Tag Archives: computational learning theory

Occam’s Razor and PAC-learning

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So far our discussion of learning theory has been seeing the definition of PAC-learning, tinkering with it, and seeing simple examples of learnable concept classes. We’ve said that our real interest is in proving big theorems about what big classes of problems can and can’t be learned. One major tool for doing this with PAC is the concept of VC-dimension, but to set the stage we’re going to prove a simpler theorem that gives a nice picture of PAC-learning when your hypothesis class is small. In short, the theorem we’ll prove says that if you have a finite set of hypotheses to work with, and you can always find a hypothesis that’s consistent with the data you’ve seen, then you can learn efficiently. It’s obvious, but we want to quantify exactly how much data you need to ensure low error. This will also give us some concrete mathematical justification for philosophical claims about simplicity, and the theorems won’t change much when we generalize to VC-dimension in a future post.

The Chernoff bound

One tool we will need in this post, which shows up all across learning theory, is the Chernoff-Hoeffding bound. We covered this famous inequality in detail previously on this blog, but the part of that post we need is the following theorem that says, informally, that if you average a bunch of bounded random variables, then the probability this average random variable deviates from its expectation is exponentially small in the amount of deviation. Here’s the slightly simplified version we’ll use:

Theorem: Let $ X_1, \dots, X_m$ be independent random variables whose values are in the range $ [0,1]$. Call $ \mu_i = \mathbf{E}[X_i]$, $ X = \sum_i X_i$, and $ \mu = \mathbf{E}[X] = \sum_i \mu_i$. Then for all $ t > 0$,

$ \displaystyle \Pr(|X-\mu| > t) \leq 2e^{-2t^2 / m}$

One nice thing about the Chernoff bound is that it doesn’t matter how the variables are distributed. This is important because in PAC we need guarantees that hold for any distribution generating data. Indeed, in our case the random variables above will be individual examples drawn from the distribution generating the data. We’ll be estimating the probability that our hypothesis has error deviating more than $ \varepsilon$, and we’ll want to bound this by $ \delta$, as in the definition of PAC-learning. Since the amount of deviation (error) and the number of samples ($ m$) both occur in the exponent, the trick is in balancing the two values to get what we want.

Realizability and finite hypothesis classes

Let’s recall the PAC model once more. We have a distribution $ D$ generating labeled examples $ (x, c(x))$, where $ c$ is an unknown function coming from some concept class $ C$. Our algorithm can draw a polynomial number of these examples, and it must produce a hypothesis $ h$ from some hypothesis class $ H$ (which may or may not contain $ c$). The guarantee we need is that, for any $ \delta, \varepsilon > 0$, the algorithm produces a hypothesis whose error on $ D$ is at most $ \varepsilon$, and this event happens with probability at least $ 1-\delta$. All of these probabilities are taken over the randomness in the algorithm’s choices and the distribution $ D$, and it has to work no matter what the distribution $ D$ is.

Let’s introduce some simplifications. First, we’ll assume that the hypothesis and concept classes $ H$ and $ C$ are finite. Second, we’ll assume that $ C \subset H$, so that you can actually hope to find a hypothesis of zero error. This is called realizability. Later we’ll relax these first two assumptions, but they make the analysis a bit cleaner. Finally, we’ll assume that we have an algorithm which, when given labeled examples, can find in polynomial time a hypothesis $ h \in H$ that is consistent with every example.

These assumptions give a trivial learning algorithm: draw a bunch of examples and output any consistent hypothesis. The question is, how many examples do we need to guarantee that the hypothesis we find has the prescribed generalization error? It will certainly grow with $ 1 / \varepsilon$, but we need to ensure it will only grow polynomially fast in this parameter. Indeed, realizability is such a strong assumption that we can prove a polynomial bound using even more basic probability theory than the Chernoff bound.

Theorem: A algorithm that efficiently finds a consistent hypothesis will PAC-learn any finite concept class provided it has at least $ m$ samples, where

$ \displaystyle m \geq \frac{1}{\varepsilon} \left ( \log |H| + \log \left ( \frac{1}{\delta} \right ) \right )$

Proof. All we need to do is bound the probability that a bad hypothesis (one with error more than $ \varepsilon$) is consistent with the given data. Now fix $ D, c, \delta, \varepsilon$, and draw $ m$ examples and let $ h$ be any hypothesis that is consistent with the drawn examples. Suppose that the bad thing happens, that $ \Pr_D(h(x) \neq c(x)) > \varepsilon$.

Because the examples are all drawn independently from $ D$, the chance that all $ m$ examples are consistent with $ h$ is

$ \displaystyle (1 – \Pr_{x \sim D}(h(x) \neq c(x)))^m < (1 – \varepsilon)^m$

What we’re saying here is, the probability that a specific bad hypothesis is actually consistent with your drawn examples is exponentially small in the error tolerance. So if we apply the union bound, the probability that some hypothesis you could produce is bad is at most $ (1 – \varepsilon)^m S$, where $ S$ is the number of hypotheses the algorithm might produce.

A crude upper bound on the number of hypotheses you could produce is just the total number of hypotheses, $ |H|$. Even cruder, let’s use the inequality $ (1 – x) < e^{-x}$ to give the bound

$ \displaystyle (1 – \varepsilon)^m |H| < e^{-\varepsilon m} |H|$

Now we want to make sure that this probability, the probability of choosing a high-error (yet consistent) hypothesis, is at most $ \delta$. So we can set the above quantity less than $ \delta$ and solve for $ m$:

$ \displaystyle e^{-\varepsilon m} |H| \leq \delta$

Taking logs and solving for $ m$ gives the desired bound.

$ \square$

An obvious objection is: what if you aren’t working with a hypothesis class where you can guarantee that you’ll find a consistent hypothesis? Well, in that case we’ll need to inspect the definition of PAC again and reevaluate our measures of error. It turns out we’ll get a similar theorem as above, but with the stipulation that we’re only achieving error within epsilon of the error of the best available hypothesis.

But before we go on, this theorem has some deep philosophical interpretations. In particular, suppose that, before drawing your data, you could choose to work with one of two finite hypothesis classes $ H_1, H_2$, with $ |H_1| > |H_2|$. If you can find a consistent hypothesis no matter which hypothesis class you use, then this theorem says that your generalization guarantees are much stronger if you start with the smaller hypothesis class.

In other words, all else being equal, the smaller set of hypotheses is better. For this reason, the theorem is sometimes called the “Occam’s Razor” theorem. We’ll see a generalization of this theorem in the next section.

Unrealizability and an extra epsilon

Now suppose that $H$ doesn’t contain any hypotheses with error less than $ \varepsilon$. What can we hope to do in this case? One thing is that we can hope to find a hypothesis whose error is within $ \varepsilon$ of the minimal error of any hypothesis in $ H$. Moreover, we might not have any consistent hypotheses for some data samples! So rather than require an algorithm to produce an $ h \in H$ that is perfectly consistent with the data, we just need it to produce a hypothesis that has minimal empirical error, in the sense that it is as close to consistent as the best hypothesis of $ h$ on the data you happened to draw. It seems like such a strategy would find you a hypothesis that’s close to the best one in $ H$, but we need to prove it and determine how many samples we need to draw to succeed.

So let’s make some definitions to codify this. For a given hypothesis, call $ \textup{err}(h)$ the true error of $ h$ on the distribution $ D$. Our assumption is that there may be no hypotheses in $ H$ with $ \textup{err}(h) = 0$. Next we’ll call the empirical error $ \hat{\textup{err}}(h)$.

Definition: We say a concept class $ C$ is agnostically learnable using the hypothesis class $ H$ if for all $ c \in C$ and all distributions $ D$ (and all $ \varepsilon, \delta > 0$), there is a learning algorithm $ A$ which produces a hypothesis $ h$ that with probability at least $ 1 – \delta$ satisfies

$ \displaystyle \text{err}(h) \leq \min_{h’ \in H} \text{err}(h’) + \varepsilon$

and everything runs in the same sort of polynomial time as for vanilla PAC-learning. This is called the agnostic setting or the unrealizable setting, in the sense that we may not be able to find a hypothesis with perfect empirical error.

We seek to prove that all concept classes are agnostically learnable with a finite hypothesis class, provided you have an algorithm that can minimize empirical error. But actually we’ll prove something stronger.

Theorem: Let $ H$ be a finite hypothesis class and $ m$ the number of samples drawn. Then for any $ \delta > 0$, with probability $ 1-\delta$ the following holds:

$ \displaystyle \forall h \in H, \hat{\text{err}}(h) \leq \text{err}(h) + \sqrt{\frac{\log |H| + \log(2 / \delta)}{2m}}$

In other words, we can precisely quantify how the empirical error converges to the true error as the number of samples grows. But this holds for all hypotheses in $ H$, so this provides a uniform bound of the difference between true and empirical error for the entire hypothesis class.

Proving this requires the Chernoff bound. Fix a single hypothesis $ h \in H$. If you draw an example $ x$, call $ Z$ the random variable which is 1 when $ h(x) \neq c(x)$, and 0 otherwise. So if you draw $ m$ samples and call the $ i$-th variable $ Z_i$, the empirical error of the hypothesis is $ \frac{1}{m}\sum_i Z_i$. Moreover, the actual error is the expectation of this random variable since $ \mathbf{E}[1/m \sum_i Z_i] = Z$.

So what we’re asking is the probability that the empirical error deviates from the true error by a lot. Let’s call “a lot” some parameter $ \varepsilon/2 > 0$ (the reason for dividing by two will become clear in the corollary to the theorem). Then plugging things into the Chernoff-Hoeffding bound gives a bound on the probability of the “bad event,” that the empirical error deviates too much.

$ \displaystyle \Pr[|\hat{\text{err}}(h) – \text{err}(h)| > \varepsilon / 2] < 2e^{-\frac{\varepsilon^2m}{2}}$

Now to get a bound on the probability that some hypothesis is bad, we apply the union bound and use the fact that $ |H|$ is finite to get

$ \displaystyle \Pr[|\hat{\text{err}}(h) – \text{err}(h)| > \varepsilon / 2] < 2|H|e^{-\frac{\varepsilon^2m}{2}}$

Now say we want to bound this probability by $ \delta$. We set $ 2|H|e^{-\varepsilon^2m/2} \leq \delta$, solve for $ m$, and get

$ \displaystyle m \geq \frac{2}{\varepsilon^2}\left ( \log |H| + \log \frac{2}{\delta} \right )$

This gives us a concrete quantification of the tradeoff between $ m, \varepsilon, \delta, $ and $ |H|$. Indeed, if we pick $ m$ to be this large, then solving for $ \varepsilon / 2$ gives the exact inequality from the theorem.

$ \square$

Now we know that if we pick enough samples (polynomially many in all the parameters), and our algorithm can find a hypothesis $ h$ of minimal empirical error, then we get the following corollary:

Corollary: For any $ \varepsilon, \delta > 0$, the algorithm that draws $ m \geq \frac{2}{\varepsilon^2}(\log |H| + \log(2/ \delta))$ examples and finds any hypothesis of minimal empirical error will, with probability at least $ 1-\delta$, produce a hypothesis that is within $ \varepsilon$ of the best hypothesis in $ H$.

Proof. By the previous theorem, with the desired probability, for all $ h \in H$ we have $ |\hat{\text{err}}(h) – \text{err}(h)| < \varepsilon/2$. Call $ g = \min_{h’ \in H} \text{err}(h’)$. Then because the empirical error of $ h$ is also minimal, we have $ |\hat{\text{err}}(g) – \text{err}(h)| < \varepsilon / 2$. And using the previous theorem again and the triangle inequality, we get $ |\text{err}(g) – \text{err}(h)| < 2 \varepsilon / 2 = \varepsilon$. In words, the true error of the algorithm’s hypothesis is close to the error of the best hypothesis, as desired.

$ \square$

Next time

Both of these theorems tell us something about the generalization guarantees for learning with hypothesis classes of a certain size. But this isn’t exactly the most reasonable measure of the “complexity” of a family of hypotheses. For example, one could have a hypothesis class with a billion intervals on $ \mathbb{R}$ (say you’re trying to learn intervals, or thresholds, or something easy), and the guarantees we proved in this post are nowhere near optimal.

So the question is: say you have a potentially infinite class of hypotheses, but the hypotheses are all “simple” in some way. First, what is the right notion of simplicity? And second, how can you get guarantees based on that analogous to these? We’ll discuss this next time when we define the VC-dimension.

Until then!

A problem that is not (properly) PAC-learnable

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In a previous post we introduced a learning model called Probably Approximately Correct (PAC). We saw an example of a concept class that was easy to learn: intervals on the real line (and more generally, if you did the exercise, axis-aligned rectangles in a fixed dimension).

One of the primary goals of studying models of learning is to figure out what is learnable and what is not learnable in the various models. So as a technical aside in our study of learning theory, this post presents the standard example of a problem that isn’t learnable in the PAC model we presented last time. Afterward we’ll see that allowing the learner to be more expressive can be helpful, and by doing so we can make this unlearnable problem learnable.

Addendum: This post is dishonest in the following sense. The original definition I presented of PAC-learning is not considered the “standard” version, precisely because it forces the learning algorithm to produce hypotheses from the concept class it’s trying to learn. As this post shows, that prohibits us from learning concept classes that should be easy to learn. So to quell any misconceptions, we’re not saying that 3-term DNF formulas (defined below) are not PAC-learnable, just that they’re not PAC-learnable under the definition we gave in the previous post. In other words, we’ve set up a straw man (or, done some good mathematics) in order to illustrate why we need to add the extra bit about hypothesis classes to the definition at the end of this post.

3-Term DNF Formulas

Readers of this blog will probably have encountered a boolean formula before. A boolean formula is just a syntactic way to describe some condition (like, exactly one of these two things has to be true) using variables and logical connectives. The best way to recall it is by example: the following boolean formula encodes the “exclusive or” of two variables.

$ \displaystyle (x \wedge \overline{y}) \vee (\overline{x} \wedge y)$

The wedge $ \wedge$ denotes a logical AND and the vee $ \vee$ denotes a logical OR. A bar above a variable represents a negation of a variable. (Please don’t ask me why the official technical way to write AND and OR is in all caps, I feel like I’m yelling math at people.)

In general a boolean formula has literals, which we can always denote by an $ x_i$ or the negation $ \overline{x_i}$, and connectives $ \wedge$ and $ \vee$, and parentheses to denote order. It’s a simple fact that any logical formula can be encoded using just these tools, but rather than try to learn general boolean formulas we look at formulas in a special form.

Definition: A formula is in three-term disjunctive normal form (DNF) if it has the form $ C_1 \vee C_2 \vee C_3$ where each $C_i$ is an AND of some number of literals.

Readers who enjoyed our P vs NP primer will recall a related form of formulas: the 3-CNF form, where the “three” meant that each clause had exactly three literals and the “C” means the clauses are connected with ANDs. This is a sort of dual normal form: there are only three clauses, each clause can have any number of variables, and the roles of AND and OR are switched. In fact, if you just distribute the $ \vee$’s in a 3-term DNF formula using DeMorgan’s rules, you’ll get an equivalent 3-CNF formula. The restriction of our hypotheses to 3-term DNFs will be the crux of the difficulty: it’s not that we can’t learn DNF formulas, we just can’t learn them if we are forced to express our hypothesis as a 3-term DNF as well.

The way we’ll prove that 3-term DNF formulas “can’t be learned” in the PAC model is by an NP-hardness reduction. That is, we’ll show that if we could learn 3-term DNFs in the PAC model, then we’d be able to efficiently solve NP-hard problems with high probability. The official conjecture we’d be violating is that RP is different from NP. RP is the class of problems that you can solve in polynomial time with randomness if you can never have false positives, and the probability of a false negative is at most 1/2. Our “RP” algorithm will be a PAC-learning algorithm.

The NP-complete problem we’ll reduce from is graph 3-coloring. So if you give me a graph, I’ll produce an instance of the 3-term DNF PAC-learning problem in such a way that finding a hypothesis with low error corresponds to a valid 3-coloring of the graph. Since PAC-learning ensures that you are highly likely to find a low-error hypothesis, the existence of a PAC-learning algorithm will constitute an RP algorithm to solve this NP-complete problem.

In more detail, an “instance” of the 3-term DNF problem comes in the form of a distribution over some set of labeled examples. In this case the “set” is the set of all possible truth assignments to the variables, where we fix the number of variables to suit our needs, along with a choice of a target 3-term DNF to be learned. Then you’d have to define the distribution over these examples.

But we’ll actually do something a bit slicker. We’ll take our graph $ G$, we’ll construct a set $ S_G$ of labeled truth assignments, and we’ll define the distribution $ D$ to be the uniform distribution over those truth assignments used in $ S_G$. Then, if there happens to be a 3-term DNF that coincidentally labels the truth assignments in $ S_G$ exactly how we labeled them, and we set the allowed error $ \varepsilon$ to be small enough, a PAC-learning algorithm will find a consistent hypothesis (and it will correspond to a valid 3-coloring of $ G$). Otherwise, no algorithm would be able to come up with a low-error hypothesis, so if our purported learning algorithm outputs a bad hypothesis we’d be certain (with high probability) that it was not bad luck but that the examples are not consistent with any 3-term DNF (and hence there is no valid 3-coloring of $ G$).

This general outline has nothing to do with graphs, and so you may have guessed that the technique is commonly used to prove learning problems are hard: come up with a set of labeled examples, and a purported PAC-learning algorithm would have to come up with a hypothesis consistent with all the examples, which translates back to a solution to your NP-hard problem.

The Reduction

Now we can describe the reduction from graphs to labeled examples. The intuition is simple: each term in the 3-term DNF should correspond to a color class, and so any two adjacent vertices should correspond to an example that cannot be true. The clauses will correspond to…

For a graph $ G$ with $ n$ nodes $ v_1, \dots, v_n$ and a set of $ m$ undirected edges $ E$, we construct a set of examples with positive labels $ S^+$ and one with negative examples $ S^-$. The examples are truth assignments to $ n$ variables, which we label $ x_1, \dots, x_n$, and we identify a truth assignment to the $ \left \{ 0,1 \right \}$-valued vector $ (x_1, x_2, \dots, x_n)$ in the usual way (true is 1, false is 0).

The positive examples $ S^+$ are simple: for each $ v_i$ add a truth assignment $ x_i = T, x_j = F$ for $ j \neq i$. I.e., the binary vector is $ (1, \dots, 1,0,1, \dots, 1)$, and the zero is in the $ i$-th position.

The negative examples $ S^-$ come from the edges. For each edge $ (v_i, v_j) \in E$, we add the example with a zero in the $ i$-th and $ j$-th components and ones everywhere else. Here is an example graph and the corresponding positive and negative examples:

PAC-reduction

Claim: $ G$ is 3-colorable if and only if the corresponding examples are consistent with some 3-term DNF formula $ \varphi$.

Again, consistent just means that $ \varphi$ is satisfied by every truth assignment in $ S^+$ and unsatisfied by every example in $ S^-$. Since we chose our distribution to be uniform over $ S^+ \cup S^-$, we don’t care what $ \varphi$ does elsewhere.

Indeed, if $ G$ is three-colorable we can fix some valid 3-coloring with colors red, blue, and yellow. We can construct a 3-term DNF that does what we need. Let $ T_R$ be the AND of all the literals $ x_i$ for which vertex $ v_i$ is not red. For each such $ i$, the corresponding example in $ S^+$ will satisfy $ T_R$, because we put a zero in the $ i$-th position and ones everywhere else. Similarly, no example in $ S^-$ will make $ T_R$ true because to do so both vertices in the corresponding edge would have to be red.

To drive this last point home say there are three vertices and your edge is $ (v_1,v_2)$. Then the corresponding negative example is $ (0,0,1)$. Unless both $ v_1$ and $ v_2$ are colored red, one of $ x_1, x_2$ will have to be ANDed as part of $ T_R$. But the example has a zero for both $ x_1$ and $ x_2$, so $ T_R$ would not be satisfied.

Doing the same thing for blue and yellow, and OR them together to get $ T_R \vee T_B \vee T_Y$. Since the case is symmetrically the same for the other colors, we a consistent 3-term DNF.

On the other hand, say there is a consistent 3-term DNF $ \varphi$. We need to construct a three coloring of $ G$. It goes in largely the same way: label the clauses $ \varphi = T_R \vee T_B \vee T_Y$ for Red, Blue, and Yellow, and then color a vertex $ v_i$ the color of the clause that is satisfied by the corresponding example in $ S^+$. There must be some clause that does this because $ \varphi$ is consistent with $ S^+$, and if there are multiple you can pick a valid color arbitrarily. Now we argue why no edge can be monochromatic. Suppose there were such an edge $ (v_i, v_j)$, and both $ v_i$ and $ v_j$ are colored, say, blue. Look at the clause $ T_B$: since $ v_i$ and $ v_j$ are both blue, the positive examples corresponding to those vertices  (with a 0 in the single index and 1’s everywhere else) both make $ T_B$ true. Since those two positive examples differ in both their $ i$-th and $ j$-th positions, $ T_B$ can’t have any of the literals $ x_i, \overline{x_i}, x_j, \overline{x_j}$. But then the negative example for the edge would satisfy $ T_B$ because it has 1’s everywhere except $ i,j$! This means that the formula doesn’t consistently classify the negative examples, a contradiction. This proves the Claim.

Now we just need to show a few more details to finish the proof. In particular, we need to observe that the number of examples we generate is polynomial in the size of the graph $ G$; that the learning algorithm would still run in polynomial time in the size of the input graph (indeed, this depends on our choice of the learning parameters); and that we only need to pick $ \delta < 1/2$ and $ \varepsilon \leq 1/(2|S^+ \cup S^-|)$ in order to enforce that an efficient PAC-learner would generate a hypothesis consistent with all the examples. Indeed, if a hypothesis errs on even one example, it will have error at least $ 1 / |S^+ \cup S^-|$, which is too big.

Everything’s not Lost

This might seem a bit depressing for PAC-learning, that we can’t even hope to learn 3-term DNF formulas. But we will give a sketch of why this is mostly not a problem with PAC but a problem with DNFs.

In particular, the difficulty comes in forcing a PAC-learning algorithm to express its hypothesis as a 3-term DNF, as opposed to what we might argue is a more natural representation. As we observed, distributing the ORs in a 3-term DNF produces a 3-CNF formula (an AND of clauses where each clause is an OR of exactly three literals). Indeed, one can PAC-learn 3-CNF formulas efficiently, and it suffices to show that one can learn formulas which are just ANDs of literals. Then you can blow up the number of variables only polynomially larger to get 3-CNFs. ANDs of literals are just called “conjunctions,” so the problem is to PAC-learn conjunctions. The idea that works is the same one as in our first post on PAC where we tried to learn intervals: just pick the “smallest” hypothesis that is consistent with all the examples you’ve seen so far. We leave a formal proof as an (involved) exercise to the reader.

The important thing to note is that a concept class $ C$ (the thing we’re trying to learn) might be hard to learn if you’re constrained to work within $ C$. If you’re allowed more expressive hypotheses (in this case, arbitrary boolean formulas), then learning $ C$ suddenly becomes tractable. This compels us to add an additional caveat to the PAC definition from our first post.

Definition: A concept class $ \mathsf{C}$ over a set $ X$ is efficiently PAC-learnable using the hypothesis class $ \mathsf{H}$ if there exists an algorithm $ A(\varepsilon, \delta)$ with access to a query function for $ \mathsf{C}$ and runtime $ O(\text{poly}(1/\varepsilon, 1/\delta))$, such that for all $ c \in \mathsf{C}$, all distributions $ D$ over $ X$, and all $ 0 < \delta , \varepsilon < 1/2$, the probability that $ A$ produces a hypothesis $ h \in \mathsf{H}$ with error at most $ \varepsilon$ is at least $ 1-\delta$.

And with that we’ll end this extended side note. The next post in this series will introduce and analyze a fascinating notion of dimension for concept classes, the Vapnik-Chervonenkis dimension.

Until then!