Solution to "Left is where the thumb is right"

Zur deutschen Version

Scarlett finds that the door she came through can no longer be opened. She is trapped in a cellar. There is an emergency exit, but in front of it there is a lattice.

If you go further to the left, you will see several numbered doors behind the lattice. A gap in the lattice, however, allows access to doors 1 and 2. You can now open one of the two doors and enter the room behind it, but it is empty - as are the rooms behind the other doors.

There is a handle on the left and right of the lattice gap. The lattice can be moved with this handle. If you move the lattice to the right so that the gap is above the exit, you have solved the puzzle and Scarlett can leave the muffy cellar.

The lattice cannot be moved any further to the left, it stops at the left side. But if you try to move it to the right, the latch of the first door blocks the lattice. So you open door no. 1 and the lattice can be moved further to the right. But if you try to move it even further to the right, the latch of door 2 blocks the lattice.

You can see the problem: The gap can only be pushed so far to the right until the latch of a door blocks the lattice. So to get all the way to the exit, all five doors must be opened.

So cheer up and get to work! Open door no. 2: Wrong! Door no. 1 blocks door no. 2. This also applies to all other doors: If the door to the left of it is open, the corresponding door cannot be opened or closed.

In summary: To move the lattice, one door must be open. But to open the door to the right of it, it must be closed! So is there a solution at all? Of course, otherwise this puzzle would be pointless!

However, the solution is rather complicated. In the following table "0" stands for a closed door and "1" for an open one. The moving of the lattice is not mentioned here, it results logically: To open or close a door, it must be in the gap of the lattice.

The moves result as follows:

So you need 21 moves (not counting moving the lattice).

Background

The binary code that results is closely related to the Gray code. As with the Gray code, only one digit can change from one line to the next. In this case, only one door can be opened or closed in one move. Logically, Scarlett is alone and can only operate one door at a time. The problem becomes twice as complex with each door, so the difficulty increases exponentially with each door.

This puzzle is closely related to the "Chinese Rings" and this puzzle.