week4 debugging more

[lambda.git] / week4.mdwn

diff --git a/week4.mdwn b/week4.mdwn
index 11eea4c..a1aa254 100644 (file)
--- a/week4.mdwn
+++ b/week4.mdwn
@@ -33,6 +33,7 @@ Y Y ≡ \T. (\x. T (x x)) (\x. T (x x)) Y
 ~~> Y (Y ((\x. Y (x x)) (\x. Y (x x))))
 ~~> Y (Y (Y (...(Y (Y Y))...)))</code></pre>
 
+
 #Q: Ouch!  Stop hurting my brain.#
 
 A: Is that a question?
@@ -52,23 +53,25 @@ 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 (succ X)
+</code></pre>
 
 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:
 
-      [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))))
+       [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))))
 
 So `succ X` looks like a numeral: it takes two arguments, `s` and `z`,
 and returns a sequence of nested applications of `s`...
@@ -85,13 +88,14 @@ represents arithmetic infinity.
 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!)
 
+
 #Q. That reminds me, what about [[evaluation order]]?#
 
 A. For a recursive function that has a well-behaved base case, such as
@@ -101,24 +105,24 @@ 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:
 
-    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
+       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
 
 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`,
@@ -131,18 +135,18 @@ 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. 
 
+
 #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:
   
-<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))
+       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>
+       let A = Y (\A m n. iszero m (succ n) (iszero n (A (pred m) 1) (A (pred m) (A m (pred n)))))
 
 For instance,
 
@@ -569,4 +573,3 @@ 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.
-