From: Mathieu Bouchard <matju@...>
Date: 2001-06-10T19:41:06+09:00
Subject: [ruby-talk:16387] Symbolic Computation II


Take the source of [ruby-talk:16383] and append the following to it. It
will allow you to isolate a variable if that variable occurs only once in
an expression.

----------------8<----------------cut-here----------------8<----------------

class Quote
	def which_side(s); @sym==s; end
end

class QuoteBin
	@inverse_of = Hash[ :+, :-, :-, :+, :*, :/, :/, :* ]
	class<<self; attr_reader :inverse_of; end
	attr_reader :l,:o,:r
	def which_side(s)
		l = @l.respond_to?(:which_side) && @l.which_side(s)
		r = @r.respond_to?(:which_side) && @r.which_side(s)
		if l and r then raise "#{s} is in both sides" end
		return l ? :l : r ? :r : nil
	end
end

class Equation < QuoteBin
	def initialize(l,r); @l,@o,@r = l,:==,r; end
	def isolate(s)
		#puts self
		case @l.which_side(s)
		when :l
			invo = QuoteBin.inverse_of[@l.o]
			nr = QuoteBin.new(invo,@r,@l.r)
			Equation.new(@l.l,nr).isolate(s)
		when :r
			nr = QuoteBin.new(@l.o,@l.l,@r)
			Equation.new(@l.r,nr).isolate(s)
		when true; self
		when nil; raise "#{s} not found"
		end
	end
end

----------------8<----------------cut-here----------------8<----------------

try this:

e=Equation.new x,0
puts e               # (((((2 + a) * 3) * b) - ((2 * a) / c)) == 0)
puts e.isolate(:b)   # (b == (((2 + a) * 3) * (0 + ((2 * a) / c))))
puts e.isolate(:c)   # (c == ((2 * a) / ((((2 + a) * 3) * b) - 0)))

note that my algorithm doesn't take care of irregularities like division
by zero, infinities, and the irreversibility of Integer#/.

matju