On Mon, Feb 17, 2003 at 12:20:36AM -0500, David Hobby wrote:
> Erik Reuter wrote:
> >
> > On Thu, Feb 13, 2003 at 05:41:26PM -0500, David Hobby wrote:
> > > 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.
> >
> > My description of X's strategy wasn't as clear as it could have been. I
> > should have explicitly indicated the "loop":
> >
> > Strategy for X:
> >
> > Initialize (in X's memory) a signal counter S = 0
> >
> > LOOP
> > IF NOT X's first time in room
> > AND state of switch B has changed from where X left it last THEN {
> > IF switch B changed from up to down THEN {yell "Shit!"; break;}
> > ELSE IF switch B changed from down to up THEN increment S
> > }
> >
> > IF S >= 22 THEN declare "Everyone has visited" [much celebration];
> >
> > TOGGLE switch B and remember its state;
> > END_LOOP
> >
>
> I think that's how I read it the first time.
> Look, when switch B is down, the warden takes one of the Rs
> into the switch room 100 times in a row. That should hold
> until the switch is down again. And X keeps toggling B,
> so it should be down again pretty soon.
You lost me. If the warden takes an R in for the first time with switch
B down, and then takes him back 100 times in a row, nothing will happen
during those hundred trips, true. But eventually X will be taken in
again, flip it up, then eventually R will see it up, then eventually X
will flip it down, then eventually R will see it down again. It has to
happen eventually.
--
"Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.net/
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