[Guest]

Imagine a small square table with an area of about 1x1 meters. The table is resting on 1 cylindrical foot attached to the floor. This allows the table to be turned clockwise or counter-clockwise freely.

The table has 4 pockets, one in each corner, like the pockets in a pool table except there are only 4, and they are larger. Each pocket has enough room to fit a small glass in it and still leave room for a hand to reach in and turn the glass around.

In each pocket there is a small glass turned either the normal way, or upside down. These are the only 2 states the glasses are allowed to be in.

You are about to play a game of "Align the glasses". The simple objective is to make sure that all 4 glasses in the corner pockets are turning the same way (either all turned normally - or all turned upside-down).

You are blindfolded by the table master. He sets a random initial position of the glasses. Then he spins the table for a random number of 1/4th turns in a random direction (clockwise or counter-clockwise). Afterwards he tells you to adjust 2 glasses. You put your hands in 2 pockets - either side-by-side or diagonally, feel the glasses, and set the position of each glass to either of the 2 legal states (up or down) - or you leave either or both as they are. The choice is yours.

After setting the 2 glasses of your choice, the table master takes control of the table again.

If all 4 glasses are turned the same way (either all up or all down) he will tell you immediately and you win the game. Otherwise he spins the table randomly again, and you get another turn of feeling/adjusting the position of 2 glasses.

This continues indefinately - or until you win the game.

Now - it is fairly obvious that you have a chance at winning the game just by going on and on until you win. However, the question for you to answer is this:

Is it possible given these rules to guarantee a victory in a certain number of turns? And if so, how many turns?

Enjoy

/Uhre

[Guest]

Well, I think my first move would be to select two diagonal pockets and align the glasses the same way (either both up or both down). Then when the table is spun again, I'll choose diagonal pockets again. If they're both aligned the same way I've already selected, then I know they're the ones I switched last time, so I'll just turn them both upside down on the off chance that they'll match the other glasses. If however on the second spin the two diagonal glasses I choose aren't in the same state that I set on the first spin, then they are the other set of diagonal glasses (they can't be the other set and also in the same state or else I would have won after the first spin). If this is the case, then I just switch them to match the other glasses and I've won. The only problem with this method is that since everything is random, I have no idea how long it will take before I select the second diagonal set of glasses.

I wouldn't be able to guarantee a victory in a certain number of turns because of the improbable (yet still possible) chance that I choose the same two glasses on a million spins in a row.

[Pickett]

Imagine a small square table with an area of about 1x1 meters. The table is resting on 1 cylindrical foot attached to the floor. This allows the table to be turned clockwise or counter-clockwise freely.

The table has 4 pockets, one in each corner, like the pockets in a pool table except there are only 4, and they are larger. Each pocket has enough room to fit a small glass in it and still leave room for a hand to reach in and turn the glass around.

In each pocket there is a small glass turned either the normal way, or upside down. These are the only 2 states the glasses are allowed to be in.

You are about to play a game of "Align the glasses". The simple objective is to make sure that all 4 glasses in the corner pockets are turning the same way (either all turned normally - or all turned upside-down).

You are blindfolded by the table master. He sets a random initial position of the glasses. Then he spins the table for a random number of 1/4th turns in a random direction (clockwise or counter-clockwise). Afterwards he tells you to adjust 2 glasses. You put your hands in 2 pockets - either side-by-side or diagonally, feel the glasses, and set the position of each glass to either of the 2 legal states (up or down) - or you leave either or both as they are. The choice is yours.

After setting the 2 glasses of your choice, the table master takes control of the table again.

If all 4 glasses are turned the same way (either all up or all down) he will tell you immediately and you win the game. Otherwise he spins the table randomly again, and you get another turn of feeling/adjusting the position of 2 glasses.

This continues indefinately - or until you win the game.

Now - it is fairly obvious that you have a chance at winning the game just by going on and on until you win. However, the question for you to answer is this:

Is it possible given these rules to guarantee a victory in a certain number of turns? And if so, how many turns?

First of all, I would say, "no" there is no way you can guarantee a victory in a certain number of turns...

Here's why:

You can assure that 3 of the 4 are all the same direction after only 2 table turns:

First turn, just feel any two glasses that are next to each other...make them the same (both up or both down)...

Second turn, feel any two glasses that are diagonal from each other...make them the same as what you chose in the first turn (one of the glasses is from your first turn)...

The third turn is the problem. Since the table is spun RANDOMLY, there is a chance that the configuration will always end up the same (but it may not)...so you can't just pick any two next to each other to find the last glass to turn the correct direction...you also can't just pick any two diagonal because you may get the same two diagonal that you've already done...

I will say that the odds are VERY VERY slim that it would take more than 4 or 5 turns, but it's definitely possible, and so as a result, I can't guarantee it...

HAHA...AI beat me to the EXACT same solution (if you read both of ours, they are almost identical)...nice...

Edited April 2, 2009 by Pickett

[Guest]

Actually, I was thinking about this problem in the bathroom and I think I came up with a way to guarantee it in 5 moves.

After the first spin, select two diagonal glasses and turn them both down.

After the second spin, select two glasses next to each other on one side. One of them is guaranteed to be one of the diagonal glasses from the first move, so it should be down. If the other glass is also down, then it means the fourth glass you haven't touched yet is up. In this case, keep both glasses turned down. If the other glass was up however, then we can't yet know what position the fourth glass is in. Turn the up glass down if this is the case and spin again. If you don't win after this spin, then you know that 3 of the glasses are down and 1 is up.

Next, select two diagonal glasses again. If one is up and one is down, then turn the up one down and you win. If they're both down however, then turn just one of them up so that you have two up glasses next to each other and two down glasses next to each other.

After the fourth spin, select two glasses next to each other on one side. If they are the same, then switch them both. This way they'll match the other two glasses. If the glasses are different, then also turn them both. This way you'll have two up glasses and two down glasses, but the similar glasses will be diagonal from each other now, and not next to each other.

Finally, spin again and this time choose diagonal glasses. They are guaranteed to be in the same orientation, so flip them both and you win!

[Guest]

Doesn't it all depend on how the glasses start initially? Say all glasses are upright and one is upside down. If, by chance, you chose to right the upside down glass on the first spin, you win. On the other hand if you feel the upside down glass and choose to flip one glass that was upright, everything changes. I don't know enough statistics to find any guarantee, but there are a lot of variables to account for.

[Guest]

Actually, I was thinking about this problem in the bathroom and I think I came up with a way to guarantee it in 5 moves.

After the first spin, select two diagonal glasses and turn them both down.

After the second spin, select two glasses next to each other on one side. One of them is guaranteed to be one of the diagonal glasses from the first move, so it should be down. If the other glass is also down, then it means the fourth glass you haven't touched yet is up. In this case, keep both glasses turned down. If the other glass was up however, then we can't yet know what position the fourth glass is in. Turn the up glass down if this is the case and spin again. If you don't win after this spin, then you know that 3 of the glasses are down and 1 is up.

Next, select two diagonal glasses again. If one is up and one is down, then turn the up one down and you win. If they're both down however, then turn just one of them up so that you have two up glasses next to each other and two down glasses next to each other.

After the fourth spin, select two glasses next to each other on one side. If they are the same, then switch them both. This way they'll match the other two glasses. If the glasses are different, then also turn them both. This way you'll have two up glasses and two down glasses, but the similar glasses will be diagonal from each other now, and not next to each other.

Finally, spin again and this time choose diagonal glasses. They are guaranteed to be in the same orientation, so flip them both and you win!

Looks good to me. Nice job

[Guest]

Doesn't it all depend on how the glasses start initially? Say all glasses are upright and one is upside down. If, by chance, you chose to right the upside down glass on the first spin, you win. On the other hand if you feel the upside down glass and choose to flip one glass that was upright, everything changes. I don't know enough statistics to find any guarantee, but there are a lot of variables to account for.

Ya, you're right about that, which is why we needed a solution to guarantee a win in the least number of moves. So even using my method you could win after the first move if you're really lucky, but since the question wants a guaranteed win, we have to assume the worst possible scenario after each move.

[Pickett]

Ya, you're right about that, which is why we needed a solution to guarantee a win in the least number of moves. So even using my method you could win after the first move if you're really lucky, but since the question wants a guaranteed win, we have to assume the worst possible scenario after each move.

Ah, well played...I agree with this...I can't think of how to guarantee it in less moves...nicely done

[Guest]

Actually, I was thinking about this problem in the bathroom and I think I came up with a way to guarantee it in 5 moves.

After the first spin, select two diagonal glasses and turn them both down.

After the second spin, select two glasses next to each other on one side. One of them is guaranteed to be one of the diagonal glasses from the first move, so it should be down. If the other glass is also down, then it means the fourth glass you haven't touched yet is up. In this case, keep both glasses turned down. If the other glass was up however, then we can't yet know what position the fourth glass is in. Turn the up glass down if this is the case and spin again. If you don't win after this spin, then you know that 3 of the glasses are down and 1 is up.

Next, select two diagonal glasses again. If one is up and one is down, then turn the up one down and you win. If they're both down however, then turn just one of them up so that you have two up glasses next to each other and two down glasses next to each other.

After the fourth spin, select two glasses next to each other on one side. If they are the same, then switch them both. This way they'll match the other two glasses. If the glasses are different, then also turn them both. This way you'll have two up glasses and two down glasses, but the similar glasses will be diagonal from each other now, and not next to each other.

Finally, spin again and this time choose diagonal glasses. They are guaranteed to be in the same orientation, so flip them both and you win!

Excellent job AI. This was the solution I was looking for. This does not necessarily mean that it is the only - or best - solution. But at least I don't know any better.

/Uhre

[Guest]

There is no minimum. There are 2 random elements 1 the spin and 2 your selection. You could keep selecting the 2 of the 3 you have turned upside down. Very unlikely to keep happening but its like red and black in a casino if its been red 20 times the chance of it being black is the same as red.

[bonanova]

Great puzzle. One that seems impossible until you think about it for a while.

This was posted a while back here.

Looks like it was solved both times.

Good job.