The same strategy works in Clojure, of course:
(defn b-of-a [a] (/ (* 1000.0 (- 500.0 a)) (- 1000.0 a)))
(defn near-int? [x] (< (- x (floor x)) 0.00001))
(first (for [a (iterate inc 1) :let [b (b-of-a a)] :when (near-int? b)]
[a b (sqrt (+ (* a a) (* b b)))]))
Stu
I like Clojure, but as a point of comparison, here's a Haskell
solution, as typed in the REPL:
Prelude> let bOf a = 1000*(500 - a)/(1000 - a)
Prelude> let nearInt x = x - fromInteger(truncate x) < 0.000001
Prelude> head [ ( a, b, sqrt(a^2 + b^2) ) | a <- [1..], b <- [bOf a],
nearInt b ]
(200.0,375.0,425.0)
The numbers in the result add up to 1000, of course. Here, I just
solved b in terms of a, which is function bOf. Predicate nearInt
detects whether its argument is an integer (or close enough). Haskell
is lazy, so even though [1..] and the big list comprehension are
infinite, head just needs the first element.
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