The Many Faces of Set Cover

A while back Peter Norvig posted a wonderful pair of articles about regex golf. The idea behind regex golf is to come up with the shortest possible regular expression that matches one given list of strings, but not the other.

“Regex Golf,” by Randall Munroe.

In the first article, Norvig runs a basic algorithm to recreate and improve the results from the comic, and in the second he beefs it up with some improved search heuristics. My favorite part about this topic is that regex golf can be phrased in terms of a problem called set cover. I noticed this when reading the comic, and was delighted to see Norvig use that as the basis of his algorithm.

The set cover problem shows up in other places, too. If you have a database of items labeled by users, and you want to find the smallest set of labels to display that covers every item in the database, you’re doing set cover. I hear there are applications in biochemistry and biology but haven’t seen them myself.

If you know what a set is (just think of the “set” or “hash set” type from your favorite programming language), then set cover has a simple definition.

Definition (The Set Cover Problem): You are given a finite set U called a “universe” and sets S_1, \dots, S_n each of which is a subset of U. You choose some of the S_i to ensure that every x \in U is in one of your chosen sets, and you want to minimize the number of S_i you picked.

It’s called a “cover” because the sets you pick “cover” every element of U. Let’s do a simple. Let U = \{ 1,2,3,4,5 \} and

\displaystyle S_1 = \{ 1,3,4 \}, S_2 = \{ 2,3,5 \}, S_3 = \{ 1,4,5 \}, S_4 = \{ 2,4 \}

Then the smallest possible number of sets you can pick is 2, and you can achieve this by picking both S_1, S_2 or both S_2, S_3. The connection to regex golf is that you pick U to be the set of strings you want to match, and you pick a set of regexes that match some of the strings in U but none of the strings you want to avoid matching (I’ll call them V). If w is such a regex, then you can form the set S_w of strings that w matches. Then if you find a small set cover with the strings w_1, \dots, w_t, then you can “or” them together to get a single regex w_1 \mid w_2 \mid \dots \mid w_t that matches all of U but none of V.

Set cover is what’s called NP-hard, and one implication is that we shouldn’t hope to find an efficient algorithm that will always give you the shortest regex for every regex golf problem. But despite this, there are approximation algorithms for set cover. What I mean by this is that there is a regex-golf algorithm A that outputs a subset of the regexes matching all of U, and the number of regexes it outputs is such-and-such close to the minimum possible number. We’ll make “such-and-such” more formal later in the post.

What made me sad was that Norvig didn’t go any deeper than saying, “We can try to approximate set cover, and the greedy algorithm is pretty good.” It’s true, but the ideas are richer than that! Set cover is a simple example to showcase interesting techniques from theoretical computer science. And perhaps ironically, in Norvig’s second post a header promised the article would discuss the theory of set cover, but I didn’t see any of what I think of as theory. Instead he partially analyzes the structure of the regex golf instances he cares about. This is useful, but not really theoretical in any way unless he can say something universal about those instances.

I don’t mean to bash Norvig. His articles were great! And in-depth theory was way beyond scope. So this post is just my opportunity to fill in some theory gaps. We’ll do three things:

  1. Show formally that set cover is NP-hard.
  2. Prove the approximation guarantee of the greedy algorithm.
  3. Show another (very different) approximation algorithm based on linear programming.

Along the way I’ll argue that by knowing (or at least seeing) the details of these proofs, one can get a better sense of what features to look for in the set cover instance you’re trying to solve. We’ll also see how set cover depicts the broader themes of theoretical computer science.

NP-hardness

The first thing we should do is show that set cover is NP-hard. Intuitively what this means is that we can take some hard problem P and encode instances of P inside set cover problems. This idea is called a reduction, because solving problem P will “reduce” to solving set cover, and the method we use to encode instance of P as set cover problems will have a small amount of overhead. This is one way to say that set cover is “at least as hard as” P.

The hard problem we’ll reduce to set cover is called 3-satisfiability (3-SAT). In 3-SAT, the input is a formula whose variables are either true or false, and the formula is expressed as an OR of a bunch of clauses, each of which is an AND of three variables (or their negations). This is called 3-CNF form. A simple example:

\displaystyle (x \vee y \vee \neg z) \wedge (\neg x \vee w \vee y) \wedge (z \vee x \vee \neg w)

The goal of the algorithm is to decide whether there is an assignment to the variables which makes the formula true. 3-SAT is one of the most fundamental problems we believe to be hard and, roughly speaking, by reducing it to set cover we include set cover in a class called NP-complete, and if any one of these problems can be solved efficiently, then they all can (this is the famous P versus NP problem, and an efficient algorithm would imply P equals NP).

So a reduction would consist of the following: you give me a formula \varphi in 3-CNF form, and I have to produce (in a way that depends on \varphi!) a universe U and a choice of subsets S_i \subset U in such a way that

\varphi has a true assignment of variables if and only if the corresponding set cover problem has a cover using k sets.

In other words, I’m going to design a function f from 3-SAT instances to set cover instances, such that x is satisfiable if and only if f(x) has a set cover with k sets.

Why do I say it only for k sets? Well, if you can always answer this question then I claim you can find the minimum size of a set cover needed by doing a binary search for the smallest value of k. So finding the minimum size of a set cover reduces to the problem of telling if theres a set cover of size k.

Now let’s do the reduction from 3-SAT to set cover.

If you give me \varphi = C_1 \wedge C_2 \wedge \dots \wedge C_m where each C_i is a clause and the variables are denoted x_1, \dots, x_n, then I will choose as my universe U to be the set of all the clauses and indices of the variables (these are all just formal symbols). i.e.

\displaystyle U = \{ C_1, C_2, \dots, C_m, 1, 2, \dots, n \}

The first part of U will ensure I make all the clauses true, and the last part will ensure I don’t pick a variable to be both true and false at the same time.

To show how this works I have to pick my subsets. For each variable x_i, I’ll make two sets, one called S_{x_i} and one called S_{\neg x_i}. They will both contain i in addition to the clauses which they make true when the corresponding literal is true (by literal I just mean the variable or its negation). For example, if C_j uses the literal \neg x_7, then S_{\neg x_7} will contain C_j but S_{x_7} will not. Finally, I’ll set k = n, the number of variables.

Now to prove this reduction works I have to prove two things: if my starting formula has a satisfying assignment I have to show the set cover problem has a cover of size k. Indeed, take the sets S_{y} for all literals y that are set to true in a satisfying assignment. There can be at most n true literals since half are true and half are false, so there will be at most n sets, and these sets clearly cover all of U because every literal has to be satisfied by some literal or else the formula isn’t true.

The reverse direction is similar: if I have a set cover of size n, I need to use it to come up with a satisfying truth assignment for the original formula. But indeed, the sets that get chosen can’t include both a S_{x_i} and its negation set S_{\neg x_i}, because there are n of the elements \{1, 2, \dots, n \} \subset U, and each i is only in the two S_{x_i}, S_{\neg x_i}. Just by counting if I cover all the indices i, I already account for n sets! And finally, since I have covered all the clauses, the literals corresponding to the sets I chose give exactly a satisfying assignment.

Whew! So set cover is NP-hard because I encoded this logic problem 3-SAT within its rules. If we think 3-SAT is hard (and we do) then set cover must also be hard. So if we can’t hope to solve it exactly we should try to approximate the best solution.

The greedy approach

The method that Norvig uses in attacking the meta-regex golf problem is the greedy algorithm. The greedy algorithm is exactly what you’d expect: you maintain a list L of the subsets you’ve picked so far, and at each step you pick the set S_i that maximizes the number of new elements of U that aren’t already covered by the sets in L. In python pseudocode:

def greedySetCover(universe, sets):
   chosenSets = set()
   leftToCover = universe.copy()
   unchosenSets = sets

   covered = lambda s: leftToCover & s

   while universe != 0:
      if len(chosenSets) == len(sets):
         raise Exception("No set cover possible")
      
      nextSet = max(unchosenSets, key=lambda s: len(covered(s)))
      unchosenSets.remove(nextSet)
      chosenSets.add(nextSet)
      leftToCover -= nextSet
      
   return chosenSets

This is what theory has to say about the greedy algorithm:

Theorem: If it is possible to cover U by the sets in F = \{ S_1, \dots, S_n \}, then the greedy algorithm always produces a cover that at worst has size O(\log(n)) \textup{OPT}, where \textup{OPT} is the size of the smallest cover. Moreover, this is asymptotically the best any algorithm can do.

One simple fact we need from calculus is that the following sum is asymptotically the same as \log(n):

\displaystyle H(n) = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} = \log(n) + O(1)

Proof. [adapted from Wan] Let’s say the greedy algorithm picks sets T_1, T_2, \dots, T_k in that order. We’ll set up a little value system for the elements of U. Specifically, the value of each T_i is 1, and in step i we evenly distribute this unit value across all newly covered elements of T_i. So for T_1 each covered element gets value 1/|T_1|, and if T_2 covers four new elements, each gets a value of 1/4. One can think of this “value” as a price, or energy, or unit mass, or whatever. It’s just an accounting system (albeit a clever one) we use to make some inequalities clear later.

In general call the value v_x of element x \in U the value assigned to x at the step where it’s first covered. In particular, the number of sets chosen by the greedy algorithm k is just \sum_{x \in U} v_x. We’re just bunching back together the unit value we distributed for each step of the algorithm.

Now we want to compare the sets chosen by greedy to the optimal choice. Call a smallest set cover C_{\textup{OPT}}. Let’s stare at the following inequality.

\displaystyle \sum_{x \in U} v_x \leq \sum_{S \in C_{\textup{OPT}}} \sum_{x \in S} v_x

It’s true because each x counts for a v_x at most once in the left hand side, and in the right hand side the sets in C_{\textup{OPT}} must hit each x at least once but may hit some x more than once. Also remember the left hand side is equal to k.

Now we want to show that the inner sum on the right hand side, \sum_{x \in S} v_x, is at most H(|S|). This will in fact prove the entire theorem: because each set S_i has size at most n, the inequality above will turn into

\displaystyle k \leq |C_{\textup{OPT}}| H(|S|) \leq |C_{\textup{OPT}}| H(n)

And so k \leq \textup{OPT} \cdot O(\log(n)), which is the statement of the theorem.

So we want to show that \sum_{x \in S} v_x \leq H(|S|). For each j define \delta_j(S) to be the number of elements in S not covered in T_1, \cup \dots \cup T_j. Notice that \delta_{j-1}(S) - \delta_{j}(S) is the number of elements of S that are covered for the first time in step j. If we call t_S the smallest integer j for which \delta_j(S) = 0, we can count up the differences up to step t_S, we get

\sum_{x \in S} v_x = \sum_{i=1}^{t_S} (\delta_{i-1}(S) - \delta_i(S)) \cdot \frac{1}{T_i - (T_1 \cup \dots \cup T_{i-1})}

The rightmost term is just the cost assigned to the relevant elements at step i. Moreover, because T_i covers more new elements than S (by definition of the greedy algorithm), the fraction above is at most 1/\delta_{i-1}(S). The end is near. For brevity I’ll drop the (S) from \delta_j(S).

\displaystyle \begin{aligned} \sum_{x \in S} v_x & \leq \sum_{i=1}^{t_S} (\delta_{i-1} - \delta_i) \frac{1}{\delta_{i-1}} \\ & \leq \sum_{i=1}^{t_S} (\frac{1}{1 + \delta_i} + \frac{1}{2+\delta_i} \dots + \frac{1}{\delta_{i-1}}) \\ & = \sum_{i=1}^{t_S} H(\delta_{i-1}) - H(\delta_i) \\ &= H(\delta_0) - H(\delta_{t_S}) = H(|S|) \end{aligned}

And that proves the claim.

\square

I have three postscripts to this proof:

  1. This is basically the exact worst-case approximation that the greedy algorithm achieves. In fact, Petr Slavik proved in 1996 that the greedy gives you a set of size exactly (\log n - \log \log n + O(1)) \textup{OPT} in the worst case.
  2. This is also the best approximation that any set cover algorithm can achieve, provided that P is not NP. This result was basically known in 1994, but it wasn’t until 2013 and the use of some very sophisticated tools that the best possible bound was found with the smallest assumptions.
  3. In the proof we used that |S| \leq n to bound things, but if we knew that our sets S_i (i.e. subsets matched by a regex) had sizes bounded by, say, B, the same proof would show that the approximation factor is \log(B) instead of \log n. However, in order for that to be useful you need B to be a constant, or at least to grow more slowly than any polynomial in n, since e.g. \log(n^{0.1}) = 0.1 \log n. In fact, taking a second look at Norvig’s meta regex golf problem, some of his instances had this property! Which means the greedy algorithm gives a much better approximation ratio for certain meta regex golf problems than it does for the worst case general problem. This is one instance where knowing the proof of a theorem helps us understand how to specialize it to our interests.
norvig-table

Norvig’s frequency table for president meta-regex golf. The left side counts the size of each set (defined by a regex)

The linear programming approach

So we just said that you can’t possibly do better than the greedy algorithm for approximating set cover. There must be nothing left to say, job well done, right? Wrong! Our second analysis, based on linear programming, shows that instances with special features can have better approximation results.

In particular, if we’re guaranteed that each element x \in U occurs in at most B of the sets S_i, then the linear programming approach will give a B-approximation, i.e. a cover whose size is at worst larger than OPT by a multiplicative factor of B. In the case that B is constant, we can beat our earlier greedy algorithm.

The technique is now a classic one in optimization, called LP-relaxation (LP stands for linear programming). The idea is simple. Most optimization problems can be written as integer linear programs, that is there you have n variables x_1, \dots, x_n \in \{ 0, 1 \} and you want to maximize (or minimize) a linear function of the x_i subject to some linear constraints. The thing you’re trying to optimize is called the objective. While in general solving integer linear programs is NP-hard, we can relax the “integer” requirement to 0 \leq x_i \leq 1, or something similar. The resulting linear program, called the relaxed program, can be solved efficiently using the simplex algorithm or another more complicated method.

The output of solving the relaxed program is an assignment of real numbers for the x_i that optimizes the objective function. A key fact is that the solution to the relaxed linear program will be at least as good as the solution to the original integer program, because the optimal solution to the integer program is a valid candidate for the optimal solution to the linear program. Then the idea is that if we use some clever scheme to round the x_i to integers, we can measure how much this degrades the objective and prove that it doesn’t degrade too much when compared to the optimum of the relaxed program, which means it doesn’t degrade too much when compared to the optimum of the integer program as well.

If this sounds wishy washy and vague don’t worry, we’re about to make it super concrete for set cover.

We’ll make a binary variable x_i for each set S_i in the input, and x_i = 1 if and only if we include it in our proposed cover. Then the objective function we want to minimize is \sum_{i=1}^n x_i. If we call our elements X = \{ e_1, \dots, e_m \}, then we need to write down a linear constraint that says each element e_j is hit by at least one set in the proposed cover. These constraints have to depend on the sets S_i, but that’s not a problem. One good constraint for element e_j is

\displaystyle \sum_{i : e_j \in S_i} x_i \geq 1

In words, the only way that an e_j will not be covered is if all the sets containing it have their x_i = 0. And we need one of these constraints for each j. Putting it together, the integer linear program is

The integer program for set cover.

The integer program for set cover.

Once we understand this formulation of set cover, the relaxation is trivial. We just replace the last constraint with inequalities.

setcoverlp

For a given candidate assignment x to the x_i, call Z(x) the objective value (in this case \sum_i x_i). Now we can be more concrete about the guarantees of this relaxation method. Let \textup{OPT}_{\textup{IP}} be the optimal value of the integer program and x_{\textup{IP}} a corresponding assignment to x_i achieving the optimum. Likewise let \textup{OPT}_{\textup{LP}}, x_{\textup{LP}} be the optimal things for the linear relaxation. We will prove:

Theorem: There is a deterministic algorithm that rounds x_{\textup{LP}} to integer values x so that the objective value Z(x) \leq B \textup{OPT}_{\textup{IP}}, where B is the maximum number of sets that any element e_j occurs in. So this gives a B-approximation of set cover.

Proof. Let B be as described in the theorem, and call y = x_{\textup{LP}} to make the indexing notation easier. The rounding algorithm is to set x_i = 1 if y_i \geq 1/B and zero otherwise.

To prove the theorem we need to show two things hold about this new candidate solution x:

  1. The choice of all S_i for which x_i = 1 covers every element.
  2. The number of sets chosen (i.e. Z(x)) is at most B times more than \textup{OPT}_{\textup{LP}}.

Since \textup{OPT}_{\textup{LP}} \leq \textup{OPT}_{\textup{IP}}, so if we can prove number 2 we get Z(x) \leq B \textup{OPT}_{\textup{LP}} \leq B \textup{OPT}_{\textup{IP}}, which is the theorem.

So let’s prove 1. Fix any j and we’ll show that element e_j is covered by some set in the rounded solution. Call B_j the number of times element e_j occurs in the input sets. By definition B_j \leq B, so 1/B_j \geq 1/B. Recall y was the optimal solution to the relaxed linear program, and so it must be the case that the linear constraint for each e_j is satisfied: \sum_{i : e_j \in S_i} x_i \geq 1. We know that there are B_j terms and they sums to at least 1, so not all terms can be smaller than 1/B_j (otherwise they’d sum to something less than 1). In other words, some variable x_i in the sum is at least 1/B_j \geq 1/B, and so x_i is set to 1 in the rounded solution, corresponding to a set S_i that contains e_j. This finishes the proof of 1.

Now let’s prove 2. For each j, we know that for each x_i = 1, the corresponding variable y_i \geq 1/B. In particular 1 \leq y_i B. Now we can simply bound the sum.

\displaystyle \begin{aligned} Z(x) = \sum_i x_i &\leq \sum_i x_i (B y_i) \\ &\leq B \sum_{i} y_i \\ &= B \cdot \textup{OPT}_{\textup{LP}} \end{aligned}

The second inequality is true because some of the x_i are zero, but we can ignore them when we upper bound and just include all the y_i. This proves part 2 and the theorem.

\square

I’ve got some more postscripts to this proof:

  1. The proof works equally well when the sets are weighted, i.e. your cost for picking a set is not 1 for every set but depends on some arbitrarily given constants w_i \geq 0.
  2. We gave a deterministic algorithm rounding y to x, but one can get the same result (with high probability) using a randomized algorithm. The idea is to flip a coin with bias y_i roughly \log(n) times and set x_i = 1 if and only if the coin lands heads at least once. The guarantee is no better than what we proved, but for some other problems randomness can help you get approximations where we don’t know of any deterministic algorithms to get the same guarantees. I can’t think of any off the top of my head, but I’m pretty sure they’re out there.
  3. For step 1 we showed that at least one term in the inequality for e_j would be rounded up to 1, and this guaranteed we covered all the elements. A natural question is: why not also round up at most one term of each of these inequalities? It might be that in the worst case you don’t get a better guarantee, but it would be a quick extra heuristic you could use to post-process a rounded solution.
  4. Solving linear programs is slow. There are faster methods based on so-called “primal-dual” methods that use information about the dual of the linear program to construct a solution to the problem. Goemans and Williamson have a nice self-contained chapter on their website about this with a ton of applications.

Additional Reading

Goemans and Williamson have a large textbook called The Design of Approximation Algorithms. One problem is that this field is like a big heap of unrelated techniques, so it’s not like the book will build up some neat theoretical foundation that works for every problem. Rather, it’s messy and there are lots of details, but there are definitely diamonds in the rough, such as the problem of (and algorithms for) coloring 3-colorable graphs with “approximately 3″ colors, and the infamous unique games conjecture.

I wrote a post a while back giving conditions which, if a problem satisfies those conditions, the greedy algorithm will give a constant-factor approximation. This is much better than the worst case \log(n)-approximation we saw in this post. Moreover, I also wrote a post about matroids, which is a characterization of problems where the greedy algorithm is actually optimal.

Set cover is one of the main tools that IBM’s AntiVirus software uses to detect viruses. Similarly to the regex golf problem, they find a set of strings that occurs source code in some viruses but not (usually) in good programs. Then they look for a small set of strings that covers all the viruses, and their virus scan just has to search binaries for those strings. Hopefully the size of your set cover is really small compared to the number of viruses you want to protect against. I can’t find a reference that details this, but that is understandable because it is proprietary software.

Until next time!

Finding the majority element of a stream

Problem: Given a massive data stream of n values in \{ 1, 2, \dots, m \} and the guarantee that one value occurs more than n/2 times in the stream, determine exactly which value does so.

Solution: (in Python)

def majority(stream):
   held = next(stream)
   counter = 1

   for item in stream:
      if item == held:
         counter += 1
      elif counter == 0:
         held = item
         counter = 1
      else:
         counter -= 1

   return held

Discussion: Let’s prove correctness. Say that s is the unknown value that occurs more than n/2 times. The idea of the algorithm is that if you could pair up elements of your stream so that distinct values are paired up, and then you “kill” these pairs, then s will always survive. The way this algorithm pairs up the values is by holding onto the most recent value that has no pair (implicitly, by keeping a count how many copies of that value you saw). Then when you come across a new element, you decrement the counter and implicitly account for one new pair.

Let’s analyze the complexity of the algorithm. Clearly the algorithm only uses a single pass through the data. Next, if the stream has size n, then this algorithm uses O(\log(n) + \log(m)) space. Indeed, if the stream entirely consists of a single value (say, a stream of all 1’s) then the counter will be n at the end, which takes \log(n) bits to store. On the other hand, if there are m possible values then storing the largest requires \log(m) bits.

Finally, the guarantee that one value occurs more than n/2 times is necessary. If it is not the case the algorithm could output anything (including the most infrequent element!). And moreover, if we don’t have this guarantee then every algorithm that solves the problem must use at least \Omega(n) space in the worst case. In particular, say that m=n, and the first n/2 items are all distinct and the last n/2 items are all the same one, the majority value s. If you do not know s in advance, then you must keep at least one bit of information to know which symbols occurred in the first half of the stream because any of them could be s. So the guarantee allows us to bypass that barrier.

This algorithm can be generalized to detect k items with frequency above some threshold n/(k+1) using space O(k \log n). The idea is to keep k counters instead of one, adding new elements when any counter is zero. When you see an element not being tracked by your k counters (which are all positive), you decrement all the counters by 1. This is like a k-to-one matching rather than a pairing.

When Greedy Algorithms are Perfect: the Matroid

Greedy algorithms are by far one of the easiest and most well-understood algorithmic techniques. There is a wealth of variations, but at its core the greedy algorithm optimizes something using the natural rule, “pick what looks best” at any step. So a greedy routing algorithm would say to a routing problem: “You want to visit all these locations with minimum travel time? Let’s start by going to the closest one. And from there to the next closest one. And so on.”

Because greedy algorithms are so simple, researchers have naturally made a big effort to understand their performance. Under what conditions will they actually solve the problem we’re trying to solve, or at least get close? In a previous post we gave some easy-to-state conditions under which greedy gives a good approximation, but the obvious question remains: can we characterize when greedy algorithms give an optimal solution to a problem?

The answer is yes, and the framework that enables us to do this is called a matroid. That is, if we can phrase the problem we’re trying to solve as a matroid, then the greedy algorithm is guaranteed to be optimal. Let’s start with an example when greedy is provably optimal: the minimum spanning tree problem. Throughout the article we’ll assume the reader is familiar with the very basics of linear algebra and graph theory (though we’ll remind ourselves what a minimum spanning tree is shortly). For a refresher, this blog has primers on both subjects. But first, some history.

History

Matroids were first introduced by Hassler Whitney in 1935, and independently discovered a little later by B.L. van der Waerden (a big name in combinatorics). They were both interested in devising a general description of “independence,” the properties of which are strikingly similar when specified in linear algebra and graph theory. Since then the study of matroids has blossomed into a large and beautiful theory, one part of which is the characterization of the greedy algorithm: greedy is optimal on a problem if and only if the problem can be represented as a matroid. Mathematicians have also characterized which matroids can be modeled as spanning trees of graphs (we will see this momentarily). As such, matroids have become a standard topic in the theory and practice of algorithms.

Minimum Spanning Trees

It is often natural in an undirected graph G = (V,E) to find a connected subset of edges that touch every vertex. As an example, if you’re working on a power network you might want to identify a “backbone” of the network so that you can use the backbone to cheaply travel from any node to any other node. Similarly, in a routing network (like the internet) it costs a lot of money to lay down cable, it’s in the interest of the internet service providers to design analogous backbones into their infrastructure.

A minimal subset of edges in a backbone like this is guaranteed to form a tree. This is simply because if you have a cycle in your subgraph then removing any edge on that cycle doesn’t break connectivity or the fact that you can get from any vertex to any other (and trees are the maximal subgraphs without cycles). As such, these “backbones” are called spanning trees. “Span” here means that you can get from any vertex to any other vertex, and it suggests the connection to linear algebra that we’ll describe later, and it’s a simple property of a tree that there is a unique path between any two vertices in the tree.

An example of a spanning tree

An example of a spanning tree

When your edges e \in E have nonnegative weights w_e \in \mathbb{R}^{\geq 0}, we can further ask to find a minimum cost spanning tree. The cost of a spanning tree T is just the sum of its edges, and it’s important enough of a definition to offset.

Definition: A minimum spanning tree T of a weighted graph G (with weights w_e \geq 0 for e \in E) is a spanning tree which minimizes the quantity

w(T) = \sum_{e \in T} w_e

There are a lot of algorithms to find minimal spanning trees, but one that will lead us to matroids is Kruskal’s algorithm. It’s quite simple. We’ll maintain a forest F in G, which is just a subgraph consisting of a bunch of trees that may or may not be connected. At the beginning F is just all the vertices with no edges. And then at each step we add to F the edge e whose weight is smallest and also does not introduce any cycles into F. If the input graph G is connected then this will always produce a minimal spanning tree.

Theorem: Kruskal’s algorithm produces a minimal spanning tree of a connected graph.

Proof. Call F_t the forest produced at step t of the algorithm. Then F_0 is the set of all vertices of G and F_{n-1} is the final forest output by Kruskal’s (as a quick exercise, prove all spanning trees on n vertices have n-1 edges, so we will stop after n-1 rounds). It’s clear that F_{n-1} is a tree because the algorithm guarantees no F_i will have a cycle. And any tree with n-1 edges is necessarily a spanning tree, because if some vertex were left out then there would be n-1 edges on a subgraph of n-1 vertices, necessarily causing a cycle somewhere in that subgraph.

Now we’ll prove that F_{n-1} has minimal cost. We’ll prove this in a similar manner to the general proof for matroids. Indeed, say you had a tree T whose cost is strictly less than that of F_{n-1} (we can also suppose that T is minimal, but this is not necessary). Pick the minimal weight edge e \in T that is not in F_{n-1}. Adding e to F_{n-1} introduces a unique cycle C in F_{n-1}. This cycle has some strange properties. First, e has the highest cost of any edge on C. For otherwise, Kruskal’s algorithm would have chosen it before the heavier weight edges. Second, there is another edge in C that’s not in T (because T was a tree it can’t have the entire cycle). Call such an edge e'. Now we can remove e' from F_{n-1} and add e. This can only increase the total cost of F_{n-1}, but this transformation produces a tree with one more edge in common with T than before. This contradicts that T had strictly lower weight than F_{n-1}, because repeating the process we described would eventually transform F_{n-1} into T exactly, while only increasing the total cost.

\square

Just to recap, we defined sets of edges to be “good” if they did not contain a cycle, and a spanning tree is a maximal set of edges with this property. In this scenario, the greedy algorithm performed optimally at finding a spanning tree with minimal total cost.

Columns of Matrices

Now let’s consider a different kind of problem. Say I give you a matrix like this one:

\displaystyle A = \begin{pmatrix} 2 & 0 & 1 & -1 & 0 \\ 0 & -4 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 7 \end{pmatrix}

In the standard interpretation of linear algebra, this matrix represents a linear function f from one vector space V to another W, with the basis (v_1, \dots, v_5) of V being represented by columns and the basis (w_1, w_2, w_3) of W being represented by the rows. Column j tells you how to write f(v_j) as a linear combination of the w_i, and in so doing uniquely defines f.

Now one thing we want to calculate is the rank of this matrix. That is, what is the dimension of the image of V under f? By linear algebraic arguments we know that this is equivalent to asking “how many linearly independent columns of A can we find”? An interesting consequence is that if you have two sets of columns that are both linearly independent and maximally so (adding any other column to either set would necessarily introduce a dependence in that set), then these two sets have the same size. This is part of why the rank of a matrix is well-defined.

If we were to give the columns of A costs, then we could ask about finding the minimal-cost maximally-independent column set. It sounds like a mouthful, but it’s exactly the same idea as with spanning trees: we want a set of vectors that spans the whole column space of A, but contains no “cycles” (linearly dependent combinations), and we want the cheapest such set.

So we have two kinds of “independence systems” that seem to be related. One interesting question we can ask is whether these kinds of independence systems are “the same” in a reasonable way. Hardcore readers of this blog may see the connection quite quickly. For any graph G = (V,E), there is a natural linear map from E to V, so that a linear dependence among the columns (edges) corresponds to a cycle in G. This map is called the incidence matrix by combinatorialists and the first boundary map by topologists.

The map is easy to construct: for each edge e = (v_i,v_j) you add a column with a 1 in the j-th row and a -1 in the i-th row. Then taking a sum of edges gives you zero if and only if the edges form a cycle. So we can think of a set of edges as “independent” if they don’t contain a cycle. It’s a little bit less general than independence over \mathbb{R}, but you can make it exactly the same kind of independence if you change your field from real numbers to \mathbb{Z}/2\mathbb{Z}. We won’t do this because it will detract from our end goal (to analyze greedy algorithms in realistic settings), but for further reading this survey of Oxley assumes that perspective.

So with the recognition of how similar these notions of independence are, we are ready to define matroids.

The Matroid

So far we’ve seen two kinds of independence: “sets of edges with no cycles” (also called forests) and “sets of linearly independent vectors.” Both of these share two trivial properties: there are always nonempty independent sets, and every subset of an independent set is independent. We will call any family of subsets with this property an independence system.

Definition: Let X be a finite set. An independence system over X is a family \mathscr{I} of subsets of X with the following two properties.

  1. \mathscr{I} is nonempty.
  2. If I \in \mathscr{I}, then so is every subset of I.

This is too general to characterize greedy algorithms, so we need one more property shared by our examples. There are a few things we do, but here’s one nice property that turns out to be enough.

Definition: A matroid M = (X, \mathscr{I}) is a set X and an independence system \mathscr{I} over X with the following property:

If A, B are in \mathscr{I} with |A| = |B| + 1, then there is an element x \in A \setminus B such that B \cup \{ a \} \in \mathscr{I}.

In other words, this property says if I have an independent set that is not maximally independent, I can grow the set by adding some suitably-chosen element from a larger independent set. We’ll call this the extension property. For a warmup exercise, let’s prove that the extension property is equivalent to the following (assuming the other properties of a matroid):

For every subset Y \subset X, all maximal independent sets contained in Y have equal size.

Proof. For one direction, if you have two maximal sets A, B \subset Y \subset X that are not the same size (say A is bigger), then you can take any subset of A whose size is exactly |B| + 1, and use the extension property to make B larger, a contradiction. For the other direction, say that I know all maximal independent sets of any Y \subset X have the same size, and you give me A, B \subset X. I need to find an a \in A \setminus B that I can add to B and keep it independent. What I do is take the subset Y = A \cup B. Now the sizes of A, B don’t change, but B can’t be maximal inside Y because it’s smaller than A (A might not be maximal either, but it’s still independent). And the only way to extend B is by adding something from A, as desired.

\square

So we can use the extension property and the cardinality property interchangeably when talking about matroids. Continuing to connect matroid language to linear algebra and graph theory, the maximal independent sets of a matroid are called bases, the size of any basis is the rank of the matroid, and the minimal dependent sets are called circuits. In fact, you can characterize matroids in terms of the properties of their circuits, which are dual to the properties of bases (and hence all independent sets) in a very concrete sense.

But while you could spend all day characterizing the many kinds of matroids and comatroids out there, we are still faced with the task of seeing how the greedy algorithm performs on a matroid. That is, suppose that your matroid M = (X, \mathscr{I}) has a nonnegative real number w(x) associated with each x \in X. And suppose we had a black-box function to determine if a given set S \subset X is independent. Then the greedy algorithm maintains a set B, and at every step adds a minimum weight element that maintains the independence of B. If we measure the cost of a subset by the sum of the weights of its elements, then the question is whether the greedy algorithm finds a minimum weight basis of the matroid.

The answer is even better than yes. In fact, the answer is that the greedy algorithm performs perfectly if and only if the problem is a matroid! More rigorously,

Theorem: Suppose that M = (X, \mathscr{I}) is an independence system, and that we have a black-box algorithm to determine whether a given set is independent. Define the greedy algorithm to iteratively adds the cheapest element of X that maintains independence. Then the greedy algorithm produces a maximally independent set S of minimal cost for every nonnegative cost function on X, if and only if M is a matroid.

It’s clear that the algorithm will produce a set that is maximally independent. The only question is whether what it produces has minimum weight among all maximally independent sets. We’ll break the theorem into the two directions of the “if and only if”:

Part 1: If M is a matroid, then greedy works perfectly no matter the cost function.
Part 2: If greedy works perfectly for every cost function, then M is a matroid.

Proof of Part 1.

Call the cost function w : X \to \mathbb{R}^{\geq 0}, and suppose that the greedy algorithm picks elements B = \{ x_1, x_2, \dots, x_r \} (in that order). It’s easy to see that w(x_1) \leq w(x_2) \leq \dots \leq w(x_r). Now if you give me any list of r independent elements y_1, y_2, \dots, y_r \in X that has w(y_1) \leq \dots \leq w(y_r), I claim that w(x_i) \leq w(y_i) for all i. This proves what we want, because if there were a basis of size r with smaller weight, sorting its elements by weight would give a list contradicting this claim.

To prove the claim, suppose to the contrary that it were false, and for some k we have w(x_k) > w(y_k). Moreover, pick the smallest k for which this is true. Note k > 1, and so we can look at the special sets S = \{ x_1, \dots, x_{k-1} \} and T = \{ y_1, \dots, y_k \}. Now |T| = |S|+1, so by the matroid property there is some j between 1 and r so that S \cup \{ y_j \} is an independent set (and y_j is not in S). But then w(y_j) \leq w(y_k) < w(x_k), and so the greedy algorithm would have picked y_j before it picks x_k (and the strict inequality means they’re different elements). This contradicts how the greedy algorithm runs, and hence proves the claim.

Proof of Part 2.

We’ll prove this contrapositively as follows. Suppose we have our independence system and it doesn’t satisfy the last matroid condition. Then we’ll construct a special weight function that causes the greedy algorithm to fail. So let A,B be independent sets with |A| = |B| + 1, but for every a \in A \setminus B adding a to B never gives you an independent set.

Now what we’ll do is define our weight function so that the greedy algorithm picks the elements we want in the order we want (roughly). In particular, we’ll assign all elements of A \cap B a tiny weight we’ll call w_1. For elements of B - A we’ll use w_2, and for A - B we’ll use w_3, with w_4 for everything else. In a more compact notation:

CodeCogsEqn

We need two things for this weight function to screw up the greedy algorithm. The first is that w_1 < w_2 < w_3 < w_4, so that greedy picks the elements in the order we want. Note that this means it’ll first pick all of A \cap B, and then all of B - A, and by assumption it won’t be able to pick anything from A - B, but since B is assumed to be non-maximal, we have to pick at least one element from X - (A \cup B) and pay w_4 for it.

So the second thing we want is that the cost of doing greedy is worse than picking any maximally independent set that contains A (and we know that there has to be some maximal independent set containing A). In other words, if we call m the size of a maximally independent set, we want

\displaystyle |A \cap B| w_1 + |B-A|w_2 + (m - |B|)w_4 > |A \cap B|w_1 + |A-B|w_3 + (m-|A|)w_4

This can be rearranged (using the fact that |A| = |B|+1) to

\displaystyle w_4 > |A-B|w_3 - |B-A|w_2

The point here is that the greedy picks too many elements of weight w_4, since if we were to start by taking all of A (instead of all of B), then we could get by with one fewer. That might not be optimal, but it’s better than greedy and that’s enough for the proof.

So we just need to make w_4 large enough to make this inequality hold, while still maintaining w_2 < w_3. There are probably many ways to do this, and here’s one. Pick some 0 < \varepsilon < 1, and set

settings

It’s trivial that w_1 < w_2 and w_3 < w_4. For the rest we need some observations. First, the fact that |A-B| = |B-A| + 1 implies that w_2 < w_3. Second, both |A-B| and |B-A| are nonempty, since otherwise the second property of independence systems would contradict our assumption that augmenting B with elements of A breaks independence. Using this, we can divide by these quantities to get

\displaystyle w_4 = 2 > 1 = \frac{|A-B|(1 + \varepsilon)}{|A-B|} - \frac{|B-A|\varepsilon}{|B-A|}

This proves the claim and finishes the proof.

\square

As a side note, we proved everything here with respect to minimizing the sum of the weights, but one can prove an identical theorem for maximization. The only part that’s really different is picking the clever weight function in part 2. In fact, you can convert between the two by defining a new weight function that subtracts the old weights from some fixed number N that is larger than any of the original weights. So these two problems really are the same thing.

This is pretty amazing! So if you can prove your problem is a matroid then you have an awesome algorithm automatically. And if you run the greedy algorithm for fun and it seems like it works all the time, then that may be hinting that your problem is a matroid. This is one of the best situations one could possibly hope for.

But as usual, there are a few caveats to consider. They are both related to efficiency. The first is the black box algorithm for determining if a set is independent. In a problem like minimum spanning tree or finding independent columns of a matrix, there are polynomial time algorithms for determining independence. These two can both be done, for example, with Gaussian elimination. But there’s nothing to stop our favorite matroid from requiring an exponential amount of time to check if a set is independent. This makes greedy all but useless, since we need to check for independence many times in every round.

Another, perhaps subtler, issue is that the size of the ground set X might be exponentially larger than the rank of the matroid. In other words, at every step our greedy algorithm needs to find a new element to add to the set it’s building up. But there could be such a huge ocean of candidates, all but a few of which break independence. In practice an algorithm might be working with X implicitly, so we could still hope to solve the problem if we had enough knowledge to speed up the search for a new element.

There are still other concerns. For example, a naive approach to implementing greedy takes quadratic time, since you may have to look through every element of X to find the minimum-cost guy to add. What if you just have to have faster runtime than O(n^2)? You can still be interested in finding more efficient algorithms that still perform perfectly, and to the best of my knowledge there’s nothing that says that greedy is the only exact algorithm for your favorite matroid. And then there are models where you don’t have direct/random access to the input, and lots of other ways that you can improve on greedy. But those stories are for another time.

Until then!

Methods of Proof — Induction

In this final post on the basic four methods of proof (but perhaps not our last post on proof methods), we consider the proof by induction.

Proving Statements About All Natural Numbers

Induction comes in many flavors, but the goal never changes. We use induction when we want to prove something is true about all natural numbers. These statements will look something like this:

For all natural numbers n, 1 + 2 + \dots + n = n(n+1)/2.

Of course, there are many ways to prove this fact, but induction relies on one key idea: if we know the statement is true for some specific number n, then that gives us information about whether the statement is true for n+1. If this isn’t true about the problem, then proof by induction is hopeless.

Let’s see how we can apply it to the italicized statement above (though we haven’t yet said what induction is, this example will pave the way for a formal description of the technique). The first thing we notice is that indeed, if we know something about the first n numbers, then we can just add n+1 to it to get the sum of the first n+1 numbers. That is,

\displaystyle 1 + \dots + n + (n+1) = (1 + \dots + n) + (n+1)

Reiterating our key point, this is how we know induction is a valid strategy: the statement written for a fixed n translates naturally into the statement about n+1. Now suppose we know the theorem is true for n. Then we can rewrite the above sum as follows:

\displaystyle 1 + \dots + n + (n+1) = \frac{n(n+1)}{2} + (n+1)

With some algebra, we can write the left-hand side as a single fraction:

\displaystyle 1 + \dots + (n+1) = \frac{n(n+1) + 2(n+1)}{2}

and factoring the numerator gives

\displaystyle 1 + \dots + (n+1) = \frac{(n+1)(n+2)}{2}

Indeed, this is precisely what we’re looking for! It’s what happens when you replace n by n+1 in the original statement of the problem.

At this point we’re very close to being finished. We proved that if the statement is true for n, then it’s true for n+1. And by the same reasoning, it will be true for n+2, n+3, and all numbers after n. But this raises the obvious question: what’s the smallest number that it’s true for?

For this problem, it’s easy to see the answer is n=1. A mathematician would say: the statement is trivially true for n=1 (here trivial means there is no thinking required to show it: you just plug in n=1 and verify). And so by our reasoning, the statement is true for n=2, n=3, and so on forever. This is the spirit of mathematical induction.

Formal Nonsense

Now that we’ve got a taste of how to use induction in practice, let’s formally write down the rules for induction. Let’s have a statement which depends on a number n, and call it p(n). This is written as a function because it actually is one (naively). It’s a function from the set of natural numbers to the set of all mathematical statements. In our example above, p(n) was the statement that the equality 1 + \dots + n = n(n+1)/2 holds.

We can plug in various numbers into this function, such as p(1) being the statement “1 = 1(1+1)/2 holds,” or p(n+1) being “1 + \dots + (n+1) = (n+1)(n+1+1)/2 holds.”

The basic recipe for induction is then very simple. Say you want to prove that p(n) is true for all n \geq 1. First you prove that p(1) is true (this is called the base case), and then you prove the implication p(n) \to p(n+1) for an arbitrary n. Each of these pieces can be proved with any method one wishes (direct, contrapositive, contradiction, etc.). Once they are proven, we get that p(n) is true for all n automatically.

Indeed, we can even prove it. A rigorous proof requires a bit of extra knowledge, but we can easily understand the argument: it’s just a proof by contradiction. Here’s a quick sketch. Let X be the set of all natural numbers n for which p(n) is false. Suppose to the contrary that X is not empty. Every nonempty set of natural numbers has a smallest element, so let’s call m the smallest number for which p(m) is false. Now m-1 < m, so p(m-1) must be true. But we proved that whenever p(n) is true then so is p(n+1), so p(m-1 + 1) = p(m) is true, a contradiction.

Besides practicing proof by induction, that’s all there is to it. One more caveat is that the base case can be some number other than 1. For instance, it is true that n! > 2^n, but only for n \geq 4. In this case, we prove p(4) is true, and p(n) \to p(n+1) with the extra assumption that n \geq 4. The same inductive result applies.

Here are some exercises the reader can practice with, and afterward we will explore some variants of induction.

  1. Prove that n! > 2^n for all n \geq 4.
  2. Prove that for all n \geq 1 the following equality holds: 1/(1 \cdot 2) + 1/(2 \cdot 3) + \dots + 1/(n \cdot (n+1)) = n/(n+1).
  3. Prove that for every natural number n, a set of n elements has 2^n subsets (including the empty subset).

This last exercise gives a hint that induction can prove more than arithmetic formulas. Indeed, if we have any way to associate a mathematical object with a number, we can potentially use induction to prove things about those objects. Unfortunately, we don’t have any mathematical objects to work with (except for sets and functions), and so we will stick primarily to proving facts about numbers.

One interesting observation about proof by induction is very relevant to programmers: it’s just recursion. That is, if we want to prove a statement p(n), it suffices to prove it for p(n-1) and do some “extra computation” to arrive at the statement for p(n). And of course, we want to make sure the recursion terminates, so we add in the known result for p(1).

Variations on Induction, and Connections to Dynamic Programming

The first variation of induction is simultaneous induction on multiple quantities. That is, we can formulate a statement p(n,m) which depends on two natural numbers independently of one another. The base case is a bit trickier, but paralleling the above remark about recursion, double-induction follows the same pattern as a two-dimensional dynamic programming algorithm. The base cases would consist of all p(1,m) and all p(n,1), and the inductive step to get p(n,m) requires p(n-1,m) and p(n,m-1) (and potentially p(n-1, m-1) or others; it depends on the problem).

Unfortunately, natural instances where double induction is useful (or anywhere close to necessary) are rare. Here is an example of a (tricky) but elementary example. Call

\displaystyle C(m,n) = \frac{(2m)!(2n)!}{m!n!(m+n)!},

where the exclamation point denotes the factorial function. We will outline a proof that C(m,n) is always an integer for all m, n \geq 0. If we look at the base cases, C(0,n), C(m,0) (recalling that 0! = 1), we get (2n!)/(n! n!), and this happens to be in the form of a binomial coefficient (here, the number of ways to choose n! objects from a collection of (2n)! objects), and binomial coefficients are known to be integers. Now the inductive step relies on the fact that C(m,n-1) and C(m+1, n-1) are both integers. If this is true then

\displaystyle C(m,n) = 4C(m,n-1) - C(m+1, n-1),

which is obviously again an integer.

If we write these values in a table, we can see how the induction progresses in a “dynamic programming” fashion:

dynamic-programming-table

In order to fill the values in the next n column (prove the statement for those values of n), we need to fill the entire n-1 column (for indeed, we rely on the inductive hypothesis for both the m+1 and m row). But since our base case was the entire n=0 column, we can fill the entire table. In fact, we have just described a dynamic programming algorithm for computing the value of C(m,n) in mn steps. The correctness of the algorithm is indeed an inductive proof of the theorem.

Perhaps uninterestingly, this is asymptotically slower than the naive algorithm of computing C(m,n) directly by computing (2n)!, (2m)!, (n+m)! directly; this would take a linear number of steps in n, assuming n > m. In passing, this author wonders if, when the numbers are really large, the lack of division and multiplication in the dynamic program (multiplying by 4 using bit shifting instead) would overtake the naive algorithm. But C(m,n) is certainly not interesting enough in its own right for anyone to care :)

At this point, we have noticed that we sometimes use induction and assume that many smaller instances of the statement are true. Indeed, why not inductively assume that the statement holds for all smaller n. This would certainly give the prover more tools to work with. Indeed, this technique is sometimes called strong induction, in the sense that we assume a stronger inductive hypothesis when we’re trying to prove p(n+1). It may not be entirely obvious (especially to one well versed in the minutiae of formal logic) that this is actually equivalent to normal induction, but it is. In fact, the concept of “strong” induction is entirely pedagogical in nature. Most working mathematicians will not mention the difference in their proofs.

The last variant we’ll mention about induction is that of transfinite induction. The concept is that if you have any set X which is well-ordered (essentially this means: allowing one to prove induction applies as we did earlier in the post), then we can perform induction its elements. In this way, we can “parameterize” statements by elements of an arbitrary well-ordered set, so that instead of p(n) being a function from natural numbers to mathematical statements, it’s a function from X to mathematical statements. One somewhat common example of when X is something besides natural numbers is when we use the so-called cardinal numbers. We have already seen two distinct infinite cardinal numbers in this series: the cardinality of the integers and the cardinality of the real numbers (indeed, “cardinal number” just means a number which is the cardinality of a set). It turns out that there are many more kinds of cardinal numbers, and you can do induction on them, but this rarely shows up outside of mathematical logic.

And of course, we should close this post on an example of when induction goes wrong. For this we point the reader to our proof gallery, and the false proof that all horses are the same color. It’s quite an amusing joke, and hopefully it will stimulate the reader’s mind to recognize the pitfalls that can occur in a proof by induction.

So those are the basic four proof techniques! Fortunately for the reader, pretty much all proofs presented on this blog follow one of these four techniques. I imagine many of my readers skip over the proofs entirely (if only I could put proofs in animated gifs, with claims illustrated by grumpy cats!). So hopefully, if you have been intimidated or confused by the structure of the proofs on this blog, this will aid you in your future mathematical endeavors.  Butchering an old phrase for the sake of a bad pun, the eating of the pudding is in the proof. :)

Until next time!