[OLD QUIZ] Pascal's Triangle

ruby-talk
1

Hey all,
my second submission to RQ. My first one is GOING TO BE fizzbuzz. Now now, now's not the time to battle my inconsistencies...

Here's my code for Pascals Triangle. Can someone help me shorten it up?

# Pascals triangle

def factorial(n)
   sum = 1
   sum.upto(n) { |i| sum *= i }
   sum
end

def nCr(n, r)
   numerator = factorial(n)
   denominator = factorial(r) * factorial(n - r)

   @combinations = numerator / denominator
end

def pascal max_row

   0.upto(max_row) {|row_num|

     holder = []
     ticker = 0
     while ticker != row_num
       result = nCr(row_num, ticker)
       ticker = ticker + 1
       holder.push(result)
     end
     puts holder.join(' ').center(80)

   }

end

puts 'How many rows do you want?'
max_row = gets.chomp.to_i
pascal max_row

Thanks,
---------------------------------------------------------------|
~Ari
"I don't suffer from insanity. I enjoy every minute of it" --1337est man alive

2

Hey all,
my second submission to RQ. My first one is GOING TO BE fizzbuzz. Now now, now's not the time to battle my inconsistencies...

Here's my code for Pascals Triangle. Can someone help me shorten it up?

# Pascals triangle

def factorial(n)
  sum = 1
  sum.upto(n) { |i| sum *= i }
  sum
end

(1..n).inject(1) { |fact, i| fact * i }

def nCr(n, r)
  numerator = factorial(n)
  denominator = factorial(r) * factorial(n - r)

  @combinations = numerator / denominator
end

def pascal max_row

  0.upto(max_row) {|row_num|

    holder =
    ticker = 0
    while ticker != row_num
      result = nCr(row_num, ticker)
      ticker = ticker + 1
      holder.push(result)
    end

That inner loop could be:

(0...row_num).each do |ticker|
   # ...
end

    puts holder.join(' ').center(80)

  }

end

puts 'How many rows do you want?'
max_row = gets.chomp.to_i
pascal max_row

James Edward Gray II

···

On Jun 2, 2007, at 9:44 AM, Ari Brown wrote:

3

Ari Brown wrote:

Hey all,
my second submission to RQ. My first one is GOING TO BE fizzbuzz. Now
now, now's not the time to battle my inconsistencies...

Here's my code for Pascals Triangle. Can someone help me shorten it up?

Most of it has already been said, anyway, here's my take based on your
approach:

class Integer
  def factorial
    (1..self).inject(1) { |fac, i| fac*i } # as James already stated
  end

  def cr(r)
    factorial/(r.factorial*(self-r).factorial)
  end
end

# IMHO number of rows is more intuitive and leaves less questions open
(like: does it start with 0 or 1?)
def pascal(rows)
  rows.times do |row_num|
    holder =
    row_num.times do |ticker|
      holder << row_num.cr(ticker)
    end
    holder # the formatting stuff etc. IMHO does not belong into this
function
end

puts 'How many rows do you want?'
max_row = gets.chomp.to_i
puts pascal(max_row).join(' ').center(80)

Regards
Stefan

···

--
Posted via http://www.ruby-forum.com/\.

4

James Gray wrote:

(0...row_num).each do |ticker|
James Edward Gray II

For those I prefer:
row_num.times { |ticker| ...

Regards
Stefan

···

--
Posted via http://www.ruby-forum.com/\.