From a7cb9446e69306f6cc533e5558f9bb323d3bdb35 Mon Sep 17 00:00:00 2001 From: jim Date: Mon, 23 Feb 2015 19:32:17 -0500 Subject: [PATCH] cleanup --- topics/_week3_eval_order.mdwn | 128 +++++++++++++++++++++--------------------- 1 file changed, 64 insertions(+), 64 deletions(-) diff --git a/topics/_week3_eval_order.mdwn b/topics/_week3_eval_order.mdwn index 26a4b04d..453bbfb4 100644 --- a/topics/_week3_eval_order.mdwn +++ b/topics/_week3_eval_order.mdwn @@ -1,31 +1,29 @@ Evaluation Strategies and Normalization ======================================= -In the assignment we asked you to reduce various expressions until it wasn't possible to reduce them any further. For two of those expressions, this was impossible to do. One of them was this: +In last week's homework, we asked you to reduce various expressions until it wasn't possible to reduce them any further. For two of those expressions, this was impossible to do. One of them was this: (\x. x x) (\x. x x) -As we saw above, each of the halves of this formula are the combinator ω; so this can also be written: +As we discuss in other notes for this week, each of the halves of this formula are the combinator `Ï`; so this can also be written: -
ω ω
+ Ï Ï -This compound expression---the self-application of ω---is named Ω. It has the form of an application of an abstract (ω) to an argument (which also happens to be ω), so it's a redex and can be reduced. But when we reduce it, we get ω ω again. So there's no stage at which this expression has been reduced to a point where it can't be reduced any further. In other words, evaluation of this expression "never terminates." (This is the standard language, however it has the unfortunate connotation that evaluation is a process or operation that is performed in time. You shouldn't think of it like that. Evaluation of this expression "never terminates" in the way that the decimal expansion of π never terminates. These are static, atemporal facts about their mathematical properties.) +This compound expression---the self-application of `Ï`---is named `Î©`. It has the form of an application of an abstract (`Ï`) to an argument (which also happens to be `Ï`), so it's a redex and can be reduced. But when we reduce it, we get `Ï Ï` again. So there's no stage at which this expression has been reduced to a point where it can't be reduced any further. In other words, evaluation of this expression "never terminates." (This is the standard language, however it has the unfortunate connotation that evaluation is a process or operation that must be performed over time. You shouldn't think of it like that. Evaluation of this expression "never terminates" in the way that the decimal expansion of π never terminates. These are static, atemporal facts about their mathematical properties.) -There are infinitely many formulas in the lambda calculus that have this same property. Ω is the syntactically simplest of them. In our meta-theory, it's common to assign such formulas a special value, , pronounced "bottom." When we get to discussing types, you'll see that this value is counted as belonging to every type. To say that a formula has the bottom value means that the computation that formula represents never terminates and so doesn't evaluate to any orthodox, computed value. +There are infinitely many formulas in the Lambda Calculus that have this same property. `Î©` is the syntactically simplest of them. In our meta-theory, it's common to assign such formulas a special meaning, , pronounced "bottom." When we get to discussing types, you'll see that expressions with this meaning are counted as belonging to every type. To say that a term "is bottom" means that the computation that term represents never terminates, and so the term doesn't evaluate to any orthodox, computed value. From a "Fregean" or "weak Kleene" perspective, if any component of an expression fails to be evaluable (to an orthodox, computed value), then the whole expression should be unevaluable as well. However, in some such cases it seems *we could* sensibly carry on evaluation. For instance, consider: -

-(\x. y) (ω ω)
-
+ (\x. y) (Ï Ï) -Should we count this as unevaluable, because the reduction of (ω ω) never terminates? Or should we count it as evaluating to `y`? +Should we count this as unevaluable, because the reduction of `(Ï Ï)` never terminates? Or should we count it as evaluating to `y`? This question highlights that there are different choices to make about how evaluation or computation proceeds. It's helpful to think of three questions in this neighborhood: -> Q1. When arguments are complex, as (ω ω) is, do we reduce them before substituting them into the abstracts to which they are arguments, or later? +> Q1. When arguments are complex, as `(Ï Ï)` is, do we reduce them before substituting them into the abstracts to which they are arguments (`(\x. y)`, in the expression displayed above), or later? > Q2. Are we allowed to reduce inside abstracts? That is, can we reduce: @@ -48,25 +46,21 @@ This question highlights that there are different choices to make about how eval With regard to Q3, it should be intuitively clear that `\x. M x` and `M` will behave the same with respect to any arguments they are given. It can also be proven that no other functions can behave -differently with respect to them (that is, no function can behave -differently depending on whether its argument is `M` versus `\x. M -x`). However, the logical system you get when eta-reduction is added -to the proof theory is importantly different from the one where only -beta-reduction is permitted. +differently with respect to them (that is, no function can interact differently with further arguments depending on whether it has been applied to `M` versus `\x. M x`). However, the logical system you get when eta-reduction is added to the proof theory is importantly different from the one where only beta-reduction is permitted. -If we answer Q2 by permitting reduction inside abstracts, and we also permit eta-reduction, then where none of y1, ..., yn occur free in M, this: +If we answer Q2 by permitting reduction inside abstracts, and we also permit eta-reduction, then where none of y1, ..., yn occur free in `M`, this: -
\x y1... yn. M y1... yn
+
\x y1...yn. M y1...yn
-(\x.I)((\x.xxx)(\x.xxx))
- |      \
- I       (\x.w)((\x.xxx)(\x.xxx)(\x.xxx))
-          /      \
-         I        (\x.w)((\x.xxx)(\x.xxx)(\x.xxx)(\x.xxx))
-                   /       \
-                  I         etc.
+(\x.I) ((\x.xxx)(\x.xxx))
+ |         \
+ |      (\x.I) ((\x.xxx)(\x.xxx)(\x.xxx))
+ |        /      \
+ |-------+    (\x.I) ((\x.xxx)(\x.xxx)(\x.xxx)(\x.xxx))
+ |              /       \
+ I ------------+         etc.

If we start by evaluating the leftmost redex, we're done in one step. @@ -151,22 +143,22 @@ rightmost redex, creating an even longer term. Clearly, in this situation, we want to prioritize reducing the leftmost redex in order to arrive at a stable result more quickly. -Indeed, it's provable that if there's *any* reduction path that delivers a value for a given expression, the normal-order evalutation strategy will terminate with that value. +Indeed, it's provable for the untyped Lambda Calculus that if there's *any* reduction path that delivers a value for a given expression, the normal-order evaluation strategy will terminate with that value. An expression is said to be in **normal form** when it's not possible to perform any more reductions (not even inside abstracts). -There's a sense in which you *can't get anything more out of* ω ω, but it's not in normal form because it still has the form of a redex. +There's a sense in which you *can't get anything more out of* `Ï Ï`, but it's not in normal form because it still has the form of a redex. A computational system is said to be **confluent**, or to have the **Church-Rosser** or **diamond** property, if, whenever there are multiple possible evaluation paths, those that terminate always terminate in the same value. In such a system, the choice of which sub-expressions to evaluate first will only matter if some of them but not others might lead down a non-terminating path. The diamond property is named after the shape of diagrams like the following:
-                     ((\x.x)((\y.y) z))
-                       /      \
-                      /        \
-                     /          \
-                    /            \
-                    ((\y.y) z)   ((\x.x) z)
+                     ((\x.x) ((\y.z) y))
+                       /        \
+                      /          \
+                     /            \
+                    /              \
+                  ((\y.z) y)   ((\x.x) z)
\             /
\           /
\         /
@@ -179,23 +171,25 @@ The diamond property is named after the shape of diagrams like the following:
Because the starting term contains more than one redex, we can imagine
reducing the leftmost redex first (the left branch of the diagram) or
else the rightmost redex (the right branch of the diagram).  But
-because the lambda calculus is confluent, no matter which lambda you
+because the Lambda Calculus is confluent, no matter which lambda you
choose to target for reduction first, you end up at the same place.
-It's like travelling in Manhattan: if you walk uptown first and then
+It's like traveling in (much of) Manhattan: if you walk uptown first and then
head east, you end up in the same place as if you walk east and then

-The untyped lambda calculus is confluent. So long as a computation terminates, it always terminates in the same way. It doesn't matter which order the sub-expressions are evaluated in.
+As we mentioned, the untyped Lambda Calculus is confluent. So long as a computation terminates, it always terminates in the same way. It doesn't matter which order the sub-expressions are evaluated in.
+
+A computational system is said to be **strongly normalizing** if *every* permitted evaluation path is guaranteed to terminate. The untyped Lambda Calculus is *not* strongly normalizing: `Ï Ï` doesn't terminate by any evaluation path; and `(\x. y) (Ï Ï)` terminates only by some evaluation paths but not by others.

-A computational system is said to be **strongly normalizing** if every permitted evaluation path is guaranteed to terminate. The untyped lambda calculus is not strongly normalizing: ω ω doesn't terminate by any evaluation path; and (\x. y) (ω ω) terminates only by some evaluation paths but not by others.
+But the untyped Lambda Calculus enjoys some compensation for this weakness. It's Turing Complete! It can represent any computation we know how to describe. (That's the cash value of being Turing Complete, not the rigorous definition. There is a rigorous definition. However, we don't know how to rigorously define "any computation we know how to describe.") And in fact, it's been proven that you can't have both. If a computational system is Turing Complete, it cannot be strongly normalizing.

-But the untyped lambda calculus enjoys some compensation for this weakness. It's Turing complete! It can represent any computation we know how to describe. (That's the cash value of being Turing complete, not the rigorous definition. There is a rigrous definition. However, we don't know how to rigorously define "any computation we know how to describe.") And in fact, it's been proven that you can't have both. If a computational system is Turing complete, it cannot be strongly normalizing.
+A computational system is said to be **weakly normalizing** if there's always guaranteed to be *at least one* evaluation path that terminates. The untyped Lambda Calculus is not weakly normalizing either, as we've seen.

-A computational system is said to be **weakly normalizing** if there's always guaranteed to be *at least one* evaluation path that terminates. The untyped lambda calculus is not weakly normalizing either, as we've seen.
+The *simply-typed* lambda calculus that linguists traditionally work with, on the other hand, is strongly normalizing. (And as a result, is not Turing Complete.) It has expressive power (concerning types) that the untyped Lambda Calculus does not natively possess, but it is also unable to represent some (terminating!) computations that the untyped Lambda Calculus can represent.

-The *typed* lambda calculus that linguists traditionally work with, on the other hand, is strongly normalizing. (And as a result, is not Turing complete.) It has expressive power (concerning types) that the untyped lambda calculus lacks, but it is also unable to represent some (terminating!) computations that the untyped lambda calculus can represent.
+(Interestingly, since the untyped Lambda Calculus *is* Turing Complete, any algorithmic manipulations we can do the simply-typed (or any other) lambda calculus can in principle be *encoded in* operations in the untyped Lambda Calculus. It then becomes an interesting question what we could mean, when we want to say things like "This typed lambda calculus has an expressive power not (natively) present in the untyped Lambda Calculus.")

-Other more-powerful type systems we'll look at in the course will also fail to be Turing complete, though they will turn out to be pretty powerful.
+Other more-powerful type systems we'll look at in the course will also fail to be Turing Complete, though they will turn out to be pretty powerful.

@@ -205,25 +199,25 @@ Further reading:
*	[[!wikipedia Strict programming language]]

* [[!wikipedia Church-Rosser theorem]] * [[!wikipedia Normalization property]] -* [[!wikipedia Turing completeness]] +* [[!wikipedia Turing Completeness]] Decidability ============ -The question whether two formulas are syntactically equal is "decidable": we can construct a computation that's guaranteed to always give us the answer. +The question whether two formulas are syntactically equal is "decidable": we can construct a computation that's guaranteed to always give us the correct answer. What about the question whether two formulas are convertible? Well, to answer that, we just need to reduce them to normal form, if possible, and check whether the results are syntactically equal. The crux is that "if possible." Some computations can't be reduced to normal form. Their evaluation paths never terminate. And if we just kept trying blindly to reduce them, our computation of what they're convertible to would also never terminate. So it'd be handy to have some way to check in advance whether a formula has a normal form: whether there's any evaluation path for it that terminates. -Is it possible to do that? Sure, sometimes. For instance, check whether the formula is syntactically equal to Ω. If it is, it never terminates. +Is it possible to do that? Sure, sometimes. For instance, check whether the formula is syntactically equal to `Î©`. If it is, it never terminates. -But is there any method for doing this in general---for telling, of any given computation, whether that computation would terminate? Unfortunately, there is not. Church proved this in 1936; Turing also essentially proved it at the same time. Geoff Pullum gives a very reader-friendly outline of the proofs here: +But is there any method for doing this in general---for telling, of any given computation, whether that computation would terminate? Unfortunately, there is not. Church proved this in 1936; Turing also essentially proved it at the same time. Geoff Pullum gives a very reader-friendly outline of Turing's proof here: * [Scooping the Loop Snooper](http://www.cl.cam.ac.uk/teaching/0910/CompTheory/scooping.pdf), a proof of the undecidability of the halting problem in the style of Dr Seuss by Geoffrey K. Pullum -Interestingly, Church also set up an association between the lambda calculus and first-order predicate logic, such that, for arbitrary lambda formulas `M` and `N`, some formula would be provable in predicate logic iff `M` and `N` were convertible. So since the right-hand side is not decidable, questions of provability in first-order predicate logic must not be decidable either. This was the first proof of the undecidability of first-order predicate logic. +Interestingly, Church also set up an association between the Lambda Calculus and first-order predicate logic, such that, for arbitrary lambda formulas `M` and `N`, some formula would be provable in predicate logic iff `M` and `N` were convertible. So since the right-hand side is not decidable, questions of provability in first-order predicate logic must not be decidable either. This was the first proof of the undecidability of first-order predicate logic. ## Efficiency @@ -233,9 +227,9 @@ course, it will have implications for the order in which side effects occur.)

-                      ((\x.w)((\y.y) z))
-                        \      \
-                         \      ((\x.w) z)
+                     ((\x.w) ((\y.z) y))
+                        \           \
+                         \   ((\x.w) z)
\       /
\     /
\   /
@@ -283,3 +277,9 @@ rightmost reducible lambda" is more efficient.

So evaluation strategies have a strong effect on how many reduction
steps it takes to arrive at a stopping point (e.g., normal form).
+
+## Side-effects
+
+In the systems we're currently working in, there are no side-effects. (Though the possibility of non-terminating computations like `Ï Ï` has some similarities to side-effects.)
+
+When we later consider systems where the evaluation of some expressions *does* have side-effects, then the choice of evaluation strategy will make a big difference to which side-effects occur, and in what order.
--
2.11.0