6 balls and a scaleTwo Sheriffs and EavesdroppersSnow White and the Secret Message (Secret Sharing Problem)Two Sheriffs and Eavesdroppers - 2The oldest wins the prize, but they won't tell their age
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6 balls and a scale
Two Sheriffs and EavesdroppersSnow White and the Secret Message (Secret Sharing Problem)Two Sheriffs and Eavesdroppers - 2The oldest wins the prize, but they won't tell their age
$begingroup$
Suppose you are given 6 balls all of which look identical. You’re told that 4 of the balls all weigh the same, but there are 2 balls that have unequal weights. Additionally, these two balls together weigh the same as any two of the 4 balls having the same weight. Determine which two balls have different weights, and which one weighs less and which one weighs more, using a balance scale with the fewest number of weighings possible.
Hint given in book: First, calculate how many questions you have to answer about the possible relations between the weight of the balls, and use this to determine how many weighings with the balance scale are needed.
I believe the mathematical way of going about it would be taking log3 of (6 choose 2), but I'm not sure.
information-theory
asked yesterday
user57753user57753
284
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user57753 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Suppose you are given 6 balls all of which look identical. You’re told that 4 of the balls all weigh the same, but there are 2 balls that have unequal weights. Additionally, these two balls together weigh the same as any two of the 4 balls having the same weight. Determine which two balls have different weights, and which one weighs less and which one weighs more, using a balance scale with the fewest number of weighings possible.
Hint given in book: First, calculate how many questions you have to answer about the possible relations between the weight of the balls, and use this to determine how many weighings with the balance scale are needed.
I believe the mathematical way of going about it would be taking log3 of (6 choose 2), but I'm not sure.
information-theory
asked yesterday
user57753user57753
284
New contributor
user57753 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Suppose you are given 6 balls all of which look identical. You’re told that 4 of the balls all weigh the same, but there are 2 balls that have unequal weights. Additionally, these two balls together weigh the same as any two of the 4 balls having the same weight. Determine which two balls have different weights, and which one weighs less and which one weighs more, using a balance scale with the fewest number of weighings possible.
Hint given in book: First, calculate how many questions you have to answer about the possible relations between the weight of the balls, and use this to determine how many weighings with the balance scale are needed.
I believe the mathematical way of going about it would be taking log3 of (6 choose 2), but I'm not sure.
information-theory
asked yesterday
user57753user57753
284
New contributor
user57753 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Suppose you are given 6 balls all of which look identical. You’re told that 4 of the balls all weigh the same, but there are 2 balls that have unequal weights. Additionally, these two balls together weigh the same as any two of the 4 balls having the same weight. Determine which two balls have different weights, and which one weighs less and which one weighs more, using a balance scale with the fewest number of weighings possible.
Hint given in book: First, calculate how many questions you have to answer about the possible relations between the weight of the balls, and use this to determine how many weighings with the balance scale are needed.
I believe the mathematical way of going about it would be taking log3 of (6 choose 2), but I'm not sure.
information-theory
information-theory
asked yesterday
user57753user57753
284
New contributor
user57753 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
user57753user57753
284
New contributor
user57753 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
user57753user57753
284
New contributor
user57753 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
user57753user57753
284
asked yesterday
user57753user57753
284
284
New contributor
user57753 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user57753 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user57753 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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1 Answer
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$begingroup$
I believe you're almost right in your assessment. Instead of
$log_3 binom62$ I believe the minimum number of weighings is $lceil log_3 [binom62 times 2] rceil$
This is because each weighing
has three possible outcomes: either the scales balance, the right side is heavier or the left side is heavier. So if the number of weighings is $X$ then the maximum possible number of different scenarios you can obtain is $3^X$.
However, there are $binom62=15$ ways to choose 2 faulty balls out of 6, and 2 ways to have one be heavier and one lighter, for a total of 30 different ways. Since we are looking at worst case scenarios, this means that if you only have three weighings (27 max scenarios), there is a way for two different arrangements to give the same outcome and thus be indistinguishable from each other.
For this reason
The minimum number of weighings is $lceil log_3 (15 times 2) rceil=4$
It's trivial to come up with a solution for 4 weighings. Call the balls ABCDEF. Weigh AB, CD, EF. If two of the weighings are balanced then the other will give the heavy and light balls. If two of the weighings are unbalanced, then for the fourth weighing weigh the two heavy sides. The heaviest side is the heavy ball and the lighter side of the weighing that didn't include the heavy ball is the light ball. I do not think it can be done in less than 4 weighings.
answered yesterday
AmorydaiAmorydai
88512
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I believe you're almost right in your assessment. Instead of
$log_3 binom62$ I believe the minimum number of weighings is $lceil log_3 [binom62 times 2] rceil$
This is because each weighing
has three possible outcomes: either the scales balance, the right side is heavier or the left side is heavier. So if the number of weighings is $X$ then the maximum possible number of different scenarios you can obtain is $3^X$.
However, there are $binom62=15$ ways to choose 2 faulty balls out of 6, and 2 ways to have one be heavier and one lighter, for a total of 30 different ways. Since we are looking at worst case scenarios, this means that if you only have three weighings (27 max scenarios), there is a way for two different arrangements to give the same outcome and thus be indistinguishable from each other.
For this reason
The minimum number of weighings is $lceil log_3 (15 times 2) rceil=4$
It's trivial to come up with a solution for 4 weighings. Call the balls ABCDEF. Weigh AB, CD, EF. If two of the weighings are balanced then the other will give the heavy and light balls. If two of the weighings are unbalanced, then for the fourth weighing weigh the two heavy sides. The heaviest side is the heavy ball and the lighter side of the weighing that didn't include the heavy ball is the light ball. I do not think it can be done in less than 4 weighings.
answered yesterday
AmorydaiAmorydai
88512
$endgroup$
add a comment |
$begingroup$
I believe you're almost right in your assessment. Instead of
$log_3 binom62$ I believe the minimum number of weighings is $lceil log_3 [binom62 times 2] rceil$
This is because each weighing
has three possible outcomes: either the scales balance, the right side is heavier or the left side is heavier. So if the number of weighings is $X$ then the maximum possible number of different scenarios you can obtain is $3^X$.
However, there are $binom62=15$ ways to choose 2 faulty balls out of 6, and 2 ways to have one be heavier and one lighter, for a total of 30 different ways. Since we are looking at worst case scenarios, this means that if you only have three weighings (27 max scenarios), there is a way for two different arrangements to give the same outcome and thus be indistinguishable from each other.
For this reason
The minimum number of weighings is $lceil log_3 (15 times 2) rceil=4$
It's trivial to come up with a solution for 4 weighings. Call the balls ABCDEF. Weigh AB, CD, EF. If two of the weighings are balanced then the other will give the heavy and light balls. If two of the weighings are unbalanced, then for the fourth weighing weigh the two heavy sides. The heaviest side is the heavy ball and the lighter side of the weighing that didn't include the heavy ball is the light ball. I do not think it can be done in less than 4 weighings.
answered yesterday
AmorydaiAmorydai
88512
$endgroup$
add a comment |
$begingroup$
I believe you're almost right in your assessment. Instead of
$log_3 binom62$ I believe the minimum number of weighings is $lceil log_3 [binom62 times 2] rceil$
This is because each weighing
has three possible outcomes: either the scales balance, the right side is heavier or the left side is heavier. So if the number of weighings is $X$ then the maximum possible number of different scenarios you can obtain is $3^X$.
However, there are $binom62=15$ ways to choose 2 faulty balls out of 6, and 2 ways to have one be heavier and one lighter, for a total of 30 different ways. Since we are looking at worst case scenarios, this means that if you only have three weighings (27 max scenarios), there is a way for two different arrangements to give the same outcome and thus be indistinguishable from each other.
For this reason
The minimum number of weighings is $lceil log_3 (15 times 2) rceil=4$
It's trivial to come up with a solution for 4 weighings. Call the balls ABCDEF. Weigh AB, CD, EF. If two of the weighings are balanced then the other will give the heavy and light balls. If two of the weighings are unbalanced, then for the fourth weighing weigh the two heavy sides. The heaviest side is the heavy ball and the lighter side of the weighing that didn't include the heavy ball is the light ball. I do not think it can be done in less than 4 weighings.
answered yesterday
AmorydaiAmorydai
88512
$endgroup$
I believe you're almost right in your assessment. Instead of
$log_3 binom62$ I believe the minimum number of weighings is $lceil log_3 [binom62 times 2] rceil$
This is because each weighing
has three possible outcomes: either the scales balance, the right side is heavier or the left side is heavier. So if the number of weighings is $X$ then the maximum possible number of different scenarios you can obtain is $3^X$.
However, there are $binom62=15$ ways to choose 2 faulty balls out of 6, and 2 ways to have one be heavier and one lighter, for a total of 30 different ways. Since we are looking at worst case scenarios, this means that if you only have three weighings (27 max scenarios), there is a way for two different arrangements to give the same outcome and thus be indistinguishable from each other.
For this reason
The minimum number of weighings is $lceil log_3 (15 times 2) rceil=4$
It's trivial to come up with a solution for 4 weighings. Call the balls ABCDEF. Weigh AB, CD, EF. If two of the weighings are balanced then the other will give the heavy and light balls. If two of the weighings are unbalanced, then for the fourth weighing weigh the two heavy sides. The heaviest side is the heavy ball and the lighter side of the weighing that didn't include the heavy ball is the light ball. I do not think it can be done in less than 4 weighings.
answered yesterday
AmorydaiAmorydai
88512
answered yesterday
AmorydaiAmorydai
88512
answered yesterday
AmorydaiAmorydai
88512
answered yesterday
AmorydaiAmorydai
88512
88512
add a comment |
add a comment |
user57753 is a new contributor. Be nice, and check out our Code of Conduct.
user57753 is a new contributor. Be nice, and check out our Code of Conduct.
user57753 is a new contributor. Be nice, and check out our Code of Conduct.
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