Recursive call map on array?

Question

I'm attempting to write a method that takes a value N and returns three integers N/2, N/3, and N/4 rounded down. It keeps taking them back until there is only zero.

def crazy_coins(n)
  coins = Array.new
  coins.push(n.to_a)
  generate = Proc.new { |x|
    temp = []
    count = 2
    while count < 5
      x = n/count
      temp << x
      count += 1    
    end
    return temp
  }
  coins.map!(&generate)
end

Outputs:

crazy_coins(5)
# => [[2, 1, 1]]

Successful output is supposed to resemble:

crazy_coins(5)

11
=> [2, 1, 1] 
=> [[1, 0, 0], [0, 0, 0], [0, 0, 0]]
=> [[[0, 0, 0], 0, 0], [0, 0, 0], [0, 0, 0]]

What might be the best way to call coins.map! again on each element (recursively perhaps) until all coins[i][j] == 0?

I tried calling coins[0].map!(&generate) but the result is [[2, 1, 1], [2, 1, 1], [2, 1, 1]] Why is it not replacing the existing values with a new array?

You've got it - the return statement in a proc will return from the context in which the proc is called, not just the proc itself. You actually don't need an explicit return statement at all - just put temp as the last statement in the proc and it will be returned. — Jon Cairns
– Jon Cairns, Commented Oct 23, 2013 at 9:16
ok nice thanks, that works but when attempting to call coins[0].map!(&generate) to try and replace the [[2, 1, 1]] values, the result is [[2, 1, 1], [2, 1, 1], [2, 1, 1]]..how come map! isnt replacing the single values with a new array? — rogueSF
– rogueSF, Commented Oct 23, 2013 at 9:26
map! is a method that changes the receiver. It isn't meant to return a new array. — vgoff
– vgoff, Commented Oct 23, 2013 at 10:44
I see so i have should make a new array from each [[2, 3, 3]]? — rogueSF
– rogueSF, Commented Oct 23, 2013 at 10:52

dbenhur · Accepted Answer · 2013-10-23 16:34:25Z

1

You just need to recurse on each generated element when it's not 0. Also, try to avoid iteration when collection composition is available.

def divisors(n)
  (2..4).map { |d| n/d }
end

def crazy_coins(a)
  Array(a).map { |e| e != 0 ? crazy_coins(divisors(e)) : e }
end

crazy_coins(5)
=> [[[[0, 0, 0], 0, 0], [0, 0, 0], [0, 0, 0]]]
crazy_coins(11)
=> [[[[[0, 0, 0], 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], 0], [[0, 0, 0], 0, 0]]]

answered Oct 23, 2013 at 16:34

dbenhur

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rogueSF · Accepted Answer · 2013-10-24 00:54:06Z

0

Thanks to dbenhur I was able to finish the problem...dbenhur showed me the recursion but the end result of my program was to return the final amount of zero coins but I couldn't make it work with two methods...

Solution:

def crazy_coins(n)
  return 1 if n == 0

      # Recursive
  return crazy_coins(n / 2) + crazy_coins(n / 3) + crazy_coins(n / 4)
end

crazy_coins(5)
=> 11
=> [[[[0, 0, 0], 0, 0], [0, 0, 0], [0, 0, 0]]]

answered Oct 24, 2013 at 0:54

rogueSF

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