On Thu, Feb 13, 2003 at 05:41:26PM -0500, David Hobby wrote:
> Erik Reuter wrote:
> >
> > I've been thinking about this in my spare moments for the past few
> > days and I think I have a solution. Hopefully there are no bugs in my
> > strategy (please let me know if there are). Here it is:
> >
> > Choose one prisoner, call X
> >
> > Rest of prisoners called R
> >
> > -------------------
> > Strategy for X:
> >
> > Initialize (in X's memory) a signal counter S = 0
> >
> > IF NOT X's first time in room THEN {
> > IF state of switch B has changed where X left it last THEN {
> > IF switch B changed from on to off THEN yell "Shit!" break;
> > ELSE IF switch B changed from off to on THEN increment S
> > }
> > }
> >
> > IF S >= 22 THEN declare "Everyone has visited" [much celebration];
> >
> > TOGGLE switch B and remember its state
> > ----------------------
> >
> > ----------------------
> > Strategy for all R's:
> >
> > Note the position of switch B
> >
> > IF you have seen switch B in an up state on a previous visit
> > AND you have NOT ever toggled switch B
> > AND switch B is currently in a down state
> > THEN TOGGLE switch B
> > ELSE TOGGLE switch A
> > ------------------------
>
> Erik--
> It seems to me that there could be Rs who never
> "checked in". Couldn't an R keep visiting the room and
> never see B in the up state? Then they would always do
> the ELSE clause, S would never get to 22, and they would
> all stay there indefinitely.
Since X always TOGGLES switch B before he leaves, it will constantly be
going on/off/on/off each time X visits. So eventually the R's must see
it on, right? Or am I missing something?
--
"Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.net/
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