Problem 25 asks us to find the index of the first Fibonacci number to have more than 1000 digits. Here’s my first solution, which boils down to making a single, human-readable line at the end.

(def fibonacci (map first (iterate (fn [[a b]] (vector b (+ a b))) [1 1])))

(defn num-digits [n] (count (str n)))

(defn first-where [condition seq]
  "Returns index of the first item in seq that satisfies condition."
  (loop [n 0
         r seq]
    (if (or (nil? r)
            (condition (first r)))
      n
      (recur (inc n) (next r)))))

(defn euler-25 []
  (inc (first-where #(>= (num-digits %) 1000) fibonacci)))

But nearly equivalent performance can be achieved using clojure-contrib.seq-utils, and if we do so we need even fewer lines of code:

(use '[clojure.contrib.seq-utils :only (indexed find-first)])

(def fibonacci (map first (iterate (fn [[a b]] (vector b (+ a b))) [1 1])))

(defn euler-25-seq-utils []
  (inc (first (find-first #(>= (count (str (second %))) 1000)
                      (indexed fibonacci)))))

As usual, proper exploitation of Clojure’s sequence abstraction gives us the shortest solution. Because first-where may be useful in the future, we could also redefine it as:

(defn first-where [condition seq]
  "Returns index of the first item in seq that satisfies condition."
  (first (find-first #(condition (second %)) (indexed seq))))

See you tomorrow!