From: Jim Pryor Date: Sun, 3 Oct 2010 18:08:42 +0000 (-0400) Subject: week4 tweaking X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff_plain;h=7d59a94d25530ed5d0bbb53f97af3bff3bc6a532 week4 tweaking Signed-off-by: Jim Pryor --- diff --git a/week4.mdwn b/week4.mdwn index a1aa2548..5238b261 100644 --- a/week4.mdwn +++ b/week4.mdwn @@ -17,7 +17,7 @@ of the equalities in the last line. #Q: How do you know that for any term `T`, `Y T` is a fixed point of `T`?# A: Note that in the proof given in the previous answer, we chose `T` -and then set `X = L L = (\x. T (x x)) (\x. T (x x))`. If we abstract over +and then set `X ≡ L L ≡ (\x. T (x x)) (\x. T (x x))`. If we abstract over `T`, we get the Y combinator, `\T. (\x. T (x x)) (\x. T (x x))`. No matter what argument `T` we feed `Y`, it returns some `X` that is a fixed point of `T`, by the reasoning in the previous answer. @@ -43,7 +43,7 @@ claimed that even `succ`---the function that added one to any number---had a fixed point. How could there be an X such that X = X+1? That would imply that - X <~~> succ X <~~> succ (succ X) <~~> succ (succ (succ (X))) <~~> succ (... (succ X)...) + X <~~> succ X <~~> succ (succ X) <~~> succ (succ (succ X)) <~~> succ (... (succ X)...) In other words, the fixed point of `succ` is a term that is its own successor. Let's just check that `X = succ X`: @@ -52,7 +52,7 @@ successor. Let's just check that `X = succ X`: let X = (\x. succ (x x)) (\x. succ (x x)) in succ X ≡ succ ( (\x. succ (x x)) (\x. succ (x x)) ) -~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x)))) +~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x)) )) ≡ succ (succ X) @@ -64,21 +64,22 @@ You should see the close similarity with `Y Y` here. A. Yes! Let's see why it makes sense to think of `Y succ` as a Church numeral: - [same definitions] - succ X - = (\n s z. s (n s z)) X - = \s z. s (X s z) - = succ (\s z. s (X s z)) ; using fixed-point reasoning - = \s z. s ([succ (\s z. s (X s z))] s z) - = \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z) - = \s z. s (s (succ (\s z. s (X s z)))) +
``````[same definitions]
+succ X
+≡ (\n s z. s (n s z)) X
+~~> \s z. s (X s z)
+<~~> succ (\s z. s (X s z)) ; using fixed-point reasoning
+~~> \s z. s ([succ (\s z. s (X s z))] s z)
+~~> \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z)
+~~> \s z. s (s (succ (\s z. s (X s z))))
+``````
So `succ X` looks like a numeral: it takes two arguments, `s` and `z`, and returns a sequence of nested applications of `s`... You should be able to prove that `add 2 (Y succ) <~~> Y succ`, -likewise for `mult`, `minus`, `pow`. What happens if we try `minus (Y -succ)(Y succ)`? What would you expect infinity minus infinity to be? +likewise for `mul`, `sub`, `pow`. What happens if we try `sub (Y +succ) (Y succ)`? What would you expect infinity minus infinity to be? (Hint: choose your evaluation strategy so that you add two `s`s to the first number for every `s` that you add to the second number.) @@ -105,35 +106,35 @@ which we have to make a choice about which beta reduction to perform next: one choice leads to a normal form, the other choice leads to endless reduction: - let prefac = \f n. iszero n 1 (mult n (f (pred n))) in - let fac = Y prefac in - fac 2 - = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2 - = [(\x.prefac(xx))(\x.prefac(xx))] 2 - = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2 - = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2 - = [(\f n. iszero n 1 (mult n (f (pred n)))) - (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2 - = [\n. iszero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2 - = iszero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2))) - = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1) - ... - = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0)) - = mult 2 (mult 1 (iszero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0)))) - = mult 2 (mult 1 1) - = mult 2 1 - = 2 +
``````let prefact = \f n. iszero n 1 (mul n (f (pred n))) in
+let fact = Y prefact in
+fact 2
+≡ [(\f. (\x. f (x x)) (\x. f (x x))) prefact] 2
+~~> [(\x. prefact (x x)) (\x. prefact (x x))] 2
+~~> [prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 2
+~~> [prefact (prefact ((\x. prefact (x x)) (\x. prefact (x x))))] 2
+≡ [ (\f n. iszero n 1 (mul n (f (pred n)))) (prefact ((\x. prefact (x x)) (\x. prefact (x x))))] 2
+~~> [\n. iszero n 1 (mul n ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred n)))] 2
+~~> iszero 2 1 (mul 2 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred 2)))
+~~> mul 2 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 1)
+...
+~~> mul 2 (mul 1 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 0))
+≡ mul 2 (mul 1 (iszero 0 1 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred 0))))
+~~> mul 2 (mul 1 1)
+~~> mul 2 1
+~~> 2
+``````
The crucial step is the third from the last. We have our choice of either evaluating the test `iszero 0 1 ...`, which evaluates to `1`, no matter what the ... contains; -or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to -produce another copy of `prefac`. If we postpone evaluting the -`iszero` test, we'll pump out copy after copy of `prefac`, and never +or we can evaluate the `Y` pump, `(\x. prefact (x x)) (\x. prefact (x x))`, to +produce another copy of `prefact`. If we postpone evaluting the +`iszero` test, we'll pump out copy after copy of `prefact`, and never realize that we've bottomed out in the recursion. But if we adopt a leftmost/call-by-name/normal-order evaluation strategy, we'll always start with the `iszero` predicate, and only produce a fresh copy of -`prefac` if we are forced to. +`prefact` if we are forced to. #Q. You claimed that the Ackerman function couldn't be implemented using our primitive recursion techniques (such as the techniques that allow us to define addition and multiplication). But you haven't shown that it is possible to define the Ackerman function using full recursion.# @@ -148,20 +149,20 @@ A. OK: let A = Y (\A m n. iszero m (succ n) (iszero n (A (pred m) 1) (A (pred m) (A m (pred n))))) -For instance, +So for instance: - A 1 2 - = A 0 (A 1 1) - = A 0 (A 0 (A 1 0)) - = A 0 (A 0 (A 0 1)) - = A 0 (A 0 2) - = A 0 3 - = 4 + A 1 2 + ~~> A 0 (A 1 1) + ~~> A 0 (A 0 (A 1 0)) + ~~> A 0 (A 0 (A 0 1)) + ~~> A 0 (A 0 2) + ~~> A 0 3 + ~~> 4 -A 1 x is to A 0 x as addition is to the successor function; -A 2 x is to A 1 x as multiplication is to addition; -A 3 x is to A 2 x as exponentiation is to multiplication--- -so A 4 x is to A 3 x as hyper-exponentiation is to exponentiation... +`A 1 x` is to `A 0 x` as addition is to the successor function; +`A 2 x` is to `A 1 x` as multiplication is to addition; +`A 3 x` is to `A 2 x` as exponentiation is to multiplication--- +so `A 4 x` is to `A 3 x` as hyper-exponentiation is to exponentiation... #Q. What other questions should I be asking?#