tag:blogger.com,1999:blog-4459360717297142573.post3883525335932334476..comments2024-03-27T20:44:56.139+11:00Comments on chessexpress: Seen in Dubai AirportShaun Presshttp://www.blogger.com/profile/00897215011002594039noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-4459360717297142573.post-68566403123221355172010-10-09T23:02:56.351+11:002010-10-09T23:02:56.351+11:00OK. I've seen a solution to this problem (as I...OK. I've seen a solution to this problem (as I eventually gave up) and while it is along the lines of what I was thinking (and what has been discussed here), it is probably 1 level of complexity beyond what I was looking at. Here is a tip: It helps if you label the balls.Shaun Presshttps://www.blogger.com/profile/00897215011002594039noreply@blogger.comtag:blogger.com,1999:blog-4459360717297142573.post-51323391562617084682010-10-05T23:03:13.487+11:002010-10-05T23:03:13.487+11:00But heavier isn't given, the different ball ma...But heavier isn't given, the different ball may be lighter.<br /><br />Thinking a long way back, I think the solution begins with dividing into three groups of four balls each.<br /><br />Weigh one group against another, if equal, then the different ball is one of the unweighed balls. Which should be easy to distinguish with the remaining weighings.<br /><br />The difficult part is finding the odd ball when it is among the two original groups. This involves some switching of balls between the two groups in the remaining weighings, but I don't recall the exact procedure nor can work it out at the moment. So, I had a won game, but resign here :)Phil Bourkehttps://www.blogger.com/profile/01984479449044393181noreply@blogger.comtag:blogger.com,1999:blog-4459360717297142573.post-34667737436951111452010-10-05T21:23:43.081+11:002010-10-05T21:23:43.081+11:00Hello Shaun.
For the 12 balls and a balance beam.
...Hello Shaun.<br />For the 12 balls and a balance beam.<br />Measure 6 a side, then retain the heavier 6.<br />Do the same for 3 balls each.<br />Out of the remaining 3, pick 2 at random. If they are equal then the heavier is the remaining ball, otherwise you know the heavier ball with the one ball on easch side.EADhttps://www.blogger.com/profile/02575727262012961734noreply@blogger.com