Level 1

You have a table with 4 corners. On each corner is a button. If you press a button it alternates between being ON or OFF, but you don't know what it is currently set as.

You can press whichever buttons you want, then end your turn. If at the end of your turn, all buttons are set to the same setting, a light turns on in the middle of the table and you win. If not, the table rotates randomly until you lose track of which button is where.

Is there a strategy you can use to guarantee that the light will turn on, regardless of what the starting arrangement was, and regardless of how the table spins? What is the maximum number of steps that it would take?


Level 2

For what other number of corners larger than 4 is this solvable? How many moves does it take to solve a larger table?

Note:
They don't all have to be On. If they're all Off you're done as well.

I don't think it is possible. Mainly because I don't think there is a strategy for just 3 buttons.

nvm

1 button - end turn. Good

2 buttons - end turn. Bad. press one button, end turn. Good

3 buttons - end turn. Bad. press one button, end turn. Bad. 2 are same, 1 isn't. Every table spin you press 1 button and have a 33% chance. Still no guarantee, unless you consider endless 33% shots a guarantee.

kinda lost with 4.

edit: with 4.

It's either XXYY or XXXY

Press two buttons.

If XXYY, you have a 2/4 x 1/3 = 2/12 = 1/6 chance of being right. Do this repeatedly until you're mathematical certain it's not XXYY.

Then start pressing 1 button.

Alternatively you could start with 1 button, then move to 2.

I assume this is solvable, but I have nfi how.

Would be very strange if it were solvable with n=2 and n=4, but not with n=3.

well, you basically have 16 combos states and 2 solutions. Since on and off are the same thing, it's really like 8 combos and 1 solution.

I can think of 3 routines:
1. press a single button
2. press 2 adjacent buttons
3. press 2 opposite buttons

Just need to come up with an order of routines that can't get back to the original state, while hitting all other 7. Something like routines 1 - 2 - 3 - 2 or something.

Pressing just one button at a time gives you a 1/4 chance for every two pushes. That might be better than the 1/6 every press, since there is a 4/7 chance it is XXXY or YYYX

We're not looking at probabilities, you need find a way that is a 100% lock

Haven't given it much thought yet, but I think there may be an important point that if you push two buttons across from each other you know you pushed one of two pairs, and if you push adjacent buttons you pushed one of 4 pairs (that are LDO distinct from the pairs above).

Other thought is that you need to switch either one or two toggles to win if you are not in a winning state.

Do we know at our first turn that we are NOT on a winning combination? So game starts, we do nothing and end our turn, can we win? (I guess this isn't that important, since we know if we go to round two we are starting without a winning combination).

I'll give some thought to this and maybe draw some pictures later.

This is a solution that works in 8 rounds. Don't know if it's the least possible guaranteed solution though. And I haven't thought about level 2 yet.

this is group theory

I have a book on group theory (because Rubik's cube) but haven't read it

My guess for Level 2 is



But coming up with a general strategy is difficult, then proving it is even harder.

chuckles I tried for the next one and failed. Can't say it's impossible but my original methods let me down.

I think my routine thing might work for figuring out how many turns it takes for level 2 but I wouldn't know how to find the order of routines you'd need to go through. If you take the number of starting states and divide it by 2, that's the max number of routines you can run in order to hit all possible combinations once.

Like for 4 lights, there's 16 combos of 0000 - 1111, with 0 being off and 1 being on. Two of those combos are the correct combo. Then you remove everything that's mirrored cus it doesn't matter if it's off or on by dividing by 2, and then you have to hit all of the remaining combinations once using all routines available.

edit: if chips' thing is right I'd also subtract 1 at the end, if iversonians is right then I wouldn't. I'm not sure if Iversonian's is achieving the all off/all on twice or not. Their routines are the same aside from the first one.

I tried to make this an iteration problem

So lets say we have 8 lights

We do a series of tests so that if 4 are lit up we win

If not, we do a series of moves that would move 2 or 6 lights to 4 lights or a win

We do a series of tests so that if 4 are lit up we win

If not we pick one button randomly and push it

We do a series of tests so that if 4 are lit up we win

If not, we do a series of moves that would move 2 or 6 lights to 4 lights or a win

We do a series of tests so that if 4 are lit up we win




This sounds great and all, but I don't know what the tests are...

I want to convert this into some sort of binary problem...

The question suggests it's solvable for N > 2 but I'm not sure.

The only scoring pattern I see is using N/2 buttons, alternating. When you use N/2 alternating and you have N/2 alternating, you win. The trick then is to transform any starting number of buttons into N/2 alternating.

With any other number or pattern of buttons than N/2 alternating, you have more than two equiv positions so you can never be sure to win.

What can you transform to N/2 alternating? When you start with N buttons total and A initially on, pushing P buttons will result in either {1,3,5,...} buttons or {0,2,4,...} buttons depending on the parity of A+P mod N. For N > 4, there's no guaranteed route to N/2 because for any A buttons on, pushing P produces multiple results. You can't force it to N/2. So I think 4 is the max you can solve. Not confident though.

This is the weak part. Maybe there is a smart way. N/2 alternating is for sure the goal though.

Guaranteed lock with 7 turns

Turn 1 - Press 2 opposite buttons. If does not solve
Turn 2 - Press 2 adjacent buttons. Will not solve at this point
Turn 3 - Press 2 opposite buttons. This method will solve ALL starting combinations of 2 on 2 off no matter how they are ordered guaranteed. If it does not solve then we know it MUST be a 1 on 3 off situation or vice versa
Turn 4 - Press any 1 light. 1 in 4 chance of solving. If does not solve then we now know we have a 2 on 2 off situation.
Turn 5, 6, 7 - Repeat Steps 1, 2, 3 and you will have a guaranteed lock by turn 7

Forgot about fact could already be set, so yeah 8 turns

You posted this one in the brainteaser thread already ibavly

Running out of material?

Lol don't remember that.

I just solved Level 2 yesterday though

First part of 2 I'm going to guess is even numbers.