[Lua] Finding What "x" Equals

--[[
Created by my_hat_stinks
Licensed under a Creative Commons Attribution 3.0 Unported License
Date of creation: 20 May 2012
--]]
--[[
****Utility functions
--]]
local function table.HasValue( tab, var )
if type(tab) ~= "table" then error("Table expected, received "..type(tab).."!n") end --Bad input
for _,v in pairs(tab) do
  if v==var then return true end --It's there
end
return false --It's not there
end
--[[
****Local functions
--]]
local operators = {"+","-","*","/","="} --TODO: Powers support, Brackets support
local function InputToTable( In )
local tab = {} --Table of values
local pos = 0 --Position in table
for i=1,string.len(In) do
  pos=pos+1

  local char = string.sub(In, i, 1) --Character by character re-formatting

  if string.find(char, "[0-9]") then --Number
   if tab[pos-1] and type( tab[pos-1] ) == "string" then
  
	tab[pos-1] = (tab[pos-1]*10) + tonumber( char ) --Shift everything left one and insert this number
  
	pos = pos-1 --No new entry
  
   else --We're the first digit of this number
	tab[pos] = tonumber( char ) --Insert into a new entry
   end
  
  elseif string.lower(char)=="x" then --What we're searching for
   tab[pos] = "X"
  
  --Mathematical operators
  elseif table.HasValue(operators, char) then
   tab[pos] = char
  
  else --Not recognised
   if char~=" " then error("Bad input or unsupported character!n") end --If not space, error
  
   pos = pos-1 --Character wasn't registered
  end
end

return tab
end
local function CheckX( var )
if type(var) == "table" then return table.HasValue(var, "X") else return var=="X" end
end
local function Simplify( tab )
--There's a good chance modifying the table mid-loop will bork it, and from past experience "break" exists all loops
--We'll use a sub-function system
local function DoMath( func )
  local result

  --Highest priority, multiply
  for k,v in ipairs(func) do
   if v=="*" then
	local bef = func[k-1]
	local aft = func[k+1]
  
	if bef==0 or aft==0 then
	 func[k-1] = 0
	 table.remove(func, k)
	 table.remove(func, k) --Twice to remove the following number, too
	
	 return func, false
	end
  
	if CheckX(bef) or CheckX(aft) then
	end
   end
  end

  return func, true
end

local done = false
while not done do
  tab, done = DoMath( tab )
end

return tab
end
local function ArrangeEquation( tab )
--Verify and simplify the equation
--[[***************************]]--
local FirstEqual = 0
local SecondEqual = nil
for k,v in ipairs(tab) do
  if v=="=" then
   if FirstEqual==0 then FirstEqual = k
   else
	SecondEqual = k
	break
   end
  end
end
if SecondEqual then --Complicates things, erase it (result should be the same)
  for i=#tab,SecondEqual,(-1) do --Everything from the end to the second equals...
   tab[i]=nil --Erase!
  end
end
--[[***************************]]--

--Find the X
--[[___________________________]]--
local XPos = {}

for k,v in pairs(tab) do
  if v=="X" then
   table.insert(XPos, k)
  end
end

if XPos == {} then error("Nothing to find!n") end
--[[___________________________]]--

Simplify( tab ) --Do maths

local AfterEqual = nil
for i=1,#XPos do
  if AfterEqual~=nil and AfterEqual~=(XPos>FirstEqual) then
   tab = MoveX(tab, XPos, FirstEqual)
   break
  end
  AfterEqual = XPos[i]
end
end
--[[
****Base function
--]]
function FindX( In )
local tab = InputToTable(In)

tab = ArrangeEquation( tab )
end
---[[ Testing (remove the first - on this line to disable)
local In = "1+x*2=20"
local x = FindX( In )
write("x is " ..tostring(x).."n")
--]]