X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=hints%2Fassignment_4_hint_3_alternate_1.mdwn;h=f75bad0efce8079169eac9a41b99cbd76c95e31f;hp=dd55e052ec797db0bb714575a94133c76cf5a304;hb=61612ba6746dc4646fb0d5bb1c9e1864bf927848;hpb=1f4f3108441cfd887eb5c9e6ee53a651a80166ea

diff --git a/hints/assignment_4_hint_3_alternate_1.mdwn b/hints/assignment_4_hint_3_alternate_1.mdwn
index dd55e052..f75bad0e 100644
--- a/hints/assignment_4_hint_3_alternate_1.mdwn
+++ b/hints/assignment_4_hint_3_alternate_1.mdwn
@@ -11,13 +11,48 @@ Alternate strategy for Y1, Y2
 
 	is implemented using regular, non-mutual recursion, like this (`u` is a variable not occurring free in `A`, `B`, or `C`):
 
-		let rec u g x = (let f = u g in A)
-		in let rec g y = (let f = u g in B)
-		in let f = u g in C
+		let rec  u g x = (let f = u g in A)  in
+		let rec    g y = (let f = u g in B)  in
+		let                   f = u g        in
+		C
 
 	or, expanded into the form we've been working with:
 
-		let u = Y (\u g x. (\f. A) (u g)) in
-		let g = Y (\g y. (\f. B) (u g)) in
-		let f = u g
+		let u =    Y (\u g. (\f x. A) (u g))  in
+		let g =    Y (  \g. (\f y. B) (u g))  in
+		let f =                        u g    in
+		C
+
+	We could abstract Y1 and Y2 combinators from this as follows:
+
+		let Yu = \ff.    Y (\u g. ff ( u      g        ) g)  in
+		let Y2 = \ff gg. Y (  \g. gg (Yu ff   g        ) g)  in
+		let Y1 = \ff gg.             (Yu ff) (Y2 ff gg)      in
+		let f  = Y1 (\f g. A) (\f g. B)  in
+		let g  = Y2 (\f g. A) (\f g. B)  in
+		C
+
+
+*	Here's the same strategy extended to three mutually-recursive functions. `f`, `g` and `h`:
+
+		let v = Y (\v g h.      (\f x. A) (v g h)) in
+		let w = Y (  \w h. (\g. (\f y. B) (v g h)) (w h)) in
+		let h = Y (    \h. (\g. (\f z. C) (v g h)) (w h)) in
+		let g =                                     w h in
+		let f =                            v g h in
+		D
+
+	<!--
+	Or in Y1of3, Y2of3, Y3of3 form:
+
+		let Yv    = \ff.       Y (\v g h.      ff ( v    g h) g h)                in
+		let Yw    = \ff gg.    Y (  \w h. (\g. gg (Yv ff g h) g h) ( w       h))  in
+		let Y3of3 = \ff gg hh. Y (    \h. (\g. hh (Yv ff g h) g h) (Yw ff gg h))  in
+		let Y2of3 = \ff gg hh.                                      Yw ff gg (Y3of3 ff gg hh)  in
+		let Y1of3 = \ff gg hh.                     Yv ff (Y2of3 ff gg hh) (Y3of3 ff gg hh)     in
+		let f = Y1of3 (\f g h. A) (\f g h. B) (\f g h. C)  in
+		let g = Y2of3 (\f g h. A) (\f g h. B) (\f g h. C)  in
+		let h = Y3of3 (\f g h. A) (\f g h. B) (\f g h. C)  in
+		D
+	-->