September 22, 1999 | Perl and the Lambda Calculus | Slide #15 |
Some expressions have no normal form:
(x.(x x) x.(x x))
Reduces to itself
Now consider (y.P (x.(x x) x.(x x)))
Could reduce immediately to P
But if you try to evalute the argument first, you lose
Is it hard to find normal forms? No.
Theorem: Left-to-right evalaution always finds the normal form if one exists
This is called normal-order evalutaion
Note: Most languages use applicative order.
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