Find a stone which is not the lightest oneWhich one is the lightest marble?Using a balance scale the minimum number of times, find the median weight personA balance with three pans, detecting the lightest pan (find the one lighter ball)A balance with three pans, detecting the lightest pan (find the one heavier ball)A balance with three pans, detecting the lightest pan (find the two heavier balls)A balance with three pans, detecting the lightest pan (find the one lighter/heavier ball, for a given number of balls)Find 2 heavy coins among 27 with a 3-pan balanceMinimum number of tries to find the balance!Lots of Gold Stacks and a Balance ScaleFive balls weighing
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Find a stone which is not the lightest one
Which one is the lightest marble?Using a balance scale the minimum number of times, find the median weight personA balance with three pans, detecting the lightest pan (find the one lighter ball)A balance with three pans, detecting the lightest pan (find the one heavier ball)A balance with three pans, detecting the lightest pan (find the two heavier balls)A balance with three pans, detecting the lightest pan (find the one lighter/heavier ball, for a given number of balls)Find 2 heavy coins among 27 with a 3-pan balanceMinimum number of tries to find the balance!Lots of Gold Stacks and a Balance ScaleFive balls weighing
$begingroup$
I've gotten this riddle and have been struggling with solving it:
Suppose you have 20 stones which have different weights [0 ... n] . You have no way of measuring the weight of any stone individually, instead you can only measure them by putting 10 on each side of a balance. After doing this you see which "group" of stones is heavier. You can do this only 10 times in total. You can also switch the stones from either side to the other as much as you want between the measurements.
Your task is to find a stone for which you can be certain is not the lightest of them all.
All the stones are marked, so you always know which is which. Their appearance or anything similar does not help you with to determine their weight, the only way you can do it is by putting it on the balance.
weighing
New contributor
$endgroup$
add a comment |
$begingroup$
I've gotten this riddle and have been struggling with solving it:
Suppose you have 20 stones which have different weights [0 ... n] . You have no way of measuring the weight of any stone individually, instead you can only measure them by putting 10 on each side of a balance. After doing this you see which "group" of stones is heavier. You can do this only 10 times in total. You can also switch the stones from either side to the other as much as you want between the measurements.
Your task is to find a stone for which you can be certain is not the lightest of them all.
All the stones are marked, so you always know which is which. Their appearance or anything similar does not help you with to determine their weight, the only way you can do it is by putting it on the balance.
weighing
New contributor
$endgroup$
1
$begingroup$
Please give credit to the author of the puzzle.
$endgroup$
– Gilles
6 hours ago
$begingroup$
I have a new and improved solution!
$endgroup$
– Brandon_J
3 hours ago
add a comment |
$begingroup$
I've gotten this riddle and have been struggling with solving it:
Suppose you have 20 stones which have different weights [0 ... n] . You have no way of measuring the weight of any stone individually, instead you can only measure them by putting 10 on each side of a balance. After doing this you see which "group" of stones is heavier. You can do this only 10 times in total. You can also switch the stones from either side to the other as much as you want between the measurements.
Your task is to find a stone for which you can be certain is not the lightest of them all.
All the stones are marked, so you always know which is which. Their appearance or anything similar does not help you with to determine their weight, the only way you can do it is by putting it on the balance.
weighing
New contributor
$endgroup$
I've gotten this riddle and have been struggling with solving it:
Suppose you have 20 stones which have different weights [0 ... n] . You have no way of measuring the weight of any stone individually, instead you can only measure them by putting 10 on each side of a balance. After doing this you see which "group" of stones is heavier. You can do this only 10 times in total. You can also switch the stones from either side to the other as much as you want between the measurements.
Your task is to find a stone for which you can be certain is not the lightest of them all.
All the stones are marked, so you always know which is which. Their appearance or anything similar does not help you with to determine their weight, the only way you can do it is by putting it on the balance.
weighing
weighing
New contributor
New contributor
edited 6 hours ago
Gilles
3,42531837
3,42531837
New contributor
asked 14 hours ago
podloga123podloga123
512
512
New contributor
New contributor
1
$begingroup$
Please give credit to the author of the puzzle.
$endgroup$
– Gilles
6 hours ago
$begingroup$
I have a new and improved solution!
$endgroup$
– Brandon_J
3 hours ago
add a comment |
1
$begingroup$
Please give credit to the author of the puzzle.
$endgroup$
– Gilles
6 hours ago
$begingroup$
I have a new and improved solution!
$endgroup$
– Brandon_J
3 hours ago
1
1
$begingroup$
Please give credit to the author of the puzzle.
$endgroup$
– Gilles
6 hours ago
$begingroup$
Please give credit to the author of the puzzle.
$endgroup$
– Gilles
6 hours ago
$begingroup$
I have a new and improved solution!
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
I have a new and improved solution!
$endgroup$
– Brandon_J
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Measure 10 and 10.
In each subsequent weighing you change one of the originals of one side for one of the originals of the other side.
At some point, the balance turns around.
The stone that you have put on the side that has come down is not the lightest, because the other is lighter than that.
If at the tenth weighing the sides have not been turned around, the only original stone left on the side that is below is not the lightest.
$endgroup$
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
13 hours ago
$begingroup$
Well, never mind my previous comment about being late; I found a better solution.
$endgroup$
– Brandon_J
2 hours ago
add a comment |
$begingroup$
Here's a solution that needs at most
5
weighings.
For convenience, I arbitrarily number each stone on side "A" (which I declare to be the lighter side for convenience) of the balance in the first weighing to be stones 1, 3, 5...19 and each stone on the heaver side B 2, 4, 6...20. The original weighing of all of the stones is taken to count toward the weighing total, but will be named "step zero" for convenience, because it's the only possible thing to do at the beginning.
Note: if at any point the scale balances, a not-lightest stone can be found in one more weighing by switching any two stones and seeing which way the balance tips.
Step one:
Swap stones 1, 3, 5, and 7 with stones 2, 4, 6, and 8. If the balance tips, you would skip to step 3, but that wouldn't be worst-case scenario (as you will see), so we won't skip ahead. We assume the balance does not tip. Total weighings: 2
Step two:
Swap stones 11, 13, 15, and 17 with stones 10, 12, 14, and 16. If the balance tips doesn't tip, you would skip to step 4, but that's not the worst case, so we will assume that the balance does tip. Total weighings: 3
Step three:
Swap four of the 8 stones that you have just swapped. Assuming that we are still following the worst case, swap stones 11 and 13 with 10 and 12. If the scale does not tip back to side B, then you know that switching 15 and 17 with 14 and 15 will. Total weighings: 4
Step four:
Regardless of which route we took to get here, we currently have four stones which we know either have tipped the scale when swapped, or would tip the scale when swapped. Swap two of them. If the scale doesn't change state when these are swapped, then the unswapped stone on the side that the scale was expected to tip away from is definitely not the lightest stone. If the scale does change state, the swapped stone that ended up on the side that the scale just tipped toward is definitely _not the lightest stone. Total weighings: 5
And there's your answer!
$endgroup$
$begingroup$
What if neither side is heavier?
$endgroup$
– noedne
2 hours ago
$begingroup$
@noedne see JaapScherpius's comment above. Also, if neither side is heavier, a not-lightest stone could be identified in two weighings, one swap.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
@noedne here's the better solution.
$endgroup$
– Brandon_J
2 hours ago
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
Measure 10 and 10.
In each subsequent weighing you change one of the originals of one side for one of the originals of the other side.
At some point, the balance turns around.
The stone that you have put on the side that has come down is not the lightest, because the other is lighter than that.
If at the tenth weighing the sides have not been turned around, the only original stone left on the side that is below is not the lightest.
$endgroup$
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
13 hours ago
$begingroup$
Well, never mind my previous comment about being late; I found a better solution.
$endgroup$
– Brandon_J
2 hours ago
add a comment |
$begingroup$
Measure 10 and 10.
In each subsequent weighing you change one of the originals of one side for one of the originals of the other side.
At some point, the balance turns around.
The stone that you have put on the side that has come down is not the lightest, because the other is lighter than that.
If at the tenth weighing the sides have not been turned around, the only original stone left on the side that is below is not the lightest.
$endgroup$
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
13 hours ago
$begingroup$
Well, never mind my previous comment about being late; I found a better solution.
$endgroup$
– Brandon_J
2 hours ago
add a comment |
$begingroup$
Measure 10 and 10.
In each subsequent weighing you change one of the originals of one side for one of the originals of the other side.
At some point, the balance turns around.
The stone that you have put on the side that has come down is not the lightest, because the other is lighter than that.
If at the tenth weighing the sides have not been turned around, the only original stone left on the side that is below is not the lightest.
$endgroup$
Measure 10 and 10.
In each subsequent weighing you change one of the originals of one side for one of the originals of the other side.
At some point, the balance turns around.
The stone that you have put on the side that has come down is not the lightest, because the other is lighter than that.
If at the tenth weighing the sides have not been turned around, the only original stone left on the side that is below is not the lightest.
answered 14 hours ago
HermesHermes
4608
4608
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
13 hours ago
$begingroup$
Well, never mind my previous comment about being late; I found a better solution.
$endgroup$
– Brandon_J
2 hours ago
add a comment |
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
13 hours ago
$begingroup$
Well, never mind my previous comment about being late; I found a better solution.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
13 hours ago
$begingroup$
You didn't explicitly mention this, but it is impossible for the two scales to ever be equal with the given set of stones. But this solution would still work even if the weights were such that the scales could be balanced, provided you know that there is at least one stone that has a unique weight.
$endgroup$
– Jaap Scherphuis
13 hours ago
$begingroup$
Well, never mind my previous comment about being late; I found a better solution.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
Well, never mind my previous comment about being late; I found a better solution.
$endgroup$
– Brandon_J
2 hours ago
add a comment |
$begingroup$
Here's a solution that needs at most
5
weighings.
For convenience, I arbitrarily number each stone on side "A" (which I declare to be the lighter side for convenience) of the balance in the first weighing to be stones 1, 3, 5...19 and each stone on the heaver side B 2, 4, 6...20. The original weighing of all of the stones is taken to count toward the weighing total, but will be named "step zero" for convenience, because it's the only possible thing to do at the beginning.
Note: if at any point the scale balances, a not-lightest stone can be found in one more weighing by switching any two stones and seeing which way the balance tips.
Step one:
Swap stones 1, 3, 5, and 7 with stones 2, 4, 6, and 8. If the balance tips, you would skip to step 3, but that wouldn't be worst-case scenario (as you will see), so we won't skip ahead. We assume the balance does not tip. Total weighings: 2
Step two:
Swap stones 11, 13, 15, and 17 with stones 10, 12, 14, and 16. If the balance tips doesn't tip, you would skip to step 4, but that's not the worst case, so we will assume that the balance does tip. Total weighings: 3
Step three:
Swap four of the 8 stones that you have just swapped. Assuming that we are still following the worst case, swap stones 11 and 13 with 10 and 12. If the scale does not tip back to side B, then you know that switching 15 and 17 with 14 and 15 will. Total weighings: 4
Step four:
Regardless of which route we took to get here, we currently have four stones which we know either have tipped the scale when swapped, or would tip the scale when swapped. Swap two of them. If the scale doesn't change state when these are swapped, then the unswapped stone on the side that the scale was expected to tip away from is definitely not the lightest stone. If the scale does change state, the swapped stone that ended up on the side that the scale just tipped toward is definitely _not the lightest stone. Total weighings: 5
And there's your answer!
$endgroup$
$begingroup$
What if neither side is heavier?
$endgroup$
– noedne
2 hours ago
$begingroup$
@noedne see JaapScherpius's comment above. Also, if neither side is heavier, a not-lightest stone could be identified in two weighings, one swap.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
@noedne here's the better solution.
$endgroup$
– Brandon_J
2 hours ago
add a comment |
$begingroup$
Here's a solution that needs at most
5
weighings.
For convenience, I arbitrarily number each stone on side "A" (which I declare to be the lighter side for convenience) of the balance in the first weighing to be stones 1, 3, 5...19 and each stone on the heaver side B 2, 4, 6...20. The original weighing of all of the stones is taken to count toward the weighing total, but will be named "step zero" for convenience, because it's the only possible thing to do at the beginning.
Note: if at any point the scale balances, a not-lightest stone can be found in one more weighing by switching any two stones and seeing which way the balance tips.
Step one:
Swap stones 1, 3, 5, and 7 with stones 2, 4, 6, and 8. If the balance tips, you would skip to step 3, but that wouldn't be worst-case scenario (as you will see), so we won't skip ahead. We assume the balance does not tip. Total weighings: 2
Step two:
Swap stones 11, 13, 15, and 17 with stones 10, 12, 14, and 16. If the balance tips doesn't tip, you would skip to step 4, but that's not the worst case, so we will assume that the balance does tip. Total weighings: 3
Step three:
Swap four of the 8 stones that you have just swapped. Assuming that we are still following the worst case, swap stones 11 and 13 with 10 and 12. If the scale does not tip back to side B, then you know that switching 15 and 17 with 14 and 15 will. Total weighings: 4
Step four:
Regardless of which route we took to get here, we currently have four stones which we know either have tipped the scale when swapped, or would tip the scale when swapped. Swap two of them. If the scale doesn't change state when these are swapped, then the unswapped stone on the side that the scale was expected to tip away from is definitely not the lightest stone. If the scale does change state, the swapped stone that ended up on the side that the scale just tipped toward is definitely _not the lightest stone. Total weighings: 5
And there's your answer!
$endgroup$
$begingroup$
What if neither side is heavier?
$endgroup$
– noedne
2 hours ago
$begingroup$
@noedne see JaapScherpius's comment above. Also, if neither side is heavier, a not-lightest stone could be identified in two weighings, one swap.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
@noedne here's the better solution.
$endgroup$
– Brandon_J
2 hours ago
add a comment |
$begingroup$
Here's a solution that needs at most
5
weighings.
For convenience, I arbitrarily number each stone on side "A" (which I declare to be the lighter side for convenience) of the balance in the first weighing to be stones 1, 3, 5...19 and each stone on the heaver side B 2, 4, 6...20. The original weighing of all of the stones is taken to count toward the weighing total, but will be named "step zero" for convenience, because it's the only possible thing to do at the beginning.
Note: if at any point the scale balances, a not-lightest stone can be found in one more weighing by switching any two stones and seeing which way the balance tips.
Step one:
Swap stones 1, 3, 5, and 7 with stones 2, 4, 6, and 8. If the balance tips, you would skip to step 3, but that wouldn't be worst-case scenario (as you will see), so we won't skip ahead. We assume the balance does not tip. Total weighings: 2
Step two:
Swap stones 11, 13, 15, and 17 with stones 10, 12, 14, and 16. If the balance tips doesn't tip, you would skip to step 4, but that's not the worst case, so we will assume that the balance does tip. Total weighings: 3
Step three:
Swap four of the 8 stones that you have just swapped. Assuming that we are still following the worst case, swap stones 11 and 13 with 10 and 12. If the scale does not tip back to side B, then you know that switching 15 and 17 with 14 and 15 will. Total weighings: 4
Step four:
Regardless of which route we took to get here, we currently have four stones which we know either have tipped the scale when swapped, or would tip the scale when swapped. Swap two of them. If the scale doesn't change state when these are swapped, then the unswapped stone on the side that the scale was expected to tip away from is definitely not the lightest stone. If the scale does change state, the swapped stone that ended up on the side that the scale just tipped toward is definitely _not the lightest stone. Total weighings: 5
And there's your answer!
$endgroup$
Here's a solution that needs at most
5
weighings.
For convenience, I arbitrarily number each stone on side "A" (which I declare to be the lighter side for convenience) of the balance in the first weighing to be stones 1, 3, 5...19 and each stone on the heaver side B 2, 4, 6...20. The original weighing of all of the stones is taken to count toward the weighing total, but will be named "step zero" for convenience, because it's the only possible thing to do at the beginning.
Note: if at any point the scale balances, a not-lightest stone can be found in one more weighing by switching any two stones and seeing which way the balance tips.
Step one:
Swap stones 1, 3, 5, and 7 with stones 2, 4, 6, and 8. If the balance tips, you would skip to step 3, but that wouldn't be worst-case scenario (as you will see), so we won't skip ahead. We assume the balance does not tip. Total weighings: 2
Step two:
Swap stones 11, 13, 15, and 17 with stones 10, 12, 14, and 16. If the balance tips doesn't tip, you would skip to step 4, but that's not the worst case, so we will assume that the balance does tip. Total weighings: 3
Step three:
Swap four of the 8 stones that you have just swapped. Assuming that we are still following the worst case, swap stones 11 and 13 with 10 and 12. If the scale does not tip back to side B, then you know that switching 15 and 17 with 14 and 15 will. Total weighings: 4
Step four:
Regardless of which route we took to get here, we currently have four stones which we know either have tipped the scale when swapped, or would tip the scale when swapped. Swap two of them. If the scale doesn't change state when these are swapped, then the unswapped stone on the side that the scale was expected to tip away from is definitely not the lightest stone. If the scale does change state, the swapped stone that ended up on the side that the scale just tipped toward is definitely _not the lightest stone. Total weighings: 5
And there's your answer!
edited 2 hours ago
answered 14 hours ago
Brandon_JBrandon_J
3,905447
3,905447
$begingroup$
What if neither side is heavier?
$endgroup$
– noedne
2 hours ago
$begingroup$
@noedne see JaapScherpius's comment above. Also, if neither side is heavier, a not-lightest stone could be identified in two weighings, one swap.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
@noedne here's the better solution.
$endgroup$
– Brandon_J
2 hours ago
add a comment |
$begingroup$
What if neither side is heavier?
$endgroup$
– noedne
2 hours ago
$begingroup$
@noedne see JaapScherpius's comment above. Also, if neither side is heavier, a not-lightest stone could be identified in two weighings, one swap.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
@noedne here's the better solution.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
What if neither side is heavier?
$endgroup$
– noedne
2 hours ago
$begingroup$
What if neither side is heavier?
$endgroup$
– noedne
2 hours ago
$begingroup$
@noedne see JaapScherpius's comment above. Also, if neither side is heavier, a not-lightest stone could be identified in two weighings, one swap.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
@noedne see JaapScherpius's comment above. Also, if neither side is heavier, a not-lightest stone could be identified in two weighings, one swap.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
@noedne here's the better solution.
$endgroup$
– Brandon_J
2 hours ago
$begingroup$
@noedne here's the better solution.
$endgroup$
– Brandon_J
2 hours ago
add a comment |
podloga123 is a new contributor. Be nice, and check out our Code of Conduct.
podloga123 is a new contributor. Be nice, and check out our Code of Conduct.
podloga123 is a new contributor. Be nice, and check out our Code of Conduct.
podloga123 is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Please give credit to the author of the puzzle.
$endgroup$
– Gilles
6 hours ago
$begingroup$
I have a new and improved solution!
$endgroup$
– Brandon_J
3 hours ago