Symmetric strategy for 100 prisoners and a light bulb problem

Question

Remark: The problem is the prisoners and lightbulb problem, but without the usual probabilistic frame (the ogre chooses as he pleases). Also, the strategy is symmetric and the dwarves don't have a sense of time. The link above does not give a solution for these hypothesis.

In more detail, the problem goes as follows: $100$ immortal dwarves are captured by an immortal ogre in order for him to play a game. The dwarves are in separate cells and never communicate, and each day the ogre chooses arbitrarily one dwarf and brings him to a room with a lightbulb which is either on or off. The dwarf can leave it as it is or switch it. On the first day, the dwarve decide on a strategy, and the lightbulb is off.

The dwarves are able to go out if one day one of them can say that all $100$ dwarves have been taken to the lightbulb (and be right about it).

Two important hypothesis:

• the choice of the ogre is arbitrary (not random, arbitrary), but the game isn't unfair so he promises that he plans to take each dwarf an infinite number of times to the lightbulb room,
• the dwarves have no sense of time (so in particular they cannot know how many dwarves went to the room before them, or if the ogre took them to the room several times in a row). Equivalently, we could say the ogre takes a dwarf to the room whenever he wants.

$ $

Question: before the game, all $100$ dwarves meet one (last?) time to decide on a strategy.

• A. Can they find a winning strategy?

• B. [Contains hint for 1...] Can they find a winning symmetric strategy (i.e. each dwarf has the same action policy) ?

---

I don't have an answer for B. I am putting this question here since I think some (very simple) combinatorics may be needed to find a proof for a symmetric strategy.

Here is my answer for A: one of the dwarf is chosen as a 'leader': only him can switch on the light, and only him can speak to the ogre. So each time he switches it on (or not if it is already) and count the number of times he switched it on. All $99$ other dwarves can only switch the light off once if they find it on (and then do nothing more). Once the leader has switched the light on $100$ times (it is off at the beginning), he tells the ogre that all dwarves have seen the lightbulb. Using the fact that after any number of days, the ogre will still bring each dwarf again in the room, it is easy to show that the dwarves will win.

Answer 1

If the dwarves are aware of the separation between rounds, then the problems are trivial, as you can see from @Arthur 's answer. Assuming that the dwarves lose their sense of time, the problems are much more interesting. A suggestion for 2): Everybody plays assuming that the light was off to begin with. The only modification in the strategy: each of the 99 non-leader dwarves turn the light off twice if they find it on. If the leader finds the light on during his first visit, then he leaves it on, and knows that this is where the game started, because he must have been the first visitor. If he finds it off, then he turns it on (according to the original strategy), and assumes that there was already zero or one visitor. In any case, after having to switch on the light for the 198th time, the leader can make the statement.

The 198th time definitely comes: even if the light was turned off by a dwarf on day one, it is turned on after each time it was turned off, which is 198 times. When the light is turned on for the 198th time, there cannot have been at most 98 dwarves in the room before that: otherwise, they would have had turned the light off 196 times, plus maybe there was one extra turn-on round by the leader (which can happen if the leader was taken to the room on day one and the light was off).

Answer 2

I believe there is no symmetric strategy for the prisoners. This problem appeared in Mathematical Puzzles, a Connoisseur's Collection by Peter Winkler. In the solution provided, he went above and beyond and gave a complete characterization of all successful strategies. Here is the exact excerpt from the book, which Winkler warns is only "sketchy."

Let us focus on one prisoner, say Alice. Her strategy can be assumed to be deterministic and based solely on the sequence of light-states she has so far observed.

Suppose that Alice's strategy calls for her (in some circumstance) to change the state (of the light) after finding it in the state in which she last left it. Then the adversary could have brought her back immediately to the room, "wasting" her previous visit; in effect, this piece of Alice's strategy can only give the adversary an extra option. We may assume, therefore, that Alice never changes the state when she finds it where she last left it.

Next, suppose Alice is required at some point to leave the state as she found it. Then we claim we can assume she will never act again! Why? Because if the adversary doesn't want her ever to act again, he can insure that she never sees a state different from the state she now finds. He can do this because if Alice did become permanently inactive, at least one of the states (on or off) will recur infinitely often; suppose it's the "off" state. Then he can schedule Alice so that she sees "off" now and at every subsequent visit, hence, by the previous argument, she will never act again. So, once more, the adversary always has the option of silencing Alice so we may assume that it is his only option.

Obviously, Alice can't then begin with the instruction to leave the state as she finds it, since in that case she is forever inactive and no one will ever know that she has visited the room.3 Say she is supposed to turn the light on if it's off, otherwise leave it on. Then she won't be doing anything until she again finds the light off, at which point she may only turn it on again or go inactive forever. Thus, she is limited to turning the light on some number of times $j$ (which may as well be constant, else the adversary has more options). We call this strategy $+j$, where $j$ is a positive integer or infinity. Similar arguments apply if she is instructed to turn the light off on first visit, leading to strategy $-j$.

The only remaining possibility is that she is instructed to change the state of the light at first visit, in which case she must proceed as above depending on whether she turned the first light on or off. This again only gives the adversary an additional option.

We are reduced to each prisoner having a strategy $+ j$ or $- j$ for various $j$. If they all turn lights only off (or only on), no one will learn anything; thus, we may assume Alice's strategy is $+j$ and Bob's is $- k$.

At this point, the question is answered. Every strategy is of the form $\pm j$ for $j\in \mathbb N\cup \{\infty\}$, and it is clear that no one of these works when all prisoners use it. Here is the rest of the solution in case anyone is curious.

If Charlie turns lights on, Alice will never be able to tell the difference between Bob and Charlie both having finished, and Bob and Charlie each having one task left. If Charlie turns lights off, it's Bob who will be "left in the dark."

Putting all this together, we have that for a prisoner to be able to determine that everyone has visited, she must turn the light on, while everyone else turns it off (or vice-versa). In fact, if her strategy is $+j_1$ and the others are $- j_2, \dots, -j_{n}$ then it's easy to check that having each $j_i$ finite, but at least $2$, and $j_1$ greater than the sum of the other $j_i$'s minus the least of them, is necessary and sufficient.

Answer 3

For (2), they can just decide that whoever gets in the the first day will leave with the light off, and then they start whatever actual strategy they are planning on the second day.

For (3), this may feel like cheating, but let the dwarf which is chosen on the second day become the leader.