16 replies to this topic

[k-man]

Spoiler for I think...

An 11 gram weight allows to solve the first problem. I haven't checked all possible combinations of stones, but I haven't found any combinations that wouldn't work yet.

 

Ahh... never mind. back to the drawing board

Edited by k-man, 23 February 2013 - 05:07 AM.

[k-man]

Spoiler for

Looks like there is no single reference weight that would allow to discover every possible combination of 12 stones. Unless I missed something here are all the weight ranges and combinations that are indistinguishable for each range:

 

 

W <= 4g - can't distinguish all 5g stones from all 6g stones

4g < W < 6g - cant distinguish all 3g stones from all 4g stones

6g <= W < 8g - can't distinguish a single 3g stone from a 4g stone among 11 6g stones

8g <= W < 9g - can't distinguish a single 1g stone from a 2g stone among 11 4g stones

9g <= W < 10g - can't distinguish a single 1g stone from a 2g stone among 11 6g stones

10g <= W < 11g - can't distinguish a single 1g stone from a 2g stone among 11 5g stones

11g <= W <= 12g - can't distinguish a single 3g stone from a 4g stone among 11 6g stones

12g < W <= 14g - can't distinguish all 5g stones from all 6g stones

W > 14g - can't distinguish a single 2g stone from a 3g stone among 11 1g stones

 

[ThunderCloud]

Spoiler for I suspect the answer is...

22. The difficult cases are the ones where all the weights are the same, or where only one of them is different. A reference weight of 22 allows one to discriminate these cases well. For example, suppose all of the weights are 6 grams except for one which is less. You can determine that all of the others are 6 by observing that 4 of them weigh more than the reference weight, and that they all weigh the same as one another. To determine the weight of the "extra" one, add it to three of the other stones and compare against the reference weight. If it is greater, the extra stone weighs 5 grams; if equal, it weighs 4; if less, combine it with the reference weight and compare against four of the other stones. If it is greater, it is 3 grams; if equal, it is 2 grams; if less, it is 1 gram. Similar reasoning seems to work with stones that are each 5 grams...

Edited by ThunderCloud, 23 February 2013 - 06:25 AM.

[Prime]

Spoiler for I suspect the answer is...

22. The difficult cases are the ones where all the weights are the same, or where only one of them is different. A reference weight of 22 allows one to discriminate these cases well. For example, suppose all of the weights are 6 grams except for one which is less. You can determine that all of the others are 6 by observing that 4 of them weigh more than the reference weight, and that they all weigh the same as one another. To determine the weight of the "extra" one, add it to three of the other stones and compare against the reference weight. If it is greater, the extra stone weighs 5 grams; if equal, it weighs 4; if less, combine it with the reference weight and compare against four of the other stones. If it is greater, it is 3 grams; if equal, it is 2 grams; if less, it is 1 gram. Similar reasoning seems to work with stones that are each 5 grams...

 

I suspect, your suspicion is correct. That's the same reference weight that I found. There is a certain criteria for it.

Rather than enumerating all different cases, I invite all participants to try and find cases, which would be indistinguishable with this reference weight.

 

K-man came close to the solution and was the first to "think outside the box",

Spoiler for noting

Noting that the reference weight does not need to be within the range of stone weights.

[k-man]

Spoiler for I suspect the answer is...

22. The difficult cases are the ones where all the weights are the same, or where only one of them is different. A reference weight of 22 allows one to discriminate these cases well. For example, suppose all of the weights are 6 grams except for one which is less. You can determine that all of the others are 6 by observing that 4 of them weigh more than the reference weight, and that they all weigh the same as one another. To determine the weight of the "extra" one, add it to three of the other stones and compare against the reference weight. If it is greater, the extra stone weighs 5 grams; if equal, it weighs 4; if less, combine it with the reference weight and compare against four of the other stones. If it is greater, it is 3 grams; if equal, it is 2 grams; if less, it is 1 gram. Similar reasoning seems to work with stones that are each 5 grams...

 

I suspect, your suspicion is correct. That's the same reference weight that I found. There is a certain criteria for it.

Rather than enumerating all different cases, I invite all participants to try and find cases, which would be indistinguishable with this reference weight.

 

K-man came close to the solution and was the first to "think outside the box",

Spoiler for noting

Noting that the reference weight does not need to be within the range of stone weights.

Spoiler for I disagree...

Larger weights solve the single lighter stone among 11 6g stones, but they don't work for one heavy stone among 11 1g stones. With 22g reference weight how will you identify the weight of a single heavy stone among 11 1g stones?

[Prime]

 

Spoiler for I suspect the answer is...

22. The difficult cases are the ones where all the weights are the same, or where only one of them is different. A reference weight of 22 allows one to discriminate these cases well. For example, suppose all of the weights are 6 grams except for one which is less. You can determine that all of the others are 6 by observing that 4 of them weigh more than the reference weight, and that they all weigh the same as one another. To determine the weight of the "extra" one, add it to three of the other stones and compare against the reference weight. If it is greater, the extra stone weighs 5 grams; if equal, it weighs 4; if less, combine it with the reference weight and compare against four of the other stones. If it is greater, it is 3 grams; if equal, it is 2 grams; if less, it is 1 gram. Similar reasoning seems to work with stones that are each 5 grams...

 

I suspect, your suspicion is correct. That's the same reference weight that I found. There is a certain criteria for it.

Rather than enumerating all different cases, I invite all participants to try and find cases, which would be indistinguishable with this reference weight.

 

K-man came close to the solution and was the first to "think outside the box",

Spoiler for noting

Noting that the reference weight does not need to be within the range of stone weights.

Spoiler for I disagree...

Larger weights solve the single lighter stone among 11 6g stones, but they don't work for one heavy stone among 11 1g stones. With 22g reference weight how will you identify the weight of a single heavy stone among 11 1g stones?

Spoiler for There is a way

When you have 11 equal stones, which put together are lighter than 22 g., you know those are 1 g. stones. Now you can use them as reference weights to measure whatever remaining heavier stone you have.

In a way, this problem is a mix of weighing problem and sort.

Edited by Prime, 23 February 2013 - 08:25 PM.

[k-man]

 

 

Spoiler for I suspect the answer is...

22. The difficult cases are the ones where all the weights are the same, or where only one of them is different. A reference weight of 22 allows one to discriminate these cases well. For example, suppose all of the weights are 6 grams except for one which is less. You can determine that all of the others are 6 by observing that 4 of them weigh more than the reference weight, and that they all weigh the same as one another. To determine the weight of the "extra" one, add it to three of the other stones and compare against the reference weight. If it is greater, the extra stone weighs 5 grams; if equal, it weighs 4; if less, combine it with the reference weight and compare against four of the other stones. If it is greater, it is 3 grams; if equal, it is 2 grams; if less, it is 1 gram. Similar reasoning seems to work with stones that are each 5 grams...

 

I suspect, your suspicion is correct. That's the same reference weight that I found. There is a certain criteria for it.

Rather than enumerating all different cases, I invite all participants to try and find cases, which would be indistinguishable with this reference weight.

 

K-man came close to the solution and was the first to "think outside the box",

Spoiler for noting

Noting that the reference weight does not need to be within the range of stone weights.

Spoiler for I disagree...

Larger weights solve the single lighter stone among 11 6g stones, but they don't work for one heavy stone among 11 1g stones. With 22g reference weight how will you identify the weight of a single heavy stone among 11 1g stones?

Spoiler for There is a way

When you have 11 equal stones, which put together are lighter than 22 g., you know those are 1 g. stones. Now you can use them as reference weights to measure whatever remaining heavier stone you have.

In a way, this problem is a mix of weighing problem and sort.

Doh!  Thanks, Prime.