Cartesian power (cartesian product with self arbitrary times)

问题

I have a need in my code to calculate the cartesian product of an array with itself a varying number of times. For example, if my array is [1,2] and I need to fill these values into three slots, the result would be:

[1,1,1]
[1,1,2]
[1,2,1]
[1,2,2]
[2,1,1]
[2,1,2]
[2,2,1]
[2,2,2]

What's the easiest way to do this?

回答1:

You are probably looking for permutation with repetition and Ruby's Array from standard library luckily implements this:

[1,2].repeated_permutation(3).to_a
# [[1, 1, 1], [1, 1, 2], [1, 2, 1], [1, 2, 2], [2, 1, 1], [2, 1, 2], [2, 2, 1], [2, 2, 2]]

回答2:

A slight variant of your answer:

class Array
  def **(n)
    product( *([self]*(n-1)) )
  end
end

[1,2]**3
  # => [[1, 1, 1], [1, 1, 2], [1, 2, 1], [1, 2, 2],
  #     [2, 1, 1], [2, 1, 2], [2, 2, 1], [2, 2, 2]] 

回答3:

Because I like monkeypatching, I put this on the array itself:

class Array
  def **(n)
    self.product( *(n-1).times.map{ self } )
  end
end

I'm not sure if there's a more elegant way to pass n-1 copies of yourself to the method, though.

来源：https://stackoverflow.com/questions/21347083/cartesian-power-cartesian-product-with-self-arbitrary-times