🕑 14 min • 👤 Thomas Graf • 📆 April 28, 2020 in Discussions • 🏷 phonology, syntax, algebra, gradience

Omer has a great post on gradience in syntax. I left a comment there that briefly touches on why gradience isn’t really that big of a deal thanks to monoids and semirings. But in a vacuum that remark might not make a lot of sense, so here’s some more background.

My central claim is that linguists’ worries about gradience are overblown because there isn’t that much of a difference between categorical systems, which only distinguish between well-formed and ill-formed, and gradient systems, which have more shades of gray than that. In particular, the difference doesn’t matter for those aspects of grammar that linguists really care about. A grammar with only a categorical distinction isn’t irredeemably impoverished, and if your formalism gets the linguistic fundamentals wrong adding gradience won’t fix that for you.

Brief note: In practice, gradient systems are usually probabilistic, but there’s no need for that. The familiar system of rating sentences as well-formed, ?, ??, ?*, and * would also be gradient. This is an important fact that’s frequently glossed over. I really wish researchers wouldn’t always jump right to probabilistic systems when they want to make something gradient. Sure, probabilities are nice because they are easy to extract from the available data, but that doesn’t mean that this is the right notion of gradience.

That said, this post will frequently use probabilistic grammars to illustrate more general points about gradience. The take-home message, though, applies equally to all gradient systems, whether they’re probabilistic or not.

## A formula for categorical grammars

Let’s start with a very simple example in the form of a strictly local grammar. SL grammars are usually negative, which means that they list all the n-grams that must not occur in a string. But for the purposes of this post, it is preferable to convert the negative grammar into an equivalent positive grammar, which lists all the n-grams that may occur in a string. For example, the positive SL-2 grammar $$G$$ below generates the language $$(ab)^+$$, which contains the strings $$\mathit{ab}$$, $$\mathit{abab}$$, $$\mathit{ababab}$$, and so on.

1. Positive SL-2 grammar for $$\mathbf{(ab)^*}$$
1. $$\mathit{\a}$$: the string may start with $$a$$
2. $$\mathit{ab}$$: $$a$$ may be followed by $$b$$
3. $$\mathit{ba}$$: $$b$$ may be followed by $$a$$
4. $$\mathit{b\}$$: the string may end with $$b$$

Now let’s consider how one actually decides whether a given string is well-formed with respect to this grammar. There’s many equivalent ways of thinking about this, but right now we want one that emphasizes the algebraic nature of grammars.

Suppose we are given the string $$\mathit{abab}$$. As always with an SL grammar, we first add edge markers to it, giving us $$\mathit{\abab\}$$. That’s just a mathematical trick to clearly distinguish the first and last symbol of the string. The SL grammar decides the well-formedness of the string $$\mathit{\abab\}$$ based on whether the bigrams that occur in it are well-formed. Those bigrams are (including repetitions)

1. $$\mathit{\a}$$,
2. $$\mathit{ab}$$,
3. $$\mathit{ba}$$,
4. $$\mathit{ab}$$,
5. $$\mathit{b\}$$.

We can write this as a single formula that doesn’t make a lick of sense at this point:

$G(\mathit{\abab\}) := f(\a) \otimes f(ab) \otimes f(ba) \otimes f(ab) \otimes f(b\)$

It sure looks fancy, but I haven’t really done anything substantial here. Let’s break this formula down into its components:

• $$G(\mathit{\abab\})$$ is the value that the grammar $$G$$ assigns to the string $$\mathit{\abab\}$$. Since $$G$$ is categorical, this can be $$1$$ for well-formed or $$0$$ for ill-formed.
• $$:=$$ means “is defined as”.
• $$f$$ is some mystery function that maps each bigram to some value.
• $$\otimes$$ is some mystery operation that combines the values produced by $$f$$.

The formula expresses in mathematical terms the most fundamental rule of SL grammars: the value that $$G$$ assigns to $$\mathit(\abab\)$$ depends on the bigrams that occur in the string. Each bigram in the string is mapped to some value, and then all these values are combined into an aggregate value for the string. The only reason the formula looks weird is because I haven’t told you what $$f$$ and $$\otimes$$ are.

The cool thing is, $$f$$ and $$\otimes$$ can be lots of things. That’s exactly what will allow us to unify categorical and gradient grammars. But let’s not get ahead of ourselves, let’s just focus on $$f$$ and $$\otimes$$ for our categorical example grammar $$G$$.

We start with $$f$$. This function maps a bigram $$b$$ to $$1$$ if it is a licit bigram according to our grammar $$G$$. If $$b$$ is not a licit bigram, $$f$$ maps it to $$0$$.

$f(b) := \begin{cases} 1 & \text{if } b \text{ is a licit bigram of } G\\ 0 & \text{otherwise} \end{cases}$

Let’s go back to the formula above and fill in the corresponding values according to $$f$$ and $$G$$.

\begin{align*} G(\mathit{\abab\}) := & f(\a) \otimes f(ab) \otimes f(ba) \otimes f(ab) \otimes f(b\)\\ = & 1 \otimes 1 \otimes 1 \otimes 1 \otimes 1\\ \end{align*}

Compare this to the formula for the illicit string $$\mathit{\abba\}$$.

\begin{align*} G(\mathit{\abba\}) := & f(\a) \otimes f(ab) \otimes f(bb) \otimes f(ba) \otimes f(a\)\\ = & 1 \otimes 1 \otimes 0 \otimes 1 \otimes 0\\ \end{align*}

Notice how we get $$1$$ or $$0$$ depending on whether the bigram is licit according to grammar $$G$$.

This only leaves us with $$\otimes$$. The job of this operation is to combine the values produced by $$f$$ such that we get $$1$$ if the string is well-formed, and $$0$$ otherwise. A string is well-formed iff it does not contain even one illicit bigram, or equivalently, iff there isn’t a single bigram that was mapped to $$0$$ by $$f$$. If there is even one $$0$$, the whole aggregate value must be $$0$$. We can replace $$\otimes$$ with any operation that satisfies this property — multiplication, for instance, will do just fine.

\begin{align*} G(\mathit{\abab\}) := & f(\a) \otimes f(ab) \otimes f(ba) \otimes f(ab) \otimes f(b\)\\ = & 1 \otimes 1 \otimes 1 \otimes 1 \otimes 1\\ = & 1 \times 1 \times 1 \times 1 \times 1\\ = & 1\\ \end{align*} \begin{align*} G(\mathit{\abba\}) := & f(\a) \otimes f(ab) \otimes f(bb) \otimes f(ba) \otimes f(a\)\\ = & 1 \otimes 1 \otimes 0 \otimes 1 \otimes 0\\ = & 1 \times 1 \times 0 \times 1 \times 0\\ = & 0\\ \end{align*}

Tada, the well-formed string gets a 1, the ill-formed string a 0, just as intended. Any string that contains at least one illicit bigram will be mapped to 0 because whenever you multiply by 0, you get 0. The only way for a string to get mapped to 1 is if only consists of well-formed bigrams. This is exactly the intuition we started out with: the well-formedness of a string is contingent on the well-formedness of its parts; in this case, bigrams.

## A formula for gradient grammars

While it’s certainly refreshing to think of a grammar as a device for multiplying $$1$$s and $$0$$s, there is a deeper purpose to this view. Here’s the crucial twist: the formula above also works for gradient SL grammars, we just have to change $$f$$ and $$\otimes$$. If we use probabilities, we can even keep $$\otimes$$ the same. The math works exactly the same for categorical and probabilistic grammars.

First, let’s turn our categorical example grammar into a probabilistic one by assigning each bigram a probability. I’ll use arbitrary numbers here, in the real world those probabilities would usually come from a corpus.

1. Probabilistic SL-2 grammar for $$\mathbf{(ab)^*}$$
1. $$\mathit{\a}$$: the probability that a string starts with $$a$$ is 100%
2. $$\mathit{ab}$$: the probability that $$a$$ is followed by $$b$$ is 100%
3. $$\mathit{ba}$$: the probability that $$b$$ is followed by $$a$$ is 75%
4. $$\mathit{b\}$$: the probability that $$b$$ is not followed by anything is 25%

Now that the grammar is probabilistic, we also have to change our formula. Except that we don’t! We keep everything the way it is and only interpret $$f$$ differently. The function $$f$$ no longer tells us whether a bigram is licit, it instead gives us the probability of the bigram according to $$G$$. The probability for bigrams that aren’t listed in the grammar is set to $$0$$.

\begin{align*} G(\mathit{\abab\}) := & f(\a) \otimes f(ab) \otimes f(ba) \otimes f(ab) \otimes f(b\)\\ = & 1 \otimes 1 \otimes .75 \otimes 1 \otimes .25\\ = & 1 \times 1 \times .75 \times 1 \times .25\\ = & .1875\\ \end{align*} \begin{align*} G(\mathit{\abba\}) := & f(\a) \otimes f(ab) \otimes f(bb) \otimes f(ba) \otimes f(a\)\\ = & 1 \otimes 1 \otimes 0 \otimes .7 \otimes 0\\ = & 1 \times 1 \times 0 \times .7 \times 0\\ = & 0\\ \end{align*}

Compare that to the formula we had for the categorical grammar — it’s exactly the same mechanism! Nothing here has changed except the values. The value of the whole is still computed from the values of the same parts.

## A trivalent SL grammar

What if we want to do a trivalent system, with well-formed, borderline, and ill-formed? Let’s modify our categorical grammar so that it marginally allows $$\mathit{bb}$$.

1. Trivalent SL-2 grammar for $$\mathbf{(ab)^*}$$
1. $$\mathit{\a}$$: the string may start with $$a$$
2. $$\mathit{ab}$$: $$a$$ may be followed by $$b$$
3. $$\mathit{ba}$$: $$b$$ may be followed by $$a$$
4. $$\mathit{b\}$$: the string may end with $$b$$
5. $$\mathit{bb}$$: $$b$$ may be marginally followed by $$b$$

The corresponding formula once again will stay the same. But instead of $$0$$ and $$1$$, we will use three values:

• $$1$$: well-formed
• $$?$$: borderline
• $$*$$: ill-formed

Instead of multiplication, $$\otimes$$ is now an operation $$\mathrm{min}$$ that always returns the least licit value, as specified in the table below.

$$\mathrm{min}$$ $$\mathbf{1}$$ $$\mathbf{?}$$ $$\mathbf{*}$$
$$\mathbf{1}$$ $$1$$ $$?$$ $$*$$
$$\mathbf{?}$$ $$?$$ $$?$$ $$*$$
$$\mathbf{*}$$ $$*$$ $$*$$ $$*$$

And here are the corresponding formulas for our familiar example strings $$\mathit{\abab\}$$ and $$\mathit{\abba\}$$

\begin{align*} G(\mathit{\abab\}) := & f(\a) \otimes f(ab) \otimes f(ba) \otimes f(ab) \otimes f(b\)\\ = & 1 \otimes 1 \otimes 1 \otimes 1 \otimes 1\\ = & 1 \mathrel{\mathrm{min}} 1 \mathrel{\mathrm{min}} 1 \mathrel{\mathrm{min}} 1 \mathrel{\mathrm{min}} 1\\ = & 1\\ \end{align*} \begin{align*} G(\mathit{\abba\}) := & f(\a) \otimes f(ab) \otimes f(bb) \otimes f(ba) \otimes f(a\)\\ = & 1 \otimes 1 \otimes ? \otimes 1 \otimes *\\ = & 1 \mathrel{\mathrm{min}} 1 \mathrel{\mathrm{min}} ? \mathrel{\mathrm{min}} 1 \mathrel{\mathrm{min}} *\\ = & *\\ \end{align*}

Note how the second string is still considered ill-formed. While the presence of the bigram $$\mathit{bb}$$ degrades it to borderline status, the presence of the illicit bigram $$\mathit{a\}$$ means that we cannot assign a higher value than $$*$$.

## Beyond acceptability

We can even use this formula to calculate aspects of the string that have nothing at all to do with well-formedness or acceptability. Suppose that $$f$$ once again maps each bigram to $$1$$ or $$0$$ depending on whether it is licit according to $$G$$. Next, we instantiate $$\otimes$$ as addition. Then we have a formula that calculates the number of licit bigrams in the string.

\begin{align*} G(\mathit{\abab\}) := & f(\a) \otimes f(ab) \otimes f(ba) \otimes f(ab) \otimes f(b\)\\ = & 1 \otimes 1 \otimes 1 \otimes 1 \otimes 1\\ = & 1 + 1 + 1 + 1 + 1\\ = & 5\\ \end{align*} \begin{align*} G(\mathit{\abba\}) := & f(\a) \otimes f(ab) \otimes f(bb) \otimes f(ba) \otimes f(a\)\\ = & 1 \otimes 1 \otimes 0 \otimes 1 \otimes 0\\ = & 1 + 1 + 0 + 1 + 0\\ = & 3\\ \end{align*}

Or maybe $$f$$ replaces each bigram $$g$$ with the singleton set $$\{g\}$$. And $$\otimes$$ will be $$\cup$$, the set union operation. Then the formula maps each string to the set of bigrams that occur in it.

\begin{align*} G(\mathit{\abab\}) := & f(\a) \otimes f(ab) \otimes f(ba) \otimes f(ab) \otimes f(b\)\\ = & \{\a\} \otimes \{ab\} \otimes \{ba\} \otimes \{ab\} \otimes \{b\\}\\ = & \{\a\} \cup \{ab\} \cup \{ba\} \cup \{ab\} \cup \{b\\}\\ = & \{\a, ab, ba, b\\}\\ \end{align*} \begin{align*} G(\mathit{\abba\}) := & f(\a) \otimes f(ab) \otimes f(bb) \otimes f(ba) \otimes f(a\)\\ = & \{\a\} \otimes \{ab\} \otimes \{bb\} \otimes \{ba\} \otimes \{a\\}\\ = & \{\a\} \cup \{ab\} \cup \{bb\} \cup \{ba\} \cup \{a\\}\\ = & \{\a, ab, bb, ba, a\\}\\ \end{align*}

Is there a point to these instantiations of $$f$$ and $$\otimes$$? They can be useful for certain computational tasks, but from a linguistic perspective there really isn’t much point to them. But, you know what, I’d say the same is true for all the other instantiations we’ve seen so far. If you’re a linguist, you shouldn’t worry at all about how $$f$$ and $$\otimes$$ are instantiated.

## Grammars combine, they don’t calculate

The general upshot is this: a grammar is a mechanism for determining the values of the whole from values of its parts. The difference between grammars is what parts they look at and how they relate them to each other.

A TSL grammar, for instance, would have a different formula. In a TSL grammar, we ignore irrelevant symbols in the string. So if we have a grammar that cares about $$a$$ but not $$b$$, the corresponding formula for the string $$\mathit{abba}$$ would be $$f(\a) \otimes f(aa) \otimes f(a\)$$. This is only a minor change because TSL grammars are very similar to SL grammars. The formula for, say, a finite-state automaton would differ by quite a bit more. That’s what linguistic analysis is all about. Linguistics is about determining the shape of the formula!

But that’s not what the categorical VS gradience divide is about. That only kicks in once you have determined the overall shape of the formula and need to define $$f$$ and $$\otimes$$. And that choice simply isn’t very crucial from a linguistic perspective.

There’s many different choices for $$f$$ and $$\otimes$$ depending on what you want to do. But the choices that are useful for a linguist will always be limited in such way that they form a particular kind of algebraic structure that’s called a monoid. I won’t bug you with the mathematical details of monoids. Whether you prefer a categorical system or a gradient system, rest assured there’s a suitable monoid for that. And that’s all that matters. That’s why linguists shouldn’t worry about the categorical VS gradience divide — linguistic insights are about the overall shape of the formula, not about calculating the result.

## From string to trees: semirings

Okay, there’s one minor complication that I’d like to cover just to cross all ts and dot all is. If you’re already worn out, just skip ahead to the wrap-up.

Beyond the pleasant valleys of string land lies the thicket of tree land. In tree land, things can get a bit more complicated depending on what your question is. Not always, though. It really depends on what kind of value you’re trying to compute.

If you just want to know whether a specific tree is well-formed, nothing really changes. Take your standard phrase structure grammar. A rewrite rule of the form S -> NP VP is a tree bigram where the mother is S and the daughters are NP and VP. Just like we can break down a string into its string bigrams, we can break down a tree into its tree bigrams. And the value of the whole tree according to a phrase structure grammar is computed by combining the values of its tree bigrams. With more expressive formalisms like MGs, things are once again more complicated, just like a finite-state automaton uses a more complicated formula in string land than the one for SL grammars above. But the general principle remains the same: once you have a formula for how the parts interact, you can plug in the operators you want. As before, we can switch between gradient and categorical systems by tweaking the values of $$f$$ and $$\otimes$$, under the condition that this still gets us a monoid.

I think this is actually enough for syntax. But perhaps you want to talk about the value of a string, rather than a tree. This is a more complex value because one string can correspond to multiple trees. For instance, in probabilistic syntax the probability of the string

1. I eat sushi with edible chopsticks.

is the sum of the probabilities of two distinct trees:

1. [I eat [sushi with edible chopsticks]]
2. [I [[eat sushi] [with edible chopsticks]]

So $$\otimes$$ by itself is not enough, there is yet another operation. For probabilistic grammars it’s $$+$$, but we may again replace it with a more general mystery operation $$\oplus$$. The job of $$\oplus$$ is to combine all the values computed by $$\otimes$$. Like $$\otimes$$, $$\oplus$$ has to yield a monoid of some kind, and the combination of $$\oplus$$ and $$\otimes$$ has to form a semiring. Again I’ll completely gloss over the math. Let’s focus only on the essential point: once again the split between categorical systems and gradient systems is not very large because either way we end up with a semiring. The nature of the grammar stays the same, only the system for computing compound values uses different functions and operators.

You might be wondering what a categorical grammar looks like from the semiring perspective. What is the mysterious operation $$\oplus$$ in that case? It can’t be addition because $$1 + 1$$ would give us $$2$$, which isn’t a possible value in a categorical system. No, with categorical systems, $$\oplus$$ behaves like logical or: it returns 1 if there is at least one 1. Suppose, then, that we want to know if some string s is well-formed according to some categorical grammar $$G$$. Here is how this would work in a very simplified manner:

1. We look at all possible trees that yield the string s, even if those strings are ill-formed according to $$G$$.
2. We use $$\otimes$$ to compute the compound value for each tree. As before, $$\otimes$$ is multiplication (but it could also be logical and, if you find that more pleasing). Well-formed tress will evaluate to $$1$$, ill-formed ones to $$0$$.
3. We then use $$\oplus$$, i.e. logical or, to combine all those compound values into a single value for the string s. Then s will get the value $$1$$, and hence be deemed well-formed, iff there is at least one well-formed tree that yields s.

Okay, that’s not how we usually think about well-formedness. We view the grammar as a system for specifying a specific set of well-formed trees, rather than a function that maps every logically conceivable tree to some value. But as you hopefully remember from your semantics intro, there is no difference between a set and its characteristic function. The procedure above treats the grammar as the characteristic function of the set of well-formed trees. Most of the time that’s not very illuminating for linguistics, but when it comes to the split between categorical and gradient it is really useful because it reveals the monoid/semiring structure of the grammar formalism.

## Wrapping up: Don’t worry, be happy

Monoids and semirings are a very abstract perspective of grammars, and I rushed through them in a (failed?) attempt to keep the post at a manageable length. But behind all that math is the simple idea that syntacticians, and linguists in general, don’t need to worry that a categorical grammar formalism is somehow irreconcilable with the fact that acceptability judgments are gradient. Even if we don’t factor out gradience as a performance phenomenon, even if we want to place it in the heart of grammar, that does not require us to completely retool our grammar formalisms. The change is largely mathematical in nature and doesn’t touch on the things that linguists care about. Linguists care about representations and how specific parts of those representations can interact. In the mathematical terms I used in this post, those issues are about the shape of the formula for computing $$G(o)$$ for some object $$o$$. It is not about the specific values or operators that appear in the formula.

In many cases, there’s actually many different operators that give the same result. We interpreted $$\otimes$$ as multiplication for categorical SL grammars, but we could have also used logical and or the min function. They all produce exactly the same values. No linguist would ever worry about which one of those functions is the right choice. The choice between categorical, probabilistic, or some other kind of gradient system isn’t all that different. Again you are needlessly worrying about the correct way of instantiating $$f$$, $$\otimes$$, and possibly $$\oplus$$.

That’s not to say that switching out, say, a categorical semiring for a probabilistic one is a trivial affair. It can create all kinds of problems. But those are mathematical problems, computational problems, they are not linguistic problems. It’s stuff like computing infinite sums of bounded reals. It’s decidedly not a linguistic issue. So don’t worry, be happy.