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=c8bb62c51fafc17b85aa54e62166fb90b6a3ff77;hb=f212354152a53c6a3ab018c7874570c600f463b9;hpb=d437f108ffa48fa141665ffb0b836a9ad53093e7

diff --git a/hints/assignment_4_hint_3_alternate_1.mdwn b/hints/assignment_4_hint_3_alternate_1.mdwn
index c8bb62c5..f75bad0e 100644
--- a/hints/assignment_4_hint_3_alternate_1.mdwn
+++ b/hints/assignment_4_hint_3_alternate_1.mdwn
@@ -18,12 +18,12 @@ Alternate strategy for Y1, Y2
 
 	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 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 abstract the Y1 and Y2 combinators from this as follows:
+	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
@@ -35,22 +35,24 @@ Alternate strategy for Y1, Y2
 
 *	Here's the same strategy extended to three mutually-recursive functions. `f`, `g` and `h`:
 
-		let v = Y (\v g h x.      (\f. A) (v g h)) in
-		let w = Y (  \w h x. (\g. (\f. B) (v g h)) (w h)) in
-		let h = Y (    \h x. (\g. (\f. C) (v g h)) (w h)) in
+		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 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
+	-->