Date: 13 Dec 1985 16:09:04 EST (Fri) From: Dan Hoey To: Alan Bawden Subject: Re: Attractive nuisances Date: Thu, 12 Dec 85 03:08:51 EST From: Alan Bawden I would have guessed that a unique solution could not fail to start with a 1... (I don't know -why-, just a feeling.) I don't know if this is a corroborative counterintuitivity, but among the 12-space length 106 ruler(s), the only one found so far is (2 3 20 12 6 16 11 15 4 9 1 7) No solution starts with 1, and this is the only one starting with 2. Unfortunately, the machine crashed, and I haven't got checkpoint restart on this yet, so there won't be anything more definite this week. ...I always wind up speculating if there is something to be gained by looking for solutions mod P. I can't make that idea go anywhere unfortunately. By looking at solutions mod 2 I was able to devise an alternate way of proving that there are no perfect solutions for -some- of the larger numbers of marks, but big deal. Using this hint, I was able to prove that the number of spaces on a perfect ruler is either one less or one greater than a perfect square. Is this your result? I think this even-odd hack has strong possibilities of cutoffs for my program. Once when debugging I saw a situation where the program was investigating a situation where a lot of even marks had been put down. There were a lot of available marks, most all of them odd. And there were a lot of available distances, most all of them odd. The result was a fairly lengthy search through a doomed space. I will look at trying to turn this into a workable heuristic. I am also a little skeptical about being able to make use of larger P. When you mentioned this, I remembered a related problem that ocurred to me when I first read about rulers. What if the ruler is a circle? It seems solutions for a circular ruler would be more regular, with the elimination of edge effects, and might give insight into the straight ruler case. Dan