By way of a puzzle, I've been trying to solve SICP exercise 2.20 ---------------------------- Using the (define (f x . args)) notation, write a procedure "same-parity" that takes one or more integers and returns a list of all the arguments that have the same even-odd parity as the first argument. For example, (same-parity 1 2 3 4 5 6 7) (1 3 5 7) (same-parity 2 3 4 5 6 7) (2 4 6) ---------------------------- using "straight recursion", that is, without using any `let` or `define` constructs. Still not managed to find the trick that will tack on the first argument as the head of the list. Anyone have a hint?
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