Sum in a sudoku with sum total of 15 and 45












2












$begingroup$


I have created a sudoku puzzle with the following restrictions:





  • Each row and column sum to $45$.

  • Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.




Is such a sudoku unique?






grid-deduction sudoku







share|improve this question









New contributor




Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    "...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
    $endgroup$
    – Hugh
    2 hours ago












  • $begingroup$
    "where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
    $endgroup$
    – Kryesec
    2 hours ago










  • $begingroup$
    @Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
    $endgroup$
    – Hugh
    1 hour ago


















2












$begingroup$


I have created a sudoku puzzle with the following restrictions:





  • Each row and column sum to $45$.

  • Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.




Is such a sudoku unique?






grid-deduction sudoku







share|improve this question









New contributor




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







$endgroup$












  • $begingroup$
    "...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
    $endgroup$
    – Hugh
    2 hours ago












  • $begingroup$
    "where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
    $endgroup$
    – Kryesec
    2 hours ago










  • $begingroup$
    @Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
    $endgroup$
    – Hugh
    1 hour ago
















2












2








2





$begingroup$


I have created a sudoku puzzle with the following restrictions:





  • Each row and column sum to $45$.

  • Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.




Is such a sudoku unique?






grid-deduction sudoku







share|improve this question









New contributor




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







$endgroup$




I have created a sudoku puzzle with the following restrictions:





  • Each row and column sum to $45$.

  • Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.




Is such a sudoku unique?






grid-deduction sudoku




grid-deduction sudoku






share|improve this question









New contributor




Aorica Ireson 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




Aorica Ireson 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




share|improve this question








edited 1 hour ago









Hugh

1,6961721




1,6961721






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asked 2 hours ago









Aorica IresonAorica Ireson

111




111




New contributor




Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





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






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












  • $begingroup$
    "...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
    $endgroup$
    – Hugh
    2 hours ago












  • $begingroup$
    "where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
    $endgroup$
    – Kryesec
    2 hours ago










  • $begingroup$
    @Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
    $endgroup$
    – Hugh
    1 hour ago




















  • $begingroup$
    "...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
    $endgroup$
    – Hugh
    2 hours ago












  • $begingroup$
    "where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
    $endgroup$
    – Kryesec
    2 hours ago










  • $begingroup$
    @Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
    $endgroup$
    – Hugh
    1 hour ago


















$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago






$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago














$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
2 hours ago




$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
2 hours ago












$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago






$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago












2 Answers
2






active

oldest

votes


















5












$begingroup$


No.




Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.






share|improve this answer









$endgroup$





















    1












    $begingroup$


    I was beaten to a very similar answer, but you can create such a square based on the magic square below:

    abc
    def
    ghi

    Turn it into this:

    abcdefghi
    defghiabc
    ghiabcdef
    bcaefdhig
    efdhigbca
    higbcaefd
    cabfdeigh
    fdeighcab
    ighcabfde

    Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).







    share|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      5












      $begingroup$


      No.




      Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.






      share|improve this answer









      $endgroup$


















        5












        $begingroup$


        No.




        Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.






        share|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$


          No.




          Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.






          share|improve this answer









          $endgroup$




          No.




          Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 1 hour ago









          Dr XorileDr Xorile

          12.1k22567




          12.1k22567























              1












              $begingroup$


              I was beaten to a very similar answer, but you can create such a square based on the magic square below:

              abc
              def
              ghi

              Turn it into this:

              abcdefghi
              defghiabc
              ghiabcdef
              bcaefdhig
              efdhigbca
              higbcaefd
              cabfdeigh
              fdeighcab
              ighcabfde

              Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).







              share|improve this answer









              $endgroup$


















                1












                $begingroup$


                I was beaten to a very similar answer, but you can create such a square based on the magic square below:

                abc
                def
                ghi

                Turn it into this:

                abcdefghi
                defghiabc
                ghiabcdef
                bcaefdhig
                efdhigbca
                higbcaefd
                cabfdeigh
                fdeighcab
                ighcabfde

                Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).







                share|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$


                  I was beaten to a very similar answer, but you can create such a square based on the magic square below:

                  abc
                  def
                  ghi

                  Turn it into this:

                  abcdefghi
                  defghiabc
                  ghiabcdef
                  bcaefdhig
                  efdhigbca
                  higbcaefd
                  cabfdeigh
                  fdeighcab
                  ighcabfde

                  Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).







                  share|improve this answer









                  $endgroup$




                  I was beaten to a very similar answer, but you can create such a square based on the magic square below:

                  abc
                  def
                  ghi

                  Turn it into this:

                  abcdefghi
                  defghiabc
                  ghiabcdef
                  bcaefdhig
                  efdhigbca
                  higbcaefd
                  cabfdeigh
                  fdeighcab
                  ighcabfde

                  Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 25 mins ago









                  NautilusNautilus

                  3,751523




                  3,751523






















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