Logic Problem that is Making Me Angry. (Hence the angry face)

[BluffTHIS!]

The purpose of the problem is to find a workable strategy, and using the switches as a means of communication is not stated to be necessary. If 2+2=4 is too easy for you, then feel free to make it 2^2=4 or 1+1+1+1=4 or whatever floats your boat.

[jason_t]

[ QUOTE ]
The purpose of the problem is to find a workable strategy, and using the switches as a means of communication is not stated to be necessary. If 2+2=4 is too easy for you, then feel free to make it 2^2=4 or 1+1+1+1=4 or whatever floats your boat.

[/ QUOTE ]

The purpose of the problem is obviously to use the switches.

[Quaalude]

All Praise Google:

Answer

Though I personally prefer something more like what BluffTHIS! mentions. The problem does not specifically state that one must use the switches, so anything is fair game. I was thinking just draw a line on the wall, or sign your name. Since the warden states that noone else enters the room, you don't have to worry about painters and the like. This method also guarantees the quickest freedom. Screw the switches.

[[censored]]

If I was the warden they would be 100% certain to get fed to the alligators

[PairTheBoard]

I read the googled link hint down to taking a long term view, a 6th grader could figure it out, and do it for 3 first. I did do it for 3 before reading the hint. 3 is easy. If the switches are both on or both off when you enter the room you have no choice but to leave them one on, one off. But if they're in the neutral position you can either leave them both on or both off. Let one person be an "off" person and the other two be "on" people. If I'm one of the "on" people then if I ever visit the room after the first time and find the lights off I'll know the "off" person has been there. If I ever visit the room after a visit where I left the lights neutral and find them on I'll know the other "on" person has been there. Either myself or the other "on" person can go to the warden after one of us knows the other two have been there.

I can't see how to extend this to more than 3 people though.

I never did understand parity very well, but what I know of it, it doesn't seem easily applied to just two switches. Also, how could a 6th grader know about parity?

I realize something might be done using the A,B labels on the switches but I don't see what.

I must sleep now.

PairTheBoard

[jason_t]

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Also, how could a 6th grader know about parity?


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It's not parity. It's counting. I don't know what made me type parity in my hint in white above.

[Jman28]

Thanks you.

I really don't like that answer, but I guess it's the right one. I wasn't thinking outside the box at all. I never considered a case where the counter (which I did have) visited the room nearly that often.

[SL__72]

Why 23 people? The only solution I could come up with doesn't change based on the number of people (other then it would take longer).

Am I missing something?

[CallMeIshmael]

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I have a solution. It's possible.

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Please post it or PM me. (although I'm about 65% sure you're just trying to piss me off more and I was right)

[/ QUOTE ]

You're terrible with probability assessments. I assure you I have found a correct solution to this problem and I hoped that by telling you that you would be able to find a solution yourself.

Hint in white below:
<font color="white">Use an idea involving parity.</font>

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Jason_t pwns logic. And, when he says he has the solution, he almost certainly has one.


You'd think he'd be better at poker

[jason_t]

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Why 23 people? The only solution I could come up with doesn't change based on the number of people (other then it would take longer).

Am I missing something?

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No, the solution has nothing to do with 23. It's not even a red herring.