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The Wires Underground


The grades of the students in a classA puzzle of trust and lies, allies and spiesSink the SubmarineThe Magic Money MachineLabeling wires in a bundleMr. Hilbert and the Problem of the Rogue ResistorThe game of Shafiqa and HabibiA way to beat the system?A Strategy Game Involving Conquering of RegionsThe Alchemist's brewSolution or Origin of logic puzzle






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








3












$begingroup$


There are ten wires that run underground across a river. Thus on each of the banks, there are ten ends of the wires. Unfortunately, given an end of a wire on this bank, one doesn't know which end on the other bank it corresponds to. Suppose that you an ohmmeter and a bunch of wires. (You can then, for example, connect two ends on side A of the river, and go to the side B and measure the resistance between every pair of ends to figure out which pair on side B corresponds to the pair that is connected on side A). What is the minimum number of crossings you need to make in order to figure out the matching ends of all the ten wires?










share|improve this question









New contributor



dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$













  • $begingroup$
    This must be a duplicate
    $endgroup$
    – Dr Xorile
    7 hours ago










  • $begingroup$
    What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
    $endgroup$
    – Helena
    7 hours ago

















3












$begingroup$


There are ten wires that run underground across a river. Thus on each of the banks, there are ten ends of the wires. Unfortunately, given an end of a wire on this bank, one doesn't know which end on the other bank it corresponds to. Suppose that you an ohmmeter and a bunch of wires. (You can then, for example, connect two ends on side A of the river, and go to the side B and measure the resistance between every pair of ends to figure out which pair on side B corresponds to the pair that is connected on side A). What is the minimum number of crossings you need to make in order to figure out the matching ends of all the ten wires?










share|improve this question









New contributor



dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    This must be a duplicate
    $endgroup$
    – Dr Xorile
    7 hours ago










  • $begingroup$
    What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
    $endgroup$
    – Helena
    7 hours ago













3












3








3





$begingroup$


There are ten wires that run underground across a river. Thus on each of the banks, there are ten ends of the wires. Unfortunately, given an end of a wire on this bank, one doesn't know which end on the other bank it corresponds to. Suppose that you an ohmmeter and a bunch of wires. (You can then, for example, connect two ends on side A of the river, and go to the side B and measure the resistance between every pair of ends to figure out which pair on side B corresponds to the pair that is connected on side A). What is the minimum number of crossings you need to make in order to figure out the matching ends of all the ten wires?










share|improve this question









New contributor



dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




There are ten wires that run underground across a river. Thus on each of the banks, there are ten ends of the wires. Unfortunately, given an end of a wire on this bank, one doesn't know which end on the other bank it corresponds to. Suppose that you an ohmmeter and a bunch of wires. (You can then, for example, connect two ends on side A of the river, and go to the side B and measure the resistance between every pair of ends to figure out which pair on side B corresponds to the pair that is connected on side A). What is the minimum number of crossings you need to make in order to figure out the matching ends of all the ten wires?







strategy






share|improve this question









New contributor



dvc 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









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share|improve this question




share|improve this question








edited 8 hours ago









athin

13.2k3 gold badges41 silver badges108 bronze badges




13.2k3 gold badges41 silver badges108 bronze badges






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asked 8 hours ago









dvcdvc

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211 bronze badge




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  • $begingroup$
    This must be a duplicate
    $endgroup$
    – Dr Xorile
    7 hours ago










  • $begingroup$
    What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
    $endgroup$
    – Helena
    7 hours ago
















  • $begingroup$
    This must be a duplicate
    $endgroup$
    – Dr Xorile
    7 hours ago










  • $begingroup$
    What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
    $endgroup$
    – Helena
    7 hours ago















$begingroup$
This must be a duplicate
$endgroup$
– Dr Xorile
7 hours ago




$begingroup$
This must be a duplicate
$endgroup$
– Dr Xorile
7 hours ago












$begingroup$
What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
$endgroup$
– Helena
7 hours ago




$begingroup$
What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
$endgroup$
– Helena
7 hours ago










2 Answers
2






active

oldest

votes


















3













$begingroup$

To solve the problem, at least I can cross




$7$ times.




And the strategy is:




- Connect two wires, says wire $1$ and $2$, then cross.

- Find the corresponding connected wires of $1$ and $2$. Connect another two wires $3$ and $4$, then cross.

- Find the corresponding connected wires of $3$ and $4$. Connect another two wires $5$ and $6$, then cross.

- Find the corresponding connected wires of $5$ and $6$. Connect another two wires $7$ and $8$, then cross.

- Find the corresponding connected wires of $7$ and $8$. You also know which two lefts are $9$ and $10$. Now, connect wires corresponding to $1$ and $3$, then cross.

- Find the corresponding connected wires of $1$ and $3$. You also know for sure about $1$, $2$, $3$, and $4$. Now, connect wires corresponding to $5$ and $7$, then cross.

- Find the corresponding connected wires of $5$ and $7$. You also know for sure about $5$, $6$, $7$, and $8$. Now, connect wires corresponding to $2$ and $9$, then cross.

- Find the corresponding connected wires of $2$ and $9$. You also know for sure about $9$ and $10$.




For improvement, if




I can pair more than once, after the fourth crossing, I can just pair $1$ and $3$, $5$ and $7$, also $2$ and $9$ simultaneously. So I can cross only $5$ times.







share|improve this answer









$endgroup$






















    2













    $begingroup$

    The absolute minimum is




    2 crossings




    First some observations:




    Notice that when you connect ends of wires together, you are forming a partition. Notice that 10 is a triangular number, that is 10 = 1 + 2 + 3 + 4




    The actual strategy:




    So, first group wires into groups of one, two, three and four. Cross the river. You will be able to identify each group by continuity.




    Next




    Take one from each group, and form a group of four that way. Then take one from the remaining 3 groups, and so on. Once again, you have formed four differently sized groups, which will be identifiable once you cross the river. Additionally, since each group is formed by members of different groups, you will be able to tell which one is which.




    Proof:




    It obviously can't be solved with one crossing, as that would mean that you can connect the wires somehow on one end, and read something different on each wire.







    share|improve this answer








    New contributor



    dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





    $endgroup$

















      Your Answer








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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3













      $begingroup$

      To solve the problem, at least I can cross




      $7$ times.




      And the strategy is:




      - Connect two wires, says wire $1$ and $2$, then cross.

      - Find the corresponding connected wires of $1$ and $2$. Connect another two wires $3$ and $4$, then cross.

      - Find the corresponding connected wires of $3$ and $4$. Connect another two wires $5$ and $6$, then cross.

      - Find the corresponding connected wires of $5$ and $6$. Connect another two wires $7$ and $8$, then cross.

      - Find the corresponding connected wires of $7$ and $8$. You also know which two lefts are $9$ and $10$. Now, connect wires corresponding to $1$ and $3$, then cross.

      - Find the corresponding connected wires of $1$ and $3$. You also know for sure about $1$, $2$, $3$, and $4$. Now, connect wires corresponding to $5$ and $7$, then cross.

      - Find the corresponding connected wires of $5$ and $7$. You also know for sure about $5$, $6$, $7$, and $8$. Now, connect wires corresponding to $2$ and $9$, then cross.

      - Find the corresponding connected wires of $2$ and $9$. You also know for sure about $9$ and $10$.




      For improvement, if




      I can pair more than once, after the fourth crossing, I can just pair $1$ and $3$, $5$ and $7$, also $2$ and $9$ simultaneously. So I can cross only $5$ times.







      share|improve this answer









      $endgroup$



















        3













        $begingroup$

        To solve the problem, at least I can cross




        $7$ times.




        And the strategy is:




        - Connect two wires, says wire $1$ and $2$, then cross.

        - Find the corresponding connected wires of $1$ and $2$. Connect another two wires $3$ and $4$, then cross.

        - Find the corresponding connected wires of $3$ and $4$. Connect another two wires $5$ and $6$, then cross.

        - Find the corresponding connected wires of $5$ and $6$. Connect another two wires $7$ and $8$, then cross.

        - Find the corresponding connected wires of $7$ and $8$. You also know which two lefts are $9$ and $10$. Now, connect wires corresponding to $1$ and $3$, then cross.

        - Find the corresponding connected wires of $1$ and $3$. You also know for sure about $1$, $2$, $3$, and $4$. Now, connect wires corresponding to $5$ and $7$, then cross.

        - Find the corresponding connected wires of $5$ and $7$. You also know for sure about $5$, $6$, $7$, and $8$. Now, connect wires corresponding to $2$ and $9$, then cross.

        - Find the corresponding connected wires of $2$ and $9$. You also know for sure about $9$ and $10$.




        For improvement, if




        I can pair more than once, after the fourth crossing, I can just pair $1$ and $3$, $5$ and $7$, also $2$ and $9$ simultaneously. So I can cross only $5$ times.







        share|improve this answer









        $endgroup$

















          3














          3










          3







          $begingroup$

          To solve the problem, at least I can cross




          $7$ times.




          And the strategy is:




          - Connect two wires, says wire $1$ and $2$, then cross.

          - Find the corresponding connected wires of $1$ and $2$. Connect another two wires $3$ and $4$, then cross.

          - Find the corresponding connected wires of $3$ and $4$. Connect another two wires $5$ and $6$, then cross.

          - Find the corresponding connected wires of $5$ and $6$. Connect another two wires $7$ and $8$, then cross.

          - Find the corresponding connected wires of $7$ and $8$. You also know which two lefts are $9$ and $10$. Now, connect wires corresponding to $1$ and $3$, then cross.

          - Find the corresponding connected wires of $1$ and $3$. You also know for sure about $1$, $2$, $3$, and $4$. Now, connect wires corresponding to $5$ and $7$, then cross.

          - Find the corresponding connected wires of $5$ and $7$. You also know for sure about $5$, $6$, $7$, and $8$. Now, connect wires corresponding to $2$ and $9$, then cross.

          - Find the corresponding connected wires of $2$ and $9$. You also know for sure about $9$ and $10$.




          For improvement, if




          I can pair more than once, after the fourth crossing, I can just pair $1$ and $3$, $5$ and $7$, also $2$ and $9$ simultaneously. So I can cross only $5$ times.







          share|improve this answer









          $endgroup$



          To solve the problem, at least I can cross




          $7$ times.




          And the strategy is:




          - Connect two wires, says wire $1$ and $2$, then cross.

          - Find the corresponding connected wires of $1$ and $2$. Connect another two wires $3$ and $4$, then cross.

          - Find the corresponding connected wires of $3$ and $4$. Connect another two wires $5$ and $6$, then cross.

          - Find the corresponding connected wires of $5$ and $6$. Connect another two wires $7$ and $8$, then cross.

          - Find the corresponding connected wires of $7$ and $8$. You also know which two lefts are $9$ and $10$. Now, connect wires corresponding to $1$ and $3$, then cross.

          - Find the corresponding connected wires of $1$ and $3$. You also know for sure about $1$, $2$, $3$, and $4$. Now, connect wires corresponding to $5$ and $7$, then cross.

          - Find the corresponding connected wires of $5$ and $7$. You also know for sure about $5$, $6$, $7$, and $8$. Now, connect wires corresponding to $2$ and $9$, then cross.

          - Find the corresponding connected wires of $2$ and $9$. You also know for sure about $9$ and $10$.




          For improvement, if




          I can pair more than once, after the fourth crossing, I can just pair $1$ and $3$, $5$ and $7$, also $2$ and $9$ simultaneously. So I can cross only $5$ times.








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 7 hours ago









          athinathin

          13.2k3 gold badges41 silver badges108 bronze badges




          13.2k3 gold badges41 silver badges108 bronze badges


























              2













              $begingroup$

              The absolute minimum is




              2 crossings




              First some observations:




              Notice that when you connect ends of wires together, you are forming a partition. Notice that 10 is a triangular number, that is 10 = 1 + 2 + 3 + 4




              The actual strategy:




              So, first group wires into groups of one, two, three and four. Cross the river. You will be able to identify each group by continuity.




              Next




              Take one from each group, and form a group of four that way. Then take one from the remaining 3 groups, and so on. Once again, you have formed four differently sized groups, which will be identifiable once you cross the river. Additionally, since each group is formed by members of different groups, you will be able to tell which one is which.




              Proof:




              It obviously can't be solved with one crossing, as that would mean that you can connect the wires somehow on one end, and read something different on each wire.







              share|improve this answer








              New contributor



              dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.





              $endgroup$



















                2













                $begingroup$

                The absolute minimum is




                2 crossings




                First some observations:




                Notice that when you connect ends of wires together, you are forming a partition. Notice that 10 is a triangular number, that is 10 = 1 + 2 + 3 + 4




                The actual strategy:




                So, first group wires into groups of one, two, three and four. Cross the river. You will be able to identify each group by continuity.




                Next




                Take one from each group, and form a group of four that way. Then take one from the remaining 3 groups, and so on. Once again, you have formed four differently sized groups, which will be identifiable once you cross the river. Additionally, since each group is formed by members of different groups, you will be able to tell which one is which.




                Proof:




                It obviously can't be solved with one crossing, as that would mean that you can connect the wires somehow on one end, and read something different on each wire.







                share|improve this answer








                New contributor



                dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





                $endgroup$

















                  2














                  2










                  2







                  $begingroup$

                  The absolute minimum is




                  2 crossings




                  First some observations:




                  Notice that when you connect ends of wires together, you are forming a partition. Notice that 10 is a triangular number, that is 10 = 1 + 2 + 3 + 4




                  The actual strategy:




                  So, first group wires into groups of one, two, three and four. Cross the river. You will be able to identify each group by continuity.




                  Next




                  Take one from each group, and form a group of four that way. Then take one from the remaining 3 groups, and so on. Once again, you have formed four differently sized groups, which will be identifiable once you cross the river. Additionally, since each group is formed by members of different groups, you will be able to tell which one is which.




                  Proof:




                  It obviously can't be solved with one crossing, as that would mean that you can connect the wires somehow on one end, and read something different on each wire.







                  share|improve this answer








                  New contributor



                  dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  $endgroup$



                  The absolute minimum is




                  2 crossings




                  First some observations:




                  Notice that when you connect ends of wires together, you are forming a partition. Notice that 10 is a triangular number, that is 10 = 1 + 2 + 3 + 4




                  The actual strategy:




                  So, first group wires into groups of one, two, three and four. Cross the river. You will be able to identify each group by continuity.




                  Next




                  Take one from each group, and form a group of four that way. Then take one from the remaining 3 groups, and so on. Once again, you have formed four differently sized groups, which will be identifiable once you cross the river. Additionally, since each group is formed by members of different groups, you will be able to tell which one is which.




                  Proof:




                  It obviously can't be solved with one crossing, as that would mean that you can connect the wires somehow on one end, and read something different on each wire.








                  share|improve this answer








                  New contributor



                  dratini0 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 answer



                  share|improve this answer






                  New contributor



                  dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.








                  answered 4 hours ago









                  dratini0dratini0

                  211 bronze badge




                  211 bronze badge




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                  New contributor




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                  Check out our Code of Conduct.

























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