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diff --git a/week3.mdwn b/week3.mdwn
index d9c53611..39e472bf 100644
--- a/week3.mdwn
+++ b/week3.mdwn
@@ -1,3 +1,12 @@
+[[!toc]]
+
+##More on evaluation strategies##
+
+Here are notes on [[evaluation order]] that make the choice of which
+lambda to reduce next the selection of a route through a network of
+links.
+
+
 ##Computing the length of a list##
 
 How could we compute the length of a list? Without worrying yet about what lambda-calculus implementation we're using for the list, the basic idea would be to define this recursively:
@@ -99,6 +108,34 @@ where this very same formula occupies the `...` position:
 
 but as you can see, we'd still have to plug the formula back into itself again, and again, and again... No dice.
 
+[At this point, some of you will recall the discussion in the first
+class concerning the conception of functions as sets of ordered pairs.
+The problem, as you will recall, was that in the untyped lambda
+calculus, we wanted a function to be capable of taking itself as an
+argument.  For instance, we wanted to be able to apply the identity
+function to itself.  And since the identity function always returns
+its argument unchanged, the value it should return in that case is
+itself:
+
+    (\x.x)(\x.x) ~~> (\x.x)
+
+If we conceive of a function as a set of ordered pairs, we would start
+off like this:
+
+    1 -> 1
+    2 -> 2
+    3 -> 3
+    ...
+    [1 -> 1, 2 -> 2, 3 -> 3, ..., [1 -> 1, 2 -> 2, 3 -> 3, ..., 
+
+Eventually, we would get to the point where we want to say what the
+identity function itself gets mapped to.  But in order to say that, we
+need to write down the identity function in the argument position as a
+set of ordered pairs.  The need to insert a copy of the entire
+function definition inside of a copy of the entire function definition
+inside of... is the same problem as the need to insert a complete
+graph of the identity function inside of the graph for the identity function.]
+
 So how could we do it? And how do OCaml and Scheme manage to do it, with their `let rec` and `letrec`?
 
 1.	OCaml and Scheme do it using a trick. Well, not a trick. Actually an impressive, conceptually deep technique, which we haven't yet developed. Since we want to build up all the techniques we're using by hand, then, we shouldn't permit ourselves to rely on `let rec` or `letrec` until we thoroughly understand what's going on under the hood.
@@ -126,7 +163,10 @@ With sufficient ingenuity, a great many functions can be defined in the same way
 
 ##However...##
 
-Some computable functions are just not definable in this way. The simplest function that *simply cannot* be defined using the resources we've so far developed is the [[!wikipedia Ackermann function]]:
+Some computable functions are just not definable in this way. We can't, for example, define a function that tells us, for whatever function `f` we supply it, what is the smallest integer `x` where `f x` is `true`.
+
+Neither do the resources we've so far developed suffice to define the 
+[[!wikipedia Ackermann function]]:
 
 	A(m,n) =
 		| when m == 0 -> n + 1
@@ -180,7 +220,8 @@ Instead of writing out a long formula twice, we could write:
 
 and the initial `(\x. x x)` is just what we earlier called the <code>&omega;</code> combinator (lower-case omega, not the non-terminating <code>&Omega;</code>). So the self-application of `H` can be written:
 
-<pre><code>&omega; (\h \lst. (isempty lst) zero (add one ((h h) (extract-tail lst))))</code></pre>
+<pre><code>&omega; (\h \lst. (isempty lst) zero (add one ((h h) (extract-tail lst))))
+</code></pre>
 
 and this will indeed implement the recursive function we couldn't earlier figure out how to define.
 
@@ -237,6 +278,8 @@ In the lambda calculus, we say a fixed point of an expression `f` is any formula
 
 What is a fixed point of the identity combinator I?
 
+What is a fixed point of the false combinator, KI?
+
 It's a theorem of the lambda calculus that every formula has a fixed point. In fact, it will have infinitely many, non-equivalent fixed points. And we don't just know that they exist: for any given formula, we can name many of them.
 
 Yes, even the formula that you're using the define the successor function will have a fixed point. Isn't that weird? Think about how it might be true.
@@ -381,7 +424,7 @@ to *the tail* of the list we were evaluating its application to at the previous
 
 ##Fixed-point Combinators Are a Bit Intoxicating##
 
-![tatoo](/y-combinator.jpg)
+![tatoo](/y-combinator-fixed.jpg)
 
 There's a tendency for people to say "Y-combinator" to refer to fixed-point combinators generally. We'll probably fall into that usage ourselves. Speaking correctly, though, the Y-combinator is only one of many fixed-point combinators.
 
@@ -396,3 +439,202 @@ then this is a fixed-point combinator:
 	L L L L L L L L L L L L L L L L L L L L L L L L L L
 
 
+##Watching Y in action##
+
+For those of you who like to watch ultra slow-mo movies of bullets
+piercing apples, here's a stepwise computation of the application of a
+recursive function.  We'll use a function `sink`, which takes one
+argument.  If the argument is boolean true (i.e., `\x y.x`), it
+returns itself (a copy of `sink`); if the argument is boolean false
+(`\x y. y`), it returns `I`.  That is, we want the following behavior:
+
+    sink false ~~> I
+    sink true false ~~> I
+    sink true true false ~~> I
+    sink true true true false ~~> I
+
+So we make `sink = Y (\f b. b f I)`: 
+
+    1. sink false 
+    2. Y (\fb.bfI) false
+    3. (\f. (\h. f (h h)) (\h. f (h h))) (\fb.bfI) false
+    4. (\h. [\fb.bfI] (h h)) (\h. [\fb.bfI] (h h)) false
+    5. [\fb.bfI] ((\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))) false
+    6. (\b.b[(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))]I)  false
+    7. false [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] I
+             --------------------------------------------
+    8. I
+
+So far so good.  The crucial thing to note is that as long as we
+always reduce the outermost redex first, we never have to get around
+to computing the underlined redex: because `false` ignores its first
+argument, we can throw it away unreduced.
+
+Now we try the next most complex example:
+
+    1. sink true false 
+    2. Y (\fb.bfI) true false
+    3. (\f. (\h. f (h h)) (\h. f (h h))) (\fb.bfI) true false
+    4. (\h. [\fb.bfI] (h h)) (\h. [\fb.bfI] (h h)) true false
+    5. [\fb.bfI] ((\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))) true false
+    6. (\b.b[(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))]I)  true false
+    7. true [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] I false
+    8. [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] false
+
+We've now arrived at line (4) of the first computation, so the result
+is again I.
+
+You should be able to see that `sink` will consume as many `true`s as
+we throw at it, then turn into the identity function after it
+encounters the first `false`. 
+
+The key to the recursion is that, thanks to Y, the definition of
+`sink` contains within it the ability to fully regenerate itself as
+many times as is necessary.  The key to *ending* the recursion is that
+the behavior of `sink` is sensitive to the nature of the input: if the
+input is the magic function `false`, the self-regeneration machinery
+will be discarded, and the recursion will stop.
+
+That's about as simple as recursion gets.
+
+##Base cases, and their lack##
+
+As any functional programmer quickly learns, writing a recursive
+function divides into two tasks: figuring out how to handle the
+recursive case, and remembering to insert a base case.  The
+interesting and enjoyable part is figuring out the recursive pattern,
+but the base case cannot be ignored, since leaving out the base case
+creates a program that runs forever.  For instance, consider computing
+a factorial: `n!` is `n * (n-1) * (n-2) * ... * 1`.  The recursive
+case says that the factorial of a number `n` is `n` times the
+factorial of `n-1`.  But if we leave out the base case, we get
+
+    3! = 3 * 2! = 3 * 2 * 1! = 3 * 2 * 1 * 0! = 3 * 2 * 1 * 0 * -1! ...
+
+That's why it's crucial to declare that 0! = 1, in which case the
+recursive rule does not apply.  In our terms,
+
+    fac = Y (\fac n. iszero n 1 (fac (predecessor n)))
+
+If `n` is 0, `fac` reduces to 1, without computing the recursive case.
+
+There is a well-known problem in philosophy and natural language
+semantics that has the flavor of a recursive function without a base
+case: the truth-teller paradox (and related paradoxes).
+
+(1)    This sentence is true.
+
+If we assume that the complex demonstrative "this sentence" can refer
+to (1), then the proposition expressed by (1) will be true just in
+case the thing referred to by *this sentence* is true.  Thus (1) will
+be true just in case (1) is true, and (1) is true just in case (1) is
+true, and so on.  If (1) is true, then (1) is true; but if (1) is not
+true, then (1) is not true.
+
+Without pretending to give a serious analysis of the paradox, let's
+assume that sentences can have for their meaning boolean functions
+like the ones we have been working with here.  Then the sentence *John
+is John* might denote the function `\x y. x`, our `true`.  
+
+Then (1) denotes a function from whatever the referent of *this
+sentence* is to a boolean.  So (1) denotes `\f. f true false`, where
+the argument `f` is the referent of *this sentence*.  Of course, if
+`f` is a boolean, `f true false <~~> f`, so for our purposes, we can
+assume that (1) denotes the identity function `I`.
+
+If we use (1) in a context in which *this sentence* refers to the
+sentence in which the demonstrative occurs, then we must find a
+meaning `m` such that `I m = I`.  But since in this context `m` is the
+same as the meaning `I`, so we have `m = I m`.  In other words, `m` is
+a fixed point for the denotation of the sentence (when used in the
+appropriate context).
+
+That means that in a context in which *this sentence* refers to the
+sentence in which it occurs, the sentence denotes a fixed point for
+the identity function.  Here's a fixed point for the identity
+function:
+
+<pre><code>Y I
+(\f. (\h. f (h h)) (\h. f (h h))) I
+(\h. I (h h)) (\h. I (h h)))
+(\h. (h h)) (\h. (h h)))
+&omega; &omega;
+&Omega
+</code></pre>
+
+Oh.  Well!  That feels right.  The meaning of *This sentence is true*
+in a context in which *this sentence* refers to the sentence in which
+it occurs is <code>&Omega;</code>, our prototypical infinite loop...
+
+What about the liar paradox?
+
+(2)  This sentence is false.
+
+Used in a context in which *this sentence* refers to the utterance of
+(2) in which it occurs, (2) will denote a fixed point for `\f.neg f`,
+or `\f l r. f r l`, which is the `C` combinator.  So in such a
+context, (2) might denote
+
+     Y C
+     (\f. (\h. f (h h)) (\h. f (h h))) I
+     (\h. C (h h)) (\h. C (h h))) 
+     C ((\h. C (h h)) (\h. C (h h)))
+     C (C ((\h. C (h h))(\h. C (h h))))
+     C (C (C ((\h. C (h h))(\h. C (h h)))))
+     ...
+
+And infinite sequence of `C`s, each one negating the remainder of the
+sequence.  Yep, that feels like a reasonable representation of the
+liar paradox.
+
+See Barwise and Etchemendy's 1987 OUP book, [The Liar: an essay on
+truth and circularity](http://tinyurl.com/2db62bk) for an approach
+that is similar, but expressed in terms of non-well-founded sets
+rather than recursive functions.
+
+##However...##
+
+You should be cautious about feeling too comfortable with
+these results.  Thinking again of the truth-teller paradox, yes,
+<code>&Omega;</code> is *a* fixed point for `I`, and perhaps it has
+some a privileged status among all the fixed points for `I`, being the
+one delivered by Y and all (though it is not obvious why Y should have
+any special status).
+
+But one could ask: look, literally every formula is a fixed point for
+`I`, since
+
+    X <~~> I X
+
+for any choice of X whatsoever.
+
+So the Y combinator is only guaranteed to give us one fixed point out
+of infinitely many---and not always the intuitively most useful
+one. (For instance, the squaring function has zero as a fixed point,
+since 0 * 0 = 0, and 1 as a fixed point, since 1 * 1 = 1, but `Y
+(\x. mul x x)` doesn't give us 0 or 1.) So with respect to the
+truth-teller paradox, why in the reasoning we've
+just gone through should we be reaching for just this fixed point at
+just this juncture?
+
+One obstacle to thinking this through is the fact that a sentence
+normally has only two truth values.  We might consider instead a noun
+phrase such as 
+
+(3)  the entity that this noun phrase refers to
+
+The reference of (3) depends on the reference of the embedded noun
+phrase *this noun phrase*.  It's easy to see that any object is a
+fixed point for this referential function: if this pen cap is the
+referent of *this noun phrase*, then it is the referent of (3), and so
+for any object.
+
+The chameleon nature of (3), by the way (a description that is equally
+good at describing any object), makes it particularly well suited as a
+gloss on pronouns such as *it*.  In the system of 
+[Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/),
+pronouns denote (you guessed it!) identity functions...
+
+Ultimately, in the context of this course, these paradoxes are more
+useful as a way of gaining leverage on the concepts of fixed points
+and recursion, rather than the other way around.