Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!tut.cis.ohio-state.edu!ukma!husc6!ogccse!blake!blake.acs.washington.edu!tom From: tom@yin.cpac.washington.edu (Tom May) Newsgroups: comp.emacs Subject: Re: Blackbox Message-ID: Date: 3 May 89 23:28:18 GMT References: <8905022249.AA02054@dsys.icst.nbs.gov> Sender: news@blake.acs.washington.edu Organization: Center for Process Analytical Chemistry, U of Wash, Seattle Lines: 65 In-reply-to: rbj@dsys.icst.nbs.GOV's message of 2 May 89 22:49:31 GMT In article <8905022249.AA02054@dsys.icst.nbs.gov> rbj@dsys.icst.nbs.GOV (Root Boy Jim) writes: ? Regarding best scores, I believe the minimum score possible in ? order to secure a sure match is equal to the number of balls in ? the box. However, as this is an extreme case, most scores will ? be at least twice the number of balls. My lowest score has been 10. I don't see how you could score 4. It is possible to score zero by guessing the position of the balls without shooting any rays at all. I believe this is possible on VMS systems, because the random number generator's low bits aren't really random. Not much fun, though. ? Ps. Is there some inherent reason why the game is limited to an ? 8 x 8 box? It might be more interesting if the size were ? adjustable as well as the number of balls. It is possible that 8x8 may be the smallest size that can uniquely determine the four balls. Four balls is the maximum that may be placed in a box arbitrarily and be uniquely determined. Consider the pattern 0 - 0 x x x 0 - 0 A fifth ball placed on any of the x squares could no be distinguished. Actually, the 8x8 grid was probably determined by play testing. A larger grid with more balls gives too much opportunity to hide balls behind other balls. A larger grid with four balls would be too easy, as the chance of multiple-ball deflections would be decreased. A smaller grid would probably require less balls, again reducing the complexity somewhat. So 8x8 seems about optimum. The above example is actually not correct. Here is the what happens to all possible rays shot into the 3x3 grid (assuming the ball is placed at the leftmost x): H R H H 0 - 0 H H 0 - - R H 0 - 0 H H R H It can be seen that the fifth ball must be on the left side. Here's a tricky setup with four balls on an 8x8. The fourth ball can be at any of the positions marked x, and the results will be the same: - - - - - - - - - - - - - - - - - - - - - - - - 0 - - - - - - - - - - - - - - - x - 0 - - - - - - - - - - - - - x - x - 0 - - - Also, in case anyone cares, I believe that blackbox has an optional prefix argument which specifies the number of balls to use. Four is the default; the commercial version suggested 4 or 5. It should be possible to use 64 balls and get a score of zero . . . -- Tom May tom@yin.cpac.washington.edu