recursion - Clojure: Write a Clojure function that returns a list of all 2^n bits strings of length n -

i want write clojure function called (nbits n) returns list of 2^n bits strings of length n.

my expected output is:

user=> (nbits -2) ()  user=> (nbits 0) ()  user=> (nbits 1) ((0) (1))  user=> (nbits 2) ((0 0) (0 1) (1 0) (1 1))  user=> (nbits 3) ((0 0 0) (0 0 1) (0 1 0) (0 1 1) (1 0 0) (1 0 1) (1 1 0) (1 1 1)) 

here try:

(defn add0 [seq]    (cond (empty? seq) 'nil         (and (seq? seq) (> (count seq) 0))            (cons (cons '0 (first seq)) (add0 (rest seq)))))  (defn add1 [seq]   (cond (empty? seq) 'nil         (and (seq? seq) (> (count seq) 0))            (cons (cons '1 (first seq)) (add1 (rest seq)))))  (defn nbits [n]   (cond (number? n)            (cond (< n 1) '()                 (= n 1) '((0) (1))                 (> n 1)                    (list (add0 (nbits (- n 1))) (add1 (nbits (- n 1)))))          :else 'nil)) 

but output not right. did go wrong? don't know how fix it.

you (lazily , iteratively) math.combinatorics library:

user=> (defn nbits [n]           (if (pos? n)             (clojure.math.combinatorics/selections [0 1] n)             '())) #'user/nbits user=> (nbits 0) () user=> (nbits 1) ((0) (1)) user=> (nbits 2) ((0 0) (0 1) (1 0) (1 1)) user=> (nbits 3) ((0 0 0) (0 0 1) (0 1 0) (0 1 1) (1 0 0) (1 0 1) (1 1 0) (1 1 1)) 

this version has advantage of leveraging iteration avoid blowing stack (which got @ n=14 original version on machine).