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Here is my solution...



rows = ( ARGV.shift || 1 ).to_i
tri = Array.new( rows ) { |i| Array.new( i+1 ) }
tri.each_with_index do |row, rowno|
    row.fill do |colno|
        ( colno == 0 or colno == rowno ) ? 1 : tri[rowno-1][colno-1] +
tri[rowno-1][colno]
    end
end

cols, col_w = tri.last.length, tri.last.max.to_s.length+1
tri.each do |row|
    row_strs = row.collect { |col| col.to_s.center( col_w ) }
    puts row_strs.join.center( cols*col_w )
end

mark

On 6/25/06, Robert Retzbach <rretzbach / googlemail.com> wrote:
>
> Hello,
>
> you can take this as negative example, cause I store the triangle and
> the formatted output :) Thus it's much slower than most solutions.
>
> kind regards
> Robert
>
>
>
> #!C:\ruby\bin\ruby.exe
>
> class PTriangle
>   def build num = 4
>     @body = Array.new
>
>     # fill the empty triangle
>     # using non-recursive algo
>     ( num + 1 ).times do | row |
>       row.times do | pos |
>         if pos == 0 or pos == row - 1
>           number = 1
>         else
>           number = @body[ row - 2 ][ pos - 1 ] + @body[ row - 2 ][ pos ]
>         end
>         @body[ row - 1 ] ||= []
>         @body[ row - 1 ][ pos ] = number
>       end
>     end
>
>     @body
>   end
>
>   def render
>     @rendered = []
>
>     # calculate maximum number length
>     # needed for alignment
>     max = @body.last[ @body.last.size / 2 ].to_s.size
>
>     # shift all lines to the right if a new line is added
>     # right justify all numbers
>     @body.each do | row |
>       @rendered.map! { | line | ' ' * max << line }
>       @rendered << row.map { | num | num.to_s.rjust( max * 2, ' ' ) }.join
>     end
>
>     # cleanup, don't know how to do it more elegant
>     # => remove not necessary indentation
>     @rendered.map! { | line | line[ ( max * 2 - 1 )..-1 ] }
>   end
> end
>
> if __FILE__ == $0
>   level = ARGV.first.to_i
>   level = 4 unless level > 0
>
>   pt = PTriangle.new
>   pt.build level
>   puts pt.render
> end
>
>
>


-- 
Mark Van Holstyn
mvette13 / gmail.com
http://lotswholetime.com

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