Is this good Ruby? (Ruby Koans' Greed Task)

Question

Having coded in Java and C# for quite some years, I'm currently learning Ruby. I'm working my way through the Ruby Koans tutorial. At some point, you are to implement a method that calculates the game-score of a dice-game called Greed.

I came up with this recursive Java/C#-like method. It passes all the supplied unit tests, so technically it's correct.

Now I'm wonderng: Is this good Ruby code? If not, how would a "Rubyist" write this method? And possibly: Why? I'm also not so happy about the amount of duplicate code but can't think of a better Rubyish way.

Thanks for your reviews!

EDIT: The updated method to review (updated upon glebm's feedback)

def score(dice)
  patterns = {[1,1,1]=>1000, [2,2,2]=>200, [3,3,3]=>300, [4,4,4]=>400,
              [5,5,5]=>500,  [6,6,6]=>600, 1=>100, 5=>50}
  sorted = dice.sort

  triple = patterns[sorted[0..2]]
  single = patterns[sorted[0]]
  if triple
    partial_score = triple
    rest = sorted[3..-1]
  elsif single
    partial_score = single
    rest = sorted[1..-1]
  else
    partial_score = 0
    rest = sorted[1..-1]
  end

  if rest
    partial_score + score(rest) 
  else
    partial_score
  end
end

The original method to review

def score(dice)   #dice is an array of numbers, i.e. [3,4,5,3,3]
  return 0 if(dice == [] || dice == nil)

  dice.sort!

  return 1000 + score(dice[3..-1]) if(dice[0..2] == [1,1,1])
  return 600 + score(dice[3..-1]) if(dice[0..2] == [6,6,6])
  return 500 + score(dice[3..-1]) if(dice[0..2] == [5,5,5])
  return 400 + score(dice[3..-1]) if(dice[0..2] == [4,4,4])
  return 300 + score(dice[3..-1]) if(dice[0..2] == [3,3,3])
  return 200 + score(dice[3..-1]) if(dice[0..2] == [2,2,2])
  return 100 + score(dice[1..-1]) if(dice[0] == 1)
  return 50 + score(dice[1..-1]) if(dice[0] == 5)
  return 0 + score(dice[1..-1]);
end

**Some background (if needed)

# Greed is a dice game where you roll up to five dice to accumulate
# points. A greed roll is scored as follows:
#
# * A set of three ones is 1000 points
#
# * A set of three numbers (other than ones) is worth 100 times the
#   number. (e.g. three fours is 400 points).
#
# * A one (that is not part of a set of three) is worth 100 points.
#
# * A five (that is not part of a set of three) is worth 50 points.
#
# * Everything else is worth 0 points.
#
#
# Examples:
#
# score([1,1,1,5,1]) => 1150 points
# score([2,3,4,6,2]) => 0 points
# score([3,4,5,3,3]) => 350 points
# score([1,5,1,2,4]) => 250 points
#
# More scoring examples are given in the tests below:


class AboutScoringProject < EdgeCase::Koan
  def test_score_of_an_empty_list_is_zero
    assert_equal 0, score([])
  end

  def test_score_of_a_single_roll_of_5_is_50
    assert_equal 50, score([5])
  end

  def test_score_of_a_single_roll_of_1_is_100
    assert_equal 100, score([1])
  end

  def test_score_of_a_single_roll_of_1_is_100
    assert_equal 200, score([1,1])
  end

  def test_score_of_multiple_1s_and_5s_is_the_sum_of_individual_scores
    assert_equal 300, score([1,5,5,1])
  end

  def test_score_of_single_2s_3s_4s_and_6s_are_zero
    assert_equal 0, score([2,3,4,6])
  end

  def test_score_of_a_triple_1_is_1000
    assert_equal 1000, score([1,1,1])
  end

  def test_score_of_other_triples_is_100x
    assert_equal 200, score([2,2,2])
    assert_equal 300, score([3,3,3])
    assert_equal 400, score([4,4,4])
    assert_equal 500, score([5,5,5])
    assert_equal 600, score([6,6,6])
  end

  def test_score_of_mixed_is_sum
    assert_equal 250, score([2,5,2,2,3])
    assert_equal 550, score([5,5,5,5])
  end

  def test_score_of_a_triple_1_is_1000A
    assert_equal 1150, score([1,1,1,5,1])
  end

  def test_score_of_a_triple_1_is_1000B
    assert_equal 350, score([3,4,5,3,3])
  end

  def test_score_of_a_triple_1_is_1000C
    assert_equal 250, score([1,5,1,2,4])
  end
end

Answer 1

No, this is not good ruby code.

Let's start with mistakes:

• Do not check for == nil. nil is not specified as a valid value for the method, therefore checking for it and returning 0 might mask other exceptions
• Do not use return statement. If you need to do series of if statements, just use if...elsif, or case
• Do not modify parameters that come into your function. I am referring to dice.sort!
• Do not use recursion when it would be a lot cleaner to do it the straight-forward way

Considering all the above, here is a cleaned up version of the code:

def score(dice)
  score = 0

  # Below is equivalent to:
  #   counts = dice.inject(Hash.new(0)) { |h, x| h[x] += 1; h }
  counts = Hash.new(0) 
  dice.each do |x|
    counts[x] += 1
  end

  (1..6).each do |i|
    if counts[i] >= 3 
      if i == 1
        score += 1000
      else
        score += 100 * i
      end

      counts[i] = [counts[i] - 3, 0].max
    end

    if i == 1
      score += 100 * counts[i]
    elsif i == 5
      score += 50 * counts[i]
    end
  end

  score
end

Answer 2

Here's my solution:

def score(dice)
  score = 0
  dice_grouped = dice.group_by { |i| i }
  dice_grouped.each do | number, all_of_number |
     if all_of_number.size >= 3
       if number == 1
         score += 1000*number
       else
         score += 100*number
       end
       all_of_number.shift(3)
     end
  end
  score += 100*dice_grouped[1].size if dice_grouped[1]
  score += 50*dice_grouped[5].size if dice_grouped[5]
  return score
end

Answer 3

A little refactoring on the second implementation;

PATTERNS = {[1,1,1]=>1000, [2,2,2]=>200, [3,3,3]=>300, [4,4,4]=>400,
            [5,5,5]=>500,  [6,6,6]=>600, [1]=>100, [5]=>50}

def score(dice)
  return 0 if dice.nil? or dice.empty?

  sorted = dice.sort    
  partial_score = PATTERNS[set=sorted.first(3)] || PATTERNS[set=sorted.first(1)] || 0    
  rest = sorted.drop(set.length)
  partial_score + score(rest) 
end

Answer 4

don't know if all bugs are ironed out but here is what i came up with while trying to learn koans.

   def score(dice)
       dice_map = Hash[1,0,2,0, 3,0, 4,0, 5,0, 6,0]
       dice.collect {|i| dice_map[i] += 1 }
       totalscore= dice_map[1]>=3 ? 1000 + (dice_map[1]-3)*100 : dice_map[1]*100
       totalscore += dice_map[5]>=3 ? 500 + (dice_map[5]-3)*50 : dice_map[5]*50
       dice_map.delete_if {|k,v| k==1 || k==5}
       dice_map.each {|k, v| v>=3 ? totalscore += k*100 : totalscore}

       return totalscore

    end

Answer 5

What about that piece:

def score(dice)
  score = 0
  1.upto(6).each { |i| score = i == 1 ? 1000 : 100*i unless dice.count(i) < 3 }
  [[1, 100], [5, 50]].each { |i,j| score += j * (dice.count(i) % 3) }
  score
end

Answer 6

This is what I just wrote:

def score(dice)

  totals = Hash.new(0)

  dice.each { |x| totals[x] += 1 }

  score = 0

  totals.each do |x, t|

    if t >= 3
      score += x == 1 ? 1000 : (x * 100)
      t -= 3;
    end

    if x == 1
      score += t * 100
    elsif x == 5
      score += t * 50
    end

  end

  score

end

Google brought me here when I decided to look for other examples, so I signed finally signed up! I create a new hash to store the totals of each rolled number, then I loop through it. First, I check if the number was rolled 3 or more times. If it was, I add 1000 if it's a 1 and x * 100 for everything else. I then minus 3 from the number of times rolled so we can properly calculate the single values to add (for example, if [1, 1, 1, 1, 1] was rolled, it should output 1,200). If a one was rolled, 100 * t is added. If a 5 was rolled, 50 * t is added.

The only difference between this and other examples is that I'm accounting for rolls above 3 times (specified in GREED_RULES.txt).

I just started learning Ruby a few days ago, so please comment if you have any optimizations. It passed the asserts in the tutorial fine.

Answer 7

def score(dice)
    sum = 0

    unless dice.empty?
        # Get score for single die
        if dice.length == 1 then
        sum = 50 if dice.first == 5
        sum = 100 if dice.first == 1
        else
            if dice.length == 3 && dice.uniq.length == 1 then
            sum = dice.first == 1 ? 1000 : dice.first * 100
            else
                if dice.all? { |d| d == 1 || d == 5 } && dice.uniq.length == 2 then
                    sum = dice.each.inject(0) {|total, d| total + score([d])}
                else
                    dice.sort!
                    sum = score([dice[-1]]) + score(dice[0...-1])
                end
            end
        end
    end

  return sum
end

Answer 8

Here's mine... somewhat different than the others.

def score(dice)
  score = 0
  dice.uniq.each{ |die| 
    count = dice.select{ |x| x == die }.count
    case die
    when 1
      score += (count / 3) * 1000
      score += count % 3 * 100
    when 5
      score += (count / 3) * 500
      score += count % 3 * 50
    else
      score += (count / 3) * (die*100)
    end
  }
  score
end