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CTK Exchange
Aishwarya
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Sep-05-04, 10:52 AM (EST) |
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"Odd coins problem"
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I am posting the solution that occured to me for the odd coins problem. I am not certain that this solution has not already been offered, but it is something that occured to me. let the coins be named 1 to 12, and the two pans be named A and B for convenience. Soln: weigh 1: A 1,2,3 B 4,5,6, If they are equal, the faulty coin lies in the other 6 coins, ie. 7,8,9,10,11,12 else, the faulty coin lies in these first 6 coins. in such a case, remove 1, 4 ie. weigh 2: A: 2,3 B: 5,6 if these are equal, the faulty coin is either 1 or 4 in such a case weigh 3: A: 1 B: 5(or any coin other than 4) if these are equal, the faulty coin is 4, else 1. consider weigh 2 in which 2,3 and 5,6 are weighed. if they are unequal,(2,3 lighter than 5,6) then exchange 3 and 5. weigh 3: A:2,5 B: 3,6 if the pan A still remains lighter, then the faulty coin is 2, else 3. similarly, if in weigh 1, we find that the pans weigh the same, then the faulty ball lies in the second 6 balls, then name them 1 to 6 and proceed from the second step. |
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jvwert
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Sep-09-04, 06:51 AM (EST) |
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1. "RE: Odd coins problem"
In response to message #0
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"Odd coins problem" In problems of this type, you will find that the solution must consider the worst case in each decision and operation. For example, you state that in the first weighing, if the balance indicates equality for the two pans containing 1,2,3 and 4,5,6, the faulty coin lies in the other 6 coins. This is the worst case for the first weighing. You have used one weighing, and gained little information as to the relative weight of the six remaining coins which includes the odd coin, and you have only two weighings left. I think you will find that that is not enough. Your reasoning for the unbalanced condition for the first weighing at least gave some information as to the relative weights of the six coins on the pans. For example in such a case (balanced for the first weighing) , and following the procedure used in your solution, you now put coins 7,8 on pan A and 11,12 on pan B. Assume they are equal. This leaves you two coins (6, and 10) and the information that the odd coin is one of them, but not the relative weight. Weighing one against one of the “good” coins can eliminate it from the problem, telling you the other one is the odd coin, but not its relative weight. I suggest you follow through the “worst case” scenario until you have exhausted all possibilities before deciding you have a solution. jvwert
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Harsh K. Mittal
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Sep-22-04, 08:10 AM (EST) |
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2. "RE: Odd coins problem"
In response to message #0
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lets have 12 coins......as coins are coins we can name them as we want.... now we compare four coins with other four coins................ out put could be either they are equal or they are not.......... so we can say like 1234>5678 1234<5678 1234 =5678 about the third case.....
we know that any of 9,10,11,12 are faulty.... in this we will weigh 9,10,11 with any good coins...like 123....... by the end of second weighing... either 9,10,11 are equal.... if they are 12 wil be faulty if not then we can find out by single more(in all three)...... after comparin 9,10...........if they are equal 11 will be faulty else which gives the same result as in second weighin is faulty..... NOW come to first or second case 1234>5678 now we will compare 1235 with 4 & three more good coins( 9,10,11)........... now if 1235>4,9,10,11 then any of 1,2 or 3 is faulty now we'll compare 1 with 2.......if 1>2 then 1 is faulty... 1<2 2 is faulty 1=2 3 is faulty.......... if 1235 <4,9,10,11 then either 4 or 5 is faulty compare 4 with any good coin....if 4 = good coin,5 is faulty else 4 is faulty........... if 1235=4,9,10,11 then 6,7 or 8 are faulty....
6=7 8 is faulty 6>7 7 is faulty 6<7 6 is faulty.............. |
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