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Apr 13, 2017 at 12:50 history edited CommunityBot
replaced http://puzzling.stackexchange.com/ with https://puzzling.stackexchange.com/
Jun 16, 2015 at 21:13 vote accept Christian Semrau
Jun 15, 2015 at 14:31 answer added user3294068 timeline score: 2
Jun 12, 2015 at 6:35 history edited Christian Semrau CC BY-SA 3.0
name the puzzle author
Jun 8, 2015 at 5:27 comment added Moti In the edit this was clarified!
Jun 7, 2015 at 18:47 history edited Christian Semrau CC BY-SA 3.0
minor rewording, hoping to clarify the rules
Jun 7, 2015 at 18:44 comment added Christian Semrau As stated in the question: The "normal" balls are all of the same weight. I'm sorry if this was not clear enough. The two heavy balls may be of different weight, but the other balls all weigh the same.
Jun 7, 2015 at 16:48 comment added Moti I believe you mean to say that ball 2 is heavier than 1, but it is heavy only if the other balls (apart from the heavy two) are equal in weight! Is this the case? In your question, you need to clearly say that the rest of the balls have the same weight! - I assume this is the case.
Jun 7, 2015 at 8:23 comment added Christian Semrau I'll give a partial example. Suppose you have 4 balls and put ball 1 on pan A, 2 on B, 3,4 on C. If A is lightest, you know that ball 2 is heavy, but do not yet know about ball 1 (it could be heavy, but lighter than 2). Similar for B being lightest. It is possible that pan C is lightest, meaning that balls 1 and 2 are both (very) heavy. If no pan is lightest, either 1 and 2 are both normal, or are both (equally) heavy. In case A, put 1 on A, 3 on B, 2,4 on C. Now only one of 1 and 3 can be heavy (because 2 is), so either pan A or B is lightest (meaning 3 or 1 is heavy), or none (4 is heavy).
Jun 7, 2015 at 8:20 comment added Christian Semrau The weighing procedure to be found has to work in all cases of weight: The two heavier balls weighing the same, or differently, being only slightly heavier or much more.
Jun 7, 2015 at 0:20 comment added Moti Your first comment is quite significant. You are not clear if the other balls are equal in weight. In any case it seems that you could only measure three balls in a time to get meaningful results, to start with.
Jun 5, 2015 at 16:09 comment added Christian Semrau The LPDR always applies: As long as there is a single lightest pan, it will go up, no matter how many balls are on each pan. If you think about it, "this is not how physics works", but I think it's a valid imaginary machine. :-)
Jun 5, 2015 at 16:04 comment added Christian Semrau All three pans are on the balance at the same time. A hanging balance with 3 arms would behave different from this LPDR balance, so instead imagine three pans in a row, sitting on a hidden weighting mechanics, like so: three-pan balance
Jun 5, 2015 at 14:47 comment added Engineer Toast Please to clarify: Does the LPDR apply no matter the weight of the balls on the scales? I.E., if there is a single lightest pan, it will go up no matter how much weight is placed upon it?
Jun 5, 2015 at 14:46 comment added Engineer Toast Please to clarify: Are all three pans on the scale at the same time (I.E., Is it a classic hanging balance with 3 arms in a triangular pattern instead of just 2) or can the scale only hold 2 pans at a time and you must change them out?
Jun 5, 2015 at 7:36 comment added Christian Semrau I do not know if it makes a difference to the puzzle, but: You may not assume that the heavier balls are just slightly heavier than a normal ball, but they may be arbitrarily heavy (even more than n times a normal ball). Imagine having an assistant move the balls, so that you never get to tough them and feel their weight.
Jun 5, 2015 at 7:28 history asked Christian Semrau CC BY-SA 3.0