Write up the instantiation rule and the new type checking algorithm in compiler series.
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@ -46,8 +46,11 @@ some of our notation from the [typechecking]({{< relref "03_compiler_typecheckin
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But using our rules so far, such a thing is impossible, since there is no way for
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\\(\text{Int}\\) to be unified with \\(\text{Bool}\\). We need a more powerful
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set of rules to describe our program's types. One such set of rules is
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the [Hindley-Milner type system](https://en.wikipedia.org/wiki/Hindley%E2%80%93Milner_type_system),
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set of rules to describe our program's types.
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### Hindley-Milner Type System
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One such powerful set of rules is the [Hindley-Milner type system](https://en.wikipedia.org/wiki/Hindley%E2%80%93Milner_type_system),
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which we have previously alluded to. In fact, the rules we came up
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with were already very close to Hindley-Milner, with the exception of two:
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__generalization__ and __instantiation__. It's been quite a while since the last time we worked on typechecking, so I'm going
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@ -74,9 +77,9 @@ Rule|Name and Description
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\{ (p_1,e_1) \ldots (p_n, e_n) \} : \tau_c }
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{{< /latex >}}| __Case__: This rule is not part of Hindley-Milner, and is specific to our language. If the expression being case-analyzed is of type \\(\\tau\\) and each branch \\((p\_i, e\_i)\\) is of the same type \\(\\tau\_c\\) when the pattern \\(p\_i\\) works with type \\(\\tau\\) producing extra bindings \\(b\_i\\), the whole case expression is of type \\(\\tau\_c\\).
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{{< latex >}}
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\frac{\Gamma \vdash e : \sigma \quad \sigma' \sqsubseteq \sigma}
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{\Gamma \vdash e : \sigma'}
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{{< /latex >}}| __Inst (New)__: If type \\(\\sigma\'\\) is an instantiation of type \\(\\sigma\\) then an expression of type \\(\\sigma\\) is also an expression of type \\(\\sigma\'\\).
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\frac{\Gamma \vdash e : \sigma' \quad \sigma' \sqsubseteq \sigma}
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{\Gamma \vdash e : \sigma}
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{{< /latex >}}| __Inst (New)__: If type \\(\\sigma\\) is an instantiation (or specialization) of type \\(\\sigma\'\\) then an expression of type \\(\\sigma\'\\) is also an expression of type \\(\\sigma\\).
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{{< latex >}}
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\frac
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{\Gamma \vdash e : \sigma \quad \alpha \not \in \text{free}(\Gamma)}
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@ -106,6 +109,155 @@ for all possible \\(a\\)". This new expression using "forall" is what we call a
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For simplicity, we only allow "forall" to be at the front of a polytype. That is, expressions like
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\\(a \\rightarrow \\forall b \\; . \\; b \\rightarrow b\\) are not valid polytypes as far as we're concerned.
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It's key to observe that only some of the typing rules in the above table use monotypes (\\(\\tau\\)). Whereas expressions
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It's key to observe that only some of the typing rules in the above table use polytypes (\\(\\sigma\\)). Whereas expressions
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consisting of a single variable can be polymorphically typed, this is not true for function applications and case expressions.
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In fact, according to our rules, there is no way to introduce a polytype anywhere into our system!
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The reason for this is that we only allow polymorphism at certain locations. In the Hindley-Milner type system,
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this is called __Let-Polymorphism__, which means that only in `let`/`in` expressions can variables or expressions
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be given a polymorphic type. We, on the other hand, do not (yet) have `let`/`in` expressions, so our polymorphism
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is further limited: only functions (and data type constructors) can be polymorphically typed.
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Let's talk about what __Inst__ does, and what "\\(\\sqsubseteq\\)" means.
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First, why don't we go ahead and write the formal inference rule for \\(\\sqsubseteq\\):
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{{< latex >}}
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\frac
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{\tau'=\{\alpha_i \mapsto \tau_i \}\tau \quad \beta_i \not \in \text{free}(\forall \alpha_1...\forall \alpha_n \; . \; \tau)}
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{\forall \alpha_1 ... \forall \alpha_n \; . \; \tau \sqsubseteq \forall \beta_1 ... \forall \beta_m \; . \; \tau'}
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{{< /latex >}}
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In my opinion, this is one of the more confusing inference rules we have to deal with in Hindley-Milner.
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In principle, though, it's not too hard to understand. \\(\\sigma\' \\sqsubseteq \\sigma\\) says "\\(\\sigma\'\\)
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is more general than \\(\\sigma\\)". Alternatively, we can interpret it as "\\(\\sigma\\) is an instance of \\(\\sigma\'\\)".
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What does it mean for one polytype to be more general than another? Intuitively, we might say that \\(\forall a \\; . \\; a \\rightarrow a\\) is
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more general than \\(\\text{Int} \\rightarrow \\text{Int}\\), because the former type can represent the latter, and more. Formally,
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we define this in terms of __substitutions__. A substitution \\(\\{\\alpha \\mapsto \\tau \\}\\) replaces a variable
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\\(\\alpha\\) with a monotype \\(\\tau\\). If we can use a substitution to convert one type into another, then the first
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type (the one on which the substitution was performed) is more general than the resulting type. In our previous example,
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since we can apply the substitution \\(\\{a \\mapsto \\text{Int}\\}\\) to get \\(\\text{Int} \\rightarrow \\text{Int}\\)
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from \\(\\forall a \\; . \\; a \\rightarrow a\\), the latter type is more general; using our notation,
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\\(\\forall a \\; . \\; a \\rightarrow a \\sqsubseteq \\text{Int} \\rightarrow \\text{Int}\\).
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That's pretty much all that the rule says, really. But what about the whole thing with \\(\\beta\\) and \\(\\text{free}\\)? The reason
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for that part of the rule is that, in principle, we can substitute polytypes into polytypes. However, we can't
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do this using \\(\\{ \\alpha \\mapsto \\sigma \\}\\). Consider, for example:
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{{< latex >}}
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\{ a \mapsto \forall b \; . \; b \rightarrow b \} \; \text{Bool} \rightarrow a \rightarrow a \stackrel{?}{=}
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\text{Bool} \rightarrow (\forall b \; . \; b \rightarrow b) \rightarrow \forall b \; . \; b \rightarrow b
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{{< /latex >}}
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Hmm, this is not good. Didn't we agree that we only want quantifiers at the front of our types? Instead, to make that substitution,
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we only substitute the monotype \\(b \\rightarrow b\\), and then add the \\(\\forall b\\) at the front. But
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to do this, we must make sure that \\(b\\) doesn't occur anywhere in the original type
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\\(\forall a \\; . \\; \\text{Bool} \\rightarrow a \\rightarrow a\\) (otherwise we can accidentally generalize
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too much). So then, our concrete inference rule is as follows:
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{{< latex >}}
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\frac
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{
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\begin{gathered}
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\text{Bool} \rightarrow (b \rightarrow b) \rightarrow b \rightarrow b =\{a \mapsto (b \rightarrow b) \} \; \text{Bool} \rightarrow a \rightarrow a \\
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b \not \in \text{free}(\forall a \; . \; \text{Bool} \rightarrow a \rightarrow a) = \varnothing
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\end{gathered}
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}
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{\forall a \; . \; \text{Bool} \rightarrow a \rightarrow a \sqsubseteq \forall b \; . \; \text{Bool} \rightarrow (b \rightarrow b) \rightarrow b \rightarrow b}
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{{< /latex >}}
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In the above rule we:
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1. Replaced \\(a\\) with \\(b \\rightarrow b\\), getting rid of \\(a\\) in the quantifier.
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2. Observed that \\(b\\) is not a free variable in the original type, and thus can be generalized.
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3. Added the generalization for \\(b\\) to the front of the resulting type.
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Now, __Inst__ just allows us to perform specialization / substitution as many times
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as we need to get to the type we want.
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### A New Typechecking Algorithm
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Alright, now we have all these rules. How does this change our typechecking algorithm?
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How about the following:
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1. To every declared function, assign the type \\(a \\rightarrow ... \\rightarrow y \\rightarrow z\\),
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where
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{{< sidenote "right" "arguments-note" "\(a\) through \(y\) are the types of the arguments to the function" >}}
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Of course, there can be more or less than 25 arguments to any function. This is just a generalization;
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we use as many input types as are needed.
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{{< /sidenote >}}, and \\(z\\) is the function's
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return type.
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2. We typecheck each declared function, using the __Var__, __Case__, __App__, and __Inst__ rules.
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3. Whatever type variables we don't fill in, we assume can be filled in with any type,
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and use the __Gen__ rule to sprinkle polymorphism where it is needed.
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Maybe this is enough. Let's go through an example. Suppose we have three functions:
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```
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defn if c t e = {
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case c of {
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True -> { t }
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False -> { e }
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}
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}
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defn testOne = { if True False True }
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defn testTwo = { if True 0 1 }
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```
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If we go through and type check them top-to-bottom, the following happens:
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1. We start by assuming \\(\\text{if} : a \\rightarrow b \\rightarrow c \\rightarrow d\\),
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\\(\\text{testOne} : e\\) and \\(\\text{testTwo} : f\\).
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2. We look at `if`. We infer the type of `c` to be \\(\\text{Bool}\\), and thus, \\(a = \\text{Bool}\\).
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We also infer that \\(b = c\\), since they occur in two branches of the same case expression.
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Finally, we infer that \\(c = d\\), since whatever the case expression returns becomes the return
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value of the function. Thus, we come out knowing that \\(\\text{if} : \\text{Bool} \\rightarrow b
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\\rightarrow b \\rightarrow b\\).
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3. Now, since we never figured out \\(b\\), we use __Gen__: \\(\\text{if} : \\forall b \\; . \\;
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\\text{Bool} \\rightarrow b \\rightarrow b \\rightarrow b\\). Like we'd want, `if` works with
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all types, as long as both its inputs are of the same type.
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4. When we typecheck the body of `testOne`, we use __Inst__ to turn the above type for `if`
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into a single, monomorphic instance. Then, type inference proceeds as before, and all is well.
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5. When we typecheck the body of `testTwo`, we use __Inst__ again, instantiating a new monotype,
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and all is well again.
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So far, so good. But what if we started from the bottom, and went to the top?
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1. We start by assuming \\(\\text{if} : a \\rightarrow b \\rightarrow c \\rightarrow d\\),
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\\(\\text{testOne} : e\\) and \\(\\text{testTwo} : f\\).
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2. We look at `testTwo`. We infer that \\(a = \\text{Bool}\\) (since
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we pass in `True` to `if`). We also infer that \\(b = \\text{Int}\\), and that \\(c = \\text{Int}\\).
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Not yet sure of the return type of `if`, this is where we stop. We are left with
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the knowledge that \\(f = d\\) (because the return type of `if` is the return type of `testTwo`),
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but that's about it. Since \\(f\\) is no longer free, we don't generalize, and conclude that \\(\text{testTwo} : f\\).
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3. We look at `testOne`. We infer that \\(a = \\text{Bool}\\) (already known). We also infer
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that \\(b = \\text{Bool}\\), and that \\(c = \\text{Bool}\\). But wait a minute! This is not right.
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We are back to where we started, with a unification error!
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What went wrong? I claim that we typechecked the functions that _used_ `if` before we typechecked `if` itself,
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which led us to infer a less-than-general type for `if`. This less-than-general type was insufficient to
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correctly check the whole program.
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To address this, we enforce a particular order of type inference on our declaration, guided by dependencies
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between functions. Haskell, which has to deal with a similar issue, has
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[a section in the 2010 report on this](https://www.haskell.org/onlinereport/haskell2010/haskellch4.html#x10-880004.5).
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In short:
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1. We find the
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{{< sidenote "right" "transitive-closure-note" "transitive closure" >}}
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A transitive closure of a vertex in a graph is the list of all vertices reachable
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from that original vertex. Check out the <a href="https://en.wikipedia.org/wiki/Transitive_closure#In_graph_theory">
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Wikipedia page on this</a>.
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{{< /sidenote >}}
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of the function dependencies. We define a function \\(f\\) to be dependent on another function \\(g\\)
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if \\(f\\) calls \\(g\\). The transitive closure will help us find functions that are related indirectly.
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For instance, if \\(g\\) also depends on \\(h\\), then the transitive closure of \\(f\\) will
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include \\(h\\), even if \\(f\\) directly doesn't use \\(h\\).
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2. We isolate groups of mutually dependent functions. If \\(f\\) depends on \\(g\\) and \\(g\\) depends \\(f\\),
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they are placed in one group. We then construct a dependency graph __of these groups__.
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3. We compute a topological order of the group graph. This helps us typecheck the dependencies
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of functions before checking the functions themselves. In our specific case, this would ensure
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we check `if` first, and only then move on to `testOne` and `testTwo`. The order of typechecking
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within a group does not matter.
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4. We typecheck the function groups, and functions within them, following the above topological order.
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To find the transitive closure of a graph, we can use [Warshall's Algorithm](https://cs.winona.edu/lin/cs440/ch08-2.pdf).
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