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Topic: Coin weighting: 8 coins with 2 defected coins (Read 3238 times) |
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wonderful
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Coin weighting: 8 coins with 2 defected coins
« on: Jul 7th, 2008, 9:28pm » |
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There are 8 identical looking coins. Two of which are defected. They could be heavier or lighter than the normal coins. Using a standard two-arm balance, what is the minimum number of scalings you need to identify these two defected coins?
To clarify the question, let's say "a" to be the weight of the regular coin; x and y to be the weights of two defected coins. The following can happen:
x<y<a ; y<x<a; x<a<y; y<a<x; a<x<y; a<x<y; x=y<a; a<x=y.
The question is how one can design an optimal scaling scheme to identify these two points weighting x, and y respectively.
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| « Last Edit: Jul 7th, 2008, 9:32pm by wonderful » |
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rmsgrey
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Re: Coin weighting: 8 coins with 2 defected coins
« Reply #2 on: Jul 8th, 2008, 10:29am » |
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What, if any, is the difference between x<y<a and y<x<a?
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towr
wu::riddles Moderator
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Re: Coin weighting: 8 coins with 2 defected coins
« Reply #3 on: Jul 8th, 2008, 11:37am » |
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We can always rename the minimum of x and y to x, and the maximum to y.
So we just have to deal with: x<y<a ; x<a<y; a<x<y; x=y<a; a<x=y.
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Wikipedia, Google, Mathworld, Integer sequence DB
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rajatiet
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Re: Coin weighting: 8 coins with 2 defected coins
« Reply #4 on: Nov 30th, 2008, 11:58pm » |
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we will divide 8 coins in groups of 2 coins each.
Lets the groups are A,B,C,D.
now compare two groups on weigh e.g. A and B.
There can be 2 possibilities . The Defective balls can be in same grp or different groups.
1) Different Group: When we will compare 2 groups , one comparison will show that 2 groups of same weights.
While the other comparison will show that groups are unequal.
Till now total scalings equal to 2.
After that we will take 1 ball from the groups of equal weights, compare with each ball of unequal group.
These comparisons will show the defective balls.
So total scaling equal to 2+4=6
2) Same Group: When we will compare 2 groups , both comparisons will show that 4 groups of same weights, this means the total weights of 2 defective coins equals to 2 good coins.
Now compare the 2 coins of each group with each other. The comparison which shows the unequal coins gives the defective coins.
So total comparisons equal to 2+4=6.
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rmsgrey
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Re: Coin weighting: 8 coins with 2 defected coins
« Reply #5 on: Dec 2nd, 2008, 7:35am » |
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on Nov 30th, 2008, 11:58pm, rajatiet wrote: we will divide 8 coins in groups of 2 coins each.
Lets the groups are A,B,C,D.
now compare two groups on weigh e.g. A and B.
There can be 2 possibilities . The Defective balls can be in same grp or different groups.
1) Different Group: When we will compare 2 groups , one comparison will show that 2 groups of same weights.
While the other comparison will show that groups are unequal.
Till now total scalings equal to 2.
After that we will take 1 ball from the groups of equal weights, compare with each ball of unequal group.
These comparisons will show the defective balls.
So total scaling equal to 2+4=6
2) Same Group: When we will compare 2 groups , both comparisons will show that 4 groups of same weights, this means the total weights of 2 defective coins equals to 2 good coins.
Now compare the 2 coins of each group with each other. The comparison which shows the unequal coins gives the defective coins.
So total comparisons equal to 2+4=6.
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If A1 and C1 are both defective, then both weighings will be uneven.
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