A robot is placed on the square A1 of a standard chessboard and has to reach H8. It understands orders Up, Down, Left, Right. On some squares, there might be a cement block; is such a case, the robot does not execute the order and continues with the next one on your list.
There always is at least one possible path.
The list of directions is finite.
The robot might reach the destination somewhere in the middle of your list.
Give the answer in the form UURRU.
(This list will work i.e. on an empty 3x3 chessboard or with a single block on A3 but will fail with a single block on B3.)

I had this solution in mind, unfortunately, it fails because of repeated situations.
Starting situation: 6 lanterns, only lantern 6 is lit.
First round:
lantern 1: no action
lantern 2: devil turns it on
remaining lanterns: angel does not do anything (If the devil turns a lantern on before the angel has turned any off, then the angel should just do nothing on that circuit. )
Situation: Lanterns 2 and 6 are lit
2nd round:
lantern 1: no action
lantern 2: angel turns it off (on each circuit, turn off all lit lanterns until the devil turns an unlit lantern on)
Now, only the lantern 6 is lit: the game ends if the lanterns return to a previously encountered state
What am I missing?

Here's an example from my code's 6 lantern output:
111110 <one lit lantern (interpret 0 as "on" and 1 as "off")
111101 <one lit lantern
You suppose the devil turns the lantern 5 on. But what if he turns the lantern 4 and not 5 on?
111110 <one lit lantern (interpret 0 as "on" and 1 as "off")
111010 your turn
P.S. I am not touchy (did not even think it could be interpreted this way), but I think things over. In the last time, I wrote a lot of nonsense, i.e. my first answer.

@EventHorizonMy post is unrelated to yours.
The number of lamps is not important, just a little bit more that those I quote.
"Let's start with one lantern lit, i.e. 33.":
This is not a binary number, it is the 33rd lantern assuming the starting position is the 1st lantern (conveniently chosen), following the lantern 32 and followed by the lantern 34.

To decide who will do the dishes, we put in a jar tickets with prime numbers up to 111. The one who draws the ticket with a larger number wins.
I look at my ticket: 3. Maybe you have a low number like 11... So I offer to exchange our tickets before showing them.
What is the probability I will do the dishes?

I get 13.
Take one coin of each chest.
a) You get two Gold, one Silver:
Silver is identified, remains to distinguish all_Gold and Gold_Silver.
Take 10 more coins from the first G:
- if they are all gold, you identified all_Gold, the remaining is Gold_Silver
- if you get a silver coin, you identified Gold_Silver, the remaining is all_Gold
b) You get one Gold, two Silver:
Same proceeding, permute Gold/Silver.

There is a strip of N (N>13) squares. In the middle is the SILVER DOLLAR.
Two players alternatively place a penny on an empty square.
Then, at each turn, a player must:
- move one coin one or more squares to the left, observing:
- the coin cannot move out of the strip
- the coin cannot jump on another coin
- the coin cannot jump over another coin
- or -
- pocket the leftmost coin
Best strategy?

@Plasmid I translated your notation into mine, as far as we are gone, we have same results (excepted some doublets).
I thought about interpolation, too, but I did not go this way estimating there is not enough data.