X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=topics%2Fweek4_more_about_fixed_point_combinators.mdwn;h=11754dde61bc6bee63675eba2ed12e2ce51f0345;hp=17d8eaba3efc477764180baa20186332f2f19327;hb=7d5250e133fdb9a566ddfa2a6a017f880c17602d;hpb=14cda93136c935ddbe8004120e4842b3808c107a diff --git a/topics/week4_more_about_fixed_point_combinators.mdwn b/topics/week4_more_about_fixed_point_combinators.mdwn index 17d8eaba..11754dde 100644 --- a/topics/week4_more_about_fixed_point_combinators.mdwn +++ b/topics/week4_more_about_fixed_point_combinators.mdwn @@ -311,61 +311,62 @@ so `A 4 x` is to `A 3 x` as hyper-exponentiation is to exponentiation... according to what ordering? How do you know it's least? Is leastness important? -## Q: I still don't fully understand the Y combinator. Can you - explain it in a different way? +## Q: I still don't fully understand the Y combinator. Can you explain it in a different way? -Sure! Here is another way to derive Y. We'll start by choosing a +Sure! Here is another way to derive `Y`. We'll start by choosing a specific goal, and at each decision point, we'll make a reasonable guess. The guesses will all turn out to be lucky, and we'll arrive at a fixed point combinator. -Given an arbitrary term f, we want to find a fixed point X such that +Given an arbitrary term `h`, we want to find a fixed point `X` such that: - X <~~> f X + X <~~> h X -Our strategy will be to seek an X such that X ~~> f X. Because X and -f X are not the same, the only way that X can reduce to f X is if X +Our strategy will be to seek an `X` such that `X ~~> h X` (this is just a guess). Because `X` and +`h X` are syntactically different, the only way that `X` can reduce to `h X` is if `X` contains at least one redex. The simplest way to satisfy this -constraint would be for the fixed point to itself be a redex: +constraint would be for the fixed point to itself be a redex (again, just a guess): - X == ((\u.M) N) ~~> f X + X â¡ ((\u. M) N) ~~> h X -The result of beta reduction on this redex will be M with some -substitutions. We know that after these substitutions, M will have -the form `f X`, since that is what the reduction arrow tells us. So we +The result of beta reduction on this redex will be `M` with some +substitutions. We know that after these substitutions, `M` will have +the form `h X`, since that is what the reduction arrow tells us. So we can refine the picture as follows: - X == ((\u.f(___)) N) ~~> f X + X â¡ ((\u. h (___)) N) ~~> h X -Here, the ___ has to be something that reduces to the fixed point X. +Here, the `___` has to be something that reduces to the fixed point `X`. It's natural to assume that there will be at least one occurrence of -"u" in the body of the head abstract: +`u` in the body of the head abstract: - X == ((\u.f(__u__)) N) ~~> f X + X â¡ ((\u. h (__u__)) N) ~~> h X After reduction of the redex, we're going to have - X == f(__N__) ~~> f X + X â¡ h (__N__) ~~> h X -Apparently, `__N__` will have to reduce to X. Therefore we should -choose a skeleton for N that is consistent with what we have decided -so far about the internal structure of X. We might like for N to -match X in its entirety, but this would require N to contain itself as -a subpart. So we'll settle for the more modest assumption that N -matches the head of X: +Apparently, `__N__` will have to reduce to `X`. Therefore we should +choose a skeleton for `N` that is consistent with what we have decided +so far about the internal structure of `X`. We might like for `N` to +syntactically match the whole of `X`, but this would require `N` to contain itself as +a subpart. So we'll settle for the more modest assumption (or guess) that `N` +matches the head of `X`: - X == ((\u.f(__u__)) (\u.f(__u__))) ~~> f X + X â¡ ((\u. h (__u__)) (\u. h (__u__))) ~~> h X At this point, we've derived a skeleton for X on which it contains two so-far identical halves. We'll guess that the halves will be exactly identical. Note that at the point at which we perform the first -reduction, `u` will get bound N, which now corresponds to a term -representing one of the halves of X. So in order to produce a full X, +reduction, `u` will get bound to `N`, which now corresponds to a term +representing one of the halves of `X`. So in order to produce a full `X`, we simply make a second copy of `u`: - X == ((\u.f(uu))(\u.f(uu))) ~~> f((\u.f(uu))(\u.f(uu))) == f X + X â¡ ((\u. h (u u)) (\u. h (u u))) + ~~> h ((\u. h (u u)) (\u. h (u u))) + â¡ h X Success. -So the function `\f.(\u.f(uu))(\u.f(uu))` maps an arbtirary function -`f` to a fixed point for `f`. +So the function `\h. (\u. h (u u)) (\u. h (u u))` maps an arbitrary term +`h` to a fixed point for `h`.