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A collection of Cassie's harebrained schemes.

Syntactic Unification Is Easy!

Published 2020-02-19

Syntactic unification is the process of trying to find an assignment of variables that makes two terms the same. For example, attempting to unify the terms in the equation (X, Y) = (1, 2) would give the substitution {X: 1, Y: 2}, but if we tried to unify (X, X) = (1, 2) it would fail since one variable can’t be equal to two different values. Unification is used to implement two things I’m very interested in: type systems, and logic programming languages.

Earlier last year I tried implementing unification in order to make a toy Prolog, and I got completely stuck on the algorithms. I eventually puzzled throught it by reading lots of papers and articles, but I don’t recall understanding it fully. You can take a look at the equations that the Wikipedia article uses to see what I was dealing with—it’s certainly intimidating!

Cut to last week, I happened to read the MicroKanren paper (which is very readable) and discovered that unification could be very simple! Let me show you how!

The Terms To Unify

We have to define what kinds of things we’re going to unify. There are a few basic things we need to have.

I’m going to write my examples in Python, so the types I’m going to work with are:

The scheme example they demonstrate in the MicroKanren article uses variables, pairs, and arbitrary objects. The Rust version I wrote just has variables, pairs, and integers. Exactly what you want to work with is completely up to you: hopefully at the end of this, you’ll be able to figure out how to do it for some custom type.

My Var type is very simple, with only one custom method to make the printing more compact:

from dataclasses import dataclass

@dataclass(frozen=True)
class Var:
name: str

def __repr__(self):
return self.name

And everything else is base Python types, so we’re done with this part.

Unifying Terms

Let’s work through unifying things. We want to unify two terms, and return a substitution that makes them equal, which I’ll call the “environment.” If they can’t possibly be equal, we’ll return None. Let’s start out simple: two things can be unified if they’re already equal.

def unify(a, b):
if a == b:
return {}
# ... other cases ...
return None

Here we return an empty environment because they’re already equal, and nothing needs to be substituted.

The variable cases can also be solved very easily. Trying to unify a variable with another term, we can just substitute that for the variable. When unifying x = 1, we get back { x: 1 }.

    if isinstance(a, Var):
return {a: b}
if isinstance(b, Var):
return {b: a}

Finally, let’s handle tuples. For this, we want to unify each individual component of the tuple. We want the union of each of these unifications, so we’ll update the dictionary each time. We have to check if the sub-unification returned None and pass that along.

    if isinstance(a, tuple) and isinstance(b, tuple):
if len(a) != len(b):
return None
env = {}
for (ax, bx) in zip(a, b):
res = unify(ax, bx)
if res is None:
return None
env.update(res)
return env

Now you might see where this could go wrong, and we’ll get to that in a moment with an example. Repeating that function all in one piece:

def unify(a, b):
if a == b:
return {}
if isinstance(a, Var):
return {a: b}
if isinstance(b, Var):
return {b: a}
if isinstance(a, tuple) and isinstance(b, tuple):
if len(a) != len(b):
return None
env = {}
for (ax, bx) in zip(a, b):
res = unify(ax, bx)
if res is None:
return None
env.update(res)
return env
return None

Now we can unify stuff:

>>> x, y = Var("x"), Var("y")
>>> unify(1, 1)
{}
>>> unify(x, 1)
{ x: 1 }
>>> unify((x, 2), (1, 2))
{ x: 1 }
>>> unify((x, y), (1, 2))
{ x: 1, y: 2 }
>>> unify((x, x), (1, 2))
{ x: 2 }

Oops! That last one is wrong! So what can we do to fix that?

Well, we need to take into account context. Let’s change our unify function to accept an environment to work inside. I’m gonna make this a pure function, so we don’t update the environment, we just make a new version of it. Some things are changed a bit when doing this. We include env in all of our outputs, instead of starting from nothing. We’ll also include

def unify(a, b, env={}):
if a == b:
return env
if isinstance(a, Var):
return {**env, a: b}
if isinstance(b, Var):
return {**env, b: a}
if isinstance(a, tuple) and isinstance(b, tuple):
if len(a) != len(b):
return None
for (ax, bx) in zip(a, b):
env = unify(a, b, env)
if env is None:
return None
return env
return None

This still doesn’t do the right thing though, because we’re not actually looking in the environment. We need to look stuff up, so let’s write a function to do that. We’ll just always call it, and if the term isn’t a key for the environment, we’ll just return it unchanged.

def walk(env, term):
if term in env:
term = env[term]
return term

Now we can add these two lines at the beginning.

    a = walk(env, a)
b = walk(env, b)

With these modifications, we correctly handle the case we were stuck on earlier.

>>> unify((x, x), (1, 2))
>>> # look ma no solution!

It’s still not perfect though. It doesn’t properly reject this case where the conflict is two steps away:

>>> unify((x, y, x), (y, 8, 9))
{ x: y, y: 9 }

Here the problem is that we have two facts around, that x = y, and that y = 8. Even so, it doesn’t see that that means x = 8 when it gets to the assertion of x = 9. Luckily this is another simple fix. We just keep walking through the environment as long as we have a binding. Just change the if to a while!

def walk(env, term):
while term in env:
term = env[term]
return term

And now this unifier should be able to handle any (finite) hard cases you throw at it!

The Whole Unifier

Here’s the whole program we wrote in one piece, in just 30 lines.

from dataclasses import dataclass

@dataclass(frozen=True)
class Var:
name: str
def __repr__(self):
return self.name

def walk(env, term):
while term in env:
term = env[term]
return term

def unify(a, b, env={}):
a = walk(env, a)
b = walk(env, b)
if a == b:
return env
if isinstance(a, Var):
return { **env, a: b }
if isinstance(b, Var):
return { **env, b: a }
if isinstance(a, tuple) and isinstance(b, tuple):
if len(a) != len(b):
return None
for (a, b) in zip(a, b):
env = unify(a, b, env)
if env is None:
return None
return env
return None

Consider writing your own in your language of choice, or adding support for new datatypes. It would be nice to be able to unify namedtuples and dictionaries and things, or do unification in the web browser. Try it out for yourself, because next time I’ll be talking about how you can go from here and use unification to build logic programs.

Postscript: Unifying Infinity

What was that earlier? About only the finite cases? Well, it turns out this unifier is good enough for implementing well-behaved logic programs, but it leaves out one complexity: the “Occurs Check.” This checks that a variable does not “occur” inside a term you unify it with, to avoid infinite solutions and non-termination. Many old Prolog implementations left out the occurs check for efficiency reasons, and we can leave it out for simplicity reasons if we want to trust our programmers to only write finite terms.

But let’s go look at the problem specifically. We can write a unification like:

>>> unify(a, (1, a))
{ a: (1, a) }

If you think through this, you can see that a = (1, (1, (1, (1, ...)))), going on infinitely deep. There are languages that work with logic on infinite terms like this, but applied naively it can end up giving us incorrect results or non-termination. For instance, with proper support for infinite terms, the following should unify:

>>> env = unify(a, (1, a))
>>> env = unify(b, (1, b), env)
>>> env = unify(a, b)

But instead, it gets stuck on the last step, descending forever. There are two ways to solve this. One is to embrace infinite terms and include fixpoints in your logic, giving you terms like a = fix x. (1, x) and b = fix x. (1, x). I don’t know how to do this, though.

The other way is to detect unifications that would lead to infinite terms, and reject them. We take our code that adds variables to the substitution, and add the occurs check there:

    if isinstance(a, Var):
if occurs(a, b, env): return None
return { **env, a: b }
if isinstance(b, Var):
if occurs(b, a, env): return None
return { **env, b: a }

If we’re using the occurs-check here, then we know a few things:

This means that we can check if the literal variable a occurs literally in walked terms and subterms of b. If there’s a variable in b that’s a synonym of a, then it will walk to a because it was the result of walking. While it’s safe to assume to that a has already been walked, note that we can’t assume that b has already been walked since we’re recursing over it. Then, the implementation looks like this:

def occurs(a, b, env={}):
b = walk(env, b)
if a == b:
return True
if isinstance(b, tuple):
return any(occurs(a, x) for x in b)
return False

Now if we try…

>>> unify(a, (1, a))
>>>

We correctly don’t unify this. We’ve got sound unification over finite terms. This is useful to clean the situtation up, but I’m still a bit sad to see infinite terms go…


Edit November 26, 2020. Thanks to Onel Harrison for giving some clarity feedback and pointing out some typos.