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OriginalGriff3-Dec-13 23:24

3-Dec-13 23:24

One of them is elected as the only person who can say "We have all visited". He is the only one who changes the left switch from down to up. As each other prisoner enters the room, if the left switch is up, and he has visited the room but not moved the left switch, he moves it down. Otherwise, he moves the right switch.
When the "special" prisoner enters the room, if the left switch is down, he changes it to up, and adds one to his mental count. Otherwise he also moves the right switch.
When his mental count reaches 23, all of the prisoners have been in the room.

Nagy Vilmos3-Dec-13 23:26

Nagy Vilmos

3-Dec-13 23:26

Almost right. What if the left switch was down to start with?

speramus in juniperus

Motor Viper3-Dec-13 23:38

Motor Viper

3-Dec-13 23:38

Shouldn't matter should it, counting to 23 is one more than required since he does not have to count himself and so will take into account the switch starting down.

Nagy Vilmos4-Dec-13 0:17

Nagy Vilmos

4-Dec-13 0:17

I'm going to retract my original answer as this may be right.

The method I have is similar but each prisoner switches the left up twice and when the caller gets to 44 he says everyone has visited, this gets over the left being up to start with. But I think that you may have beaten the maths brains of IBM[^] Big Grin | :-D

speramus in juniperus

OriginalGriff4-Dec-13 0:30

4-Dec-13 0:30

The trouble with that approach is that 2 prisoners could visit 44 times, and 21 of them sit on their hands waiting to die... Laugh | :laugh:

Nagy Vilmos4-Dec-13 0:35

Nagy Vilmos

4-Dec-13 0:35

No it doesn't, each prisoner only switches up TWICE, so your two visitors would only raise the switch FOUR TIMES.

speramus in juniperus

OriginalGriff4-Dec-13 0:37

4-Dec-13 0:37

Maybe I need another coffee!

CBadger4-Dec-13 0:41

CBadger

4-Dec-13 0:41

Griff was close but is foresee another problem.

I will agree they choose one person that will change a specific switch to a specific value and no one else ever will. So now lets say they choose that the prisoner counting changes the left switch ON. Now when anyone enters the room they can switch the Left switch OFF only 2 times. So when it is ON they switch it OFF until they did that 2, only the counting prisoner will ever switch the Left switch ON, and every time he does that he counts the number of times he does that. Now being that there is 23 prisoners (- 1 as the counter is excluded), the counter should switch the left switch ON 41 times. Because under worst case scenario if the first person were to switch OFF the left switch before the counter's first visit, when he counts to 41, most would have switched it twice, if not it would mean that once person only switched it once. $Shucks | :-\$

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Motor Viper4-Dec-13 0:52

Motor Viper

4-Dec-13 0:52

I still think Griff is correct. The first time the counter goes in he effectively resets the problem to the one where the left switch starts in the up position, however he does not know if anyone has been in before him so he would not start counting until next time he visits, which is effectively the same as starting the count immediately but adding an extra 1.

CBadger4-Dec-13 1:34

CBadger

4-Dec-13 1:34

but the dilemma remains in the counting, that is why I say he has to count to double the amount of prisoners excluding the counter times two. Safer and more secure letting everyone do it twice at so that worst case will be when someone went in and changed it before the counter's first visit, the worst case scenario would be that one person only changed it once instead of twice. Now if you are only going to do it once you will run into the dilemma of miscounting one...

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Freak304-Dec-13 1:41

Freak30

4-Dec-13 1:41

But if the left switch starts in the up position, the "counting" prisoner will never reach 23, because the other 22 prisoners will only move the switch down once each.

I think this is the reason to let everyone switch two times at most. This way you have a safe margin of 1.
- If the switch was down at the beginning and the "counting" prisoner enters first, he will reach 44 after 1 prisoner has moved the switch down once and the others twice.
- If the switch was up at the beginning and the "counting" prisoner enters first, he will reach 44 after all other prisoners have moved the switch down twice.
- If the switch was down at the beginning and another prisoner enters first, the "counter" will reach 44 after all other prisoners have moved the switch down twice.
- If the switch was up at the beginning and another prisoner enters first, the "counter" will reach 44 after all other prisoners have moved the switch down twice.

The good thing about pessimism is, that you are always either right or pleasently surprised.

CBadger4-Dec-13 2:08

CBadger

4-Dec-13 2:08

Exactly

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Mohamad M. Mohamad4-Dec-13 2:12

Mohamad M. Mohamad

4-Dec-13 2:12

Use a pen to mark the count each time one enter the room write on wall and add the counter, once the counter reach 23 they are all visited.

My Name in English[^] Translation done By OriginalGriff

Dave has decorated for X-mas, I see...

PIEBALDconsult4-Dec-13 4:10

PIEBALDconsult

4-Dec-13 4:10

They all agree to call the same lawyer and tell him they were there.

Edit:
0) The warden would likely lose his job if he carried out the promise.
1) There is no guarantee that the warden will ever have any of the prisoners enter the room.
2) There is no guarantee that the warden will have them enter in a time frame that precludes the death of any of the prisoners.
3) There is no guarantee that the warden (or others) won't mess with the switches.

"The only way to win is not to play." -- WOPR

Johnny J.3-Dec-13 22:22

Johnny J.

3-Dec-13 22:22

Proof[^]

Anything that is unrelated to elephants is irrelephant
Anonymous
-----
Do not argue with an idiot. He will drag you down to his level and beat you with experience
Greg King
-----
I always wanted to be somebody, but now I realize I should have been more specific.
Lily Tomlin, Actress

Re: Dave has decorated for X-mas, I see...

Dalek Dave3-Dec-13 22:40

Dalek Dave

3-Dec-13 22:40

That is so cool!

Michelle would go mad if I did that though.

---------------------------------
Obscurum per obscurius.
Ad astra per alas porci.
Quidquid latine dictum sit, altum videtur.

Re: Dave has decorated for X-mas, I see...

Johnny J.3-Dec-13 23:05

Johnny J.

3-Dec-13 23:05

Speaking of Michelle[^] Wink | ;)

Re: Dave has decorated for X-mas, I see...

Johnny J.4-Dec-13 2:04

Johnny J.

4-Dec-13 2:04

More Dalek X-mas[^]

CCC 4/12/13

Dalek Dave3-Dec-13 22:01

Dalek Dave

3-Dec-13 22:01

"Watery, airy, fiery, earthy and straightforward."(10)

---------------------------------
Obscurum per obscurius.
Ad astra per alas porci.
Quidquid latine dictum sit, altum videtur.

OriginalGriff3-Dec-13 22:09

3-Dec-13 22:09

Watery, airy, fiery, earthy

and

straightforward

ELEMENTARY my dear Watson! Laugh | :laugh:

DaveAuld3-Dec-13 22:16

DaveAuld

3-Dec-13 22:16

What about the 5th Element......Milla, or should I say LeeLoo Rose | [Rose]

Dave
Find Me On: Web|Facebook|Twitter|LinkedIn|GitHub

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OriginalGriff3-Dec-13 22:27

3-Dec-13 22:27

Ooo - I'd forgotten her! Now I have to dig the carpet fluff off the back of my tongue! Laugh | :laugh: