Ruby script to solve approximate equation with least squares method.

least_squares_method.rb

#! /usr/local/bin/ruby
# ***************************************
# 最小二乗法
# ***************************************
#
class LeastSquaresMethod
  N, M = 7, 5                    # データ数, 予測曲線の次数
  X = [-3, -2, -1,  0, 1, 2, 3]  # 測定データ x
  Y = [ 5, -2, -3, -1, 1, 4, 5]  # 測定データ y

  def initialize
    # a, s, t データ
    @a = Array.new(M + 1).map{ Array.new(M + 2, 0.0) }
    @s = Array.new(2 * M + 1, 0.0)
    @t = Array.new(M + 1, 0.0)
  end

  # 実行
  def exec
    # s, t 計算
    calc_st
    # a に s, t 代入
    ins_st
    # 掃き出し
    sweap_out
    # y 値計算＆結果出力
    display
  end

  private

  # s, t 計算
  def calc_st
    N.times do |i|
      (2 * M + 1).times { |j| @s[j] += (X[i] ** j) }
      (M + 1).times { |j| @t[j] += (X[i] ** j) * Y[i] }
    end
  rescue => e
    raise
  end

  # a に s, t 代入
  def ins_st
    (M + 1).times do |i|
      (M + 1).times { |j| @a[i][j] = @s[i + j] }
      @a[i][M + 1] = @t[i]
    end
  rescue => e
    raise
  end

  # 掃き出し
  def sweap_out
    (M + 1).times do |k|
      p = @a[k][k]
      k.upto(M + 1) { |j| @a[k][j] /= p }
      (M + 1).times do |i|
        unless i == k
          d = @a[i][k]
          k.upto(M + 1) { |j| @a[i][j] -= d * @a[k][j] }
        end
      end
    end
  rescue => e
    raise
  end

  # y 値計算＆結果出力
  def display
    (M + 1).times { |k| puts "a%d = %10.6f" % [k, @a[k][M + 1]] }
    puts "    x    y"
    -3.step(3, 0.5) do |px|
      p = 0
      (M + 1).times { |k| p += @a[k][M + 1] * (px ** k) }
      puts "%5.1f%5.1f" % [px, p]
    end
  rescue => e
    raise
  end
end

exit 0 unless __FILE__ == $0

begin
  # インスタンス化＆実行
  LeastSquaresMethod.new.exec
rescue => e
  $stderr.puts "[#{e.class}] #{e.message}"
  e.backtrace.each{ |tr| $stderr.puts "\t#{tr}" }
end