week4 tweaking

[lambda.git] / week4.mdwn

diff --git a/week4.mdwn b/week4.mdwn
index e1dcc4e..f92ea16 100644 (file)
--- a/week4.mdwn
+++ b/week4.mdwn
@@ -8,7 +8,8 @@ TX` (that's what it means to *have* a fixed point).
 
 <pre><code>let L = \x. T (x x) in
 let X = L L in
 
 <pre><code>let L = \x. T (x x) in
 let X = L L in

-X &equiv; L L &equiv; (\x. T (x x)) L ~~> T (L L) &equiv; T X</code></pre>
+X &equiv; L L &equiv; (\x. T (x x)) L ~~> T (L L) &equiv; T X
+</code></pre>

 
 Please slow down and make sure that you understand what justified each
 of the equalities in the last line.
 
 Please slow down and make sure that you understand what justified each
 of the equalities in the last line.


@@ -16,7 +17,7 @@ of the equalities in the last line.
 #Q: How do you know that for any term `T`, `Y T` is a fixed point of `T`?#
 
 A: Note that in the proof given in the previous answer, we chose `T`
 #Q: How do you know that for any term `T`, `Y T` is a fixed point of `T`?#
 
 A: Note that in the proof given in the previous answer, we chose `T`

-and then set `X = L L = (\x. T (x x)) (\x. T (x x))`.  If we abstract over
+and then set <code>X &equiv; L L &equiv; (\x. T (x x)) (\x. T (x x))</code>.  If we abstract over

 `T`, we get the Y combinator, `\T. (\x. T (x x)) (\x. T (x x))`.  No matter
 what argument `T` we feed `Y`, it returns some `X` that is a fixed point
 of `T`, by the reasoning in the previous answer.
 `T`, we get the Y combinator, `\T. (\x. T (x x)) (\x. T (x x))`.  No matter
 what argument `T` we feed `Y`, it returns some `X` that is a fixed point
 of `T`, by the reasoning in the previous answer.


@@ -26,11 +27,14 @@ of `T`, by the reasoning in the previous answer.
 A: Right:
 
 <pre><code>let Y = \T. (\x. T (x x)) (\x. T (x x)) in
 A: Right:
 
 <pre><code>let Y = \T. (\x. T (x x)) (\x. T (x x)) in

-Y Y &equiv; \T. (\x. T (x x)) (\x. T (x x)) Y
-       ~~> (\x. Y (x x)) (\x. Y (x x))
-       ~~> Y ((\x. Y (x x)) (\x. Y (x x)))
-       ~~> Y (Y ((\x. Y (x x)) (\x. Y (x x))))
-       ~~> Y (Y (Y (...(Y (Y Y))...)))</code></pre>
+Y Y
+&equiv;   \T. (\x. T (x x)) (\x. T (x x)) Y
+~~> (\x. Y (x x)) (\x. Y (x x))
+~~> Y ((\x. Y (x x)) (\x. Y (x x)))
+~~> Y (Y ((\x. Y (x x)) (\x. Y (x x))))
+~~> Y (Y (Y (...(Y (Y Y))...)))
+</code></pre>
+

 
 #Q: Ouch!  Stop hurting my brain.#
 
 
 #Q: Ouch!  Stop hurting my brain.#
 


@@ -41,7 +45,7 @@ claimed that even `succ`---the function that added one to any
 number---had a fixed point.  How could there be an X such that X = X+1?
 That would imply that
 
 number---had a fixed point.  How could there be an X such that X = X+1?
 That would imply that
 

-    X <~~> succ X <~~> succ (succ X) <~~> succ (succ (succ (X))) <~~> succ (... (succ X)...)
+    X <~~> succ X <~~> succ (succ X) <~~> succ (succ (succ X)) <~~> succ (... (succ X)...)

 
 In other words, the fixed point of `succ` is a term that is its own
 successor.  Let's just check that `X = succ X`:
 
 In other words, the fixed point of `succ` is a term that is its own
 successor.  Let's just check that `X = succ X`:


@@ -49,32 +53,35 @@ successor.  Let's just check that `X = succ X`:
 <pre><code>let succ = \n s z. s (n s z) in
 let X = (\x. succ (x x)) (\x. succ (x x)) in
 succ X 
 <pre><code>let succ = \n s z. s (n s z) in
 let X = (\x. succ (x x)) (\x. succ (x x)) in
 succ X 

-  &equiv; succ ( (\x. succ (x x)) (\x. succ (x x)) ) 
-  ~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x))))
-  &equiv; succ (succ X)</code></pre>
+&equiv;   succ ( (\x. succ (x x)) (\x. succ (x x)) ) 
+~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x)) ))
+&equiv;   succ (succ X)
+</code></pre>

 
 You should see the close similarity with `Y Y` here.
 
 
 You should see the close similarity with `Y Y` here.
 

+

 #Q. So `Y` applied to `succ` returns a number that is not finite!#
 
 A. Yes!  Let's see why it makes sense to think of `Y succ` as a Church
 numeral:
 
 #Q. So `Y` applied to `succ` returns a number that is not finite!#
 
 A. Yes!  Let's see why it makes sense to think of `Y succ` as a Church
 numeral:
 

-      [same definitions]
-      succ X
-      = (\n s z. s (n s z)) X 
-      = \s z. s (X s z)
-      = succ (\s z. s (X s z)) ; using fixed-point reasoning
-      = \s z. s ([succ (\s z. s (X s z))] s z)
-      = \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z)
-      = \s z. s (s (succ (\s z. s (X s z))))
+<pre><code>[same definitions]
+succ X
+&equiv;    (\n s z. s (n s z)) X 
+~~>  \s z. s (X s z)
+<~~> succ (\s z. s (X s z)) ; using fixed-point reasoning
+&equiv;    (\n s z. s (n s z)) (\s z. s (X s z))
+~~>  \s z. s ((\s z. s (X s z)) s z)
+~~>  \s z. s (s (X s z))
+</code></pre>

 
 So `succ X` looks like a numeral: it takes two arguments, `s` and `z`,
 and returns a sequence of nested applications of `s`...
 
 You should be able to prove that `add 2 (Y succ) <~~> Y succ`,
 
 So `succ X` looks like a numeral: it takes two arguments, `s` and `z`,
 and returns a sequence of nested applications of `s`...
 
 You should be able to prove that `add 2 (Y succ) <~~> Y succ`,

-likewise for `mult`, `minus`, `pow`.  What happens if we try `minus (Y
-succ)(Y succ)`?  What would you expect infinity minus infinity to be?
+likewise for `mul`, `sub`, `pow`.  What happens if we try `sub (Y
+succ) (Y succ)`?  What would you expect infinity minus infinity to be?

 (Hint: choose your evaluation strategy so that you add two `s`s to the
 first number for every `s` that you add to the second number.)
 
 (Hint: choose your evaluation strategy so that you add two `s`s to the
 first number for every `s` that you add to the second number.)
 


@@ -84,13 +91,14 @@ represents arithmetic infinity.
 It's important to bear in mind the simplest term in question is not
 infinite:
 
 It's important to bear in mind the simplest term in question is not
 infinite:
 

-     Y succ = (\f. (\x. f (x x)) (\x. f (x x))) (\n s z. s (n s z))
+       Y succ = (\f. (\x. f (x x)) (\x. f (x x))) (\n s z. s (n s z))

 
 The way that infinity enters into the picture is that this term has
 no normal form: no matter how many times we perform beta reduction,
 there will always be an opportunity for more beta reduction.  (Lather,
 rinse, repeat!)
 
 
 The way that infinity enters into the picture is that this term has
 no normal form: no matter how many times we perform beta reduction,
 there will always be an opportunity for more beta reduction.  (Lather,
 rinse, repeat!)
 

+

 #Q. That reminds me, what about [[evaluation order]]?#
 
 A. For a recursive function that has a well-behaved base case, such as
 #Q. That reminds me, what about [[evaluation order]]?#
 
 A. For a recursive function that has a well-behaved base case, such as


@@ -100,63 +108,63 @@ which we have to make a choice about which beta reduction to perform
 next: one choice leads to a normal form, the other choice leads to
 endless reduction:
 
 next: one choice leads to a normal form, the other choice leads to
 endless reduction:
 

-    let prefac = \f n. iszero n 1 (mult n (f (pred n))) in
-    let fac = Y prefac in
-    fac 2
-       = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2
-       = [(\x.prefac(xx))(\x.prefac(xx))] 2
-       = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2
-       = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
-       = [(\f n. iszero n 1 (mult n (f (pred n))))
-          (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
-       = [\n. iszero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2
-       = iszero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2)))
-       = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1)
-       ...
-       = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0))
-       = mult 2 (mult 1 (iszero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0))))
-       = mult 2 (mult 1 1)
-       = mult 2 1
-       = 2
+<pre><code>let prefact = \f n. iszero n 1 (mul n (f (pred n))) in
+let fact = Y prefact in
+fact 2
+&equiv;   [(\f. (\x. f (x x)) (\x. f (x x))) prefact] 2
+~~> [(\x. prefact (x x)) (\x. prefact (x x))] 2
+~~> [prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 2
+~~> [prefact (prefact ((\x. prefact (x x)) (\x. prefact (x x))))] 2
+&equiv;   [ (\f n. iszero n 1 (mul n (f (pred n)))) (prefact ((\x. prefact (x x)) (\x. prefact (x x))))] 2
+~~> [\n. iszero n 1 (mul n ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred n)))] 2
+~~> iszero 2 1 (mul 2 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred 2)))
+~~> mul 2 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 1)
+...
+~~> mul 2 (mul 1 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 0))
+&equiv;   mul 2 (mul 1 (iszero 0 1 (mul 1 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred 0)))))
+~~> mul 2 (mul 1 1)
+~~> mul 2 1
+~~> 2
+</code></pre>

 
 The crucial step is the third from the last.  We have our choice of
 either evaluating the test `iszero 0 1 ...`, which evaluates to `1`,
 no matter what the ... contains;
 
 The crucial step is the third from the last.  We have our choice of
 either evaluating the test `iszero 0 1 ...`, which evaluates to `1`,
 no matter what the ... contains;

-or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to
-produce another copy of `prefac`.  If we postpone evaluting the
-`iszero` test, we'll pump out copy after copy of `prefac`, and never
+or we can evaluate the `Y` pump, `(\x. prefact (x x)) (\x. prefact (x x))`, to
+produce another copy of `prefact`.  If we postpone evaluting the
+`iszero` test, we'll pump out copy after copy of `prefact`, and never

 realize that we've bottomed out in the recursion.  But if we adopt a
 leftmost/call-by-name/normal-order evaluation strategy, we'll always
 start with the `iszero` predicate, and only produce a fresh copy of
 realize that we've bottomed out in the recursion.  But if we adopt a
 leftmost/call-by-name/normal-order evaluation strategy, we'll always
 start with the `iszero` predicate, and only produce a fresh copy of

-`prefac` if we are forced to. 
+`prefact` if we are forced to. 
+

 
 #Q.  You claimed that the Ackerman function couldn't be implemented using our primitive recursion techniques (such as the techniques that allow us to define addition and multiplication).  But you haven't shown that it is possible to define the Ackerman function using full recursion.#
 
 
 #Q.  You claimed that the Ackerman function couldn't be implemented using our primitive recursion techniques (such as the techniques that allow us to define addition and multiplication).  But you haven't shown that it is possible to define the Ackerman function using full recursion.#
 

+

 A. OK:
   
 A. OK:
   

-<pre>
-A(m,n) =
-    | when m == 0 -> n + 1
-    | else when n == 0 -> A(m-1,1)
-    | else -> A(m-1, A(m,n-1))
-
-let A = Y (\A m n. iszero m (succ n) (iszero n (A (pred m) 1) (A (pred m) (A m (pred n))))) in
-</pre>
-
-For instance,
-
-    A 1 2
-    = A 0 (A 1 1)
-    = A 0 (A 0 (A 1 0))
-    = A 0 (A 0 (A 0 1))
-    = A 0 (A 0 2)
-    = A 0 3
-    = 4
-
-A 1 x is to A 0 x as addition is to the successor function;
-A 2 x is to A 1 x as multiplication is to addition;
-A 3 x is to A 2 x as exponentiation is to multiplication---
-so A 4 x is to A 3 x as hyper-exponentiation is to exponentiation...
+       A(m,n) =
+               | when m == 0 -> n + 1
+               | else when n == 0 -> A(m-1,1)
+               | else -> A(m-1, A(m,n-1))
+
+       let A = Y (\A m n. iszero m (succ n) (iszero n (A (pred m) 1) (A (pred m) (A m (pred n)))))
+
+So for instance:
+
+       A 1 2
+       ~~> A 0 (A 1 1)
+       ~~> A 0 (A 0 (A 1 0))
+       ~~> A 0 (A 0 (A 0 1))
+       ~~> A 0 (A 0 2)
+       ~~> A 0 3
+       ~~> 4
+
+`A 1 x` is to `A 0 x` as addition is to the successor function;
+`A 2 x` is to `A 1 x` as multiplication is to addition;
+`A 3 x` is to `A 2 x` as exponentiation is to multiplication---
+so `A 4 x` is to `A 3 x` as hyper-exponentiation is to exponentiation...

 
 #Q. What other questions should I be asking?#
 
 
 #Q. What other questions should I be asking?#
 


@@ -568,4 +576,3 @@ detail](http://okmij.org/ftp/Streams.html#enumerator-stream).
 3.     To extract tails efficiently, too, it'd be nice to fuse the apparatus developed
        in these v5 lists with the ideas from [v4](/advanced/#index1h1) lists.
        But that also is left as an exercise.
 3.     To extract tails efficiently, too, it'd be nice to fuse the apparatus developed
        in these v5 lists with the ideas from [v4](/advanced/#index1h1) lists.
        But that also is left as an exercise.

-