A new mind twister
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This doesn't work because.... "take lighter or heavier N" Well, since you don't know if the odd-ball is lighter or heavier, you don't know which group to take. Tim Smith I know what you're thinking punk, you're thinking did he spell check this document? Well, to tell you the truth I kinda forgot myself in all this excitement. But being this here's CodeProject, the most powerful forums in the world and would blow your head clean off, you've got to ask yourself one question, Do I feel lucky? Well do ya punk?
Tim Smith wrote: Well, since you don't know if the odd-ball is lighter or heavier, you don't know which group to take. I was under the impression that you tested for one or the other, not both at the same time. :confused: Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country. -
6 and 6 take lighter or heavier 6 3 and 3 take lighter or heavier 3 1 and 1 take lighter or heavier if they balance it's the ball thats left ;) This Sig For Rent
Corky -
6 and 6 take lighter or heavier 6 3 and 3 take lighter or heavier 3 1 and 1 take lighter or heavier if they balance it's the ball thats left ;) This Sig For Rent
CorkyWon't work. What happens at stage two if the 3 & 3 balance? Paresh Solanki There is no substitute for genuine lack of preparation.
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You have 12 balls - one of which is either lighter or heavier than the others. Using a balance scale determine the odd ball and whether it is heavy or light in 3 weighings. This is much more difficult than it first appears. Have fun. Dave Huff 'tis a silly place!
Its the green one! No seriously, I have a solution that works for some of the cases depending on whether you randomly pick the odd ball, but not for the cases when you don't :( (yet) Roger Allen Sonork 100.10016 If I had a quote, it would be a very good one.
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Remember it is a balance scale so 6 vs 6 will have one side up and one side down but gives no indication if one side has a heavy ball or the other side has a light ball. Try again. Dave Huff 'tis a silly place!
But it will because the other eleven are all assumed to weigh the same. If each ball weighed a pound with the heavier one weighing two pounds, you'd be weighing 6 to 7. We know the heavier one is on the side that goes down. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country. -
You have 12 balls - one of which is either lighter or heavier than the others. Using a balance scale determine the odd ball and whether it is heavy or light in 3 weighings. This is much more difficult than it first appears. Have fun. Dave Huff 'tis a silly place!
Ok, I got it now. The special step is to not compare the last 6 balls against either other once you've got down to 6! I won't post the full solution. The others can sweat it out. Roger Allen Sonork 100.10016 If I had a quote, it would be a very good one.
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You have 12 balls - one of which is either lighter or heavier than the others. Using a balance scale determine the odd ball and whether it is heavy or light in 3 weighings. This is much more difficult than it first appears. Have fun. Dave Huff 'tis a silly place!
Measure four against four. Possible results: 1) weigh the same - take three balls from a measured four and three of the remaining and weigh: if the same measure remaining ball is the one - measure it against known ball and you have whether it is heavy or light. if different you know whether it is heavy or light because you know the three balls already measured are the same. Measure one of the three unknows against another. If the same the last ball is the odd man. If not the same the heavier or lighter ball can be determined from previous step. This is the easy part. What to do if the 4 vs 4 do not weigh the same is where it gets hard. Dave Huff 'tis a silly place!
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But it will because the other eleven are all assumed to weigh the same. If each ball weighed a pound with the heavier one weighing two pounds, you'd be weighing 6 to 7. We know the heavier one is on the side that goes down. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country. -
But it will because the other eleven are all assumed to weigh the same. If each ball weighed a pound with the heavier one weighing two pounds, you'd be weighing 6 to 7. We know the heavier one is on the side that goes down. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country.What if this odd ball weighs half a pound? We do not know whether the odd ball is lighter or heavier. Regards Thomas Finally with Sonork id: 100.10453 Thömmi
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Won't work. What happens at stage two if the 3 & 3 balance? Paresh Solanki There is no substitute for genuine lack of preparation.
It can't because only one ball is a different weight. 6 and 6 will never balance because the heaviest is on one side. Taking the 6 with the different weighted ball and splitting it guarantees one will be a different weight; therefore, they won't balance. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country. -
But it will because the other eleven are all assumed to weigh the same. If each ball weighed a pound with the heavier one weighing two pounds, you'd be weighing 6 to 7. We know the heavier one is on the side that goes down. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country.But the heavier one could actually be lighter (read the question) so its doesn't help Roger Allen Sonork 100.10016 If I had a quote, it would be a very good one.
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Ok, I got it now. The special step is to not compare the last 6 balls against either other once you've got down to 6! I won't post the full solution. The others can sweat it out. Roger Allen Sonork 100.10016 If I had a quote, it would be a very good one.
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But the heavier one could actually be lighter (read the question) so its doesn't help Roger Allen Sonork 100.10016 If I had a quote, it would be a very good one.
Are we testing for either circumstance or one or the other at a time? If it's either, well I'm going back to the drawing board. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country. -
Are we testing for either circumstance or one or the other at a time? If it's either, well I'm going back to the drawing board. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country. -
It can't because only one ball is a different weight. 6 and 6 will never balance because the heaviest is on one side. Taking the 6 with the different weighted ball and splitting it guarantees one will be a different weight; therefore, they won't balance. Jeremy L. Falcon "If you eat the cookie..." Homepage : Sonork = 100.16311
I care to not care about the care of not caring for the caring of a caring one's country.How? you do not know whether the different ball is lighter or heavier doing the 6:6 will tell you that something is different, but lets say you take the lighter group. If Odball is lighter, your plan will work, but if odball is heavier, then the 3:3 split will balance. you are stuck Paresh Solanki There is no substitute for genuine lack of preparation.
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You have 12 balls - one of which is either lighter or heavier than the others. Using a balance scale determine the odd ball and whether it is heavy or light in 3 weighings. This is much more difficult than it first appears. Have fun. Dave Huff 'tis a silly place!
- Put four balls in each scale to find a pile that contains normal balls. (If equal, then both weigthed piles, if not, the one not weigthted). 2) If eigth balls are found normal it is easy to find the odd one in two weighings. If not: Remove two ball from first scale, one from second scale and swap two of the remaining ones from scale to scale. Add one normal ball to have equal number of balls on each scale. Case 1: Scale is in balance - one of the three balls removed is the odd one. Case 2: Scale shifts - one of the two moved balls is odd. Case 3: Scale does not shift - one of the three unmoved balls are odd. 3.1) Place two of the odd balls from each scale on one of the scales. Balance with two normal balls on the other scale. Case 1: Scale in balance - The remaining ball is odd and it's status can be revealed from weighting 2). Case 2: The scale with the unknown balls are heavier/lighter - which one is odd can be determined from weighting 2) 3.2) Determine which is odd by comparing with a normal ball. 3.3) Do as 3.1) Is that the correct solution? /moliate
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Measure four against four. Possible results: 1) weigh the same - take three balls from a measured four and three of the remaining and weigh: if the same measure remaining ball is the one - measure it against known ball and you have whether it is heavy or light. if different you know whether it is heavy or light because you know the three balls already measured are the same. Measure one of the three unknows against another. If the same the last ball is the odd man. If not the same the heavier or lighter ball can be determined from previous step. This is the easy part. What to do if the 4 vs 4 do not weigh the same is where it gets hard. Dave Huff 'tis a silly place!
I already got that far:confused: Paresh Solanki There is no substitute for genuine lack of preparation.
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- Put four balls in each scale to find a pile that contains normal balls. (If equal, then both weigthed piles, if not, the one not weigthted). 2) If eigth balls are found normal it is easy to find the odd one in two weighings. If not: Remove two ball from first scale, one from second scale and swap two of the remaining ones from scale to scale. Add one normal ball to have equal number of balls on each scale. Case 1: Scale is in balance - one of the three balls removed is the odd one. Case 2: Scale shifts - one of the two moved balls is odd. Case 3: Scale does not shift - one of the three unmoved balls are odd. 3.1) Place two of the odd balls from each scale on one of the scales. Balance with two normal balls on the other scale. Case 1: Scale in balance - The remaining ball is odd and it's status can be revealed from weighting 2). Case 2: The scale with the unknown balls are heavier/lighter - which one is odd can be determined from weighting 2) 3.2) Determine which is odd by comparing with a normal ball. 3.3) Do as 3.1) Is that the correct solution? /moliate
moliate wrote: 2) If eigth balls are found normal it is easy to find the odd one in two weighings. If not: Remove two ball from first scale, one from second scale and swap two of the remaining ones from scale to scale. Add one normal ball to have equal number of balls on each scale. Case 1: Scale is in balance - one of the three balls removed is the odd one. At this point you have three balls and you still do not know whether they are heavier or lighter with only one weighing remaining. But its closer. Dave Huff 'tis a silly place!
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moliate wrote: 2) If eigth balls are found normal it is easy to find the odd one in two weighings. If not: Remove two ball from first scale, one from second scale and swap two of the remaining ones from scale to scale. Add one normal ball to have equal number of balls on each scale. Case 1: Scale is in balance - one of the three balls removed is the odd one. At this point you have three balls and you still do not know whether they are heavier or lighter with only one weighing remaining. But its closer. Dave Huff 'tis a silly place!
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Assuming you keep the two balls removed from the first scale separated from the one removed from the second one you could do it in one weighing... Or am I not thinking clearly here :confused: /moliate