Re: [QUIZ] pp Pascal (#84)

Hi everybody,

here's my solution, next to few examples.
The result might look pretty poor displayed on an email client :-(, but the 
script does print a perfectly equilateral triangle.

Thanks for the interesting quiz!


$ ./pascal.rb 12
                          1
                        1   1
                      1   2   1
                    1   3   3   1
                  1   4   6   4   1
                1   5  10  10   5   1
              1   6  15  20  15   6   1
            1   7  21  35  35  21   7   1
          1   8  28  56  70  56  28   8   1
        1   9  36  84 126 126  84  36   9   1
      1  10  45 120 210 252 210 120  45  10   1
    1  11  55 165 330 462 462 330 165  55  11   1
  1  12  66 220 495 792 924 792 495 220  66  12   1


$ ./pascal.rb 12 -1
  1
  1   1
  1   2   1
  1   3   3   1
  1   4   6   4   1
  1   5  10  10   5   1
  1   6  15  20  15   6   1
  1   7  21  35  35  21   7   1
  1   8  28  56  70  56  28   8   1
  1   9  36  84 126 126  84  36   9   1
  1  10  45 120 210 252 210 120  45  10   1
  1  11  55 165 330 462 462 330 165  55  11   1

$ ./pascal.rb 12 1
                                              1
                                          1   1
                                      1   2   1
                                  1   3   3   1
                              1   4   6   4   1
                          1   5  10  10   5   1
                      1   6  15  20  15   6   1
                  1   7  21  35  35  21   7   1
              1   8  28  56  70  56  28   8   1
          1   9  36  84 126 126  84  36   9   1
      1  10  45 120 210 252 210 120  45  10   1
  1  11  55 165 330 462 462 330 165  55  11   1

$ ./pascal.rb 12 0.7

                                   1
                                1   1
                             1   2   1
                          1   3   3   1
                       1   4   6   4   1
                    1   5  10  10   5   1
                 1   6  15  20  15   6   1
              1   7  21  35  35  21   7   1
           1   8  28  56  70  56  28   8   1
        1   9  36  84 126 126  84  36   9   1
     1  10  45 120 210 252 210 120  45  10   1
  1  11  55 165 330 462 462 330 165  55  11   1


By tweaking a bit the script you can easily obtain a Sierpinski triangle, as 
mentioned in a previous post:

./sierpinski.rb 32


                               *

                             * * *
                              * *
                           *   *   *

                         * * * * * * *
                          * * * * * *
                       *   * * * * *   *
                            * * * *
                     * * *   * * *   * * *
                      * *     * *     * *
                   *   *   *   *   *   *   *

                 * * * * * * * * * * * * * * *
                  * * * * * * * * * * * * * *
               *   * * * * * * * * * * * * *   *
                    * * * * * * * * * * * *
             * * *   * * * * * * * * * * *   * * *
              * *     * * * * * * * * * *     * *
           *   *   *   * * * * * * * * *   *   *   *
                        * * * * * * * *
         * * * * * * *   * * * * * * *   * * * * * * *
          * * * * * *     * * * * * *     * * * * * *
       *   * * * * *   *   * * * * *   *   * * * * *   *
            * * * *         * * * *         * * * *
     * * *   * * *   * * *   * * *   * * *   * * *   * * *
      * *     * *     * *     * *     * *     * *     * *
   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *







#! /usr/bin/ruby
#
############### Pascal Triangle by Eric DUMINIL ###############
#   
# This program is free software; you can redistribute it
# and/or modify it under the terms of the GNU General Public
# License as published by the Free Software Foundation;
# either version 2 of the License, or (at your option) any
# later version.
# 
# How-to use it?
#       pascal.rb height excentricity
# with:      
#   height, pretty self-explanatory
#   excentricity, float between -1 and 1.
#   
#               height set to 5 and excentricity set to -1:
#                          
#               1
#               1 1
#               1 2 1
#               1 3 3 1
#               1 4 6 4 1
#               
#               height set to 5 and excentricity set to 1 :
#              
#                      1
#                    1 1
#                  1 2 1
#                1 3 3 1
#              1 4 6 4 1
#              
#              
#              By default, excentricity is set to 0:      
#              
#                  1
#                 1 1
#                1 2 1
#               1 3 3 1
#              1 4 6 4 1            
#


#Just to make sure we can calculate C(n,n/2)
#without having to build the whole tree
class Fixnum
    def fact
        return 1 if self<2
        self*(self-1).fact
    end
    
    def cnp(p)
        self.fact/(p.fact*(self-p).fact)
    end
end


class PascalTriangle  
    def initialize (height=15,excentricity=0)
        @height=height
        @excentricity=excentricity
        #maxLength should be odd, so that the alignment is preserved
        @maxLength=(height-1).cnp((height-1)/2).to_s.length|1
        createAndShow
    end
    
    attr_reader :height, :maxLength, :excentricity
    
    def createAndShow
        previous=[1]
        current=Array.new
        height.times{|i|
            current[0]=current[i]=1
            #Taking care of the symetry
            1.upto(i/2){|j|
                current[j]=current[i-j]=previous[j]+previous[j-1]   
            }
            puts " "*((maxLength+1)*(excentricity+1)/2)*(height-i-1)+
                 current.map{|number|
                    number.to_s.rjust(maxLength)
                 }.join(" ")
            #No need to remember the whole triangle,
	    #the previous row will be enough
            previous.replace(current)
        }
    end
end

PascalTriangle.new(ARGV[0].to_i,ARGV[1].to_f)

###############################################################################

Bye,

Eric




Le Vendredi 23 Juin 2006 15:31, Ruby Quiz a écrit :
> The three rules of Ruby Quiz:
>
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> until 48 hours have passed from the time on this message.
>
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>
> 3.  Enjoy!
>
> Suggestion:  A [QUIZ] in the subject of emails about the problem helps
> everyone on Ruby Talk follow the discussion.  Please reply to the original
> quiz message, if you can.
>
> -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
>=-=-=
>
> by Dirk Meijer
>
> I recently showed a friend what an amazing language Ruby was, by quickly
> programming up a script to calculate Fibonacci's Sequence, and his first
> response was: "Can you do Pascal's Triangle?"  So I did, which proved
> harder than expected.
>
> For those not familiar with Pascal's Triangle, it is very similar to
> Fibonacci's Sequence. It's a pyramid of numbers. The outside of the pyramid
> is all ones, the other numbers are the sum of the two numbers above, like
> this:
>
>       1
>      1 1
>     1 2 1
>    1 3 3 1
>   1 4 6 4 1
>
> The input and output should be as follows:
>
>   $ ruby pp_pascal.rb 10
>                               1
>                            1     1
>                         1     2     1
>                      1     3     3     1
>                   1     4     6     4     1
>                1     5    10    10     5     1
>             1     6    15    20    15     6     1
>          1     7    21    35    35    21     7     1
>       1     8    28    56    70    56    28     8     1
>    1     9    36    84    126   126   84    36     9     1
>
> A number should be given as command-line argument.  This is the amount of
> rows the triangle has. For the output, there should be spacing between the
> numbers based on the size of the numbers with the most digits, so it will
> look like a proper triangle.
>
> Good luck!
>
> [Editor's Note:  If you are working through Chris Pine's Learn to Program,
> you can do this problem using only things you learned in the first eight
> chapters. Since he doesn't teach how to grab the row count in those pages
> though, just add this as the first line of your program:  `rows =
> ARGV[0].to_i]`.  After that, the rows variable will hold the number of rows
> to print.  --JEG2]