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The Robot Chess Convention

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#1 unreality


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Posted 17 August 2007 - 05:58 PM

here's a problem i made up:

There are a bunch of robots playing chess in a convention. It takes infinite amount of time for two equal robots to beat each other. It cannot happen. (All the robots are perfect logicians, by the way.) Anyway, the Central Robot (who isnt playing chess), who gives the orders to the other robots, makes an announcement that some (at least one) of the robots playing are of the Lesser Enemy Robot, also called LER's. However the LER's arent playing each other, their goal and the reason they are there is to beat the normal robots at chest. So no two LER's are playing each other.

Instantly all the normal robots can pick out the LER's from the normal robots, EXCEPT IN THE CASE OF THEIR PARTNER, because they are hidden from their partner for fair play, so the robots cannot tell if their own partner is a LER or normal.

However if you DO know your opponent is an LER, they are Lesser, thus the normal robots can easily pull a checkmate on them.

However nobody knows if their own partner is an LER, and discussion between robots is illegal in the Chess Convention. Even the Central Robot doesnt know, all he knows is there's at least one. So, being the cleverest robot of all, he tells all the robots: "If you know your partner is an LER, checkmate him at the end of every hour-and-a-half timer buzz."

Thirty-four point five hours go by of utter silence. Then, suddenly, multiple checkmates happen at once. The Central Robot grins (if a robot can grin...?). How many LER's had snuck into the Chess Convention?
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#2 Riddari


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Posted 17 August 2007 - 08:12 PM

This puzzle is extremely similar to the one about Queen Henrietta from Atlantis, entitled Josephine.

If I did not mess up my logic the answer is 23.

Each robot can see all of the LERs except the one they might be playing. So, each robot knows that there are either as many LERs as they can see or one more than that. They also know that every other robot recognizes this fact.
If there were only one LER, his opponent would not see any other LERs and would know that his opponent is an LER, since he was told there was at least one, and would buzz on the first hour-and-a-half timer.
If there were two LERs, the two LER opponents would expect the other to buzz on the first hour-and-a-half timer unless they could also see an LER, which would mean there own opponent was an LER.
Etc. Etc until the 22rd hour-and-a-half timer passes with no robots buzzing, meaning that none of the robots see only 21 LERs implying that their own opponent is an LER. Then all the robots who can see 22 LERs know that the other robots with LER opponents must also see 22 LERs, meaning there are 23 LERs.
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#3 unreality


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Posted 18 August 2007 - 05:58 PM


good job!
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