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blind bartender


Best Answer plasmid, 30 September 2013 - 04:34 AM

Very neat problem. The answer is certainly unintuitive.

Spoiler for Checkmate in 7

Edit: renamed states for clarity Go to the full post


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#1 BMAD

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Posted 28 September 2013 - 06:30 AM

The following game is played between a customer and a bartender. The customer places four glasses on a revolving tray, arranged in a circle. Each glass is either right-side-up or upside down. The bartender is blindfolded and cannot see which way the glasses are placed, but the goal is to turn all the glasses the same direction.
In each round, the tray is spun, and the bartender is allowed to touch only two glasses, turning over either or both of them. But the bartender does not know the orientation when he touches his glasses.  After each round, the bartender is told if all glasses are oriented the same and the game is over. What is the best strategy for the bartender? Is there a maximum number of moves, after which the bartender can be certain all the glasses are identically oriented?

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#2 superprismatic

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Posted 29 September 2013 - 04:55 PM

Spoiler for well....


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#3 phil1882

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Posted 29 September 2013 - 05:11 PM

Spoiler for you sure?

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#4 phil1882

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Posted 29 September 2013 - 05:43 PM

Spoiler for my suggestion

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#5 superprismatic

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Posted 29 September 2013 - 06:33 PM

Spoiler for you sure?

I took the statement " Each glass is either right-side-up or up side down"  to mean that all the glasses were all in the same orientation to start with.    I thought it must have been there to emphasize that no glasses where placed on their sides, which would, of course, be stupid.  But, that makes it too easy, I suppose.  I haven't yet considered the case where of an arbitrary beginning setup.  Thanks for pointing that out, phil.


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#6 BMAD

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Posted 30 September 2013 - 12:12 AM

 

Spoiler for you sure?

I took the statement " Each glass is either right-side-up or up side down"  to mean that all the glasses were all in the same orientation to start with.    I thought it must have been there to emphasize that no glasses where placed on their sides, which would, of course, be stupid.  But, that makes it too easy, I suppose.  I haven't yet considered the case where of an arbitrary beginning setup.  Thanks for pointing that out, phil.

 

If all the glasses were already in the same orientation (all up or all down), the game would be over before it started.


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#7 superprismatic

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Posted 30 September 2013 - 12:32 AM

 

 

Spoiler for you sure?

I took the statement " Each glass is either right-side-up or up side down"  to mean that all the glasses were all in the same orientation to start with.    I thought it must have been there to emphasize that no glasses where placed on their sides, which would, of course, be stupid.  But, that makes it too easy, I suppose.  I haven't yet considered the case where of an arbitrary beginning setup.  Thanks for pointing that out, phil.

 

If all the glasses were already in the same orientation (all up or all down), the game would be over before it started.

 

But I assumed that the bartender had to change at least one glass on his first move.  Then the spin would cause a problem for him.


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#8 plasmid

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Posted 30 September 2013 - 04:34 AM   Best Answer

Very neat problem. The answer is certainly unintuitive.
Spoiler for Checkmate in 7

Edit: renamed states for clarity

Edited by plasmid, 30 September 2013 - 04:39 AM.

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#9 DeGe

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Posted 30 September 2013 - 04:01 PM

Spoiler for Possible strategy


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#10 radio1

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Posted 01 October 2013 - 04:09 PM

Very neat problem. The answer is certainly unintuitive.

Spoiler for Checkmate in 7

Edit: renamed states for clarity

Hmmm... does your "checkmate in 7" work if all four glasses are initially in the same state, or would that make it a "checkmate in 8", or ??


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