The following is a function which, when called without arguments, creates an incrementing sequence of numbers where iterating from n to n+1 takes about 2^n seconds.
(defn slow-nums
([]
(slow-nums 0 0))
([n sleep-for]
(Thread/sleep (* sleep-for 1000))
(cons n (lazy-seq (slow-nums (inc n) (Math/pow 2 n))))))Example
;; takes about 5 seconds (2^0 + 2^1 + 2^2)
(take 4 (slow-nums))
;; => (0 1 2 3)
;; print each number and the time it was read forever
(doseq [n (slow-nums)]
(println n (System/currentTimeMillis)))
;; => 0 1423690218544
;; 1 1423690218544
;; 2 1423690219549
;; 3 1423690223553
;; 4 1423690232558
;; ...
;; print each number in pairs and the time they were read forever
(doseq [numbers (partition 2 (slow-nums))]
(println numbers (System/currentTimeMillis)))
;; => (0 1) 1423690273680
;; (2 3) 1423690278684
;; (4 5) 1423690303691
;; (6 7) 1423690364694
;; ...Problem 1
Write a function that iterates through the sequence and paritions each number by the minute at which it was read. Then prints the minute and the numbers read during that minute.
Example output (results may vary based on the second you started)
2015-02-11 16:34 (0 1 2 3 4 5)
2015-02-11 16:35 (6)
2015-02-11 16:36 (7 8)
2015-02-11 16:38 (9)
2015-02-11 16:39 (10)
2015-02-11 16:41 (11)
2015-02-11 16:44 (12)
Problem 2
Write a function that, every minute, prints the minute and numbers read that were read minute, including empty sets of numbers.
Example output (results may vary based on the second you started)
2015-02-11 16:34 (0 1 2 3 4 5)
2015-02-11 16:35 (6)
2015-02-11 16:36 (7 8)
2015-02-11 16:37 ()
2015-02-11 16:38 (9)
2015-02-11 16:39 (10)
2015-02-11 16:40 ()
2015-02-11 16:41 (11)
2015-02-11 16:42 ()
2015-02-11 16:43 ()
2015-02-11 16:44 (12)
For both problems, strive to write the most "clojure-esque" solution.