GC16/3011 Functional Programming
Lecture 6
Recursion and the Lambda
Calculus
4/7/2015
1
Contents
•
•
•
Introduction
Recursive functions in Miranda
Recursive functions in the lambda calculus
•
•
•
•
Binding for function name
Fixed point
Fixed point operator
self-application
4/7/2015
2
Introduction
Recursive function f
Intermediate function h
Fixed points and the
fixpoint operator Y
Definition of f in terms of Y
How to define Y in the
lambda calculus
Final lambda-calculus definition of f
4/7/2015
3
Recursion and the lambda
calculus
•
Consider:
f x = 3, if (x = 0)
= 11, otherwise
•
In the lambda calculus this is:
l x.(if (x = 0) 3 11)
4/7/2015
4
Recursion and the lambda
calculus
•
However:
f x = 3, if (x = 0)
= 1 + f (x - 1), otherwise
•
What’s wrong with this?:
l x.(if (x = 0) 3 (1 + (f (x - 1))))
4/7/2015
5
Recursion and the lambda
calculus
•
Whole program:
f x = 3, if (x = 0)
= 1 + f (x - 1), otherwise
main = f 7
•
Whole program?:
l x.(if (x = 0) 3 (1 + (f (x - 1)))) 7
4/7/2015
6
Recursion and the lambda
calculus
•
1
SO how can we represent a recursive
function in the lambda calculus?
First define a new function (e.g. call it “h”)
as follows:
h f x = 3, if (x = 0)
= 1 + f (x-1), otherwise
4/7/2015
7
Recursion and the lambda
calculus
2
Now the following lambda expression for “h” is
fine, because “f” is bound:
l f. (l x.(if (x = 0) 3 (1 + (f (x - 1)))))
BUT “h” is not “f”, so we haven’t solved the
problem yet!
3
However, notice that (h f) gives the same
expression as f, so (h f) = f
(this is an equality, not a definition)
4/7/2015
8
Recursion and the lambda
calculus
•
A “fixed point” (or “fixpoint”) of a function f is a
value x from the input domain of f such that (f x)
returns the same value x
•
Example:
•
•
•
•
id is called the “identity function”
Every value in the input domain of id is a fixed point of id
Example:
•
id x = x
three x = if (x=3) 3 else (x+1)
The input value 3 is the only fixed point of three
Note that f is a fixpoint of h !!!!
4/7/2015
9
Recursion and the lambda
calculus
•
There is a special function that we can use (called the “fixpoint
operator”) which will return the fixed-point of any function.
•
The fixpoint operator is often denoted “Y”
•
It takes a function as its argument and returns the fixed point of that
function
• If the fixed-point is a function, it returns the “least” (i.e. most
general) function that is a fixed-point
•
So now (Y h) gives the least fixpoint of h, which we know is f
f=Yh
4/7/2015
10
Recursion and the lambda
calculus
•
Y can easily be expressed operationally as:
Y g = g (Y g)
•
|| not valid Miranda!
Now, for example (f is fixpoint of h):
f 3 = (Y h) 3
= (h (Y h)) 3
= (l x.(if (x = 0) 3 (1 + ((Y h) (x - 1))))) 3
= 1 + ((Y h) 2)
= 1 + ((h (Y h)) 2)
4/7/2015
because f = Y h
because Y g = g (Y g)
loop!
11
Recursion and the lambda
calculus
•
Now we have a lambda expression that defines “f”
Y (l f. (l x.(if (x = 0) 3 (1 + (f (x - 1))))))
•
But haven’t we really just shifted the
problem - how do we define “Y” in the
lambda calculus?
4/7/2015
12
Recursion and the lambda
calculus
The real magic: self-application
lq.( (lx.(q (x x))) (lx.(q (x x))) )
4/7/2015
13
Recursion and the lambda
calculus
Example:
Y h = lq.( (lx.(q (x x))) (lx.(q (x x))) ) h
= (lx.(h (x x))) (lx.(h (x x)))
= h ((lx.(h (x x))) (lx.(h (x x))))
= h (Y h)
4/7/2015
14
Recursion and the lambda
calculus
f = (Y h)
= (Y (l f. (l x.(if (x = 0) 3 (1 + (f (x - 1)))))))
= ((lq.((lx.(q (x x)))(lx.(q (x x)))))
(l f. (l x.(if (x = 0) 3 (1 + (f (x - 1)))))))
4/7/2015
15
Summary
•
•
Recursive functions in Miranda
Recursive functions in the lambda calculus
•
•
•
•
Binding for function name
Fixed point
Fixed point operator
Self-application
4/7/2015
16