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        riddles >> easy >> Russian Table Problem
        
(Message started by: Ace_T on Mar 7^th, 2005, 5:21pm)

Title: Russian Table Problem
Post by Ace_T on Mar 7^th, 2005, 5:21pm

(I originally heard of this by this title)

Background: You have a (very thick) table that has four holes cut into the corners, each deep enough to hold a glass (but not cut through the table!). In each hole there is placed a glass either right side up or upside down. The holes are then covered so you can't see the glasses. The table can rotate around its centre.

Problem: Your task is to ensure that the glasses are either all right side up or all upside down. You do this in 'turns'. For any one turn the table is first spun a random number of times (you don't know how many). You can then pick any two corners and can look at the glasses in them. You can replace them in the holes any way you wish (either both right side up, both upside down, or one up and one down, or perhaps 'reverse' the way they were). The holes are then covered again and the table is again spun a random number of times and you are on to the next turn. If at any time you get all four glasses either upright or upside down a buzzer will sound automatically and you win.

QUESTION: How many turns does it take to ensure (100%) that you win?

Note (and a bit of a hint): Since you don't know how much the table is spun, there are really only two choices you can make at any turn - 'adjacent' or 'opposite' - to describe when you pick two corners that are side-by-side, or two corners that are across from each other respectively.

Title: Re: Russian Table Problem
Post by SWF on Mar 7^th, 2005, 7:08pm

This problem is already on the board at Arrange hidden coins on a table (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1096746708) and Blind Bartender (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1028574222)