I'm working my way through the Ruby Koans in order to try and learn Ruby, and so far, so good. I've gotten to the greed koan, which at the time of this writing is 183. I've got a working solution, but I feel like I've cobbled together just a bunch of if/then logic and that I'm not embracing Ruby patterns.
In the following code, are there ways you would point me to more fully embracing Ruby patterns? (My code is wrapped in "MY CODE [BEGINS|ENDS] HERE" comments.
# Greed is a dice game where you roll up to five dice to accumulate
# points. The following "score" function will be used 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.
# MY CODE BEGINS HERE
def score(dice)
# set up basic vars to handle total points and count of each number
total = 0
count = [0, 0, 0, 0, 0, 0]
# for each die, make sure we've counted how many occurrencess there are
dice.each do |die|
count[ die - 1 ] += 1
end
# iterate over each, and handle points for singles and triples
count.each_with_index do |count, index|
if count == 3
total = doTriples( index + 1, total )
elsif count < 3
total = doSingles( index + 1, count, total )
elsif count > 3
total = doTriples( index + 1, total )
total = doSingles( index + 1, count % 3, total )
end
end
# return the new point total
total
end
def doTriples( number, total )
if number == 1
total += 1000
else
total += ( number ) * 100
end
total
end
def doSingles( number, count, total )
if number == 1
total += ( 100 * count )
elsif number == 5
total += ( 50 * count )
end
total
end
# MY CODE ENDS HERE
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
Thanks so much to any help you can give as I try to get my head around Ruby.
Wow! There are a lot of really cool approaches being done here. I like everybody's creativity. However, I have a pedagogical problem with all of the answers presented here. ("Pedagogy is the study of … the process of teaching." -- Wikipedia)
It is obvious from the first several koans (back in about_asserts.rb) that the Path to Enlightenment does not require any prior/outside knowledge of Ruby. It also seems fairly clear that the Path doesn't even require prior programming experience. So from an educational standpoint, this koan must be answerable using only the methods, language constructs, and programming techniques taught in earlier koans. That means:
no Enumerable#each_with_index
no Enumerable#count
no Enumerable#sort
no Hash.new(0) specifying a default value
no Numeric#abs
no Numeric#divmod
no recursion
no case when
etc
Now, I'm not saying that you are not allowed to use these things in your implementation, but the koan mustn't require using them. There must be a solution that only uses constructs introduced by prior koans.
Also, since the template was just
def score(dice)
# You need to write this method
end
it seemed implied that the solution should not define other methods or classes. That is, you should only replace the # You need to write this method line.
Here is a solution that fits my philosophical requirements:
def score (dice)
sum = 0
(1..6).each do |i|
idice = dice.select { |d| d == i }
count = idice.size
if count >= 3
sum += (i==1 ? 1000 : i*100)
end
sum += (count % 3) * 100 if i == 1
sum += (count % 3) * 50 if i == 5
end
sum
end
The methods/constructs here are introduced in the following koan files:
Enumerable#each about_iteration.rb
Enumerable#select about_iteration.rb
Array#size about_arrays.rb
a ? b : c about_control_statements.rb
% about_control_statements.rb
Related StackOverflow Questions:
Ruby Koans 182. Refactor help
https://codereview.stackexchange.com/questions/423/is-this-good-ruby-ruby-koans-greed-task
A student asked Joshu, "How can I write an algorithm to calculate the scores for a dice game?"
Joshu struck the student with his stick and said: "Use a calculator."
def score(dice)
score = [0, 100, 200, 1000, 1100, 1200][dice.count(1)]
score += [0, 50, 100, 500, 550, 600][dice.count(5)]
[2,3,4,6].each do |num|
if dice.count(num) >= 3 then score += num * 100 end
end
score
end
I went through and passed each of the tests one at a time. Not sure this is a very "ruby" solution, but I do like that it's obvious what each section is doing and that there are no excess declarations of values
def score(dice)
## score is set to 0 to start off so if no dice, no score
score = 0
## setting the 1000 1,1,1 rule
score += 1000 if (dice.count(1) / 3) == 1
## taking care of the single 5s and 1s here
score += (dice.count(5) % 3) * 50
score += (dice.count(1) % 3) * 100
## set the other triples here
[2, 3, 4, 5, 6].each do |num|
score += num * 100 if (dice.count(num) / 3 ) == 1
end
score
end
Looks OK. I might have written some things slightly differently, say:
def do_triples number, total
total + (number == 1 ? 1000 : number * 100)
end
If you want to do something that few languages other than Ruby can do, I suppose the following might be justifiable under DIE and DRY, on alternate Tuesdays, but I don't think those Ruby maxims were really intended to apply to common subexpression elimination. Anyway:
def do_triples number, total
total +
if number == 1
1000
else
number * 100
end
end
def do_triples number, total
if number == 1
1000
else
number * 100
end + total
end
Here's what I did. Looks pretty similar to a few of the older replies. I would love to find some ingenious usage of inject for this one (mikeonbike's one is niiiice).
def score(dice)
total = 0
# handle triples scores for all but '1'
(2..6).each do |num|
total += dice.count(num) / 3 * num * 100
end
# non-triple score for '5'
total += dice.count(5) % 3 * 50
# all scores for '1'
total += dice.count(1) / 3 * 1000 + dice.count(1) % 3 * 100
total
end
You can condense this down to fewer lines but the readability of the algorithm gets lost so I ended up with this:
def score(dice)
result = 0;
(1..6).each do |die|
multiplier = die == 1 ? 1000 : 100
number_of_triples = dice.count(die) / 3
result += die * multiplier * number_of_triples
end
result += 100 * (dice.count(1) % 3)
result += 50 * (dice.count(5) % 3)
end
And if you're using 1.8.6, you'll have to use backports or add the count method to Array yourself:
class Array
def count(item)
self.select { |x| x == item }.size
end
end
Here is the answer I went with after about four iterations and trying to take advantage of the Ruby constructs I'm learning doing the koans:
def score(dice)
total = 0
(1..6).each { |roll| total += apply_bonus(dice, roll)}
return total
end
def apply_bonus(dice, roll, bonus_count = 3)
bonus = 0
bonus = ((roll == 1 ? 1000 : 100) * roll) if (dice.count(roll) >= bonus_count)
bonus += 50 * (dice.count(5) % bonus_count) if (roll == 5)
bonus += 100 * (dice.count(1) % bonus_count) if (roll == 1)
return bonus
end
Yet another answer :)
def score(dice)
score = 0
for num in 1..6
occurrences = dice.count {|dice_num| dice_num == num}
score += 1000 if num == 1 and occurrences >= 3
score += 100 * (occurrences % 3) if num == 1
score += 100 * num if num != 1 and occurrences >= 3
score += 50 * (occurrences % 3) if num == 5
end
score
end
This is the simplest and most readable solution that I came up with. This also accounts for a few situations not in the tests, such as a roll of six 5's or six 1's.
def score(dice)
score = 0
(1..6).each { |d|
count = dice.find_all { |a| a == d }
score = ( d == 1 ? 1000 : 100 ) * d if count.size >= 3
score += (count.size - 3) * 50 if (count.size >= 4) && d == 5
score += (count.size - 3) * 100 if (count.size >= 4) && d == 1
score += count.size * 50 if (count.size < 3) && d == 5
score += count.size * 100 if (count.size < 3) && d == 1
}
score
end
I opted to use the size method instead of the count method as count isn't supported by all versions of Ruby and the koans had not tested count up to this test.
def score(dice)
total = 0
sets = dice.group_by{|num| num }
sets.each_pair do |num, values|
number_of_sets, number_of_singles = values.length.divmod(3)
number_of_sets.times { total += score_set(num) }
number_of_singles.times { total += score_single(num) }
end
total
end
def score_set(num)
return 1000 if num == 1
num * 100
end
def score_single(num)
return 100 if num == 1
return 50 if num == 5
0
end
This was my eventual solution after initially having a similar if/then/else mess on my first attempt.
def score(dice)
score = 0
dice.uniq.each do |roll|
score += dice.count(roll) / 3 * (roll == 1 ? 1000 : 100*roll)
score += dice.count(roll) % 3 * (roll == 1 ? 100 : (roll == 5 ? 50 : 0))
end
score
end
I'd say you have it looking very Ruby-like already. The only thing that doesn't look very Rubyish to me would be the use of camelCase method names instead of snake_case, but of course that's a personal convention and I haven't read the koans myself.
Other than that, your example wouldn't be improved much by using case/when or any other solution for that matter. Aim for anything less than 3 elseif operations, anything more than that and you'd probably want to hunt for a better solution.
You could shorten [0, 0, 0, 0, 0, 0] to [0] * 6 but aside from the camelCase #injekt mentioned it looks fine to me. I'd be quite happy to see this in a code review.
Also I suppose your doTriples and doSingles don't really need their temporary variables.
def doTriples( number, total )
if number == 1
total + 1000
else
total + ( number ) * 100 # be careful with precedence here
end
end
You may want to change
# for each die, make sure we've counted how many occurrencess there are
dice.each do |die|
count[ die - 1 ] += 1
end
into a hash, such as
count = Hash.new(0)
dice.each do |die|
count[die] += 1
end
or even
count = {} # Or Hash.new(0)
grouped_by_dots = dice.group_by {|die| die}
1.upto(6) do |dots| # Or grouped_by_dots.each do |dots, dice_with_those_dots|
dice_with_those_dots = grouped_by_dots.fetch(dots) {[]}
count_of_that_dots = dice_with_those_dots.length
count[dots] = count_of_that_dots
end
That way you don't have to have index + 1 littered throughout your code.
It'd be nice if Ruby had a count_by method built in.
My 2 cents. Having new methods for singles/doubles seems like a roundabout way of doing something very simple.
def score(dice)
#fill initial throws
thrown = Hash.new(0)
dice.each do |die|
thrown[die]+=1
end
#calculate score
score = 0
faces.each do |face,amount|
if amount >= 3
amount -= 3
score += (face == 1 ? 1000 : face * 100)
end
score += (100 * amount) if (face == 1)
score += (50 * amount) if (face == 5)
end
score
end
Well,
Here's my solution:
def score(dice)
total = 0
#Iterate through 1-6, and add triples to total if found
(1..6).each { |roll| total += (roll == 1 ? 1000 : 100 * roll) if dice.count(roll) > 2 }
#Handle Excess 1's and 5's
total += (dice.count(1) % 3) * 100
total += (dice.count(5) % 3) * 50
end
Once I found the "count" method for an array, this exercise was pretty easy.
Here is my answer. I do not know if it is good or not, but at least, it looks clear :)
RULEHASH = {
1 => [1000, 100],
2 => [100,0],
3 => [100,0],
4 => [100,0],
5 => [100,50],
6 => [100,0]
}
def score(dice)
score = 0
RULEHASH.each_pair do |i, rule|
mod = dice.count(i).divmod(3)
score += mod[0] * rule[0] * i + mod[1] * rule[1]
end
score
end
My solution is not ruby-like style. Just for fun and shortest code. We can set rules through hash p.
def score(dice)
p = Hash.new([100,0]).merge({1 => [1000,100], 5 => [100,50]})
dice.uniq.inject(0) { |sum, n| sum + dice.count(n) / 3 * n * p[n][0] + dice.count(n) % 3 * p[n][1] }
end
My answer uses a "lookup table" approach...
def score(dice)
tally = (1..6).inject(Array.new(7,0)){|a,i| a[i] = dice.count(i); a}
rubric = {1 => [0,100,200,1000,1100,1200], 5 => [0,50,100,500,550,600]}
score = rubric[1][tally[1]] + rubric[5][tally[5]]
[2,3,4,6].each do |i| score += 100 * i if dice.count(i) >= 3 end
score
end
Mine was similar to a couple of others posted here.
score = 0
[1,2,3,4,5,6].each {|d|
rolls = dice.count(d)
score = (d==1 ? 1000 : 100)*d if rolls >= 3
score += 100*(rolls % 3) if d == 1
score += 50*(rolls % 3) if d == 5
}
score
Me and my girlfriend were going through these rubykoans this weekend and I had quite a bit of fun golfing on this and trying many different solutions. Here is a reasonably short data-driven solution:
SCORES = [[1000, 100], [200, 0], [300, 0], [400, 0], [500, 50], [600, 0]]
def score(dice)
counts = dice.group_by(&:to_i).map { |i, j| [i-1, j.length] }
counts.inject(0) do |score, (i, count)|
sets, singles = count.divmod 3
score + sets * SCORES[i][0] + singles * SCORES[i][1]
end
end
Here is my obligatory one-liner (and perhaps FP version):
SCORES = [[1000, 100], [200, 0], [300, 0], [400, 0], [500, 50], [600, 0]]
def score(dice)
dice.group_by(&:to_i).inject(0) {|s,(i,j)| s + j.size / 3 * SCORES[i-1][0] + j.size % 3 * SCORES[i-1][1]}
end
I also went some weird routes as well:
SCORES = [[1000, 100], [200, 0], [300, 0], [400, 0], [500, 50], [600, 0]]
def score(dice)
dice.group_by(&:to_i).inject(0) do |s, (i,j)|
s + j.size.divmod(3).zip(SCORES[i-1]).map {|a,b| a*b }.reduce(:+)
end
end
All programmers should be screwing around with little problems like this...It is like performing morning stretches :)
def score(dice)
result = 0
result += 1000 * (dice.find_all{|e| e == 1}).length.divmod(3)[0]
result += 100 * (dice.find_all{|e| e == 1}).length.divmod(3)[1]
result += 50 * (dice.find_all{|e| e == 5}).length.divmod(3)[1]
(2..6).each {|value| result += value*100 * (dice.find_all{|e| e == value}).length.divmod(3)[0]}
return result
end
And what about this solution?
Thanks for the feedback!
def score(dice)
count = Hash.new(0)
dice.each do |die|
count[die] += 1
end
total = 0
count.each_pair { |die, set| total += set < 3 ? single_value(die,set) : triple_value(die,set)}
total
end
def single_value(die,set)
value = 0
value += (set * 100) if die == 1
value += (set * 50) if die == 5
value
end
def triple_value(die,set)
value = 0
diff = set - 3
value += single_value(die,diff)
value += die == 1 ? 1000 : die * 100
value
end
I used a slightly different method to others here, and (naturally) it's one that I see as preferable. It's very DRY and uses ruby methods fairly extensively to avoid manual loops and branches as much as possible. Should be relatively obvious, but essentially what is happening is we loop through each unique dice roll, and use iterative erosion of the number of occurences of that roll to add the appropriate points to an aggregate total score.
def score(dice)
score = 0 # An initial score of 0.
throw_scores = { 1 => 10, 2 => 2, 3 => 3, 4 => 4, 5 => 5, 6 => 6 }
# A hash to store the scores for each dice throw
dice.uniq.each { |throw| # for each unique dice value present in the "hand"
throw_count = (dice.select { |item| item == throw }).count
# use select to store the number of times this throw occurs
while throw_count > 0
# iteratively erode the throw count, accumulating
# points as appropriate along the way.
if throw_count >= 3
score += throw_scores[throw] * 100
throw_count -= 3
elsif throw == 1 || throw == 5
score += throw_scores[throw] * 10
throw_count -= 1
else
throw_count -= 1
end
end
}
return score
end
And another one, just for the fun of it:
def score(dice)
result = 0
dice.uniq.each { |k|
result += ((dice.count(k) / 3) * 1000 + (dice.count(k) % 3) * 100) if k == 1
result += ((dice.count(k) / 3) * 100 * k + (dice.count(k) % 3) * ( k == 5 ? 50 : 0 )) if k != 1
}
result
end
Here's my opinion. All other solutions here try to be clever. There's a place for learning clever tricks, but it's even more important to learn to write clear and maintainable code. The main problem I see with all of these solutions is that it's very difficult to discern the scoring rules from the code. Can you read your solution and make sure that it's correct in your head? Then imagine someone asks you to add a new scoring rule, or remove one. Can you quickly point to the place where the rule must be added or removed?
Here's my solution. I'm sure it can be improved, but look at the shape of the "score" function. This is the sort of code that I would not mind to maintain.
class Array
def occurrences_of(match)
self.select{ |number| match == number }.size
end
def delete_one(match)
for i in (0..size)
if match == self[i]
self.delete_at(i)
return
end
end
end
end
def single_die_rule(match, score, dice)
dice.occurrences_of(match) * score
end
def triple_rule(match, score, dice)
return 0 if dice.occurrences_of(match) < 3
3.times { dice.delete_one match }
score
end
def score(dice)
triple_rule(1, 1000, dice) +
triple_rule(2, 200, dice) +
triple_rule(3, 300, dice) +
triple_rule(4, 400, dice) +
triple_rule(5, 500, dice) +
triple_rule(6, 600, dice) +
single_die_rule(1, 100, dice) +
single_die_rule(5, 50, dice)
end
I'm gonna have to go with:
def score(dice)
# some checks
raise ArgumentError, "input not array" unless dice.is_a?(Array)
raise ArgumentError, "invalid array size" unless dice.size <= 5
raise ArgumentError, "invalid dice result" if dice.any? { |x| x<1 || x>6 }
# setup (output var, throws as hash)
out = 0
freqs = dice.inject(Hash.new(0)) { |m,x| m[x] += 1; m }
# 3-sets
1.upto(6) { |i| out += freqs[i]/3 * (i == 1 ? 10 : i) * 100 }
# one not part of 3-set
out += (freqs[1] % 3) * 100
# five not part of 3-set
out += (freqs[5] % 3) * 50
out
end
Because most of the solutions presented so far lack basic checks. And some of them are fairly unreadable (in my book) and not very idiomatic.
Granted, the 3-set condition could be made more readable by splitting into two clauses:
# 3-sets of ones
out += freqs[1]/3 * 1_000
# 3-sets of others
2.upto(6) { |i| out += freqs[i]/3 * i * 100 }
but that's IMO mostly about personal preference.
Coming from Perl, my instinct is to use a hash:
def score(dice)
# You need to write this method
score = 0
count = Hash.new(0)
for die in dice
count[die] += 1
is_triple = (count[die] % 3 == 0)
if die == 1 then
score += is_triple ? 800 : 100
elsif die == 5 then
score += is_triple ? 400 : 50
elsif is_triple
score += 100 * die
end
end
return score
end
This has the advantage that it makes a single pass over dice. I could probably have used an Array in place of the Hash.
I grouped the dice by face, then looped over these groups, first scoring the threes, then individual dice. This is how I would score the game were I playing IRL
def score(dice)
points = 0
dice.group_by {|face| face}.each do |face,group|
while group.size >= 3
if face == 1
# A set of three ones is 1000 points
points += 1000
else
# A set of three numbers (other than ones) is worth 100 times the number.
points += 100 * face
end
group.pop(3)
end
group.each do |x|
# A one (that is not part of a set of three) is worth 100 points.
points += 100 if x==1
# A five (that is not part of a set of three) is worth 50 points.
points += 50 if x==5
end
end
return points
end
That's how I roll
Late to the party, but wanted to take a shot at answering only using knowledge introduced thus far in the Koans. Specifically, I don't use Enumerable#count like most others have.
This seems very straightforward to me, but if anyone happens along, I'd be happy to hear about an optimizations you may have.
And what can I say? I like taking advantage of array indexing.
def score(dice)
return 0 if dice.empty? # Immediately recognize an empty roll
# Create an array to hold the scores for each die face
totals = []
7.times { totals << 0 }
# Handle each roll and calculate new score
dice.each do |roll|
if roll == 5
# If we have seen two 5s thus far, make the score 500 for 5s, otherwise add 50
totals[roll] == 100 ? totals[roll] = 500 : totals[roll] += 50
elsif roll == 1
# If we have seen two 1s thus far, make the score 1000 for 5s, otherwise add 100
totals[roll] == 200 ? totals[roll] = 1000 : totals[roll] += 100
else
# If we see any other number three times, score is the number times 100
totals[roll] == 2 ? totals[roll] = roll * 100 : totals[roll] += 1
end
end
# Count up the scores for each die face; if score is less than 50, then it's just zero
return totals.inject(0) { |sum, points| points >= 50 ? sum += points : sum }
end
Related
I'm learning ruby and practicing with codewars, and I've come to a challenge that I feel I mainly understand (rudimentarily) but I'm unable to figure out how to continue looping over the method until I reach the result I'm looking for.
The challenge is asking to reduce a number, by multiplying its digits, until the multiplication results in a single digit. In the end it wants you to return the number of times you had to multiply the number until you arrived at a single digit. Example -> given -> 39; 3 * 9 = 27, 2 * 7 = 14, 1 * 4 = 4; answer -> 3
Here's my code :
def persistence(n)
if n < 10
return 0
end
arr = n.to_s.split("")
sum = 1
count = 0
arr.each do |num|
sum *= num.to_i
if num == arr[-1]
count += 1
end
end
if sum < 10
return count
else
persistence(sum)
end
end
Thanks for your help!
Your function is looking great with recursion but you are reseting the count variable to 0 each time the loop runs, I think if you use an auxiliar method it should run ok:
this is in base of your code with minor improvements:
def persistence(n)
return 0 if n < 10
count = 0
multiply_values(n, count)
end
def multiply_values(n, count)
arr = n.to_s.chars
sum = 1
arr.each do |num|
sum *= num.to_i
if num == arr[-1]
count += 1
end
end
if sum < 10
return count
else
multiply_values(sum, count)
end
end
a shorter solution could be to do:
def persistence(n)
return 0 if n < 10
multiply_values(n, 1)
end
def multiply_values(n, count)
sum = n.to_s.chars.map(&:to_i).reduce(&:*)
return count if sum < 10
multiply_values(sum, count + 1)
end
and without recursion:
def persistence(n)
return 0 if n < 10
count = 0
while n > 10
n = n.to_s.chars.map(&:to_i).reduce(&:*)
count += 1
end
count
end
Let's look at a nicer way to do this once:
num = 1234
product = num.to_s.split("").map(&:to_i).reduce(&:*)
Breaking it down:
num.to_s.split("")
As you know, this gets us ["1", "2", "3", "4"]. We can easily get back to [1, 2, 3, 4] by mapping the #to_i method to each string in that array.
num.to_s.split("").map(&:to_i)
We then need to multiply them together. #reduce is a handy method. We can pass it a block:
num.to_s.split("").map(&:to_i).reduce { |a, b| a * b }
Or take a shortcut:
num.to_s.split("").map(&:to_i).reduce(&:*)
As for looping, you could employ recursion, and create product_of_digits as a new method for Integer.
class Integer
def product_of_digits
if self < 10
self
else
self.to_s.split("").map(&:to_i).reduce(&:*).product_of_digits
end
end
end
We can now simply call this method on any integer.
1344.product_of_digits # => 6
I am currently working through the Ruby Koans and I am stuck on the scoring project. First of all, I am having a hard time evaluating the instructions and laying it out in terms of what I am suppose to do. Secondly, I am not sure if I am on the right track with the method score I wrote below. My questions is - Is there a way to understand these instructions better? Also, with the score method I wrote, I am still not passing the first test. I think I must understand what I need to do first but I can't figure it out. Any help and simple explanations or direction is appreciated.
Thank you.
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)
(1..6).each do |num|
amount = dice.count(num)
if amount >= 3
100 * num
elsif num == 1
100 * amount
elsif num == 5
50 * amount
else
0
end
end
end
# test code for method
class AboutScoringProject < Neo::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])
assert_equal 1100, score([1,1,1,1])
assert_equal 1200, score([1,1,1,1,1])
assert_equal 1150, score([1,1,1,5,1])
end
end
This is what I have done:
def score(dice)
score = 0
return score if dice == nil || dice == []
quantity = dice.inject(Hash.new(0)) {|result,element| result[element] +=1; result}
score += quantity[1] >= 3 ? 1000 + ((quantity[1] - 3) * 100) : quantity[1] * 100
score += quantity[5] >= 3 ? 500 + ((quantity[5] - 3) * 50) : quantity[5] * 50
[2,3,4,6].each {|x| score += x * 100 if quantity[x] >= 3 }
score
end
You can use hash for score instead of writing switch-case
def score(dice)
score_map = {
1 => 100,
5 => 50
}
cluster = dice.inject(Hash.new(0)) {|hash, number| hash[number] += 1; hash}
cluster.inject(0) do |sum, (num, count)|
set_count = count / 3
sum += num == 1 ? 1000 * set_count : num * 100 * set_count
sum + (score_map[num] || 0) * (count % 3) # use 0 if score of num dosn't exist
end
end
I've built a Yahtzee game for Ruby Quiz #19. I have the game up and running, however, there are 2 bugs.
When a player chooses to use a roll of >=3 of a kind as a "small straight" (sequence of 4 dice) to "scratch" (score it as zero) that section an error occurs. Here is the code for the small straight:
def sm_straight
#roll = #roll.sort.uniq
if (0..1).any? {|x| (#roll[x+3] - #roll[x+2] == 1) && (#roll[x+2] - #roll[x+1] == 1) && (#roll[x+1] - #roll[x] == 1)}
#scorecard["sm. straight"] = 30
else
puts "Your roll is not a sm. straight! Please select another section or type scratch to score 0 for this section."
scratch = gets.chomp
if scratch == "scratch"
#scorecard["sm. straight"] = "scratch"
elsif #scorecard.has_key?(scratch)
#turn -= 1
section_to_score(scratch)
else
sm_straight
end
end
end
This is the error:
NoMethodError: undefined method -' for nil:NilClass
from Yahtzee_test.rb:209:inblock in sm_straight'
Line 209 is the "if statement" line
When a player incorrectly enters which dice to keep. I am trying to figure out a better way to ask the player how to enter the dice to keep or catch the error and have them re-enter the numbers with the current system. Here is the code"
def roll_again
puts "Which dice would you like to keep from this roll? (1, 2, 3, 4, 5)"
dice_to_keep = gets.chomp.split(',').map {|x| (x.to_i) - 1}.map {|x| #roll[x]}
new_roll = Array.new(5 - dice_to_keep.size) {rand(6) + 1}
#roll = new_roll + dice_to_keep
p #roll
#roll_count += 1
puts "That was roll number #{#roll_count}, you have #{3-#roll_count} remaining."
if #roll_count < 3
more_rolls?
else
section(#roll)
end
end
Any advice on how to write this code better and make it bug free would be greatly appreciated!
To check for a straight of at least 4 out of 5 dice, you could replace:
#roll = #roll.sort.uniq
if (0..1).any? {|x| (#roll[x+3] - #roll[x+2] == 1) && (#roll[x+2] - #roll[x+1] == 1) && (#roll[x+1] - #roll[x] == 1)}
with this:
if has_straight(roll, 4)
And define has_straight:
def has_straight( roll, need )
num = 1
roll = roll.sort.uniq
roll.each_with_index do |e, i|
if i < roll.length-1 then
if (roll[i+1] - roll[i]) > 1 then
break if num >= need
num = 1
end
num += 1
end
end
num >= need
end
There may be a slightly more clever Ruby-ism that will do this, but it fixes your array out-of-bounds issue.
Yesterday I completed an exercise (9.5) from the book "Learn to Program" from Chris Pine.
It's an integer to Old-school Roman numerals converter.
This is how I did it:
def old_roman_numeral number
roman_number = ""
while number != 0
if number % 1000 == 0
number -= 1000
roman_number += "M"
next
elsif number % 500 == 0
number -= 500
roman_number += "D"
next
elsif number % 100 == 0
number -= 100
roman_number += "C"
next
elsif number % 50 == 0
number -= 50
roman_number += "L"
next
elsif number % 10 == 0
number -= 10
roman_number += "X"
next
elsif number % 5 == 0
number -= 5
roman_number += "V"
next
else
number -= 1
roman_number += "I"
end
end
roman_number.reverse
end
puts "Please enter any number and I convert it to"
puts "Old-school Roman numerals."
puts
num = gets.chomp.to_i
puts "Your number #{num} converted to Old-school Roman is:"
puts (old_roman_numeral num)
When I run the script, it outputs the correct Roman numerals.
For example 1200 => MCC
But, when I woke up today the first thing I thought was, this can't be right!
The remainder of 1200 % 1000 is 200 and not 0!
But why is the output MCC and not CCCCCCCCCCCC???
If you trace through the program, it's actually matching the % 100 == 0 twice first, getting CC, and leaving you with 1000. Then it matches % 1000 == 0, leaving CCM. Finally it reverses the string, leaving you with MCC.
Side comment: Interesting approach to the problem as I probably would have used a bunch of >= comparisons building the string in forward order with special cases for the 'subtraction' parts (IV or IX). Although on second read, this solution appears to output IIII and not IV, so the special cases are moot.
the code calculates the digits in reverse order. i.e.
you get a C first, then another C and in the third iteration of the loop, you get the M.
in the end, this line of code:
roman_number.reverse
reverses CCM to MCC and thus you get the result you actually get.
to better understand what happens, you can change your code as follows:
if number % 1000 == 0
number -= 1000
roman_number += "M"
next
becomes:
if number % 1000 == 0
number -= 1000
roman_number += "M"
puts "number " + number.to_s
puts "roman_number " + roman_number
next
do this for every if-block. that way you will see what happens in each step.
Not an answer, but just for note, here is a method with similar purpose, taken from my personal library:
class Numeric
RomanNumerals = {
1000 => "m", 900 => "cm", 500 => "d", 400 => "cd",
100 => "c", 90 => "xc", 50 => "l", 40 => "xl",
10 => "x", 9 => "ix", 5 => "v", 4 => "iv", 1 => "i"
}
def roman
return to_s unless 0 < self and self < 4000 # Cannot be romanized
s, r = "", self
RomanNumerals.each{|d, c| q, r = r.divmod(d); s.concat(c*q)}
s
end
end
I'm working through the Ruby Koans, and I'm having a bit of trouble figuring out what is going wrong with a method I've written. I'm in about_scoring_project.rb, and I've written the score method for the dice game:
def score(dice)
return 0 if dice == []
sum = 0
rolls = dice.inject(Hash.new(0)) { |result, element| result[element] += 1; result; }
rolls.each { |key, value|
# special condition for rolls of 1
if key == 1
sum += 1000 | value -= 3 if value >= 3
sum += 100*value
next
end
sum += 100*key | value -= 3 if value >= 3
sum += 50*value if key == 5 && value > 0
}
return sum
end
For those unfamiliar with the exercise:
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.
I run into trouble when I try to run the last test in the file: assert_equal 550, score([5,5,5,5])
For some reason I am returning 551 instead of 550. Thanks for your help!
Here is my approach:
def score(dice)
# Count how many what
clusters = dice.reduce(Hash.new(0)) {|hash, num| hash[num] += 1; hash }
# Since 1's are special, handle them first
ones = clusters.delete(1) || 0
score = ones % 3 * 100 + ones / 3 * 1000
# Then singular 5's
score += clusters[5] % 3 * 50
# Then the triples other than triple-one
clusters.reduce(score) {|s, (num, count)| s + count / 3 * num * 100 }
end
My approach uses two lookup tables - one containing the scores for triples, the other for singles. I work out the score for each number using the tables, and accumulate the total using inject:
def score(dice)
triple_scores = [1000, 200, 300, 400, 500, 600]
single_scores = [100, 0, 0, 0, 50, 0]
(1..6).inject(0) do |score, number|
count = dice.count(number)
score += triple_scores[number - 1] * (count / 3)
score += single_scores[number - 1] * (count % 3)
end
end
I went with
def score(dice)
dice.uniq.map do |die|
count = dice.count die
if count > 2
count -= 3
die == 1 ? 1000 : 100 * die
else 0
end + case die
when 1 then count * 100
when 5 then count * 50
else 0
end
end.inject(:+) || 0
end
This is because you're really adding the result of a | operator (Bitwise OR) to the total score:
sum += 100*key | value -= 3 if value >= 3 # This is 501 in your case
Proof:
irb(main):004:0> value = 4
=> 4
irb(main):005:0> 100 * 5 | value -= 3 # This should be read as (500) | 1 which is 501
=> 501
So rewrite it like this:
if value >= 3
sum += 100 * key
value -= 3
end
My approach was:
def score(dice)
calculator = ->(no, group_multipler, individual_multipler) { (no / 3 * group_multipler) + (no % 3 * individual_multipler) }
dice.group_by {|i| i % 7 }.inject(0) do |total, (value, scores)|
group_multipler, individual_multipler = case value
when 1
[1000, 100]
when 5
[500, 50]
else
[value * 100, 0]
end
total += calculator.call(scores.size, group_multipler, individual_multipler)
end
end
My approach:
def score(dice)
score = 0
score += dice.count(1) >= 3? (1000+ (dice.count(1) -3)*100): dice.count(1) * 100
score += dice.count(5) >= 3 ? (500 + (dice.count(5) -3)*50): dice.count(5) * 50
[2,3,4,6].each {|x| dice.count(x) >=3? score+= x*100:0}
return score
end
Here's my answer:
def score(dice)
frequency = dice.inject(Hash.new(0)) do |h, el|
h[el] += 1
h
end
score_triples = { 1 => 1000 }
score_singles = { 1 => 100, 5 => 50 }
score = 0
frequency.each do |k, v|
score += v / 3 * score_triples.fetch(k, 100 * k)
score += v % 3 * score_singles.fetch(k, 0)
end
score
end
My approach used integer division and modulus division:
def score(dice)
points = 1000 * (dice.count(1) / 3)
points += 100 * (dice.count(1) % 3)
points += 50 * (dice.count(5) % 3)
(2..6).each do |i|
points += (100 * i) * (dice.count(i) / 3)
end
points
end
This was the first piece of code I ever wrote by myself (With a ton of help of stackoverflow, of course.) After watching all other answers I realize it is way overkill specially because it works for a 9 numbers dice (does that exist?)
def score(dice)
if dice.empty?
return 0
end
var_score = 0
conteo = (0..9).to_a.each.map { |x| dice.count(x)}
#Evaluating 1
if ( conteo[1] / 3 ) >= 0
multiplier1 = conteo[1]/3
var_score += multiplier1 * 1000
end
if ( conteo[1] % 3 ) != 0
var_score += (conteo[1] % 3)*100
end
#Evaluating 5
if ( conteo[5] % 3 ) != 0
var_score += (conteo[5] % 3)* 50
end
#Evaluating numbers x 3
if (conteo[2..9].count { |x| x >= 3 }) > 0
triplets = conteo[2..9].map {|x| x / 3}
array_multiplicator = triplets.each_with_index.select {|num,index| (num > 0)}.map {|x| x[0]}
product_triplets = triplets.each_with_index.select {|num,index| (num > 0)}.map {|x| x[1]}.map {|x| (x+2)*100}
var_score += array_multiplicator.zip(product_triplets).map{|x| x.inject(&:*)}.sum
end
var_score
end
It took 29 lines, but this is my first Ruby
def score(dice)
return 0 if dice == []
sums = Array.new # To hold number of occurrences 1 - 6
for i in 0..6 # Initialize to 0... note [0] is not used
sums[i] = 0
end
total = 0 # To hold total
dice.each do |dots| # Number of dots showing on dice
sums[dots] += 1 # Increment the array members 1 - 6
end
if sums[1] > 2 then # If 3 1's
total += 1000
sums[1] -= 3 # Remove the 3 you took, in case there's more
end
if sums[2] > 2 then total += 200 # If 3 2's
end
if sums[3] > 2 then total += 300 #If 3 3's
end
if sums[4] > 2 then total += 400 #If 3 4's
end
if sums[5] > 2 then total += 500 #If 3 5's
sums[5] -= 3 #Remove the 5's you took
end
if sums[6] > 2 then total += 600 #If 3 6's
end
total += (sums[1] * 100) # If any ones are left
total += (sums[5] * 50) # Same for fives
return total
end
This is my solutions.
def score(dice)
score = 0
# grab all the numbers and their amounts
number_amounts = dice.reduce(Hash.new(0)) { |hash, numb| hash[numb] += 1; hash }
# iterate through each pair
number_amounts.each do |key, value|
# case with number 1
score += (value % 3) * 100 + value / 3 * 1000 if (key == 1)
# case with number 5
score += (value % 3) * 50 + value / 3 * key * 100 if (key == 5)
# all numbers except 1 and 5
score += (value / 3) * key * 100 if (key != 1 && key != 5)
end
score
end
def score(dice)
# Set up rules Hash
rules = { 1 => {:triples => 1000, :singles => 100}, 5 => {:triples => 100, :singles => 50} }
[2,3,4,6].each {|i| rules[i] = {:triples => 100, :singles => 0} }
# Count all ocourencies
counts = dice.each_with_object(Hash.new(0)) {|e, h| h[e] += 1}
#calculate total
total = 0
counts.each_pair{ | key, value |
total += value >= 3? (rules[key][:triples]*key + (value -3)*rules[key][:singles]): value * rules[key][:singles]
}
return total
end
I used the new ruby enumerable method tally
def score(dice)
return 0 if dice.empty?
ans = dice.tally.map do |k,v|
case k
when 1
three = (k * 1000) * (v/3)
val = (v%3) * 100
val + three
when 5
three = (k * 100) * (v/3)
val = (v%3) * 50
val + three
else
(k * 100) * (v/3)
end
end
ans.reduce(0, :+)
end
My attempt, feedback and refactoring suggestions most welcome:
def score(dice)
score = 0
score_array = [[0, 100, 200, 1000, 1100, 1200], [0, 0, 0, 200, 200, 200], [0, 0, 0, 300, 300, 300], [0, 0, 0, 400, 400, 400], [0, 50, 100, 500, 550, 600], [0, 0, 0, 600, 600, 600]]
tally_hash = {1=>0, 2=>0, 3=>0, 4=>0, 5=>0, 6=>0}
dice.sort.tally.each do |key, value|
tally_hash[key] += value
end
tally_hash.each do |key, value|
score += score_array[key -1][value]
end
return score
end
I used hash for score
def score(dice)
score_map = {
1 => 100,
5 => 50
}
cluster = dice.inject(Hash.new(0)) {|hash, num| hash[num] += 1; hash}
cluster.inject(0) do |sum, (num, count)|
set_count = count / 3
sum += num == 1 ? 1000 * set_count : num * 100 * set_count
sum + (score_map[num] || 0) * (count % 3)
end
end
def score(dice)
score = 0
dice.uniq.each do |number|
count = dice.count number
weight = if number == 1 then 10 else number end
if count >= 3
score += weight * 100
count -= 3
end
if count > 0 and number == 1 or number == 5
score += count * weight * 10
end
end
score
end
def score(dice)
ones = fives = rest = 0
one_count = dice.count(1)
if one_count > 2
ones = 1000
one_count -= 3
end
ones += one_count * 100
five_count = dice.count(5)
if five_count > 2
fives = 500
five_count -= 3
end
fives += five_count * 50
[2,3,4,6].each do |num|
if dice.count(num) > 2
rest += num * 100
end
end
return ones + fives + rest
end