moved fixed point for succ to exercises

[lambda.git] / topics / _week4_fixed_point_combinator.mdwn

diff --git a/topics/_week4_fixed_point_combinator.mdwn b/topics/_week4_fixed_point_combinator.mdwn
index bc714f3..c93a16b 100644 (file)
--- a/topics/_week4_fixed_point_combinator.mdwn
+++ b/topics/_week4_fixed_point_combinator.mdwn
@@ -19,15 +19,15 @@ How could we compute the length of a list? Without worrying yet about what lambd
 
 In OCaml, you'd define that like this:
 
 
 In OCaml, you'd define that like this:
 

-       let rec get_length = fun lst ->
-               if lst == [] then 0 else 1 + get_length (tail lst)
-       in ... (* here you go on to use the function "get_length" *)
+       let rec length = fun lst ->
+               if lst == [] then 0 else 1 + length (tail lst)
+       in ... (* here you go on to use the function "length" *)

 
 In Scheme you'd define it like this:
 
 
 In Scheme you'd define it like this:
 

-       (letrec [(get_length
-                               (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )]
-               ... ; here you go on to use the function "get_length"
+       (letrec [(length
+                               (lambda (lst) (if (null? lst) 0 [+ 1 (length (cdr lst))] )) )]
+               ... ; here you go on to use the function "length"

        )
 
 Some comments on this:
        )
 
 Some comments on this:


@@ -36,67 +36,67 @@ Some comments on this:
 
 2. `cdr` is function that gets the tail of a Scheme list. (By definition, it's the function for getting the second member of an ordered pair. It just turns out to return the tail of a list because of the particular way Scheme implements lists.)
 
 
 2. `cdr` is function that gets the tail of a Scheme list. (By definition, it's the function for getting the second member of an ordered pair. It just turns out to return the tail of a list because of the particular way Scheme implements lists.)
 

-3.     I use `get_length` instead of the convention we've been following so far of hyphenated names, as in `make-list`, because we're discussing OCaml code here, too, and OCaml doesn't permit the hyphenated variable names. OCaml requires variables to always start with a lower-case letter (or `_`), and then continue with only letters, numbers, `_` or `'`. Most other programming languages are similar. Scheme is very relaxed, and permits you to use `-`, `?`, `/`, and all sorts of other crazy characters in your variable names.
+3.     I use `length` instead of the convention we've been following so far of hyphenated names, as in `make-list`, because we're discussing OCaml code here, too, and OCaml doesn't permit the hyphenated variable names. OCaml requires variables to always start with a lower-case letter (or `_`), and then continue with only letters, numbers, `_` or `'`. Most other programming languages are similar. Scheme is very relaxed, and permits you to use `-`, `?`, `/`, and all sorts of other crazy characters in your variable names.

 
 4.     I alternate between `[ ]`s and `( )`s in the Scheme code just to make it more readable. These have no syntactic difference.
 
 
 The main question for us to dwell on here is: What are the `let rec` in the OCaml code and the `letrec` in the Scheme code?
 
 
 4.     I alternate between `[ ]`s and `( )`s in the Scheme code just to make it more readable. These have no syntactic difference.
 
 
 The main question for us to dwell on here is: What are the `let rec` in the OCaml code and the `letrec` in the Scheme code?
 

-Answer: These work like the `let` expressions we've already seen, except that they let you use the variable `get_length` *inside* the body of the function being bound to it---with the understanding that it will there refer to the same function that you're then in the process of binding to `get_length`. So our recursively-defined function works the way we'd expect it to. In OCaml:
+Answer: These work like the `let` expressions we've already seen, except that they let you use the variable `length` *inside* the body of the function being bound to it---with the understanding that it will there refer to the same function that you're then in the process of binding to `length`. So our recursively-defined function works the way we'd expect it to. In OCaml:

 
 

-       let rec get_length = fun lst ->
-               if lst == [] then 0 else 1 + get_length (tail lst)
-       in get_length [20; 30]
+       let rec length = fun lst ->
+               if lst == [] then 0 else 1 + length (tail lst)
+       in length [20; 30]

        (* this evaluates to 2 *)
 
 In Scheme:
 
        (* this evaluates to 2 *)
 
 In Scheme:
 

-       (letrec [(get_length
-                               (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )]
-                       (get_length (list 20 30)))
+       (letrec [(length
+                               (lambda (lst) (if (null? lst) 0 [+ 1 (length (cdr lst))] )) )]
+                       (length (list 20 30)))

        ; this evaluates to 2
 
 If you instead use an ordinary `let` (or `let*`), here's what would happen, in OCaml:
 
        ; this evaluates to 2
 
 If you instead use an ordinary `let` (or `let*`), here's what would happen, in OCaml:
 

-       let get_length = fun lst ->
-               if lst == [] then 0 else 1 + get_length (tail lst)
-       in get_length [20; 30]
+       let length = fun lst ->
+               if lst == [] then 0 else 1 + length (tail lst)
+       in length [20; 30]

        (* fails with error "Unbound value length" *)
 
 Here's Scheme:
 
        (* fails with error "Unbound value length" *)
 
 Here's Scheme:
 

-       (let* [(get_length
-                               (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )]
-                       (get_length (list 20 30)))
-       ; fails with error "reference to undefined identifier: get_length"
+       (let* [(length
+                               (lambda (lst) (if (null? lst) 0 [+ 1 (length (cdr lst))] )) )]
+                       (length (list 20 30)))
+       ; fails with error "reference to undefined identifier: length"

 
 Why? Because we said that constructions of this form:
 
 
 Why? Because we said that constructions of this form:
 

-       let get_length = A
+       let length = A

                in B
 
 really were just another way of saying:
 
                in B
 
 really were just another way of saying:
 

-       (\get_length. B) A
+       (\length. B) A

 
 

-and so the occurrences of `get_length` in A *aren't bound by the `\get_length` that wraps B*. Those occurrences are free.
+and so the occurrences of `length` in A *aren't bound by the `\length` that wraps B*. Those occurrences are free.

 
 

-We can verify this by wrapping the whole expression in a more outer binding of `get_length` to some other function, say the constant function from any list to the integer 99:
+We can verify this by wrapping the whole expression in a more outer binding of `length` to some other function, say the constant function from any list to the integer 99:

 
 

-       let get_length = fun lst -> 99
-       in let get_length = fun lst ->
-                       if lst == [] then 0 else 1 + get_length (tail lst)
-       in get_length [20; 30]
+       let length = fun lst -> 99
+       in let length = fun lst ->
+                       if lst == [] then 0 else 1 + length (tail lst)
+       in length [20; 30]

        (* evaluates to 1 + 99 *)
 
        (* evaluates to 1 + 99 *)
 

-Here the use of `get_length` in `1 + get_length (tail lst)` can clearly be seen to be bound by the outermost `let`.
+Here the use of `length` in `1 + length (tail lst)` can clearly be seen to be bound by the outermost `let`.

 
 

-And indeed, if you tried to define `get_length` in the lambda calculus, how would you do it?
+And indeed, if you tried to define `length` in the lambda calculus, how would you do it?

 
 

-       \lst. (isempty lst) zero (add one (get_length (extract-tail lst)))
+       \lst. (isempty lst) zero (add one (length (extract-tail lst)))

 
 

-We've defined all of `isempty`, `zero`, `add`, `one`, and `extract-tail` in earlier discussion. But what about `get_length`? That's not yet defined! In fact, that's the very formula we're trying here to specify.
+We've defined all of `isempty`, `zero`, `add`, `one`, and `extract-tail` in earlier discussion. But what about `length`? That's not yet defined! In fact, that's the very formula we're trying here to specify.

 
 What we really want to do is something like this:
 
 
 What we really want to do is something like this:
 


@@ -116,14 +116,14 @@ So how could we do it? And how do OCaml and Scheme manage to do it, with their `
 
 2.     If you tried this in Scheme:
 
 
 2.     If you tried this in Scheme:
 

-               (define get_length
-                               (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )
+               (define length
+                               (lambda (lst) (if (null? lst) 0 [+ 1 (length (cdr lst))] )) )

 
 

-               (get_length (list 20 30))
+               (length (list 20 30))

 
        You'd find that it works! This is because `define` in Scheme is really shorthand for `letrec`, not for plain `let` or `let*`. So we should regard this as cheating, too.
 
 
        You'd find that it works! This is because `define` in Scheme is really shorthand for `letrec`, not for plain `let` or `let*`. So we should regard this as cheating, too.
 

-3.     In fact, it *is* possible to define the `get_length` function in the lambda calculus despite these obstacles. This depends on using the "version 3" implementation of lists, and exploiting its internal structure: that it takes a function and a base value and returns the result of folding that function over the list, with that base value. So we could use this as a definition of `get_length`:
+3.     In fact, it *is* possible to define the `length` function in the lambda calculus despite these obstacles. This depends on using the "version 3" implementation of lists, and exploiting its internal structure: that it takes a function and a base value and returns the result of folding that function over the list, with that base value. So we could use this as a definition of `length`:

 
                \lst. lst (\x sofar. successor sofar) zero
 
 
                \lst. lst (\x sofar. successor sofar) zero
 


@@ -134,9 +134,9 @@ lists and numbers. The reason we can do this is that our "version 3,"
 fold-based implementation of lists, and Church's implementations of
 numbers, have a internal structure that *mirrors* the common recursive
 operations we'd use lists and numbers for.  In a sense, the recursive
 fold-based implementation of lists, and Church's implementations of
 numbers, have a internal structure that *mirrors* the common recursive
 operations we'd use lists and numbers for.  In a sense, the recursive

-structure of the `get_length` operation is built into the data
+structure of the `length` operation is built into the data

 structure we are using to represent the list.  The non-recursive
 structure we are using to represent the list.  The non-recursive

-version of get_length exploits this embedding of the recursion into
+version of length exploits this embedding of the recursion into

 the data type.
 
 This is one of the themes of the course: using data structures to
 the data type.
 
 This is one of the themes of the course: using data structures to


@@ -183,9 +183,9 @@ But functions like the Ackermann function require us to develop a more general t
 
 ###Fixed points###
 
 
 ###Fixed points###
 

-In general, we call a **fixed point** of a function f any value *x*
-such that f <em>x</em> is equivalent to *x*. For example,
-consider the squaring function `sqare` that maps natural numbers to their squares.
+In general, a **fixed point** of a function `f` is any value `x`
+such that `f x` is equivalent to `x`. For example,
+consider the squaring function `square` that maps natural numbers to their squares.

 `square 2 = 4`, so `2` is not a fixed point.  But `square 1 = 1`, so `1` is a
 fixed point of the squaring function.
 
 `square 2 = 4`, so `2` is not a fixed point.  But `square 1 = 1`, so `1` is a
 fixed point of the squaring function.
 


@@ -193,7 +193,7 @@ There are many beautiful theorems guaranteeing the existence of a
 fixed point for various classes of interesting functions.  For
 instance, imainge that you are looking at a map of Manhattan, and you
 are standing somewhere in Manhattan.  The the [[!wikipedia Brouwer
 fixed point for various classes of interesting functions.  For
 instance, imainge that you are looking at a map of Manhattan, and you
 are standing somewhere in Manhattan.  The the [[!wikipedia Brouwer

-fixed point]] guarantees that there is a spot on the map that is
+fixed-point theorem]] guarantees that there is a spot on the map that is

 directly above the corresponding spot in Manhattan.  It's the spot
 where the blue you-are-here dot should be.
 
 directly above the corresponding spot in Manhattan.  It's the spot
 where the blue you-are-here dot should be.
 


@@ -204,13 +204,13 @@ attention to the natural numbers, then this function has no fixed
 point.  (See the discussion below concerning a way of understanding
 the successor function on which it does have a fixed point.)
 
 point.  (See the discussion below concerning a way of understanding
 the successor function on which it does have a fixed point.)
 

-In the lambda calculus, we say a fixed point of an expression `f` is any formula `X` such that:
+In the lambda calculus, we say a fixed point of a term `f` is any term `X` such that:

 
        X <~~> f X
 
 You should be able to immediately provide a fixed point of the
 
        X <~~> f X
 
 You should be able to immediately provide a fixed point of the

-identity combinator I.  In fact, you should be able to provide a whole
-bunch of distinct fixed points.
+identity combinator I.  In fact, you should be able to provide a
+whole bunch of distinct fixed points.

 
 With a little thought, you should be able to provide a fixed point of
 the false combinator, KI.  Here's how to find it: recall that KI
 
 With a little thought, you should be able to provide a fixed point of
 the false combinator, KI.  Here's how to find it: recall that KI


@@ -227,13 +227,13 @@ a fixed point. In fact, it will have infinitely many, non-equivalent
 fixed points. And we don't just know that they exist: for any given
 formula, we can explicit define many of them.
 
 fixed points. And we don't just know that they exist: for any given
 formula, we can explicit define many of them.
 

-Yes, even the formula that you're using the define the successor
-function will have a fixed point. Isn't that weird? Think about how it
-might be true.  We'll return to this point below.
+Yes, as we've mentioned, even the formula that you're using the define
+the successor function will have a fixed point. Isn't that weird?
+Think about how it might be true.  We'll return to this point below.

 
 

-###How fixed points help definie recursive functions###
+###How fixed points help define recursive functions###

 
 

-Recall our initial, abortive attempt above to define the `get_length` function in the lambda calculus. We said "What we really want to do is something like this:
+Recall our initial, abortive attempt above to define the `length` function in the lambda calculus. We said "What we really want to do is something like this:

 
        \list. if empty list then zero else add one (... (tail lst))
 
 
        \list. if empty list then zero else add one (... (tail lst))
 


@@ -249,29 +249,34 @@ it's not complete, since we don't know what value to use for the
 symbol `length`.  Technically, it has the status of an unbound
 variable.
 
 symbol `length`.  Technically, it has the status of an unbound
 variable.
 

-Imagine now binding the mysterious variable:
+Imagine now binding the mysterious variable, and calling the resulting
+function `h`:

 
        h := \length \list . if empty list then zero else add one (length (tail list))
 
 Now we have no unbound variables, and we have complete non-recursive
 definitions of each of the other symbols.
 
 
        h := \length \list . if empty list then zero else add one (length (tail list))
 
 Now we have no unbound variables, and we have complete non-recursive
 definitions of each of the other symbols.
 

-Let's call this function `h`.  Then `h` takes an argument, and returns
-a function that accurately computes the length of a list---as long as
-the argument we supply is already the length function we are trying to
-define.  (Dehydrated water: to reconstitute, just add water!)
+So `h` takes an argument, and returns a function that accurately
+computes the length of a list---as long as the argument we supply is
+already the length function we are trying to define.  (Dehydrated
+water: to reconstitute, just add water!)

 
 

-But this is just another way of saying that we are looking for a fixed point.
-Assume that `h` has a fixed point, call it `LEN`.  To say that `LEN`
-is a fixed point means that
+Here is where the discussion of fixed points becomes relevant.  Saying
+that `h` is looking for an argument (call it `LEN`) that has the same
+behavior as the result of applying `h` to `LEN` is just another way of
+saying that we are looking for a fixed point for `h`.

 
     h LEN <~~> LEN
 
 
     h LEN <~~> LEN
 

-But this means that
+Replacing `h` with its definition, we have

 
     (\list . if empty list then zero else add one (LEN (tail list))) <~~> LEN
 
 
     (\list . if empty list then zero else add one (LEN (tail list))) <~~> LEN
 

-So at this point, we are going to search for fixed point.
+If we can find a value for `LEN` that satisfies this constraint, we'll
+have a function we can use to compute the length of an arbitrary list.
+All we have to do is find a fixed point for `h`.
+

 The strategy we will present will turn out to be a general way of
 finding a fixed point for any lambda term.
 
 The strategy we will present will turn out to be a general way of
 finding a fixed point for any lambda term.
 


@@ -285,22 +290,24 @@ work, but examining the way in which it fails will lead to a solution.
 
     h h <~~> \list . if empty list then zero else 1 + h (tail list)
 
 
     h h <~~> \list . if empty list then zero else 1 + h (tail list)
 

-There's a problem.  The diagnosis is that in the subexpression `h
-(tail list)`, we've applied `h` to a list, but `h` expects as its
-first argument the length function.
+The problem is that in the subexpression `h (tail list)`, we've
+applied `h` to a list, but `h` expects as its first argument the
+length function.

 
 So let's adjust h, calling the adjusted function H:
 
     H = \h \list . if empty list then zero else one plus ((h h) (tail list))
 
 
 So let's adjust h, calling the adjusted function H:
 
     H = \h \list . if empty list then zero else one plus ((h h) (tail list))
 

-This is the key creative step.  Since `h` is expecting a
-length-computing function as its first argument, the adjustment
-tries supplying the closest candidate avaiable, namely, `h` itself.
+This is the key creative step.  Instead of applying `h` to a list, we
+apply it first to itself.  After applying `h` to an argument, it's
+ready to apply to a list, so we've solved the problem just noted.
+We're not done yet, of course; we don't yet know what argument to give
+to `H` that will behave in the desired way.

 
 

-We now reason about `H`.  What exactly is H expecting as its first
-argument?  Based on the excerpt `(h h) (tail l)`, it appears that `H`'s
-argument, `h`, should be a function that is ready to take itself as an
-argument, and that returns a function that takes a list as an
+So let's reason about `H`.  What exactly is H expecting as its first
+argument?  Based on the excerpt `(h h) (tail l)`, it appears that
+`H`'s argument, `h`, should be a function that is ready to take itself
+as an argument, and that returns a function that takes a list as an

 argument.  `H` itself fits the bill:
 
     H H <~~> (\h \list . if empty list then zero else 1 + ((h h) (tail list))) H
 argument.  `H` itself fits the bill:
 
     H H <~~> (\h \list . if empty list then zero else 1 + ((h h) (tail list))) H


@@ -322,11 +329,9 @@ finite lambda term with no undefined symbols.
 
 Since `H H` turns out to be the length function, we can think of `H`
 by itself as half of the length function (which is why we called it
 
 Since `H H` turns out to be the length function, we can think of `H`
 by itself as half of the length function (which is why we called it

-`H`, of course).  Given the implementation of addition as function
-application for Church numerals, this (H H) is quite literally H + H.
-Can you think up a recursion strategy that involves "dividing" the
-recursive function into equal thirds `T`, such that the length
-function <~~> T T T?
+`H`, of course).  Can you think up a recursion strategy that involves
+"dividing" the recursive function into equal thirds `T`, such that the
+length function <~~> T T T?

 
 We've starting with a particular recursive definition, and arrived at
 a fixed point for that definition.
 
 We've starting with a particular recursive definition, and arrived at
 a fixed point for that definition.


@@ -352,9 +357,67 @@ That is, Yh is a fixed point for h.
 
 Works!
 
 
 Works!
 

+Let's do one more example to illustrate.  We'll do `K`, since we
+wondered above whether it had a fixed point.
+
+Before we begin, we can reason a bit about what the fixed point must
+be like.  We're looking for a fixed point for `K`, i.e., `\xy.x`.  `K`
+ignores its second argument.  That means that no matter what we give
+`K` as its first argument, the result will ignore the next argument
+(that is, `KX` ignores its first argument, no matter what `X` is).  So
+if `KX <~~> X`, `X` had also better ignore its first argument.  But we
+also have `KX == (\xy.x)X ~~> \y.X`.  This means that if `X` ignores
+its first argument, then `\y.X` will ignore its first two arguments.
+So once again, if `KX <~~> X`, `X` also had better ignore at least its
+first two arguments.  Repeating this reasoning, we realize that `X`
+must be a function that ignores an infinite series of arguments.  
+Our expectation, then, is that our recipe for finding fixed points
+will build us a function that somehow manages to ignore an infinite
+series of arguments.
+
+    h := \xy.x
+    H := \f.h(ff) == \f.(\xy.x)(ff) ~~> \fy.ff
+    H H := (\fy.ff)(\fy.ff) ~~> \y.(\fy.ff)(\fy.ff)
+
+Let's check that it is in fact a fixed point:
+
+    K(H H) == (\xy.x)((\fy.ff)(\fy.ff)
+           ~~> \y.(\fy.ff)(\fy.ff)
+
+Yep, `H H` and `K(H H)` both reduce to the same term.  
+
+To see what this fixed point does, let's reduce it a bit more:
+
+    H H == (\fy.ff)(\fy.ff)
+        ~~> \y.(\fy.ff)(\fy.ff)
+        ~~> \yy.(\fy.ff)(\fy.ff)
+        ~~> \yyy.(\fy.ff)(\fy.ff)
+    
+Sure enough, this fixed point ignores an endless, infinite series of
+arguments.  It's a write-only memory, a black hole.
+
+Now that we have one fixed point, we can find others, for instance,
+
+    (\fy.fff)(\fy.fff) 
+    ~~> \y.(\fy.fff)(\fy.fff)(\fy.fff)
+    ~~> \yy.(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)
+    ~~> \yyy.(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)
+
+Continuing in this way, you can now find an infinite number of fixed
+points, all of which have the crucial property of ignoring an infinite
+series of arguments.

 
 ##What is a fixed point for the successor function?##
 
 
 ##What is a fixed point for the successor function?##
 

+As we've seen, the recipe just given for finding a fixed point worked
+great for our `h`, which we wrote as a definition for the length
+function.  But the recipe doesn't make any assumptions about the
+internal structure of the function it works with.  That means it can
+find a fixed point for literally any function whatsoever.
+
+In particular, what could the fixed point for the
+successor function possibly be like?
+

 Well, you might think, only some of the formulas that we might give to the `successor` as arguments would really represent numbers. If we said something like:
 
        successor make-pair
 Well, you might think, only some of the formulas that we might give to the `successor` as arguments would really represent numbers. If we said something like:
 
        successor make-pair


@@ -363,18 +426,20 @@ who knows what we'd get back? Perhaps there's some non-number-representing formu
 
 Yes! That's exactly right. And which formula this is will depend on the particular way you've implemented the successor function.
 
 
 Yes! That's exactly right. And which formula this is will depend on the particular way you've implemented the successor function.
 

-Moreover, the recipes that enable us to name fixed points for any
-given formula aren't *guaranteed* to give us *terminating* fixed
-points. They might give us formulas X such that neither `X` nor `f X`
-have normal forms. (Indeed, what they give us for the square function
-isn't any of the Church numerals, but is rather an expression with no
-normal form.) However, if we take care we can ensure that we *do* get
-terminating fixed points. And this gives us a principled, fully
-general strategy for doing recursion. It lets us define even functions
-like the Ackermann function, which were until now out of our reach. It
-would also let us define arithmetic and list functions on the "version
-1" and "version 2" implementations, where it wasn't always clear how
-to force the computation to "keep going."
+One (by now obvious) upshot is that the recipes that enable us to name
+fixed points for any given formula aren't *guaranteed* to give us
+*terminating* fixed points. They might give us formulas X such that
+neither `X` nor `f X` have normal forms. (Indeed, what they give us
+for the square function isn't any of the Church numerals, but is
+rather an expression with no normal form.) However, if we take care we
+can ensure that we *do* get terminating fixed points. And this gives
+us a principled, fully general strategy for doing recursion. It lets
+us define even functions like the Ackermann function, which were until
+now out of our reach. It would also let us define arithmetic and list
+functions on the "version 1" and "version 2" implementations, where it
+wasn't always clear how to force the computation to "keep going."
+
+###Varieties of fixed-point combinators###

 
 OK, so how do we make use of this?
 
 
 OK, so how do we make use of this?