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- or go back to the newest paste.
| 1 | local efficiency = 8 --the amount of items each fuel item can smelt | |
| 2 | local furnaces = 6 --the amount of furnaces the chef has (MAX 8) | |
| 3 | local amounts = {}
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| 4 | function selectNext() --selects the next non-empty slot in the turtle and returns true, else returns false | |
| 5 | for i = 1, 16 do | |
| 6 | if turtle.getItemCount(i) ~= 0 then | |
| 7 | turtle.select(i) | |
| 8 | return true | |
| 9 | end | |
| 10 | end | |
| 11 | return false | |
| 12 | end | |
| 13 | while turtle.suck() do | |
| 14 | for i = 2, furnaces do --get's the ores out of the chest, until chest is empty or has [furnaces] stacks | |
| 15 | table.insert(amounts, 0) | |
| 16 | if not turtle.suck() then break end | |
| 17 | end | |
| 18 | for k, v in pairs(amounts) do --puts the ores it has collected into the furnaces | |
| 19 | turtle.back() | |
| 20 | selectNext() | |
| 21 | amounts[k] = turtle.getItemCount() | |
| 22 | turtle.dropDown() | |
| 23 | end | |
| 24 | while turtle.forward() do end --resets to the chest location | |
| 25 | turtle.select(1) | |
| 26 | turtle.down() | |
| 27 | for k, v in pairs(amounts) do | |
| 28 | turtle.suck(math.ceil(v/efficiency)) --sucks out of the coal chest how much fuel it needs | |
| 29 | end | |
| 30 | turtle.down() | |
| 31 | local biggestStack = 0 | |
| 32 | for k, v in pairs(amounts) do | |
| 33 | if v > biggestStack then --calculates the size of the biggest stack | |
| 34 | biggestStack = v | |
| 35 | end | |
| 36 | turtle.back() | |
| 37 | selectNext() | |
| 38 | turtle.dropUp(math.ceil(v/efficiency)) --puts the correct amount of fuel into each furnace | |
| 39 | end | |
| 40 | sleep(biggestStack*10) --waits for the biggest stack to be done | |
| 41 | for k, v in pairs(amounts) do --collects all the items | |
| 42 | turtle.suckUp() | |
| 43 | turtle.forward() | |
| 44 | end | |
| 45 | while selectNext() do turtle.drop() end | |
| 46 | turtle.up() | |
| 47 | turtle.up() | |
| 48 | turtle.select(1) | |
| 49 | end |
