Repeat this GCD operation

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Problem A3 from the 2008 Putnam competition says:




Start with a finite sequence $a_1, a_2, dots, a_n$ of positive integers. If possible, choose two indices $j < k$ such that $a_j$ does not divide $a_k$, and replace $a_j$ and $a_k$ by $gcd(a_j, a_k)$ and $textlcm(a_j, a_k)$, respectively. Prove that if this process is repeated, it must eventually stop and the final sequence does not depend on the choices made.




Your goal in this challenge is to take a finite sequence of positive integers as input, and output the result of repeating this process until no further progress is possible. (That is, until every number in the resulting sequence divides all the numbers that come after it.) You don't need to solve the Putnam problem.



This is code-golf: the shortest solution in every programming language wins.



Test cases



[1, 2, 4, 8, 16, 32] => [1, 2, 4, 8, 16, 32]
[120, 24, 6, 2, 1, 1] => [1, 1, 2, 6, 24, 120]
[97, 41, 48, 12, 98, 68] => [1, 1, 2, 4, 12, 159016368]
[225, 36, 30, 1125, 36, 18, 180] => [3, 9, 18, 90, 180, 900, 4500]
[17, 17, 17, 17] => [17, 17, 17, 17]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] => [1, 1, 1, 1, 1, 2, 2, 6, 60, 2520]





code-golf array-manipulation number-theory







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    up vote
    4
    down vote

    favorite












    Problem A3 from the 2008 Putnam competition says:




    Start with a finite sequence $a_1, a_2, dots, a_n$ of positive integers. If possible, choose two indices $j < k$ such that $a_j$ does not divide $a_k$, and replace $a_j$ and $a_k$ by $gcd(a_j, a_k)$ and $textlcm(a_j, a_k)$, respectively. Prove that if this process is repeated, it must eventually stop and the final sequence does not depend on the choices made.




    Your goal in this challenge is to take a finite sequence of positive integers as input, and output the result of repeating this process until no further progress is possible. (That is, until every number in the resulting sequence divides all the numbers that come after it.) You don't need to solve the Putnam problem.



    This is code-golf: the shortest solution in every programming language wins.



    Test cases



    [1, 2, 4, 8, 16, 32] => [1, 2, 4, 8, 16, 32]
    [120, 24, 6, 2, 1, 1] => [1, 1, 2, 6, 24, 120]
    [97, 41, 48, 12, 98, 68] => [1, 1, 2, 4, 12, 159016368]
    [225, 36, 30, 1125, 36, 18, 180] => [3, 9, 18, 90, 180, 900, 4500]
    [17, 17, 17, 17] => [17, 17, 17, 17]
    [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] => [1, 1, 1, 1, 1, 2, 2, 6, 60, 2520]





    code-golf array-manipulation number-theory







    share|improve this question

























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      Problem A3 from the 2008 Putnam competition says:




      Start with a finite sequence $a_1, a_2, dots, a_n$ of positive integers. If possible, choose two indices $j < k$ such that $a_j$ does not divide $a_k$, and replace $a_j$ and $a_k$ by $gcd(a_j, a_k)$ and $textlcm(a_j, a_k)$, respectively. Prove that if this process is repeated, it must eventually stop and the final sequence does not depend on the choices made.




      Your goal in this challenge is to take a finite sequence of positive integers as input, and output the result of repeating this process until no further progress is possible. (That is, until every number in the resulting sequence divides all the numbers that come after it.) You don't need to solve the Putnam problem.



      This is code-golf: the shortest solution in every programming language wins.



      Test cases



      [1, 2, 4, 8, 16, 32] => [1, 2, 4, 8, 16, 32]
      [120, 24, 6, 2, 1, 1] => [1, 1, 2, 6, 24, 120]
      [97, 41, 48, 12, 98, 68] => [1, 1, 2, 4, 12, 159016368]
      [225, 36, 30, 1125, 36, 18, 180] => [3, 9, 18, 90, 180, 900, 4500]
      [17, 17, 17, 17] => [17, 17, 17, 17]
      [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] => [1, 1, 1, 1, 1, 2, 2, 6, 60, 2520]





      code-golf array-manipulation number-theory







      share|improve this question















      Problem A3 from the 2008 Putnam competition says:




      Start with a finite sequence $a_1, a_2, dots, a_n$ of positive integers. If possible, choose two indices $j < k$ such that $a_j$ does not divide $a_k$, and replace $a_j$ and $a_k$ by $gcd(a_j, a_k)$ and $textlcm(a_j, a_k)$, respectively. Prove that if this process is repeated, it must eventually stop and the final sequence does not depend on the choices made.




      Your goal in this challenge is to take a finite sequence of positive integers as input, and output the result of repeating this process until no further progress is possible. (That is, until every number in the resulting sequence divides all the numbers that come after it.) You don't need to solve the Putnam problem.



      This is code-golf: the shortest solution in every programming language wins.



      Test cases



      [1, 2, 4, 8, 16, 32] => [1, 2, 4, 8, 16, 32]
      [120, 24, 6, 2, 1, 1] => [1, 1, 2, 6, 24, 120]
      [97, 41, 48, 12, 98, 68] => [1, 1, 2, 4, 12, 159016368]
      [225, 36, 30, 1125, 36, 18, 180] => [3, 9, 18, 90, 180, 900, 4500]
      [17, 17, 17, 17] => [17, 17, 17, 17]
      [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] => [1, 1, 1, 1, 1, 2, 2, 6, 60, 2520]





      code-golf array-manipulation number-theory




      code-golf array-manipulation number-theory






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      edited 47 mins ago

























      asked 56 mins ago









      Misha Lavrov

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          2 Answers
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          Jelly, 9 bytes



          ÆEz0Ṣ€ZÆẸ


          Try it online!



          How it works



          ÆEz0Ṣ€ZÆẸ Main link. Argument: A (array)

          ÆE For each n in A, compute the exponents of n's prime factorization.
           E.g., 2000000 = 2⁷3⁰5⁶ gets mapped to [7, 0, 6].
           z0 Zip 0; append 0's to shorter exponent arrays to pad them to the same
           length, then read the resulting matrix by columns.
           Ṣ€ Sort the resulting arrays (exponents that correspond to the same prime).
           Z Zip; read the resulting matrix by columns, re-grouping the exponents by
           the integers they represent.
           ÆẸ Unexponents; map the exponent arrays back to integers.





          share|improve this answer





























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            0
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            Retina, 65 bytes



            d+
            *
            +`b((_+)(2)+)b(.*)b(?!1+b)(2+)b
            $2$4$5$#3*$5
            _+
            $.&


            Try it online! Link includes the faster test cases. Explanation:



            d+
            *


            Convert to unary.



            +`b((_+)(2)+)b(.*)b(?!1+b)(2+)b


            Repeatedly match: any number with a factor, then a later number that is not divisible by the first number but is divisible by the factor.



            $2$4$5$#3*$5


            $1 is the first number. $2 is the factor. Because regex is greedy this is the largest factor i.e the gcd. $4 is the part of the match between the original numbers. $5 is the second number. $#3 (in decimal rather than unary) is one less than $1 divided by $2, since it doesn't include the original $2. This means that to calculate the lcm we need to multiply $5 by one more than $#3 which is most succintly written as the sum of $5 and the product of $#3 and $5.



            _+
            $.&


            Convert to decimal.





            share




















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              2 Answers
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              active

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              active

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              up vote
              5
              down vote














              Jelly, 9 bytes



              ÆEz0Ṣ€ZÆẸ


              Try it online!



              How it works



              ÆEz0Ṣ€ZÆẸ Main link. Argument: A (array)

              ÆE For each n in A, compute the exponents of n's prime factorization.
               E.g., 2000000 = 2⁷3⁰5⁶ gets mapped to [7, 0, 6].
               z0 Zip 0; append 0's to shorter exponent arrays to pad them to the same
               length, then read the resulting matrix by columns.
               Ṣ€ Sort the resulting arrays (exponents that correspond to the same prime).
               Z Zip; read the resulting matrix by columns, re-grouping the exponents by
               the integers they represent.
               ÆẸ Unexponents; map the exponent arrays back to integers.





              share|improve this answer


























                up vote
                5
                down vote














                Jelly, 9 bytes



                ÆEz0Ṣ€ZÆẸ


                Try it online!



                How it works



                ÆEz0Ṣ€ZÆẸ Main link. Argument: A (array)

                ÆE For each n in A, compute the exponents of n's prime factorization.
                 E.g., 2000000 = 2⁷3⁰5⁶ gets mapped to [7, 0, 6].
                 z0 Zip 0; append 0's to shorter exponent arrays to pad them to the same
                 length, then read the resulting matrix by columns.
                 Ṣ€ Sort the resulting arrays (exponents that correspond to the same prime).
                 Z Zip; read the resulting matrix by columns, re-grouping the exponents by
                 the integers they represent.
                 ÆẸ Unexponents; map the exponent arrays back to integers.





                share|improve this answer
























                  up vote
                  5
                  down vote










                  up vote
                  5
                  down vote










                  Jelly, 9 bytes



                  ÆEz0Ṣ€ZÆẸ


                  Try it online!



                  How it works



                  ÆEz0Ṣ€ZÆẸ Main link. Argument: A (array)

                  ÆE For each n in A, compute the exponents of n's prime factorization.
                   E.g., 2000000 = 2⁷3⁰5⁶ gets mapped to [7, 0, 6].
                   z0 Zip 0; append 0's to shorter exponent arrays to pad them to the same
                   length, then read the resulting matrix by columns.
                   Ṣ€ Sort the resulting arrays (exponents that correspond to the same prime).
                   Z Zip; read the resulting matrix by columns, re-grouping the exponents by
                   the integers they represent.
                   ÆẸ Unexponents; map the exponent arrays back to integers.





                  share|improve this answer















                  Jelly, 9 bytes



                  ÆEz0Ṣ€ZÆẸ


                  Try it online!



                  How it works



                  ÆEz0Ṣ€ZÆẸ Main link. Argument: A (array)

                  ÆE For each n in A, compute the exponents of n's prime factorization.
                   E.g., 2000000 = 2⁷3⁰5⁶ gets mapped to [7, 0, 6].
                   z0 Zip 0; append 0's to shorter exponent arrays to pad them to the same
                   length, then read the resulting matrix by columns.
                   Ṣ€ Sort the resulting arrays (exponents that correspond to the same prime).
                   Z Zip; read the resulting matrix by columns, re-grouping the exponents by
                   the integers they represent.
                   ÆẸ Unexponents; map the exponent arrays back to integers.






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 17 mins ago

























                  answered 27 mins ago









                  Dennis♦

                  183k32293725




                  183k32293725




















                      up vote
                      0
                      down vote














                      Retina, 65 bytes



                      d+
                      *
                      +`b((_+)(2)+)b(.*)b(?!1+b)(2+)b
                      $2$4$5$#3*$5
                      _+
                      $.&


                      Try it online! Link includes the faster test cases. Explanation:



                      d+
                      *


                      Convert to unary.



                      +`b((_+)(2)+)b(.*)b(?!1+b)(2+)b


                      Repeatedly match: any number with a factor, then a later number that is not divisible by the first number but is divisible by the factor.



                      $2$4$5$#3*$5


                      $1 is the first number. $2 is the factor. Because regex is greedy this is the largest factor i.e the gcd. $4 is the part of the match between the original numbers. $5 is the second number. $#3 (in decimal rather than unary) is one less than $1 divided by $2, since it doesn't include the original $2. This means that to calculate the lcm we need to multiply $5 by one more than $#3 which is most succintly written as the sum of $5 and the product of $#3 and $5.



                      _+
                      $.&


                      Convert to decimal.





                      share
























                        up vote
                        0
                        down vote














                        Retina, 65 bytes



                        d+
                        *
                        +`b((_+)(2)+)b(.*)b(?!1+b)(2+)b
                        $2$4$5$#3*$5
                        _+
                        $.&


                        Try it online! Link includes the faster test cases. Explanation:



                        d+
                        *


                        Convert to unary.



                        +`b((_+)(2)+)b(.*)b(?!1+b)(2+)b


                        Repeatedly match: any number with a factor, then a later number that is not divisible by the first number but is divisible by the factor.



                        $2$4$5$#3*$5


                        $1 is the first number. $2 is the factor. Because regex is greedy this is the largest factor i.e the gcd. $4 is the part of the match between the original numbers. $5 is the second number. $#3 (in decimal rather than unary) is one less than $1 divided by $2, since it doesn't include the original $2. This means that to calculate the lcm we need to multiply $5 by one more than $#3 which is most succintly written as the sum of $5 and the product of $#3 and $5.



                        _+
                        $.&


                        Convert to decimal.





                        share






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote










                          Retina, 65 bytes



                          d+
                          *
                          +`b((_+)(2)+)b(.*)b(?!1+b)(2+)b
                          $2$4$5$#3*$5
                          _+
                          $.&


                          Try it online! Link includes the faster test cases. Explanation:



                          d+
                          *


                          Convert to unary.



                          +`b((_+)(2)+)b(.*)b(?!1+b)(2+)b


                          Repeatedly match: any number with a factor, then a later number that is not divisible by the first number but is divisible by the factor.



                          $2$4$5$#3*$5


                          $1 is the first number. $2 is the factor. Because regex is greedy this is the largest factor i.e the gcd. $4 is the part of the match between the original numbers. $5 is the second number. $#3 (in decimal rather than unary) is one less than $1 divided by $2, since it doesn't include the original $2. This means that to calculate the lcm we need to multiply $5 by one more than $#3 which is most succintly written as the sum of $5 and the product of $#3 and $5.



                          _+
                          $.&


                          Convert to decimal.





                          share













                          Retina, 65 bytes



                          d+
                          *
                          +`b((_+)(2)+)b(.*)b(?!1+b)(2+)b
                          $2$4$5$#3*$5
                          _+
                          $.&


                          Try it online! Link includes the faster test cases. Explanation:



                          d+
                          *


                          Convert to unary.



                          +`b((_+)(2)+)b(.*)b(?!1+b)(2+)b


                          Repeatedly match: any number with a factor, then a later number that is not divisible by the first number but is divisible by the factor.



                          $2$4$5$#3*$5


                          $1 is the first number. $2 is the factor. Because regex is greedy this is the largest factor i.e the gcd. $4 is the part of the match between the original numbers. $5 is the second number. $#3 (in decimal rather than unary) is one less than $1 divided by $2, since it doesn't include the original $2. This means that to calculate the lcm we need to multiply $5 by one more than $#3 which is most succintly written as the sum of $5 and the product of $#3 and $5.



                          _+
                          $.&


                          Convert to decimal.






                          share











                          share


                          share










                          answered 5 mins ago









                          Neil

                          77.2k744174




                          77.2k744174



























                               

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