
Re: Please help me with the following question
Posted:
Mar 9, 2013 2:38 AM


That is a solution. What was the strategy.
Bob Hansen
On Mar 8, 2013, at 7:53 PM, GS Chandy <gs_chandy@yahoo.com> wrote:
> Responding to Robert Hansen's (RH's) post of Mar 8, 2013 9:15 PM (pasted below my signature), Written mainly at around 0400 hours; to be sent much later, whenever Internet becomes available again  right now it's not accessible: > > SPOILER! > === > I've not yet looked at the solution that Joe Niederberger seems to have posted. However, in what you (RH) have posted below, you have got it right that the 12 coins are divided into 3 groups of 4 each (A,B,C), and two groups (say A, B) are weighed against each other. > > That is 'Weighing I'  Groups A and B only. > > If A and B are equal in weight, then the 'odd' coin must be in group C  and the problem is reduced to finding the odd coin out of 4 in 2 weighings. See later. > > If groups A and B are NOT equal in weight, then the 'odd' coin is in either A or B  the difficulty arises because you don't know whether the 'odd' coin is heavier or lighter. > > But you DO know that all the coins in group 'C' are of the same weight! It is ESSENTIAL to use this bit of information. > > Back to groups A and B: > Suppose group A is lighter. Then you know the following: > > i) IF the 'odd' coin happens to be lighter, it will be in group A. > ii) IF the 'odd' coin is heavier, it HAS to be in group B. > iii) ALL coins in group C are of equal weight  and we shall use these as 'reference coins'. > > It is ESSENTIAL to use these bits of information as well. > > The crucial 'idea' is simply to take out 3 coins from group A (the lighter group), and 3 coins from group B (the heavier group) and put one of these sets of 3 coins aside. One must keep in mind whether you are putting aside coins from the 'heavier' or the 'lighter' group. > > Say we put 3 coins from group B (from the heavier scale pan) aside  call this group 'D' (3 coins only). If the 'odd' coin happens to be heavier, then it may be in D, or it may be the one coin left from group B. > > Now put the 3 coins from group A (lighter group) into the scale pan of group B. Put 3 coins from group C (all equal) into scale pan of group A. > > Weigh again  this is 'Weighing II'. > > Now, I have some preparation to do for a workshop (for which purpose I had risen early  it's about 4 AM Indian Standard Time now). Also I'm a bit bored. I believe it should now be possible to work out the rest of the problem with no hassles at all. The neat 'trick' (what in fact gave me the "EUREKA!" moment) was to take three coins out from each of groups 'A' 'B' and, keep one such set aside and put the other set into the other group's scale pan. > > Once that 'trick' is discovered, all of the rest of the problem more or less solves itself  you DO need to remember to use the reference weights of the coins in group C. > > Yes, there is the small problem left over of discovering the odd coin in a group of 4 coins (in group C), in case the coins in A and B are all of equal weight. This should pose no problems at all  am now in a hurry. > > The underlying point is that it requires a little 'outofthebox' thinking to get to that little trick. Reference to one of RH's other posts: I did NOT spend 2 whole days doing the problem  I did all my other stuff as usual; yes, the problem was always 'in mind', so to speak, or at the back of the mind. It did take about 2 days (or so I seem to recall) for that little trick to 'come to mind'. That was the 'EUREKA!' moment. > > There probably are several other ways to do this problem. Perhaps shall try and discover them later. > > The following had been written earlier: > > No, my father didn't show me the trick. He was delighted that I had solved it on my own. > > The real issue is that we ALL do have these powers. After all, we all learn how to recognize our parents, sit up, stand up, walk, talk  ANY of these skills is INFINITELY more complex than anything taught at MIT or Harvard. Our conventional systems (of thought and work) severely constrain us from accessing these powers in a natural way. > > I claim the OPMS can quite significantly help us: > > a) recognise that we all do possess these powers; > b) access them quite often; understand them a bit better; > c) not get frustrated because of the barriers and disappointments we all do encounter, most of the time in conventional circumstances. Frustration is the danger  because then the mind stops working in that crucial 'questionasking frame of mind'; > d) keep going to try and discover the way to overcome barriers; > e) do MANY other things essential in problemsolving situations (in particular, to retain the 'questionasking frame of mind' which I believe is the heart of the matter). > > No, of course I did not have OPMS in solving that little trick problem  OPMS came much, MUCH later in my life. However, the moment I thought of the OPMS as a 'system structure', I did recognize that it was a practical means enable us all to put the systems approach to work on issues of interest  exactly as I recognized the couple of little 'tricks' needed to solve the 12coins problem. > > As noted in one of my other posts on this thread, OPMS will NOT be **directly** useful to solve this problem. > > As mentioned in several other posts of mine, OPMS just helps us simply not lose that 'questionasking frame of mind', which is crucial. Barriers encountered can often lead us to become frustrated  which I believe severely detracts from or even totally destroys the 'questionasking frame of mind'. > > No  OPMS does NOT mean PERT Charts at all as RH very wrongly believes (though we certainly can develop PERT Charts using OPMS). I personally have very rarely EVER needed to create a PERT  because PERT is fundamentally insignificant from the point of view of enabling us to understand the 'systems' under consideration, whatever they may be. > > What IS important in a system (from the human point of view) is to arrive at an adequate understanding of how things done from day to day (including generation of relevant ideas) may 'CONTRIBUTE TO' accomplishment of the chosen Mission, WHATEVER it may be. > > GSC > ("Still Shoveling!") > Robert Hansen posted Mar 8, 2013 9:15 PM: >> >> On Mar 8, 2013, at 8:55 AM, GS Chandy >> <gs_chandy@yahoo.com> wrote: >> >>> It was successfully done in 3 steps by Joe >> Niederberger (JN) when he was around 10 years old. >> And by the underigned when he was 10 or 11 years >> s old. I guarantee you neither JN nor GSC used >> "binary search" or "exhaustive search" or anything of >> the sort. Those terms were not even imagined (or >> imaginable) by either of us at the time. What was >> actually used was the 'questionasking frame of mind' >> that is inherent in every real learner. >> >> I am just telling you the strategy I used. I don't >> know what your strategies were, both of you seem >> incapable of sharing them. "3 steps" is not a >> strategy. "Question asking" is not a strategy. Also, >> I am a software engineer by trade and work with other >> software engineers and when I use a phrase like >> "binary search" off the cuff, I do it out of habit >> because I am used to talking to practiced people. The >> last thing they would do or have to do is look up >> definitions in text books or on google. They just >> connect it to the experience like practiced people >> do. >> >> The problem BEGGED of a tree approach. After I >> realized that we did not know the relative weight of >> the target coin I refactored. >> >> 1. Break the coins into 3 groups of 4, A, B, and C. >> 2. WEIGH A v B, if they are the same then C has the >> coin, finish off C (trivial). >> 3. Break the remaining 8 coins into 3 groups of 3, 3 >> and 2, D, E and F. >> 4. WEIGH D v E, if they are the same then F has the >> coin, finish off F (trivial). >> 5. WEIGH D v {3 coins from C}. If D < C then D has >> the coin and it is lighter, finish off D in 1 step >> (trivial). if D = C then E has the coin and it is >> heavier, finish off E in one step (trivial). >> >> The issue of course is that step 5 requires 2 >> weighings, so the trick is to figure out how to make >> the comparison in step 4 and be left with 3 or fewer >> coins to test. You need step 4 because there is no >> strategy to insure that you have only 2 coins left, >> and if you have 3 coins left then you need to know >> which is heavier or lighter, as in the end of my >> first solution. I tried to close that gap for a >> couple of hours, and after looking up the answer, >> realized that I was pretty close. >> >> If a problem takes me more than 30 minutes to solve, >> I consider it pretty hard. You said it took you a >> couple days. Surely, after a couple of days you must >> have some sort of strategy to share with us? Or was >> it hit and miss? aka, exhaustive search. >> >> Bob Hansen

