s(a) adjacent locations from 1 to D.
  • No two items overlap the same slot during the time they are in storage. In other words, if two items a and a’ are mapped to the same slot in D, the must not have any overlap between their arrival and departure times.

Example: Here’s a simple set of items:

Item Number Arrival Departure Size
1 1 2 4
2 2 3 4
3 1 3 2

If D=6, we can store these items by using slots 1-4 to hold both items 1 and 2 (notice that they don’t overlap in time, and having one item arrive right as the other departs is ok), and slots 5-6 to hold item 3.

Reduction: The reference to Stockmeyer in G&J is to a private communication.  I tried working out my own reduction from 3-Partition, but couldn’t make it work.  My approach was to make the sizes of the elements in the 3-Parttion instance map to times in this problem, since G&J give the hint that you can make all sizes 1 or 2.  But I couldn’t figure out how to make it work.  I sort of expect there to be 3 possible sizes for a 3-partition problem, instead of 2.

Eventually, I found a paper by Lim that uses regular Partition, using the storage allocation problem as a special case of a problem involving berthing ships.   (The ship problem adds extra complications like each ship needing a specified clearance between it and other ships).  He starts with a set A of elements, and defines T to be the sum of all of the element sizes.  He then creates one item in the storage allocation problem for each element in S.  For a given s(a) in A, the new item has size s(a), arrival time 2, departure time 3 (so exist for just one time duration) .  He also adds 9 new items that have the effect of creating only two sequences of storage slots that can hold the items from s, each of size= T/2. We can place the items in these slots if and only if there is a partition of S.

Difficulty: 7.  I don’t think the idea is too hard to understand, but the 9 sets that are created are hard to come up with (even if you can understand what their purpose is, coming up with the sets that actually get that purpose accomplished is pretty hard).

Ratio Clique

Last week it was pointed out to me that my reduction for Balanced Complete Bipartite Subgraph was wrong, and in my searches to fix it, I found that the real reduction (by Johnson) used a variant of Clique that said (without proof)) that Clique is NP-Complete even if K was fixed to be |V|/2.  I looked up the Clique problem in G&J, and they say in the comments that it is NP-Complete for K = any fixed ratio of V.

I thought this was a neat easy problem that fit in the 3-6 difficulty range I mentioned last week and decided it was worth a post.  But thinking about this brings up some subtle issues relating to ratios and constants that are common sources of errors among students.  I’ll talk about that at the end.

The problem: I don’t know if there is an official name, so I’m calling it “Ratio Clique”.  It is mentioned in the comments to GT19 (Clique).

The description: For any fixed number r, 0< r < 1, does G have a clique of size r*|V| or more?

Example:  Here’s a graph we’ve used for a previous problem:

maximum fixed-length disjoint paths

If r = .5, then r*|V| = 3.5.  So we’re asking if a clique of 3.5 or more vertices exists (which really means a clique of 4 or more vertices).  It does not exist in this graph.  If r ≤ \frac{3}{7}, then we would be looking for a clique of size 3, which does exist in this graph (vertices b, c, and t)

The reduction: We will be reducing from the regular Clique problem.  Since we want to show this “for any fixed value of r”, we can’t change r inside our reduction.

So we’re given a graph G=(V, E) and a K as our instance of Clique. We need to build a graph G’=(V’, E’) that has a fixed K’ = ⌈r*|V’|⌉.

G’ will start with G, and will add new vertices to the graph.  The vertices we add depend on the ratio s of K to |V|    (K = ⌈s*|V|⌉).  K’ is initially K, but may change as vertices are added to the graph.

If r > s, then we need to add vertices to V’ that will connect to each other vertex in V’, and will increase K’ by 1.  This increases the ratio of \frac{K'}{|V'|}, and we keep adding vertices until that ratio is at least r.

If G has a clique of size K, then the extra vertices in K’ can be added to the clique to form a larger clique (since these new vertices connect to every other vertex)

If G’ has a clique of size K’, notice that it must contain at least K vertices that were initially in G. (We only added K’-K new vertices).  These vertices that exist in G are all connected to each other and so will form a clique in G.

If r < s, then we will add vertices to V’ that are isolated (have no edges connecting to them).  K’ will stay equal to K.  Each vertex we add will reduce the ratio of \frac{K'}{|V'|}, and we keep adding vertices until  K=⌈r*|V’|⌉.

Since these new vertices can not be part of any clique in G’, any clique in G’ must consist only of vertices from G.  Since K=K’, this gives us a clique of size K in both graphs.

It is probably also worth mentioning just how many vertices need to get added to the graph in each case, to make sure that we are adding a polynomial number.  If r>s, we will be adding w vertices to satisfy the equation: ⌈s*|V|⌉ + w = ⌈r*(|V|+w)⌉

(These are both ways of expressing K’)

Dropping the ceiling function (since it only leads to a difference of at most one vertex) Solving for w gets us w = \frac{(s|V|-r|V|)}{(r-1)}.  Since r > s, both sides of that division are negative, so w ends up being positive, and polynomial in |V|.

If r < s, we will be adding w vertices to satisfy the equation:

⌈s*|V|⌉ = ⌈r(|V|+w)⌉

(These are both ways of expressing K)

This can similarly be solved to w = s|V|-r|V|.  Since s > v, this is also a positive (and polynomial) number of new vertices.

A possible source of mistakes: I’m pretty sure this reduction works, but we need to be careful that there is a difference between “for any fixed ratio r of |V|” and “for any fixed K”.  Because for a fixed K (say, K=7) solving the “Does this graph have a 7-Clique?” problem can be solved in polynomial (by enumerating all subgraphs of size 7, for example.  There are n \choose 7 subgraphs, which is O(N^7)).  By choosing a ratio instead of a constant K, we gain the ability to scale the size of K’ along with the size of the graph and avoid this issue.  But it is worth mentioning this to students as a possible pitfall.  It’s very easy to do things in a way that effectively is treating r|V| as a constant K, which won’t work.

Difficulty: 3, but if you’re going to make students to the algebra to show the number of vertices that are added, bump it up to a 4.

Bin Packing Take 2

[So WordPress’s search function has failed me.  A search for posts on Bin Packing didn’t turn up this post, so I went ahead and wrote a whole second post for this problem.  Since this time my reduction uses 3-Partition instead of Partition (and so is a little less trivial for use as a homework problem), I figured I’d leave it for you as an alternate reduction.

I have been thinking off and on about whether it would be useful when I’m done with this project (years from now) to go back and try to find reductions that can be done easier (or harder) than what I’ve shown here, to give more options that are in the 3-6 difficulty range that I think is best for homework problems.  I’m not sure how feasible that task would be, but it’s something I’ll try to keep in mind as I go forward.

Anyway, here’s my post that talks about Bin Packing again:]

On to a new chapter! A4- “Storage and Retrieval”

This first one is a classic problem that I guess I haven’t done yet.

The problem: Bin Packing.  This is problem SR1 in the appendix.

The description: Given a finite set U of items, each with a positive integer size, and positive integers B and K.  Can we split U into k disjoint sets such that the sum of the elements in each set is B or less?

Example: Suppose U was {1,2,3,4,5,6}, K=4, and B= 6.  We want 4 disjoint sets that each sum to 6 or less.  For example:

  • {1,5}
  • {2,4}
  • {3}
  • {6}

Note that if K = 3, we would need 3 sets instead of 4, and this wouldn’t be solvable.

The simple reduction: G&J on page 124 say that Bin Packing contains 3-Partition as a special case.  So let’s try reducing from there. Recall the definition of 3-Partition:

Given a set A of 3M elements and an integer B such that the sizes of each element are between B/4 and B/2 and the sum of all of the elements is m*B, can we split A into m disjoint sets that each sum to exactly B?

Keeping in mind that the bounds on the elements in A mean that there are exactly 3 elements in each set in the partition, we can see how this maps easily to the Bin Packing problem:

  • U = A
  • K = m
  • Use the same B

While it is true that the Bin Packing problem allows the sums to be B or less, and the 3-Parittion problem forces the sets to sum to exactly B, the fact that all of the sets have to contain 3 elements and the fact that the sum of all of the element in U is m*B means that if any set in the Bin Packing answer is < B some other set will necessarily be more than B.

Difficulty: 3.  It is basically the same problem, but I think there is enough work needed to justify the reduction that it makes sense as a good easy homework problem.

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).

3-Matroid Intersection

This is a problem that was really hard for me to understand and explain.  It may be less so or other people, though.

The problem: 3-Matroid Intersection.  This is problem SP11 in the appendix.

The description: Given three matroids (E,F1), (E, F2), (E,F3), and a positive integer K.  Can we find a subset E’ of E with K elements that is in the intersection of F1, F2, and F3?

A matroid (E,F) is a set of elements E, and a family F of subsets of E with the following properties:

  • Some set S in the family F implies that all subsets of S are also in F.
  • If I have two sets S and S’ in F, and |S| = |S’|+1, then there is some element e that is in S but not S’, and adding e to S’ gets us a set also in F.

A common example of matroids comes from graph theory: Given a graph G=(V,E), the matroid based on this graph is (E,F), where E is the set of edges in the graph, and F is the collection of edges that have no cycles (i.e, forests)  This follows the two rules above:

  • If S is a collection of edges that do not form a cycle, then any subset of S is also a collection of edges that do not form a cycle.
  • If I have two collections of edges S and S’, and S is larger than S’ by one element, we need to find an edge in S that connects two connected components  (trees in the forest) in S’.  Since each tree on K vertices has exactly K-1 edges, S has at least one edge that is not part of any tree in S’ (adding an edge to a tree causes a cycle).  That edge must either connect two trees together, or connect a vertex in some tree to a vertex not used in S’, or connect two vertices not used in S’.  Either way, adding this edge to S’ gives us another forest.

Example:  Since the reduction we’re going to use involves building three matroids from a basic graph, let’s do that.  Here is a graph:

3matroidintersection1

We’re going to use this graph to induce 3 matroids.  The first will be the graph matroid (using the rules defined above) on the undirected version of this graph:

3matroidintersection2

So F1 is the set of all forests in the above graph.

F2 is collections of sets of edges in the directed graph where each vertex has at most one incoming edge.  So, sets like {(s,a), (a,t)} or {(a,b), (b,c), (c,t)}.  The exception is that vertex s must have zero incoming edges in any set in the family.

F3 is built similarly, except we count outgoing edges, and t (instead of s) is the vertex restricted to have 0 outgoing edges.

If K=4, then the following set of edges is in all three families: {(s,a}, (a,b), (b,c), (c,t)}.  It’s acyclic, so it’s in F1.  No vertex in that set has indegree or outdegree of more than 1 (and s has indegree 0 and t has outdegree o), so it’s in F2 and F3.

Just to clarify, because it’s confusing to me.  The sets E that are the “elements” of the matroid are edges in the graph, and the families F are collections of edges.  We just found a collection of 4 edges that will be in all three families.

The reduction: G&J don’t give a reference to this problem, but suggest using 3DM. In my searches, I found that Wikipedia and this set of lecture notes from MIT both want to use Hamiltonian path between two vertices.   (Notice that the “corrected” reduction for HP actually specifies the two vertices to connect between, so the HP reduction also proves this variant).  Here is my explanation of the reduction in the MIT lecture notes.  The class was taught by Michael Goemans, and I got the notes from his web site.

So, we start with an instance of HP- a possibly directed graph G=(V,E), and two special vertices s and t that we want to see if there is a Hamiltonian Path between.  We build the three matroids described in the examples based on using the edge set E: One based on forests in E, one based on collections of edges that have indegree at most 1 on each vertex, and one based on collections of edges that have outdegree at most 1 on each vertex.  Set K = |V|-1.

Notice that an intersection of these three matroids has to be a collection of paths that do not encounter the same vertex twice.  If we can create a collection that has K edges in it, the path will start at s (because F2 ensures that no edges going into s will be part of the intersection), encounter each vertex once (because F1 ensures that we have no cycles, and thus hit each vertex at most once), and end at t (because F3 ensures that no edges leaving t will be part of the intersection).  Thus, we have an intersection of K edges if and only if V has a Hamiltonian cycle.

Difficulty: It’s an 8 for me.  Since this is taught in a relatively small part of one lecture of a class (yes, a class at MIT, but still just a class), presumably it may be less for other people.  But I have a lot of trouble thinking in terms of matroids, and even now, I’m not really convinced that the three families of edges we create are polynomial in size.

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