1 <!-- λ Λ ∀ ≡ α β ω Ω -->
5 # Reasoning about evaluation order in Combinatory Logic
7 We've discussed [[evaluation order|topics/week3_evaluation_order]]
8 before, primarily in connection with the untyped lambda calculus.
9 Whenever a term contains more than one redex, we have to choose which
10 one to reduce, and this choice can make a difference. For instance,
13 Ω == ωω == (\x.xx)(\x.xx), so
15 ((\x.I)Ω) == ((\x.I)((\x.xx)(\x.xx)))
18 There are two redexes in this term; we've marked the operative lambdas
19 with a star. If we reduce the leftmost redex first, the term reduces
20 to the normal form `I` in one step. But if we reduce the rightmost
21 redex instead, the "reduced" form is `(\x.I)Ω` again, and we are in
22 danger of entering an infinite loop.
24 Thanks to the introduction of sum types (disjoint union) in the last lecture, we
25 are now in a position to gain a deeper understanding of evaluation
26 order by writing a program that allows us to reasoning explicitly about evaluation.
28 One thing we'll see is that it is all too easy for the evaluation
29 order properties of an evaluator to depend on the evaluation order
30 properties of the programming language in which the evaluator is
31 written. We would like to write an evaluator in which the order of
32 evaluation is insensitive to the evaluator language. The goal is to
33 find an order-insensitive way to reason about evaluation order. We
34 will not fully succeed in this first attempt, but we will make good
37 The first evaluator we will develop will evaluate terms in Combinatory
38 Logic. This significantly simplifies the discussion, since we won't
39 need to worry about variables or substitution. As we develop and
40 extend our evaluator in future weeks, we'll switch to lambdas, but for
41 now, working with the simplicity of Combinatory Logic will make it
42 easier to highlight evaluation order issues.
44 A brief review of Combinatory Logic: a term in CL is the combination
45 of three basic expressions, `S`, `K`, and `I`, governed by the
46 following reduction rules:
52 where `a`, `b`, and `c` stand for an arbitrary term of CL. We've seen
53 how to embed the untyped lambda calculus in CL, so it's no surprise
54 that evaluation order issues arise in CL. To illustrate, we'll use
55 the following definition:
58 Skomega = skomega skomega == SII(SII)
62 We'll use the same symbol, `Ω`, for Omega and Skomega: in a lambda
63 term, `Ω` refers to Omega, but in a CL term, `Ω` refers to Skomega as
66 Just as in the corresponding term in the lambda calculus, CL terms can
67 contain more than one redex:
72 we can choose to reduce the leftmost redex by applying the reduction
73 rule for `K`, in which case the term reduces to the normal form `I` in
74 one step; or we can choose to reduce the Skomega part, by applying the
75 reduction rule `S`, in which case we do not get a normal form, and
76 we're headed towards an infinite loop.
78 With sum types, we can define CL terms in OCaml as follows:
80 type term = I | K | S | App of (term * term)
82 let skomega = App (App (App (S, I), I), App (App (S, I), I))
84 This type definition says that a term in CL is either one of the three
85 simple expressions (`I`, `K`, or `S`), or else a pair of CL
86 expressions. `App` stands for Functional Application. With this type
87 definition, we can encode skomega, as well as other terms whose
88 reduction behavior we want to try to control.
90 Using pattern matching, it is easy to code the one-step reduction
93 let reduce_one_step (t:term):term = match t with
95 | App(App(K,a),b) -> a
96 | App(App(App(S,a),b),c) -> App(App(a,c),App(b,c))
99 # reduce_one_step (App(App(K,S),I));;
101 # reduce_one_step skomega;;
102 - : term = App (App (I, App (App (S, I), I)), App (I, App (App (S, I), I)))
104 The definition of `reduce_one_step` explicitly says that it expects
105 its input argument `t` to have type `term`, and the second `:term`
106 says that the type of the output it delivers as a result will be of
109 The type constructor `App` obscures things a bit, but it's still
110 possible to see how the one-step reduction function is just the
111 reduction rules for CL. The OCaml interpreter shows us that the
112 function faithfully recognizes that `KSI ~~> S`, and `skomega ~~>
115 We can now say precisely what it means to be a redex in CL.
117 let is_redex (t:term):bool = not (t = reduce_one_step t)
121 # is_redex (App(K,I));;
123 # is_redex (App(App(K,I),S));;
128 Warning: this definition relies on the accidental fact that the
129 one-step reduction of a CL term is never identical to the original
130 term. This would not work for the untyped lambda calculus, since
131 `((\x.xx)(\x.xx)) ~~> ((\x.xx)(\x.xx))` in one step.
133 Note that in order to decide whether two terms are equal, OCaml has to
134 recursively compare the elements of complex CL terms. It is able to
135 figure out how to do this because we provided an explicit definition
136 of the datatype `term`.
138 As you would expect, a term in CL is in normal form when it contains
139 no redexes (analogously for head normal form, weak head normal form, etc.)
141 In order to fully reduce a term, we need to be able to reduce redexes
142 that are not at the top level of the term.
143 Because we need to process subparts, and because the result after
144 processing a subpart may require further processing, the recursive
145 structure of our evaluation function has to be somewhat subtle. To
146 truly understand, you will need to do some sophisticated thinking
147 about how recursion works.
149 We'll develop our full reduction function in two stages. Once we have
150 it working, we'll then consider a variant.
152 let rec reduce_stage1 (t:term):term =
153 if (is_redex t) then reduce_stage1 (reduce_one_step t)
156 If the input is a redex, we ship it off to `reduce_one_step` for
157 processing. But just in case the result of the one-step reduction is
158 itself a redex, we recursively call `reduce_stage1`. The recursion
159 will continue until the result is no longer a redex. We're aiming at
160 allowing the evaluator to recognize that
162 I (I K) ~~> I K ~~> K
164 When trying to understand how recursive functions work, it can be
165 extremely helpful to examining an execution trace of inputs and
168 # #trace reduce_stage1;;
169 reduce_stage1 is now traced.
170 # reduce_stage1 (App (I, App (I, K)));;
171 reduce_stage1 <-- App (I, App (I, K))
172 reduce_stage1 <-- App (I, K)
179 In the trace, "`<--`" shows the input argument to a call to
180 `reduce_stage1`, and "`-->`" shows the output result.
182 Since the initial input (`I(IK)`) is a redex, the result after the
183 one-step reduction is `IK`. We recursively call `reduce_stage1` on
184 this input. Since `IK` is itself a redex, the result after one-step
185 reduction is `K`. We recursively call `reduce_stage1` on this input. Since
186 `K` is not a redex, the recursion bottoms out, and we return the
189 But this function doesn't do enough reduction. We want to recognize
190 the following reduction path:
194 But the reduction function as written above does not deliver this result:
196 # reduce_stage1 (App (App (I, I), K));;
197 - : term = App (App (I, I), K)
199 Because the top-level term is not a redex to start with,
200 `reduce_stage1` returns it without any evaluation. What we want is to
201 evaluate the subparts of a complex term. We'll do this by evaluating
202 the subparts of the top-level expression.
204 let rec reduce (t:term):term = match t with
209 let t' = App (reduce a, reduce b) in
210 if (is_redex t') then reduce 2 (reduce_one_step t')
213 Since we need access to the subterms, we do pattern matching on the
214 input. If the input is simple (the first three `match` cases), we
215 return it without further processing. But if the input is complex, we
216 first process the subexpressions, and only then see if we have a redex
217 at the top level. To understand how this works, follow the trace
220 # reduce (App(App(I,I),K));;
221 reduce <-- App (App (I, I), K)
223 reduce <-- K ; first main recursive call
226 reduce <-- App (I, I) ; second main recursive call
240 Ok, there's a lot going on here. Since the input is complex, the
241 first thing the function does is construct `t'`. In order to do this,
242 it must reduce the two main subexpressions, `II` and `K`.
244 There are three recursive calls to the `reduce` function, each of
245 which gets triggered during the processing of this example. They have
246 been marked in the trace.
248 The don't quite go in the order in which they appear in the code,
249 however! We see from the trace that it begins with the right-hand
250 expression, `K`. We didn't explicitly tell it to begin with the
251 right-hand subexpression, so control over evaluation order is starting
252 to spin out of our control. (We'll get it back later, don't worry.)
254 In any case, in the second main recursive call, we evaluate `II`. The
257 At this point, we have constructed `t' == App(I,K)`. Since that's a
258 redex, we ship it off to reduce_one_step, getting the term `K` as a
259 result. The third recursive call checks that there is no more
260 reduction work to be done (there isn't), and that's our final result.
262 You can see in more detail what is going on by tracing both reduce
263 and reduce_one_step, but that makes for some long traces.
265 So we've solved our first problem: reduce recognizes that `IIK ~~>
268 Because the OCaml interpreter evaluates each subexpression in the
269 course of building `t'`, however, it will always evaluate the right
270 hand subexpression, whether it needs to or not. And sure enough,
272 # reduce (App(App(K,I),skomega));;
275 Running the evaluator with this input leads to an infinite loop, and
276 the only way to get out is to kill the interpreter with control-c.
278 Instead of performing the leftmost reduction first, and recognizing
279 that this term reduces to the normal form `I`, we get lost endlessly
280 trying to reduce skomega.
282 ## Laziness is hard to overcome
284 To emphasize that our evaluation order here is at the mercy of the
285 evaluation order of OCaml, here is the exact same program translated
286 into Haskell. We'll put them side by side to emphasize the exact parallel.
290 ========================================================== =========================================================
292 type term = I | S | K | App of (term * term) data Term = I | S | K | App Term Term deriving (Eq, Show)
294 let skomega = App (App (App (S,I), I), App (App (S,I), I)) skomega = (App (App (App S I) I) (App (App S I) I))
296 reduce_one_step :: Term -> Term
297 let reduce_one_step (t:term):term = match t with reduce_one_step t = case t of
298 App(I,a) -> a App I a -> a
299 | App(App(K,a),b) -> a App (App K a) b -> a
300 | App(App(App(S,a),b),c) -> App(App(a,c),App(b,c)) App (App (App S a) b) c -> App (App a c) (App b c)
303 is_redex :: Term -> Bool
304 let is_redex (t:term):bool = not (t = reduce_one_step t) is_redex t = not (t == reduce_one_step t)
306 reduce :: Term -> Term
307 let rec reduce (t:term):term = match t with reduce t = case t of
311 | App (a, b) -> App a b ->
312 let t' = App (reduce a, reduce b) in let t' = App (reduce a) (reduce b) in
313 if (is_redex t') then reduce (reduce_one_step t') if (is_redex t') then reduce (reduce_one_step t')
317 There are some differences in the way types are made explicit, and in
318 the way terms are specified (`App(a,b)` for Ocaml versus `App a b` for
319 Haskell). But the two programs are essentially identical.
321 Yet the Haskell program finds the normal form for `KIΩ`:
323 *Main> reduce (App (App K I) skomega)
326 Woa! First of all, this is wierd. Haskell's evaluation strategy is
327 called "lazy". Apparently, Haskell is so lazy that even after we've
328 asked it to construct t' by evaluating `reduce a` and `reduce b`, it
329 doesn't bother computing `reduce b`. Instead, it waits to see if we
330 ever really need to use the result.
332 So the program as written does NOT fully determine evaluation order
333 behavior. At this stage, we have defined an evaluation order that
334 still depends on the evaluation order of the underlying interpreter.
336 There are two questions we could ask:
338 * Can we adjust the OCaml evaluator to exhibit lazy behavior?
340 * Can we adjust the Haskell evaluator to exhibit eager behavior?
342 The answer to the first question is easy and interesting, and we'll
343 give it right away. The answer to the second question is also
344 interesting, but not easy. There are various tricks available in
345 Haskell we could use (such as the `seq` operator, or the `deepseq`
346 operator), but a fully general, satisifying resolution will have to
347 wait until we have Continuation Passing Style transforms.
349 The answer to the first question (Can we adjust the OCaml evaluator to
350 exhibit lazy behavior?) is quite simple:
353 let rec reduce_lazy (t:term):term = match t with
358 let t' = App (reduce_lazy a, b) in
359 if (is_redex t') then reduce_lazy (reduce_one_step t')
363 There is only one small difference: instead of setting `t'` to `App
364 (reduce a, reduce b)`, we omit one of the recursive calls, and have
365 `App (reduce a, b)`. That is, we don't evaluate the right-hand
366 subexpression at all. Ever! The only way to get evaluated is to
367 somehow get into functor position.
369 # reduce3 (App(App(K,I),skomega));;
374 The evaluator now has no trouble finding the normal form for `KIΩ`,
375 but evaluating skomega still gives an infinite loop.
377 We can now clarify the larger question at the heart of
380 *How can we can we specify the evaluation order of a computational
381 system in a way that is completely insensitive to the evaluation order
382 of the specification language?*
384 As a final note, we should mention that the evaluators given here are
385 absurdly inefficient computationally. Some computer scientists have
386 trouble even looking at code this inefficient, but the emphasis here
387 is on getting the concepts across as simply as possible.