返回n个布尔的所有组合的函数？ [英] Function to return all combinations of n booleans?

问题描述

我试图实现一个函数，它接受一个数字n，并返回一个包含所有可能的n布尔组合的布尔列表的列表。例如， （make-bools 3）应该像

I am trying to implement a function that takes a number n and returns a list of lists of booleans that contains all possible combinations of n booleans. The output of e.g. (make-bools 3) should look like

[[false false false]
 [false false true ]
 [false true  false]
 [false true  true ]
 [true  false false]
 [true  false true ]
 [true  true  false]
 [true  true  true ]]

数字从0到（2 ^ n）-1到二进制格式，并使用 bit-test 创建一个布尔列表，最后链接所有这些列表。但对我来说似乎很笨拙，我想必须有一个更优雅的解决方案。

I thought of converting the numbers from 0 to (2^n) - 1 to a binary format and use bit-test to make a list of booleans, finally chaining all those lists. But that seems quite clumsy to me and I suppose there must be a more elegant solution.

推荐答案

我不知道是否被认为是欺骗使用现有库，而不是响应算法细节你的问题，但Clojure-contrib有一组常用的组合函数，用于计算排列：

I don't know if it's considered 'cheating' to use an existing library, rather than respond to the algorithmic details of your question, but Clojure-contrib has a set of common combinatorial functions that are useful for calculating permutations:

(require '[clojure.contrib.combinatorics :as comb])

(defn make-bools [n]
  (apply comb/cartesian-product (repeat n [true false])))

http://richhickey.github.com/clojure-contrib/combinatorics-api.html

这篇关于返回n个布尔的所有组合的函数？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！