* I hear that `Y` delivers the/a *least* fixed point. Least
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?
+
+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 `h`, we want to find a fixed point `X` such that:
+
+ X <~~> h 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 (again, just a guess):
+
+ X ≡ ((\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 `h X`, since that is what the reduction arrow tells us. So we
+can refine the picture as follows:
+
+ X ≡ ((\u. h (___)) N) ~~> h 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:
+
+ X ≡ ((\u. h (__u__)) N) ~~> h X
+
+After reduction of the redex, we're going to have
+
+ X ≡ 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
+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. 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 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. h (u u)) (\u. h (u u)))
+ ~~> h ((\u. h (u u)) (\u. h (u u)))
+ ≡ h X
+
+Success.
+
+So the function `\h. (\u. h (u u)) (\u. h (u u))` maps an arbitrary term
+`h` to a fixed point for `h`.