Largest smallest value in sudoku like puzzle












1












$begingroup$


In this post, a sudoku like math puzzle is proposed.



The grid must be filled while respecting a unique constraint : the sum of all $3times3$ sub-squares must equal $2019$.



It is not that difficult to complete the grid, and there are many different solutions. I have noticed empirically that in every solution, the smallest value (among all cells) is at most $4$.



Question : is there an algebraic reason for this ?



The grid is in the picture below.



enter image description here



Note : the statement is easy to prove with a linear solver. Solving the following problem shows that the largest smallest possible value is indeed $4$.
$$
max left{ min{ {x_{ij}} } right}
$$
subject to
$$
sum_{k=i}^{i+2}sum_{ell=j}^{j+2} x_{kell} = 2019 quad forall i,j = 1,...,5
$$
where $x_{ij}$ is the value in cell $(i,j)$.






combinatorics number-theory recreational-mathematics puzzle







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just to be clear; your observation is that there is always a number less than $5$?
    $endgroup$
    – Servaes
    Dec 3 '18 at 9:12










  • $begingroup$
    @Servaes : Yes exactly.
    $endgroup$
    – Kuifje
    Dec 3 '18 at 9:13






  • 2




    $begingroup$
    The number in middle cell of the bottom row is $4$. So the smallest value of any solution has to be at most $4$.
    $endgroup$
    – achille hui
    Dec 3 '18 at 9:16










  • $begingroup$
    Also just to be clear; all entries must be positive integers?
    $endgroup$
    – Servaes
    Dec 3 '18 at 9:17










  • $begingroup$
    @achillehui : aaaaa yes....of course. Thanks :) But how do you know another cell does not take value $3$ for example ?
    $endgroup$
    – Kuifje
    Dec 3 '18 at 9:18


















1












$begingroup$


In this post, a sudoku like math puzzle is proposed.



The grid must be filled while respecting a unique constraint : the sum of all $3times3$ sub-squares must equal $2019$.



It is not that difficult to complete the grid, and there are many different solutions. I have noticed empirically that in every solution, the smallest value (among all cells) is at most $4$.



Question : is there an algebraic reason for this ?



The grid is in the picture below.



enter image description here



Note : the statement is easy to prove with a linear solver. Solving the following problem shows that the largest smallest possible value is indeed $4$.
$$
max left{ min{ {x_{ij}} } right}
$$
subject to
$$
sum_{k=i}^{i+2}sum_{ell=j}^{j+2} x_{kell} = 2019 quad forall i,j = 1,...,5
$$
where $x_{ij}$ is the value in cell $(i,j)$.






combinatorics number-theory recreational-mathematics puzzle







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just to be clear; your observation is that there is always a number less than $5$?
    $endgroup$
    – Servaes
    Dec 3 '18 at 9:12










  • $begingroup$
    @Servaes : Yes exactly.
    $endgroup$
    – Kuifje
    Dec 3 '18 at 9:13






  • 2




    $begingroup$
    The number in middle cell of the bottom row is $4$. So the smallest value of any solution has to be at most $4$.
    $endgroup$
    – achille hui
    Dec 3 '18 at 9:16










  • $begingroup$
    Also just to be clear; all entries must be positive integers?
    $endgroup$
    – Servaes
    Dec 3 '18 at 9:17










  • $begingroup$
    @achillehui : aaaaa yes....of course. Thanks :) But how do you know another cell does not take value $3$ for example ?
    $endgroup$
    – Kuifje
    Dec 3 '18 at 9:18
















1












1








1





$begingroup$


In this post, a sudoku like math puzzle is proposed.



The grid must be filled while respecting a unique constraint : the sum of all $3times3$ sub-squares must equal $2019$.



It is not that difficult to complete the grid, and there are many different solutions. I have noticed empirically that in every solution, the smallest value (among all cells) is at most $4$.



Question : is there an algebraic reason for this ?



The grid is in the picture below.



enter image description here



Note : the statement is easy to prove with a linear solver. Solving the following problem shows that the largest smallest possible value is indeed $4$.
$$
max left{ min{ {x_{ij}} } right}
$$
subject to
$$
sum_{k=i}^{i+2}sum_{ell=j}^{j+2} x_{kell} = 2019 quad forall i,j = 1,...,5
$$
where $x_{ij}$ is the value in cell $(i,j)$.






combinatorics number-theory recreational-mathematics puzzle







share|cite|improve this question











$endgroup$




In this post, a sudoku like math puzzle is proposed.



The grid must be filled while respecting a unique constraint : the sum of all $3times3$ sub-squares must equal $2019$.



It is not that difficult to complete the grid, and there are many different solutions. I have noticed empirically that in every solution, the smallest value (among all cells) is at most $4$.



Question : is there an algebraic reason for this ?



The grid is in the picture below.



enter image description here



Note : the statement is easy to prove with a linear solver. Solving the following problem shows that the largest smallest possible value is indeed $4$.
$$
max left{ min{ {x_{ij}} } right}
$$
subject to
$$
sum_{k=i}^{i+2}sum_{ell=j}^{j+2} x_{kell} = 2019 quad forall i,j = 1,...,5
$$
where $x_{ij}$ is the value in cell $(i,j)$.






combinatorics number-theory recreational-mathematics puzzle




combinatorics number-theory recreational-mathematics puzzle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 9:16







Kuifje

















asked Dec 3 '18 at 9:06









KuifjeKuifje

7,1302725




7,1302725












  • $begingroup$
    Just to be clear; your observation is that there is always a number less than $5$?
    $endgroup$
    – Servaes
    Dec 3 '18 at 9:12










  • $begingroup$
    @Servaes : Yes exactly.
    $endgroup$
    – Kuifje
    Dec 3 '18 at 9:13






  • 2




    $begingroup$
    The number in middle cell of the bottom row is $4$. So the smallest value of any solution has to be at most $4$.
    $endgroup$
    – achille hui
    Dec 3 '18 at 9:16










  • $begingroup$
    Also just to be clear; all entries must be positive integers?
    $endgroup$
    – Servaes
    Dec 3 '18 at 9:17










  • $begingroup$
    @achillehui : aaaaa yes....of course. Thanks :) But how do you know another cell does not take value $3$ for example ?
    $endgroup$
    – Kuifje
    Dec 3 '18 at 9:18




















  • $begingroup$
    Just to be clear; your observation is that there is always a number less than $5$?
    $endgroup$
    – Servaes
    Dec 3 '18 at 9:12










  • $begingroup$
    @Servaes : Yes exactly.
    $endgroup$
    – Kuifje
    Dec 3 '18 at 9:13






  • 2




    $begingroup$
    The number in middle cell of the bottom row is $4$. So the smallest value of any solution has to be at most $4$.
    $endgroup$
    – achille hui
    Dec 3 '18 at 9:16










  • $begingroup$
    Also just to be clear; all entries must be positive integers?
    $endgroup$
    – Servaes
    Dec 3 '18 at 9:17










  • $begingroup$
    @achillehui : aaaaa yes....of course. Thanks :) But how do you know another cell does not take value $3$ for example ?
    $endgroup$
    – Kuifje
    Dec 3 '18 at 9:18


















$begingroup$
Just to be clear; your observation is that there is always a number less than $5$?
$endgroup$
– Servaes
Dec 3 '18 at 9:12




$begingroup$
Just to be clear; your observation is that there is always a number less than $5$?
$endgroup$
– Servaes
Dec 3 '18 at 9:12












$begingroup$
@Servaes : Yes exactly.
$endgroup$
– Kuifje
Dec 3 '18 at 9:13




$begingroup$
@Servaes : Yes exactly.
$endgroup$
– Kuifje
Dec 3 '18 at 9:13




2




2




$begingroup$
The number in middle cell of the bottom row is $4$. So the smallest value of any solution has to be at most $4$.
$endgroup$
– achille hui
Dec 3 '18 at 9:16




$begingroup$
The number in middle cell of the bottom row is $4$. So the smallest value of any solution has to be at most $4$.
$endgroup$
– achille hui
Dec 3 '18 at 9:16












$begingroup$
Also just to be clear; all entries must be positive integers?
$endgroup$
– Servaes
Dec 3 '18 at 9:17




$begingroup$
Also just to be clear; all entries must be positive integers?
$endgroup$
– Servaes
Dec 3 '18 at 9:17












$begingroup$
@achillehui : aaaaa yes....of course. Thanks :) But how do you know another cell does not take value $3$ for example ?
$endgroup$
– Kuifje
Dec 3 '18 at 9:18






$begingroup$
@achillehui : aaaaa yes....of course. Thanks :) But how do you know another cell does not take value $3$ for example ?
$endgroup$
– Kuifje
Dec 3 '18 at 9:18












1 Answer
1






active

oldest

votes


















0












$begingroup$

fleablood's answer to the previous question fills in some more cells. Labelling the remaining ones with variables we get $$begin{matrix}10 & a_{12} & a_{13} & 8 & a_{15} & a_{16} & 11 \
a_{21} & a_{22} & a_{23} & a_{24} & a_{25} & a_{26} & a_{27} \
a_{31} & a_{32} & a_{33} & a_{34} & a_{35} & a_{36} & a_{37} \
7 & a_{42} & a_{43} & 5 & a_{45} & a_{46} & 8 \
a_{51} & a_{52} & a_{53} & a_{54} & a_{55} & a_{56} & a_{57} \
a_{61} & a_{62} & a_{63} & a_{64} & a_{65} & a_{66} & a_{67} \
6 & a_{72} & a_{73} & 4 & a_{75} & a_{76} & 7end{matrix}$$
Then the constraints on the 3x3 subsquares give 25 0-1 linear equations, each involving 8 unknowns, and there are 40 unknowns in total so the system is very underdetermined. Write them all out as a matrix, perform row reduction, and some rows drop out as linear combinations of others, leaving 21 rows in row-reduced echelon form, many of which are the original constraints. The simpler equations relate pairs: e.g. $a_{21} + a_{31} + 2 = a_{24} + a_{34}$. But none of the remaining unknowns is determined, and there's nothing stopping them all from being approximately $frac{2000}{8} = 250$. For example, working by hand with one particular rref filling in 250s where not otherwise constrained I get a solution



$$begin{matrix}10 & 246 & 250 & 8 & 246 & 250 & 11 \
251 & 262 & 250 & 253 & 262 & 250 & 250 \
250 & 250 & 250 & 250 & 250 & 250 & 250 \
7 & 249 & 250 & 5 & 249 & 250 & 8 \
251 & 262 & 250 & 253 & 262 & 250 & 250 \
250 & 250 & 250 & 250 & 250 & 250 & 250 \
6 & 250 & 250 & 4 & 250 & 250 & 7end{matrix}$$






share|cite|improve this answer









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    $begingroup$

    fleablood's answer to the previous question fills in some more cells. Labelling the remaining ones with variables we get $$begin{matrix}10 & a_{12} & a_{13} & 8 & a_{15} & a_{16} & 11 \
    a_{21} & a_{22} & a_{23} & a_{24} & a_{25} & a_{26} & a_{27} \
    a_{31} & a_{32} & a_{33} & a_{34} & a_{35} & a_{36} & a_{37} \
    7 & a_{42} & a_{43} & 5 & a_{45} & a_{46} & 8 \
    a_{51} & a_{52} & a_{53} & a_{54} & a_{55} & a_{56} & a_{57} \
    a_{61} & a_{62} & a_{63} & a_{64} & a_{65} & a_{66} & a_{67} \
    6 & a_{72} & a_{73} & 4 & a_{75} & a_{76} & 7end{matrix}$$
    Then the constraints on the 3x3 subsquares give 25 0-1 linear equations, each involving 8 unknowns, and there are 40 unknowns in total so the system is very underdetermined. Write them all out as a matrix, perform row reduction, and some rows drop out as linear combinations of others, leaving 21 rows in row-reduced echelon form, many of which are the original constraints. The simpler equations relate pairs: e.g. $a_{21} + a_{31} + 2 = a_{24} + a_{34}$. But none of the remaining unknowns is determined, and there's nothing stopping them all from being approximately $frac{2000}{8} = 250$. For example, working by hand with one particular rref filling in 250s where not otherwise constrained I get a solution



    $$begin{matrix}10 & 246 & 250 & 8 & 246 & 250 & 11 \
    251 & 262 & 250 & 253 & 262 & 250 & 250 \
    250 & 250 & 250 & 250 & 250 & 250 & 250 \
    7 & 249 & 250 & 5 & 249 & 250 & 8 \
    251 & 262 & 250 & 253 & 262 & 250 & 250 \
    250 & 250 & 250 & 250 & 250 & 250 & 250 \
    6 & 250 & 250 & 4 & 250 & 250 & 7end{matrix}$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      fleablood's answer to the previous question fills in some more cells. Labelling the remaining ones with variables we get $$begin{matrix}10 & a_{12} & a_{13} & 8 & a_{15} & a_{16} & 11 \
      a_{21} & a_{22} & a_{23} & a_{24} & a_{25} & a_{26} & a_{27} \
      a_{31} & a_{32} & a_{33} & a_{34} & a_{35} & a_{36} & a_{37} \
      7 & a_{42} & a_{43} & 5 & a_{45} & a_{46} & 8 \
      a_{51} & a_{52} & a_{53} & a_{54} & a_{55} & a_{56} & a_{57} \
      a_{61} & a_{62} & a_{63} & a_{64} & a_{65} & a_{66} & a_{67} \
      6 & a_{72} & a_{73} & 4 & a_{75} & a_{76} & 7end{matrix}$$
      Then the constraints on the 3x3 subsquares give 25 0-1 linear equations, each involving 8 unknowns, and there are 40 unknowns in total so the system is very underdetermined. Write them all out as a matrix, perform row reduction, and some rows drop out as linear combinations of others, leaving 21 rows in row-reduced echelon form, many of which are the original constraints. The simpler equations relate pairs: e.g. $a_{21} + a_{31} + 2 = a_{24} + a_{34}$. But none of the remaining unknowns is determined, and there's nothing stopping them all from being approximately $frac{2000}{8} = 250$. For example, working by hand with one particular rref filling in 250s where not otherwise constrained I get a solution



      $$begin{matrix}10 & 246 & 250 & 8 & 246 & 250 & 11 \
      251 & 262 & 250 & 253 & 262 & 250 & 250 \
      250 & 250 & 250 & 250 & 250 & 250 & 250 \
      7 & 249 & 250 & 5 & 249 & 250 & 8 \
      251 & 262 & 250 & 253 & 262 & 250 & 250 \
      250 & 250 & 250 & 250 & 250 & 250 & 250 \
      6 & 250 & 250 & 4 & 250 & 250 & 7end{matrix}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        fleablood's answer to the previous question fills in some more cells. Labelling the remaining ones with variables we get $$begin{matrix}10 & a_{12} & a_{13} & 8 & a_{15} & a_{16} & 11 \
        a_{21} & a_{22} & a_{23} & a_{24} & a_{25} & a_{26} & a_{27} \
        a_{31} & a_{32} & a_{33} & a_{34} & a_{35} & a_{36} & a_{37} \
        7 & a_{42} & a_{43} & 5 & a_{45} & a_{46} & 8 \
        a_{51} & a_{52} & a_{53} & a_{54} & a_{55} & a_{56} & a_{57} \
        a_{61} & a_{62} & a_{63} & a_{64} & a_{65} & a_{66} & a_{67} \
        6 & a_{72} & a_{73} & 4 & a_{75} & a_{76} & 7end{matrix}$$
        Then the constraints on the 3x3 subsquares give 25 0-1 linear equations, each involving 8 unknowns, and there are 40 unknowns in total so the system is very underdetermined. Write them all out as a matrix, perform row reduction, and some rows drop out as linear combinations of others, leaving 21 rows in row-reduced echelon form, many of which are the original constraints. The simpler equations relate pairs: e.g. $a_{21} + a_{31} + 2 = a_{24} + a_{34}$. But none of the remaining unknowns is determined, and there's nothing stopping them all from being approximately $frac{2000}{8} = 250$. For example, working by hand with one particular rref filling in 250s where not otherwise constrained I get a solution



        $$begin{matrix}10 & 246 & 250 & 8 & 246 & 250 & 11 \
        251 & 262 & 250 & 253 & 262 & 250 & 250 \
        250 & 250 & 250 & 250 & 250 & 250 & 250 \
        7 & 249 & 250 & 5 & 249 & 250 & 8 \
        251 & 262 & 250 & 253 & 262 & 250 & 250 \
        250 & 250 & 250 & 250 & 250 & 250 & 250 \
        6 & 250 & 250 & 4 & 250 & 250 & 7end{matrix}$$






        share|cite|improve this answer









        $endgroup$



        fleablood's answer to the previous question fills in some more cells. Labelling the remaining ones with variables we get $$begin{matrix}10 & a_{12} & a_{13} & 8 & a_{15} & a_{16} & 11 \
        a_{21} & a_{22} & a_{23} & a_{24} & a_{25} & a_{26} & a_{27} \
        a_{31} & a_{32} & a_{33} & a_{34} & a_{35} & a_{36} & a_{37} \
        7 & a_{42} & a_{43} & 5 & a_{45} & a_{46} & 8 \
        a_{51} & a_{52} & a_{53} & a_{54} & a_{55} & a_{56} & a_{57} \
        a_{61} & a_{62} & a_{63} & a_{64} & a_{65} & a_{66} & a_{67} \
        6 & a_{72} & a_{73} & 4 & a_{75} & a_{76} & 7end{matrix}$$
        Then the constraints on the 3x3 subsquares give 25 0-1 linear equations, each involving 8 unknowns, and there are 40 unknowns in total so the system is very underdetermined. Write them all out as a matrix, perform row reduction, and some rows drop out as linear combinations of others, leaving 21 rows in row-reduced echelon form, many of which are the original constraints. The simpler equations relate pairs: e.g. $a_{21} + a_{31} + 2 = a_{24} + a_{34}$. But none of the remaining unknowns is determined, and there's nothing stopping them all from being approximately $frac{2000}{8} = 250$. For example, working by hand with one particular rref filling in 250s where not otherwise constrained I get a solution



        $$begin{matrix}10 & 246 & 250 & 8 & 246 & 250 & 11 \
        251 & 262 & 250 & 253 & 262 & 250 & 250 \
        250 & 250 & 250 & 250 & 250 & 250 & 250 \
        7 & 249 & 250 & 5 & 249 & 250 & 8 \
        251 & 262 & 250 & 253 & 262 & 250 & 250 \
        250 & 250 & 250 & 250 & 250 & 250 & 250 \
        6 & 250 & 250 & 4 & 250 & 250 & 7end{matrix}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 10:27









        Peter TaylorPeter Taylor

        8,80312341




        8,80312341






























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