From: Farrel Lifson <farrel.lifson@...>
Date: 2006-06-26T16:31:43+09:00
Subject: Re: [QUIZ] pp Pascal (#84)

Here's my solution. It's not very efficient but it works:

numberOfRows = ARGV[0].to_i



# Handles the case where the number of rows asked for is 1

numberOfRows.eql?(1) ? (puts("1");exit) : nil



# Genereate Pascal's Triangle

rows = [[1],[1,1]]

2.upto(numberOfRows-1) do |currentRowIndex|

  rows[currentRowIndex] = [1]

  1.upto(currentRowIndex-1) do |elementIndex|

    rows[currentRowIndex] << rows[currentRowIndex-1][elementIndex-1] +
rows[currentRowIndex-1][elementIndex]

  end

  rows[currentRowIndex] << 1

end



# Get  the length in characters  of the largest element

maxElementLength = rows[numberOfRows - 1][numberOfRows/2].to_s.length



# Format and ouput the triangle

puts(rows.map do |row|

  ' '*maxElementLength*(numberOfRows - row.length) +

  row.map do |element|

    element.to_s +

    ' '*(maxElementLength-element.to_s.length) +

    ' '*maxElementLength

    end.join

end.join("\n"))