lua code to solve maze 24 and any target though + - * /. Original source:http://blog.henix.info/dig/reverse-polish-notation-catalan-number/_.html

maze24.lua

--[[
24点解法：4个数字的逆波兰式形式只有以下5种。1代表数字，+代表运算符，括号隐藏在逆波兰式的形式中
1111+++
111+1++
111++1+
11+11++
11+1+1+
]]

local target = 24

local num4 = { 8, 4, 2, 24 }

local op = { '+', '-', '*', '/' }

--计算一种逆波兰式的值
function evalPo(e)
	local s = {}	--临时存储的栈
	local str = {}	--正常中缀表达式
	local mask = {}	--str中各项表达式的符号
	local a, b, re, s1, s2, m1, m2
	for i = 1, #e do
		if type(e[i]) == "number" then
			table.insert(s, e[i])
			if e[i] < 0 then
				table.insert(str, "("..e[i]..")")
			else
				table.insert(str, e[i])
			end
			table.insert(mask, 1)
		elseif type(e[i]) == "string" then
			if #s < 2 then
				error("Po statement wrong!")
			end
			b = table.remove(s)
			a = table.remove(s)
			s2 = table.remove(str)
			s1 = table.remove(str)
			m2 = table.remove(mask)
			m1 = table.remove(mask)
			if e[i] == '+' then
				re = a + b
				table.insert(mask, 0)
			elseif e[i] == '-' then
				re = a - b
				table.insert(mask, 0)
			elseif e[i] == '*' then
				re = a * b
				table.insert(mask, 1)
			elseif e[i] == '/' then
				if b == 0 then
					--除0的直接返回-1吧
					return -1
				end
				re = a / b
				table.insert(mask, 1)
			else
				error("Po wrong op: "..e[i])
			end
			if mask[#mask] == 1 then
				if m1 == 0 then
					s1 = '('..s1..')'
				end
				if m2 == 0 then
					s2 = '('..s2..')'
				end
			end
			table.insert(str, s1..e[i]..s2)
			table.insert(s, re)
		end
	end
	if #s ~= 1 then
		error("Po statement not enough op!")
	end
	return s[1], str[1]
end

--local test = { 5, 3, 14, 2, '/', '+', '*' }
--print(evalPo(test))

local exp = {}
local pos = {
	[1] = { 1, 2, 3, 4, 5, 6, 7 },	--1111+++
	[2] = { 1, 2, 3, 5, 4, 6, 7 },	--111+1++
	[3] = { 1, 2, 3, 6, 4, 5, 7 },	--111++1+
	[4] = { 1, 2, 4, 5, 3, 6, 7 },	--11+11++
	[5] = { 1, 2, 4, 6, 3, 5, 7 },	--11+1+1+
}


for a = 1, 4 do
	for b = 1, 4 do
		if b ~= a then
			for c = 1, 4 do
				if c ~= a and c ~= b then
					for d = 1, 4 do
						if d ~= a and d ~= b and d ~= c then
							for op1 = 1, 4 do
								for op2 = 1, 4 do
									for op3 = 1, 4 do
										for model = 1, 5 do
											exp[pos[model][1]] = num4[a]
											exp[pos[model][2]] = num4[b]
											exp[pos[model][3]] = num4[c]
											exp[pos[model][4]] = num4[d]
											exp[pos[model][5]] = op[op1]
											exp[pos[model][6]] = op[op2]
											exp[pos[model][7]] = op[op3]
											local value, str = evalPo(exp)
											if value == target then
												io.write(str.."="..value, '\n')
												os.exit()
											end
										end
									end
								end
							end
						end
					end
				end
			end
		end
	end
end

print("maze can't solve!")