Show that every sequence of $2^n$ numbers taken from $A$ contains a consecutive block of numbers whose...












3












$begingroup$


Let $A$ be a set of $n$ positive integers. Show that every sequence of $2^n$ numbers taken from $A$ contains a consecutive block of numbers whose product is square.(For instance, {2,5,3,2,5,2,3,5} contains the block 5,3,2,5,2,3 .)



I think this has something to do with the pigeon-hole principle but apart from that I have no idea how to proceed any further.



Any hint guys?



Thank You






combinatorics







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$endgroup$

















    3












    $begingroup$


    Let $A$ be a set of $n$ positive integers. Show that every sequence of $2^n$ numbers taken from $A$ contains a consecutive block of numbers whose product is square.(For instance, {2,5,3,2,5,2,3,5} contains the block 5,3,2,5,2,3 .)



    I think this has something to do with the pigeon-hole principle but apart from that I have no idea how to proceed any further.



    Any hint guys?



    Thank You






    combinatorics







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let $A$ be a set of $n$ positive integers. Show that every sequence of $2^n$ numbers taken from $A$ contains a consecutive block of numbers whose product is square.(For instance, {2,5,3,2,5,2,3,5} contains the block 5,3,2,5,2,3 .)



      I think this has something to do with the pigeon-hole principle but apart from that I have no idea how to proceed any further.



      Any hint guys?



      Thank You






      combinatorics







      share|cite|improve this question









      $endgroup$




      Let $A$ be a set of $n$ positive integers. Show that every sequence of $2^n$ numbers taken from $A$ contains a consecutive block of numbers whose product is square.(For instance, {2,5,3,2,5,2,3,5} contains the block 5,3,2,5,2,3 .)



      I think this has something to do with the pigeon-hole principle but apart from that I have no idea how to proceed any further.



      Any hint guys?



      Thank You






      combinatorics




      combinatorics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Oct 29 '11 at 22:30









      RashmirathiRashmirathi

      32037




      32037






















          2 Answers
          2






          active

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          3












          $begingroup$

          Hint: You have a list of $2^n$ vectors of length $n$ over $mathbb{Z}_2$. Show that there is a consecutive block of vectors that sum to zero.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Length $leq n$?
            $endgroup$
            – zyx
            Oct 30 '11 at 2:12










          • $begingroup$
            @zyx, yes because we can pretend wlog that the numbers in $A$ are distinct primes. Then the criterion is just that the consecutive block must contain an even number of instances of each element.
            $endgroup$
            – Henning Makholm
            Oct 30 '11 at 6:59










          • $begingroup$
            @Henning, that observation was the point of my question. All numbers relatively prime is the worst case. In the easier case such as powers of one prime, the mod 2 vector space is of smaller dimension so one can either improve the statement of the problem (one does not quite need 2^n numbers from A but 2^(dimension of the mod 2 space for A), or the wording of the answer should be modified slightly. Assuming that you and I and Yuval are discussing the same vector space...
            $endgroup$
            – zyx
            Oct 30 '11 at 7:07












          • $begingroup$
            @zyx, I thought you were asking why $le n$ would always be enough. Even if a smaller space can sometimes work, it is easier always to use the same size and just pad with zeroes if not all dimensions are needed. It's just an existence proof we're after, after all.
            $endgroup$
            – Henning Makholm
            Oct 30 '11 at 7:12



















          0












          $begingroup$

          First I'm sorry for bad English. But there is a solution.



          Define sets for i = 1 to 2^n :



          M(i) = {x ; x is number with odd occurence in subsequence from position 1 to i}
          M(0) = {}



          For every i M(i) is subset of A.



          {M(0), M(1), …, M(2^n)} is set of 2^n + 1 set. But number of all subsets of A i 2^n. So there must exists j and k that M(j) = M(k). Subsequence from position j + 1 to position k is then a perfect square.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Hint: You have a list of $2^n$ vectors of length $n$ over $mathbb{Z}_2$. Show that there is a consecutive block of vectors that sum to zero.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Length $leq n$?
              $endgroup$
              – zyx
              Oct 30 '11 at 2:12










            • $begingroup$
              @zyx, yes because we can pretend wlog that the numbers in $A$ are distinct primes. Then the criterion is just that the consecutive block must contain an even number of instances of each element.
              $endgroup$
              – Henning Makholm
              Oct 30 '11 at 6:59










            • $begingroup$
              @Henning, that observation was the point of my question. All numbers relatively prime is the worst case. In the easier case such as powers of one prime, the mod 2 vector space is of smaller dimension so one can either improve the statement of the problem (one does not quite need 2^n numbers from A but 2^(dimension of the mod 2 space for A), or the wording of the answer should be modified slightly. Assuming that you and I and Yuval are discussing the same vector space...
              $endgroup$
              – zyx
              Oct 30 '11 at 7:07












            • $begingroup$
              @zyx, I thought you were asking why $le n$ would always be enough. Even if a smaller space can sometimes work, it is easier always to use the same size and just pad with zeroes if not all dimensions are needed. It's just an existence proof we're after, after all.
              $endgroup$
              – Henning Makholm
              Oct 30 '11 at 7:12
















            3












            $begingroup$

            Hint: You have a list of $2^n$ vectors of length $n$ over $mathbb{Z}_2$. Show that there is a consecutive block of vectors that sum to zero.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Length $leq n$?
              $endgroup$
              – zyx
              Oct 30 '11 at 2:12










            • $begingroup$
              @zyx, yes because we can pretend wlog that the numbers in $A$ are distinct primes. Then the criterion is just that the consecutive block must contain an even number of instances of each element.
              $endgroup$
              – Henning Makholm
              Oct 30 '11 at 6:59










            • $begingroup$
              @Henning, that observation was the point of my question. All numbers relatively prime is the worst case. In the easier case such as powers of one prime, the mod 2 vector space is of smaller dimension so one can either improve the statement of the problem (one does not quite need 2^n numbers from A but 2^(dimension of the mod 2 space for A), or the wording of the answer should be modified slightly. Assuming that you and I and Yuval are discussing the same vector space...
              $endgroup$
              – zyx
              Oct 30 '11 at 7:07












            • $begingroup$
              @zyx, I thought you were asking why $le n$ would always be enough. Even if a smaller space can sometimes work, it is easier always to use the same size and just pad with zeroes if not all dimensions are needed. It's just an existence proof we're after, after all.
              $endgroup$
              – Henning Makholm
              Oct 30 '11 at 7:12














            3












            3








            3





            $begingroup$

            Hint: You have a list of $2^n$ vectors of length $n$ over $mathbb{Z}_2$. Show that there is a consecutive block of vectors that sum to zero.






            share|cite|improve this answer









            $endgroup$



            Hint: You have a list of $2^n$ vectors of length $n$ over $mathbb{Z}_2$. Show that there is a consecutive block of vectors that sum to zero.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 29 '11 at 22:38









            Yuval FilmusYuval Filmus

            48.8k472146




            48.8k472146












            • $begingroup$
              Length $leq n$?
              $endgroup$
              – zyx
              Oct 30 '11 at 2:12










            • $begingroup$
              @zyx, yes because we can pretend wlog that the numbers in $A$ are distinct primes. Then the criterion is just that the consecutive block must contain an even number of instances of each element.
              $endgroup$
              – Henning Makholm
              Oct 30 '11 at 6:59










            • $begingroup$
              @Henning, that observation was the point of my question. All numbers relatively prime is the worst case. In the easier case such as powers of one prime, the mod 2 vector space is of smaller dimension so one can either improve the statement of the problem (one does not quite need 2^n numbers from A but 2^(dimension of the mod 2 space for A), or the wording of the answer should be modified slightly. Assuming that you and I and Yuval are discussing the same vector space...
              $endgroup$
              – zyx
              Oct 30 '11 at 7:07












            • $begingroup$
              @zyx, I thought you were asking why $le n$ would always be enough. Even if a smaller space can sometimes work, it is easier always to use the same size and just pad with zeroes if not all dimensions are needed. It's just an existence proof we're after, after all.
              $endgroup$
              – Henning Makholm
              Oct 30 '11 at 7:12


















            • $begingroup$
              Length $leq n$?
              $endgroup$
              – zyx
              Oct 30 '11 at 2:12










            • $begingroup$
              @zyx, yes because we can pretend wlog that the numbers in $A$ are distinct primes. Then the criterion is just that the consecutive block must contain an even number of instances of each element.
              $endgroup$
              – Henning Makholm
              Oct 30 '11 at 6:59










            • $begingroup$
              @Henning, that observation was the point of my question. All numbers relatively prime is the worst case. In the easier case such as powers of one prime, the mod 2 vector space is of smaller dimension so one can either improve the statement of the problem (one does not quite need 2^n numbers from A but 2^(dimension of the mod 2 space for A), or the wording of the answer should be modified slightly. Assuming that you and I and Yuval are discussing the same vector space...
              $endgroup$
              – zyx
              Oct 30 '11 at 7:07












            • $begingroup$
              @zyx, I thought you were asking why $le n$ would always be enough. Even if a smaller space can sometimes work, it is easier always to use the same size and just pad with zeroes if not all dimensions are needed. It's just an existence proof we're after, after all.
              $endgroup$
              – Henning Makholm
              Oct 30 '11 at 7:12
















            $begingroup$
            Length $leq n$?
            $endgroup$
            – zyx
            Oct 30 '11 at 2:12




            $begingroup$
            Length $leq n$?
            $endgroup$
            – zyx
            Oct 30 '11 at 2:12












            $begingroup$
            @zyx, yes because we can pretend wlog that the numbers in $A$ are distinct primes. Then the criterion is just that the consecutive block must contain an even number of instances of each element.
            $endgroup$
            – Henning Makholm
            Oct 30 '11 at 6:59




            $begingroup$
            @zyx, yes because we can pretend wlog that the numbers in $A$ are distinct primes. Then the criterion is just that the consecutive block must contain an even number of instances of each element.
            $endgroup$
            – Henning Makholm
            Oct 30 '11 at 6:59












            $begingroup$
            @Henning, that observation was the point of my question. All numbers relatively prime is the worst case. In the easier case such as powers of one prime, the mod 2 vector space is of smaller dimension so one can either improve the statement of the problem (one does not quite need 2^n numbers from A but 2^(dimension of the mod 2 space for A), or the wording of the answer should be modified slightly. Assuming that you and I and Yuval are discussing the same vector space...
            $endgroup$
            – zyx
            Oct 30 '11 at 7:07






            $begingroup$
            @Henning, that observation was the point of my question. All numbers relatively prime is the worst case. In the easier case such as powers of one prime, the mod 2 vector space is of smaller dimension so one can either improve the statement of the problem (one does not quite need 2^n numbers from A but 2^(dimension of the mod 2 space for A), or the wording of the answer should be modified slightly. Assuming that you and I and Yuval are discussing the same vector space...
            $endgroup$
            – zyx
            Oct 30 '11 at 7:07














            $begingroup$
            @zyx, I thought you were asking why $le n$ would always be enough. Even if a smaller space can sometimes work, it is easier always to use the same size and just pad with zeroes if not all dimensions are needed. It's just an existence proof we're after, after all.
            $endgroup$
            – Henning Makholm
            Oct 30 '11 at 7:12




            $begingroup$
            @zyx, I thought you were asking why $le n$ would always be enough. Even if a smaller space can sometimes work, it is easier always to use the same size and just pad with zeroes if not all dimensions are needed. It's just an existence proof we're after, after all.
            $endgroup$
            – Henning Makholm
            Oct 30 '11 at 7:12











            0












            $begingroup$

            First I'm sorry for bad English. But there is a solution.



            Define sets for i = 1 to 2^n :



            M(i) = {x ; x is number with odd occurence in subsequence from position 1 to i}
            M(0) = {}



            For every i M(i) is subset of A.



            {M(0), M(1), …, M(2^n)} is set of 2^n + 1 set. But number of all subsets of A i 2^n. So there must exists j and k that M(j) = M(k). Subsequence from position j + 1 to position k is then a perfect square.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              First I'm sorry for bad English. But there is a solution.



              Define sets for i = 1 to 2^n :



              M(i) = {x ; x is number with odd occurence in subsequence from position 1 to i}
              M(0) = {}



              For every i M(i) is subset of A.



              {M(0), M(1), …, M(2^n)} is set of 2^n + 1 set. But number of all subsets of A i 2^n. So there must exists j and k that M(j) = M(k). Subsequence from position j + 1 to position k is then a perfect square.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                First I'm sorry for bad English. But there is a solution.



                Define sets for i = 1 to 2^n :



                M(i) = {x ; x is number with odd occurence in subsequence from position 1 to i}
                M(0) = {}



                For every i M(i) is subset of A.



                {M(0), M(1), …, M(2^n)} is set of 2^n + 1 set. But number of all subsets of A i 2^n. So there must exists j and k that M(j) = M(k). Subsequence from position j + 1 to position k is then a perfect square.






                share|cite|improve this answer









                $endgroup$



                First I'm sorry for bad English. But there is a solution.



                Define sets for i = 1 to 2^n :



                M(i) = {x ; x is number with odd occurence in subsequence from position 1 to i}
                M(0) = {}



                For every i M(i) is subset of A.



                {M(0), M(1), …, M(2^n)} is set of 2^n + 1 set. But number of all subsets of A i 2^n. So there must exists j and k that M(j) = M(k). Subsequence from position j + 1 to position k is then a perfect square.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 19 '18 at 18:15









                RajkoRajko

                1




                1






























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