Weighing Problem Resurected

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You have a dozen (12) stones weighing a whole number of grams between 1 and 6 each. You can obtain one reference weight of your choosing.

What reference weight can you choose to be able to figure out the individual weights of the 12 stones using a balance device for any possibility that may exist therein?

For an encore: what is the maximum weight range of stones (1 to N) that you could solve using 2 reference weights of your choice? Provided you can have as many stones as you need.

I don't believe, I have solved this one myself. We could make it a community project after the first question is answered.

HISTORICAL NOTE:

This problem originated on Brain Den. I constructed it based on Bonanova's problem Weighty Thoughts: http://brainden.com/forum/index.php/topic/4932--/?p=84107 few years ago.

Back then limited number of people participated. The solution found was for specific numbers in that problem (range 1 to 5) – not general. I'd like to give it another try.

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Posted (edited) · Report post

I thought I had this one... but found a flaw in my reasoning. [solution withdrawn] :blush:

Edited by ThunderCloud
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Posted · Report post

First part: do we know that all weights between 1-6 are represented in the 12 stones? Or is it possible that we have, say, 12 stones each waiting 6?

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Posted · Report post

seems to me you don't even need a reference weight. you can rank the weights using any standard sorting algorithm, and determine the mass by comparing multiple amounts of the same weight.

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Posted · Report post

You have a dozen (12) stones weighing a whole number of grams between 1 and 6 each. You can obtain one reference weight of your choosing.

What reference weight can you choose to be able to figure out the individual weights of the 12 stones using a balance device for any possibility that may exist therein?

For an encore: what is the maximum weight range of stones (1 to N) that you could solve using 2 reference weights of your choice? Provided you can have as many stones as you need.

I don't believe, I have solved this one myself. We could make it a community project after the first question is answered.

HISTORICAL NOTE:

This problem originated on Brain Den. I constructed it based on Bonanova's problem Weighty Thoughts: http://brainden.com/forum/index.php/topic/4932--/?p=84107 few years ago.

Back then limited number of people participated. The solution found was for specific numbers in that problem (range 1 to 5) – not general. I'd like to give it another try.

Using a reference weight of 5 grams, it would be easy to distinguish stones weighing 6 grams and 5 grams from the rest. Stones of lower weight could be sorted according to like weight, and then combined to see which combinations yield totals of 5 or 6+. The only difficulty is distinguishing a case where every stone weighs 3 grams from a case where every stone weighs 4 grams; either way, each is less than the reference weight, each is identical to every other, and any two are greater than the reference weight. The only method i can conceive to decide between these cases is by adding some other unknown but finely granulated weight to the scale (such as sand) to balance it... using this trick, one could determine which is the greater difference: one stone from the reference weight or the reference weight from two stones. If the weight difference between the reference weight and one stone is larger than between two stones and the reference, then the stones all weigh 3 grams; else they weigh 4.

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Posted · Report post

First part: do we know that all weights between 1-6 are represented in the 12 stones? Or is it possible that we have, say, 12 stones each waiting 6?

There is no guaranty what weights are present/absent in your collection. The only guaranty is that each of the stones weighs a whole number of grams between 1 and 6.

12 stones each weighing 6 grams is one of the variations we must account for.

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Posted · Report post

I like T-cloud's cheat. It's not a reference weight, but just a found object that represents a difference. Good outside the box thinking!

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Part 2: you can have as many stones as you want. Which stones? Reference stones of my two chosen denominations? Or as many unknown stones?

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Posted · Report post

Part 2: you can have as many stones as you want. Which stones? Reference stones of my two chosen denominations? Or unknown stones?

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Part 2: you can have as many stones as you want. Which stones? Reference stones of my two chosen denominations? Or as many unknown stones?

For the part 2, you can have 2 reference weights of your choice, as many stones as you want in the weight range from 1 to N. (Must find the largest N and the 2 reference weights.)

Let's solve the first part first. No tricks, no cheating, no different interpretations of the OP. If there is an ambiguity, I'll clarify it.

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Posted (edited) · Report post

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
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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

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Posted (edited) · Report post

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
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Posted · Report post

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",

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

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Posted · Report post

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",

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

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?

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Posted (edited) · Report post

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",

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

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?

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
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Posted · Report post

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",

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

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?

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! :duh: Thanks, Prime.

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