Generating binary numbers of n digits in clojure

Question

I'd like to generate binary numbers of `n` digits from 0 to 2^n-1. For example of 3 digits, "000", "001", "010", ..., "111" (0 to 7 in decimal). The way I used is to use `java.lang.Integer.toBinaryString()` method and add zeros if necessary like the following: ``` (defn pad-zero [s n] (str (reduce str (repeat (- n (count s)) "0")) s)) (defn binary-permutation [n] (map (fn [s] (pad-zero s n)) (map #(Integer/toBinaryString %) (range 0 (Math/pow 2 n))))) ``` With this code, I can generate what I want like this. For 3 digits: ``` (binary-permutation 3) => ("000" "001" "010" "011" "100" "101" "110" "111") ``` But this codes look a little verbose. Aren't there any ways better or more clojure way to do this?

Accepted Answer

You can simplify the formatting using cl-format from clojure.pprint:

```
(defn binary-permutation [n]
  (map (partial cl-format nil "~v,'0B" n) (range 0 (Math/pow 2 n))))
```

You may also be interested to know that `(Math/pow 2 n)` is equivalent to `(bit-shift-left 1 n)`.

Another way to express this would be in term of selections from clojure.math.combinatorics:

```
(defn binary-permutation [n]
  (map (partial apply str) (selections [0 1] n)))
```

Generating binary numbers of n digits in clojure

Solution

Problem