So, you would recombine one of the good piles(PileA or PileB) with PileC to make 8 coins total and retest?
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So, you would recombine one of the good piles(PileA or PileB) with PileC to make 8 coins total and retest?
This is a second solution I've found, one which is much more dynamic than the first which I sent to Yves. I'll start you off: 1234v5678. If balanced, you can weigh 9(10)(11)v123 and go from there. If imbalanced, you can weigh 125v369 and go from there. See if you can finish my thought. Yves, msg me if you want me to send you this second solution (although I'll probably forget it by then...;)).
Well, it looks like your second solution is a bit closer to my original pen and paper one :)
Actually it's true that your solution to weigh 9,(10),(11) in one pile is actually nicer than mine where I weigh 1,9 vs (10),(11). But the net result is the same, you split the possibilities into 3, 3 and 2 :)
Nono :) Maybe my notation was misleading ;)Quote:
So, you would recombine one of the good piles(PileA or PileB) with PileC to make 8 coins total and retest?
A1 is the first coin from pile A. C1 is the first coin from pile C etc.
So I would weigh A1, C1 versus C2, C3 and leave C4 aside. I would only test 4 coins and not eight :)
But it's true that it's also possible by recombining to make 8 coins. After all, adding known correct coins on each side doesn't matter at all. This is what solarflare's original solution relies on (which is actually pretty nice, I have to say :) )
Thanks, Solarflare and Yves!
I understand it now:D
Theoretically, each weighing can have three outcomes: left, balanced, or right. There are originally 24 possibilities for the fake coin (12 coins, one of which is either heavier or lighter). You have three weighings, giving the possibility of narrowing it down 3^3(3 weighings, 3 possibilities each time)=27. Since 24<=27, the coin can be determined. It seems as though one of 13 coins could be determined using this information alone (26 possibilities < 27 possible outcomes) but don't be decieved. The first weighing presents a difficulty. Ideally, it should divide 26 possibilities into 8,9,or 9 possibilities. But the scale does not allow this: if balanced, the possibilities are either lighter or heavier for the coins not weighed, giving an even number of leftover possibilities. If unbalanced, there are an even number of coins on the scale (otherwise the scale would not work- same number on each side), and therefore an even number of possibilities leftover. So, the closest to 8,9,9 we can come with this scenario is 8,8,10. Since 10 isn't <= 3^2 (2 weighings left), this approach cannot always work. The only way it could work is if you had a special scale in which you could weigh 5 coins vs 4 coins, and it would tip one way if a coin on that side were heavier than average or a coin on the other side were lighter than average. This type of scale would not be a balance scale, however. In a balance scale, the side with 5 coins would always tip the scale, regardless of the location of the anomalous coin. Therefore, this problem cannot work with 13 coins unless the weighing rules were changed slightly. But it is still easily possible with 12 coins.
Hey Yves, I'm really beginning to like this problem, in case you haven't noticed ;).
I like your unbalanced scale idea ;)
Of course, you'd have to know the weight of a good coin first....
No you wouldn't, the scale would do all the work. That's why it doesn't work, because there's no such scale that does that.Quote:
Originally posted by gjs368
Of course, you'd have to know the weight of a good coin first....
Yves, got any more "interesting logical tasks" where that came from?
I know sort of a similar riddle. I think I'll post it here just for fun...
One day a big giant came down from his land to the land of the dwarves. There he captured x dwarves and took them back to his land to eat them. The giant, fortunately for the dwarves, was in a good mood and decided to give them 1 chance on freedom. Now the dwarves all have hats on but they don't know which color it is because their wives put them on, fortunately for you the dwarves only like blue and red so they only have blue and red hats. The giant told them: All of you stand in a line, you can't communicate with eachother and you don't know the colors of your hat, but, if all blue hats manage to step 1 step forward all at once and the red hats all stand still I'll let you all go free, I'll come look every morning if you'll do that, if you don't do anything at all I'll go back home and come back the next morning. If you do it wrong I'll eat you all.
Now the dwarves all went free. One day the giant came out and all the blue hats did a step forward. 2 things you need to know, the dwarves have to figure out for themselves which color their head is, no telepathy, kicking and stuff like that. Also, they all think the same solution, otherwise this won't work :-)
How did they do it?
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I forgot some stuff... Atleast both colors occur once :-) now the numbers don't manner pick any you like but if it helps you let's say 11 dwarves 6 blue ones and 5 red
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can they see each other? and if not, what do they know?
yea they can see eachother.
they know that if they make a wrong move they die, they all think the same solution, they all know there's atleast 1 of each color and they know that each day they do nothing, nothing happens. They do get free though.
Ha, I know this one, I'll PM you the solution :)
By the way, dimm_coder has (nearly) solved the twelve coins riddle without looking at the solutions solarflare and I provided here in the thread :) That is of course if we believe him ;)
Well, dimm_coder is the godfather of this thread. Show some respect. :cool:
Ok, so we believe him then ;)