[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 599

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Mathemat:
This solution is correct if the light is initially off. But if they are on, there's a problem. This is where I get stuck.

then let it wait not 9 but 10 turns on :)

 

is it possible to construct a closure if a gating changes state on two occasions --

1. the first time

2. when it sees two state changes (e.g. went in for the first time turned on, went in again turned off, went in again turned on turned off)

?

 
I suspect not, then you just have to calculate the maximum.
 
PapaYozh:

They must choose one, let's call him "The Chosen One".

The Favourite will count the number of times the switch is in the ON position and make sure it is in the OFF position when they visit the room.

Each of the remaining 9 will only turn the switch ON once, it will never turn it OFF.

Therefore, once a Chosen One counts nine ONs, everyone has been in the room.

Clarification.

Each of the remaining 9 moves the switch to the ON position only twice, never to the OFF position.

Accordingly, once the Chosen One counts eighteen ONs - the room is guaranteed to have been visited by everyone.
 
PapaYozh:

Clarification.

Each of the 9 remaining switches only turns the switch to ON twice, never to OFF.

Accordingly, once the Chosen One counts eighteen ON's - the room is guaranteed to be visited by all.

That's what you didn't mention the first time. Yes, there is such an option. I haven't considered it yet. And that's even if the lamp was initially on?

sergeev: let it wait not 9 but 10 times :)

The 10th one won't wait: all 9 will already be counted (if the normal one switches on once).

P.S. Imagine that the Meter entered the room first (he doesn't know this) and the light was on initially. No one was there yet. What should he do? Adding one is wrong because no one was there.

 
Mathemat:

That's what you didn't mention the first time. Yes, there is such an option. I haven't considered it yet. And this is even if the lamp was initially on?


If the switch is in the ON position when you start, the moment the Chosen One counts to 18, one of the 9 will only be in the room once, and the others will be in the room twice.
 
PapaYozh: If the switch is in the ON position when you start, the moment the Chosen One counts to 18, one of the nine will only be in the room once, and the others will be in the room twice.

I mean, maybe twice, but only once... Right?

So it means that even if the Counter entered the room first (without knowing it), and the light was originally on, it still adds one, even though no one was there yet?

 
Mathemat:

So it turns out that even if the Counter entered the room first (without knowing it) and the lights were originally on, it still adds one, even though no one was there yet?

Of course it does.
 
Mathemat:

I mean, maybe twice, but only once... Right?

I meant that it is guaranteed to visit, i.e. it will leave a mark as a light.
 
Sometimes I look at this thread and I get the feeling that people who often solve complex problems are cycling through simple ones.
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