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CTK Exchange
Edward
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Oct-12-09, 07:05 AM (EST) |
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"question about 8 balls, 3 weighings"
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There are many websit's which detail the 12 balls, 3 weighings problem. (one ball of a different weight to the other 11, but not known whether lighter or heavier). But I can't find a single website which gives the answer for 8 balls. Would this be right: NB, the following is my own work, and just for personal interest. It'seems to work to me, but I would appreciate a professional eye to tell me if I've missed something obvious. 1st attempt: 123 v 456 If balanced, then weigh 7 against 1 in the second attempt, or 8 against 1 in the third attempt. 2nd attempt: If 123 goes up: either one of 123 is light, or one of 456 is heavy. Therefore weigh 14 v 25. If balanced: either 3 is light or 6 is heavy. Easy to determine in final weighing (and by deduction if balanced). If scale tips in same direction as first time: bad ball must be among those two which didn't move (ie, either 1 light or 5 heavy). Easy to determine in final weighing (and by deduction if balanced). If scale tips in opposite direction: bad ball must be among those two which moved (ie either 4 heavy or 2 light). Easy to determine in final weighing (and by deduction if balanced). 3rd attempt: (as detailed above) .... Many thanks, if you have 60 seconds to cast an eye over this, and confirm whether it's right!
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alexb
Charter Member
2443 posts |
Oct-12-09, 07:18 AM (EST) |
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1. "RE: question about 8 balls, 3 weighings"
In response to message #0
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I can confirm that your solution achieves its goal: finding a defective ball out of 8 in 3 weighings. I must admit that my first reaction to your problem was that of astonishment. Surely, if one can solve the problem for 12 balls, he must be able to solve it for a smaller number, 8 in particular. Why to bother? Just get a few good balls to complete a dozen and procede to solve the 12 balls problem. On second thought, as obviously, even if there were supply of balls to complete a dozen, knowing that all of them good is a kind of presumption that was not explicitly assumed in the problem. Finding extra good balls will take the precious weighings ... This reasoning makes the 8 balls problem into a valid subject for investigation. But then it drags on the 9, 10, and 11 balls problems. If indeed a problem with fewer balls is simper (in some sense) than a problem with more balls, the three problems (for 9, 10, and 11 balls) must have a solution in three weighings. Have you tried these?
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Edward
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Oct-12-09, 03:18 PM (EST) |
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2. "RE: question about 8 balls, 3 weighings"
In response to message #1
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Thank you for your kind reply. Of course with the 8 ball problem there can be no further supply of known good balls! (Otherwise it would no longer be an "8 ball" problem...) And please NB, my problem was not merely to find a "defective ball" (as you commented), but also to identify whether it was light or heavy. So, that said, does my proposed solution definitively solve the 8 ball problem? By the way, I believe that 12 is the highest number of balls that can be definitively solved (ie without chancing to hit lucky) in just 3 weighings. |
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