On Mon, Feb 17, 2003 at 09:13:55AM -0500, David Hobby wrote: > I'm not sure what you mean by the loop, since you don't say > how it relates to the event 'X goes to the room'. (Also, it doesn't > seem to actually be a loop since I see no sign that it gets executed > repeatedly. Maybe you could rewrite this with 'on(roomvisit) do...', > or somesuch? : ) )
That's what I meant. > But I do read it as saying that X toggles switch B on every > visit. Yes. > So if the one R needs to visit the room again to catch up, To catch up? Catch up what? > the warden sends X to the room until switch B is down. Then R can go > 100 times in a row, or whatever. If the people are chosen randomly and go repeatedly to the room, then eventually > (We are trying for a guaranteed solution. A probabilistic one > is not good enough.) This is vague. Obviously, if the warden knew the strategy he could always thwart it if he wished. The warden must be assumed to be using a random-number generator to have any hope of having the problem be well-defined. And at random, eventually every R must see the B switch in the up and down position because X keeps getting chosen randomly to go in and flips B every time. Can you focus on why you think my solution is probabilistic and yours is not? It seems to me your objection applies equally to your solution. If one particular guy gets led in 100 times in a row, no progress is made until the random choosing of prisoners resumes, in either case. My solution is not probabilistic in the sense that when X declares everyone has visited, (if no one screwed up) then there is exactly 0 chance he is incorrect. And it must happen eventually, if the prisoners are chosen at random. It seems one of us has a mental block or is not understanding the other. Can someone else contribute here? -- "Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.net/ _______________________________________________ http://www.mccmedia.com/mailman/listinfo/brin-l
