How to find a smallest value without trying all posibilities - math

Question

I am trying to write a program (in Lua, but I guess this is more of a mathematical problem) that calculates the total distance between a set of numbers and there are two or three different possibilities for some of these numbers. 
For example a set of numbers is: 2, 5, 0, 1. The sum of distances in this case is 9. (5-2 + 5-0 + 1-0)
The first row with alternatives is: 5, -, 3, 2.
The second row with alternatives is: 3, -, -, 3
The combination between the two rows with the smallest sum of distances is: 5, 5, 3, 3 with a sum of distances of 2.
My first attempt was to write a program that tries all possible iterations, but with a set of about 40 numbers, there are so many possibilities that my computer crashed...
Here is the code for that version, it first creates all the different possibilities and then calculates the differences and places them in column 0. Afterwards I can find the smallest value easily and see also the combination of numbers which results in that value.
local table1 = {2, 5, 0 ,1}
local table2 = {5, nil, 3, 2}
local imax = 1
local solution = {}
local answer = table1[1]
for x = 1,#table1 do
solution[x]={}
for i= 1, (2^imax)/2 do
solution[x][i] = table1[x]
end
if table2[x] ~= nil then -- there is an alternative number
for y = 1, x-1 do -- copy all the previous table entries except the last one
for j = ((2^imax)/2)+1, 2^imax do -- the number of new rows increases exponentially
solution[y][j] = solution[y][j-imax]
end
end
for j = ((2^imax)/2)+1, 2^imax do -- create the new table entry with the alternative number
solution[x][j] = table2[x]
end
imax = imax + 1 -- this number is to remind how many alternative numbers where found
end
end
solution[0]={}
for x = 1, #table1 do
for i = 1, (2^imax)/2 do
if x < #table1 then answer = math.sqrt((solution[x+1][i]-solution[x][i])^2) else answer = 0 end
if solution[0][i] == nil then solution[0][i] = answer else solution[0][i] = solution[0][i] + answer end
end
end
After reading about dynamic programming, I wrote a new version of this program. It calculates the smallest sum of differences, but I also want to know the path (the combination of numbers) to that sum... Still work to do...
local table1 = {2, 5, 0 ,1}
local table2 = {5, nil, 3, 2}
local solution = {}
local smallestsolution = {}
solution[1]={}
solution[2]={}
solution[3]={}
solution[4]={}
for i = 1, (#table1-1) do
solution[1][i] = math.sqrt((table1[i+1]-table1[i])^2)
if table2[i] ~= nil then solution[2][i] = math.sqrt((table1[i+1]-table2[i])^2) end
if table2[i+1] ~= nil then solution[3][i] = math.sqrt((table2[i+1]-table1[i])^2) end
if table2[i] ~= nil and table2[i+1] ~= nil then solution[4][i] = math.sqrt((table2[i+1]-table2[i])^2) end
end
for i = 1, (#table1-1) do
smallestsolution[i]=100000
for j = 1, 4 do
if solution[j][i] ~= nil and solution[j][i] < smallestsolution[i] then smallestsolution[i]=solution[j][i] end
end
end
local smallestsum = 0
for i = 1, (#table1-1) do
smallestsum = smallestsum + smallestsolution[i]
end
Thanks,
Emile

Answer 1

I managed to solve it myself! The hint to dynamic programming by #EgorSkriptunoff did the trick!
local table1 = {2, 5, 0 ,1}
local table2 = {5, nil, 3, 2}
local solution = {}
local smallestsolution = {}
solution[1]={}
solution[2]={}
solution[3]={}
solution[4]={}
local path = {}
local temp = {}
for i = 1, (#table1-1) do
solution[1][i] = math.sqrt((table1[i+1]-table1[i])^2)
if table2[i] ~= nil then solution[2][i] = math.sqrt((table1[i+1]-table2[i])^2) end
if table2[i+1] ~= nil then solution[3][i] = math.sqrt((table2[i+1]-table1[i])^2) end
if table2[i] ~= nil and table2[i+1] ~= nil then solution[4][i] = math.sqrt((table2[i+1]-table2[i])^2) end
end
for i = 1, (#table1-1) do
smallestsolution[i]=100000
temp[i] = 0
for j = 1, 4 do
if solution[j][i] ~= nil and solution[j][i] < smallestsolution[i] then smallestsolution[i]=solution[j][i] temp[i] = j end
end
end
local smallestsum = 0
for i = 1, (#table1-1) do
smallestsum = smallestsum + smallestsolution[i]
end
for i = 1, (#table1) do -- find the path belonging to the smallest sum of differences
if temp[i] == 1 then path[i] = table1[i] end
if temp[i] == 2 then path[i] = table2[i] end
if temp[i] == 3 then path[i] = table1[i] end
if temp[i] == 4 then path[i] = table2[i] end
if i == (#table1) then
if temp[i-1] == 1 then path[i] = table1[i] end
if temp[i-1] == 2 then path[i] = table1[i] end
if temp[i-1] == 3 then path[i] = table2[i] end
if temp[i-1] == 4 then path[i] = table2[i] end
end
end