6 balls and a scale












4












$begingroup$


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.






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    4












    $begingroup$


    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







    share|improve this question







    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.







    $endgroup$















      4












      4








      4





      $begingroup$


      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







      share|improve this question







      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






      share|improve this question







      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.











      share|improve this question







      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.









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      asked 23 hours ago









      user57753user57753

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          $begingroup$

          I believe you're almost right in your assessment. Instead of




          $log_3 binom{6}{2}$ I believe the minimum number of weighings is $lceil log_3 [binom{6}{2} times 2] rceil$




          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 $binom{6}{2}=15$ 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.







          share|improve this answer









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            $begingroup$

            I believe you're almost right in your assessment. Instead of




            $log_3 binom{6}{2}$ I believe the minimum number of weighings is $lceil log_3 [binom{6}{2} times 2] rceil$




            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 $binom{6}{2}=15$ 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.







            share|improve this answer









            $endgroup$


















              4












              $begingroup$

              I believe you're almost right in your assessment. Instead of




              $log_3 binom{6}{2}$ I believe the minimum number of weighings is $lceil log_3 [binom{6}{2} times 2] rceil$




              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 $binom{6}{2}=15$ 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.







              share|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                I believe you're almost right in your assessment. Instead of




                $log_3 binom{6}{2}$ I believe the minimum number of weighings is $lceil log_3 [binom{6}{2} times 2] rceil$




                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 $binom{6}{2}=15$ 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.







                share|improve this answer









                $endgroup$



                I believe you're almost right in your assessment. Instead of




                $log_3 binom{6}{2}$ I believe the minimum number of weighings is $lceil log_3 [binom{6}{2} times 2] rceil$




                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 $binom{6}{2}=15$ 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.








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                answered 20 hours ago









                AmorydaiAmorydai

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