COMP4630: [fg]structure-Calculus - 4. Standardisation · If term Mis in hnf, then it will look ......

COMP4630: λ-Calculus4. Standardisation

Michael [email protected]

Canberra Research Lab., NICTA

Semester 2, 2015

mailto:[email protected]

Last Time

Confluence

I The property that divergent evaluations can rejoin one another

Proof

I Diamond properties

I Uses parallel reduction (=⇒β); and

I Many inductions

Consequences

I Soundness of λ

I (With an analogous proof) Soundness of λη

I Incompleteness of λ

I (See end of lecture 2)

Today

Introduction

Head ReductionWeak Head Reduction

The ProofFailing ApproachesThe Right ApproachConsequences

Conclusion

Objective

From last time, we know that a term M has at most one normalform.

Unfortunately, we also know that not all evaluation strategies willlead to that normal form.

I This is not inconsistent with confluence.

I Why?

Introduction 4/27

Evaluation Strategies

An evaluation strategy is basically a way of answering the question:

Where (i.e., in which sub-term) should I do my nextreduction?

Or:

Where should I do the next bit of work? Some languages (e.g., Java) do not allow for choices to be madeat all. They specify a precise evaluation order.Why would they do that?

Introduction 5/27

Evaluation Strategy #1: Applicative Order

Evaluate everything from the bottom up.

I I.e., in (λv.M) N work will start with M, passing to N andperforming the top-level β-reduction last

A function’s arguments (and the function itself) will be evaluatedbefore the argument is passed to the function.

Also known as strict evaluation.(Used in Fortran, C, Pascal, Ada, SML, Java. . . )

Causes (λx. y) Ω to go into an infinite loop.

Introduction 6/27

Evaluation Strategy #2: Normal Order

Evaluate top-down, left-to-right.

I With (λv.M) N, start by performing the β-reduction,producing M[v := N]

I Find the top-most, left-most β-redex and reduce it.

I Keep going

This strategy is behind the lazy evaluation of languages likeHaskell.

Normal order evaluation will always terminate with a normal formif a term has one. (Proof to come. . . )

Introduction 7/27

Evaluation Strategy Trade-offs

An evaluation strategy might

1. be guaranteed to find normal forms; or

2. aim to perform the least number of β-reductions

Naıvely,

I normal order reduction does 1;

I applicative order “sort of” achieves 2, but gives up on 1

(In fact, optimal reduction is very difficult to get right.)

Introduction 8/27

Proving Normal Order Evaluation

Our focus is in showing that normal order evaluation is guaranteedto find normal forms.

I (That’s why it’s called normal order. . . )

Here are the rules:

(λv.M) N −→n M[v := N]

M −→n M′

(λv.M) −→n (λv.M ′)

M −→n M′ M not an abstraction

M N −→n M′ N

N −→n N′ M not an abstraction M in β-nf

M N −→n M N ′

Introduction 9/27

Head Reduction

We can divide normal order reduction into two different sorts ofreduction.

First, normal order reduction:

(λv.M) N −→n M[v := N]

M −→n M′

(λv.M) −→n (λv.M ′)

M −→n M′ M not an abstraction

M N −→n M′ N


M N −→n M N ′

When head reducing, you never reduce to the right of anapplication

Head Reduction 10/27

Head Reduction

We can divide normal order reduction into two different sorts ofreduction.

Then, head reduction:

(λv.M) N −→h M[v := N]

M −→h M′

(λv.M) −→h (λv.M ′)

M −→h M′ M not an abstraction

M N −→h M′ N

N −→h N′ M not an abstraction M in β-nf

M N −→h M N ′

When head reducing, you never reduce to the right of anapplication


Hence, Head Normal Forms

Head Normal Form is a reasonable stopping place.

Rules again:

(λv.M) N h−→M[v := N]

M h−→M ′

(λv.M) h−→ (λv.M ′)

M h−→M ′ M not an abstraction

M N h−→M ′ N

Examples:

I v is in hnf

I (λv. v) is in hnf

I (λvw. v (λu.M) N) is in hnf

I (λuwz. v ((λu.M) N)) is in hnf


Head Normal Forms, Generally

If term M is in hnf, then it will look like:

(λ~v. uM1 · · ·Mn)

The vector ~v may be empty, u may be free or bound, andthe number of extra arguments, n, may be 0.

Once a term is in hnf, its top-level structure can’t change.


After Head Reductions

Once a term is in hnf, its top-level structure can’t change.

Any further reductions (of any sort) inside

(λ~v. uM1 · · ·Mn)

will be reductions within an Mi.

Each argument will evolve independently, and the number ofarguments can’t change.

These are internal reductions i−→.

I If the term is (λ~v. (λu.M)N1N2 . . . ) (not in hnf) and Mreduces, then that is an internal reduction too

I All reductions are either head or internal


Normal Order Reduction Splits in Two

The last rule of normal order reduction(which we deleted to get head reduction):


M N −→n M N ′

If M −→∗n N, and the reductions aren’t all head, then there mustbe a first head normal form P, such that

M h−→∗ P i−→∗ NWe want to show the same sort of split for arbitraryβ-reduction (−→∗β)


Interlude: Weak Head Reduction

Thanks to:M h−→M ′

(λv.M) h−→ (λv.M ′)

head reduction proceeds inside function bodies.

If you take this rule out, you get weak head reduction.

Weak head normal forms are

I head normal forms, or

I abstractions.

Weak head reduction is used in implementations of functionalprogramming languages.

Head Reduction Weak Head Reduction 15/27

Basic Strategy

We want to know that, if by some path:

M −→∗β Nwith N a normal form, then normal order reduction willtake M to N too.

Will do this by showing a more general result.

That for any N, if M −→∗β N, then there exists a P such that

M h−→∗ P i−→∗ N

The Proof 16/27

We Want to Commute Steps

If we had

M

N

P

i

h

N ′

h

∗

i

∗

we’d like to know that there was a N ′ that we could get to viahead reduction, and from which we could make internal reductionsto get to P (maybe with multiple steps?).

Maybe this would allow head and internal steps to be“bubble-sorted” so that all head steps came first.

The Proof Failing Approaches 17/27


If we had

M

N

P

i

h

N ′

h

∗

i

∗

we’d like to know that there was a N ′ that we could get to viahead reduction, and from which we could make internal reductionsto get to P

(maybe with multiple steps?).




If we had

M

N

P

i

h

N ′

h

∗

i∗

we’d like to know that there was a N ′ that we could get to viahead reduction, and from which we could make internal reductionsto get to P (maybe with multiple steps?).



But Direct Commuting is Hard

Commuting does require multiple steps:

(λx. f x x) ((λy. y z) u)

(λx. f x x) (u z)

f (u z) (u z)

h

i

f ((λy. y z) u) ((λy. y z) u)

f (u z) ((λy. y z) u)

f (u z) (u z)

i

i

f ((λy. y z) u) (u z)

f (u z) (u z)

i

i

h

This example requires multiple (2) internal reductions.


But Direct Commuting is Hard

Commuting does require multiple steps:

(λu. (λv. v u z) f) N

(λu. f u z) N

f N z

h

i

(λv. v N z) f

f N z

h

h

This example requires multiple head reductions (and no internals).


Commuting with Multiple Steps Isn’t Good Enough

This is a theorem:

M i−→ N h−→ P

=⇒∃N ′. M h−→∗ N ′ i−→∗ P

But it’s not good enough.

Our examples show us that we can have

ih → hi2

ih → h2


The Bubble-Sort That Never Ends

Our examples show us that we can have

ih → hi2

ih → h2

Start with a reduction sequence ihihi:

ihihi → hi3hi→ hi2h2i→ hihi2hi→ · · ·

This is not progress: we still have two head reductions that haven’tbeen “sorted” to the start of the sequence.


A Better Lemma

Though commuting internal and head reductions can result in

multiple internal reductions, the latter are all parallel (write i=⇒ ).

So, prove instead:

M

P

i

N

h

P ′

h

∗

i

If a internal parallel reduction is followed by a head reduction,

there is an alternative route where head reductions come first, andthere is one internal parallel reduction afterwards.

The Proof The Right Approach 21/27

A Better Lemma

Though commuting internal and head reductions can result in

multiple internal reductions, the latter are all parallel (write i=⇒ ).

So, prove instead:

M

P

i

N

h

P ′

h

∗

i

If a internal parallel reduction is followed by a head reduction,there is an alternative route where head reductions come first, andthere is one internal parallel reduction afterwards.


Proof in More Detail

Have

M i=⇒ P h−→ N

As P head-reduces it is (λ~v. (λu. P0)P1 P2 · · ·Pn), with n ≥ 1

And M is of form (λ~v. (λu.M0)M1M2 · · ·Mn), with Mi =⇒β Pi

SoM h−→ (λ~v.M0[u :=M1]M2 · · ·Mn)

=⇒β N

Last transition is =⇒β, not i=⇒ because M0’s reduction may beat top level, making it head.

A little more work is still required(decomposing =⇒β into head and internal parts).


The Last Big Lemma

If M =⇒β N, then there are Mi such that

M h−→M1h−→M2 · · · h−→Mn

i=⇒ N

and eachMi =⇒β N

This gives:

M N

P

β

h

∗

i


Putting the Lemmas Together

Proving: M −→∗β N =⇒ ∃P. M h−→∗ P i−→∗ NHave M −→∗β N, and so also M =⇒∗β N.

(Base case of zero steps trivial.)

By Last Big Lemma, also have P1 s.t. M h−→∗ P1 i=⇒ M ′

By inductive hypothesis, have P2 s.t. M ′ h−→∗ P2 i−→∗ NI.e.,

M h−→∗ P1 i=⇒ M ′ h−→∗ P2 i−→∗ NNow, we can “bubble” head reductions after M ′ up over the i=⇒ ,using the Better Lemma.




So, assume M =⇒β M′ =⇒∗β N.










M h−→∗ P1 i=⇒ M ′ h−→∗ P2 i−→∗ N

Now, we can “bubble” head reductions after M ′ up over the i=⇒ ,using the Better Lemma.


Consequences: Standardisation

Have shown: M −→∗β N =⇒ ∃P. M h−→∗ P i−→∗ NIt’s obviously possible to order the internal reductions so that theyoccur left-to-right.

I By induction.The internal terms within N are all smaller than N itself,so the internal reductions within each Ni can themselves besorted appropriately.

Gives Standardisation:

If M −→∗β N is possible, then N can be reached from M

in a “standard” way (doing reductions in left to rightorder)

The Proof Consequences 25/27

Consequences: Normal Order Evaluation Works

Recall that normal order evaluation is a “standard” evaluationstrategy that does all possible reductions.

If M can reduce to N, a β-normal form, then there is a standardreduction that does the same.

If a standard reduction terminates in a β-normal form, it has doneall possible reductions.

And so that standard reduction was a normal order reduction.

So, normal order evaluation finds normal forms if they exist.

The Proof Consequences 26/27

Summary

An involved proof.

Lesson #1: The λ-calculus is a plausible programming language

I there is an algorithm for turning λ-terms into values

I (when those terms have values at all)

Lesson #2: Evaluation Orders are a Design QuestionI Do you want to guarantee as much termination as possible?

I Use normal order

I Or, do you want more speed, and unnecessarynon-terminations?

I Use applicative order (like C, Java etc)

Next time: adding plausibility.Numbers, Pairs and Lists for the λ-calculus.

Conclusion 27/27

COMP4630: [fg]structure-Calculus - 4. Standardisation · If term Mis in hnf, then it will look ......

Documents

Transcript of COMP4630: [fg]structure-Calculus - 4. Standardisation · If term Mis in hnf, then it will look ......