My colleague Matt wrote about a recent Ruby Quiz which emphasised short, elegant solutions. He was particularly impressed with a solution which used a property of Ruby hashes to generate a wondrous number sequence for any given integer in a single line of code.
That got me thinking about the solution in Scheme, a language I find very appealing, and Haskell, which I had started picking up a few months ago but had to abandon due to lack of time. Even so I managed to remember enough to get this basic solution within a few minutes:
wondrous :: Int -> [Int] wondrous 1 =  wondrous x | even x = x : wondrous (x `div` 2) | otherwise = x : wondrous (3*x+1)
Here is the most obvious implementation in Scheme:
(define wondrous (lambda (x) (cond ((= x 1) ‘(1)) ((= (modulo x 2) 0) (cons x (wondrous (quotient x 2)))) (else (cons x (wondrous (+ (* 3 x) 1)))) ) ) )
As you can see both of these functional languages are really suited to expressing such algorithms, particularly Haskell with its excellent pattern matching expressions. I have no doubts that there are far more elegant solutions in both languages; I’ll post them if I get suitably inspired.