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July 18, 2011 05:29
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About Scoring from Ruby Koans
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require File.expand_path(File.dirname(__FILE__) + '/edgecase') | |
# Greed is a dice game where you roll up to five dice to accumulate | |
# points. The following "score" function will be used to calculate the | |
# score of a single roll of the dice. | |
# | |
# 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 fives is 500 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: | |
# | |
# Your goal is to write the score method. | |
def score(dice) | |
score = 0 | |
return score if dice.empty? | |
counts = [0,0,0,0,0,0] | |
dice.each {|d| counts[d-1] += 1 } | |
ones, twos, threes, fours, fives, sixes = counts | |
if ones >= 3 then | |
score += 1000 + ((ones-3) * 100) | |
else | |
score += (ones * 100) | |
end | |
if fives >= 3 then | |
score += 500 + ((fives-3) * 50) | |
else | |
score += fives * 50 | |
end | |
score += 200 if twos >= 3 | |
score += 300 if threes >= 3 | |
score += 400 if fours >= 3 | |
score += 600 if sixes >= 3 | |
score | |
end | |
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_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 | |
end |
Interesting approach ...
Most of the solutions I have seen to this, including yours, uses multiple enumerations - I think I count 4 in yours - in mine I made a conscious choice to eliminate looping over anything more than the once necessary to count the elements
Not sure if my approach is less "Rubyish" because of that
I also made a conscious effort to keep my algorithm close to the definition of the problem - so for example, I explicit use "1000" as the score for 3 ones, and add the extras on above that (where as your solution, and many others I have seen, pre-calculates the sum for 3, minus the figure for each.
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here was my take on this.
But I didnt refactor it after the comment left by Corey. I will soon have time again to dig more into ruby :)
https://gist.github.com/906429