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Is it true that if a Sudoku puzzle has the following features there will be no repetitions in rows, columns and $3 \times 3$ subsquares?

  • The sum of each row must be $45$
  • The sum of each column must be $45$
  • The sum of each $3 \times 3$ subsquare must be $45$

If so, why? Is there a mathematical proof? If not, why? Is there a case where these conditions are satisfied, but is there at least one repetition?

Thanks!

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    $\begingroup$ Uh, no.... Of course not. ... Consider a suduko grid consisting of 81 5s. The other way, if there are no reptitions in rows, columns and quatrants then each row, column and quadrant add s up to 45 but that is obvious. $\endgroup$ – fleablood Oct 9 '16 at 17:43
  • $\begingroup$ A $9 \times 9$ Sudoku matrix has no such repetitions by definition. $\endgroup$ – Rodrigo de Azevedo Oct 20 '16 at 11:24
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No. For instance, this "sudoku" fulfills your conditions, but has some repetitions: $$ \begin{array}{|ccc|ccc|ccc|} \hline 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ \hline 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ \hline 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ \hline \end{array} $$

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  • $\begingroup$ Yes, it does have "some" repetitions. I know that it is inappropriate to reply with "LOL" on a place like this, but that made me laugh. $\endgroup$ – Aaron Oct 9 '16 at 16:26
  • $\begingroup$ Thanks to all!! But the number 5 is the only case? $\endgroup$ – Antonio Franco Oct 9 '16 at 16:26
  • $\begingroup$ @AntonioFranco It has to balance around $5$ because the sum is supposed to be $45$, and you have nine entries. But it's possible to do with, for instance, $3,4,4,4,6,6,6,6,6$ instead of $5,5,5,5,5,5,5,5,5$. It probably takes some fiddling to make sure every row, column and box has the right amount of numbers, but it's definitely doable. $\endgroup$ – Arthur Oct 9 '16 at 16:31
  • $\begingroup$ Clear, but if I put in their place the numbers that I know, there may be many cases with ripeture figures? $\endgroup$ – Antonio Franco Oct 9 '16 at 16:35
  • $\begingroup$ There many millions of counter examples. There are more counter examples that dont work than examples that do work. Any combination that adds to 45 adds to 45. No reason that the need to be distinct. That the other columns and rows must also not be distinct is not a significant restriction. $\endgroup$ – fleablood Oct 9 '16 at 17:53
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If all the cells are distinct 1 through 9 then the sum is 1+2+....+9 =45. But there is utterly no reason earth to assume the converse, that is $a+b+.... +i = 45$ then they are all distinct.

For any $b,...,h =N$ we can have $a$ be any $1 \le a \le 45-N $ and $i = 45-N-a$. And we can determine values for the other rows and columns. Yes, it takes a bit of thought to actually work this out but there is no reason that that keeping them distinct will be a requirement.

Let's suppose for instance we have a grid labeled A1....A9..... I1.... I9 where every row, column and quadrant add up to 45. Then lets say we replace mk (where $A \le m \le I$ and $1\le k \le 9$) with mk + 1. Then we replace mj in the same column and quadrant with mk - 1$, replace nk in the same column and quadrant with nk-1 and nj with nj + 1. Then all the quadrants, columns and rows still add to 45 but one or the other or both grids are no longer distinct.

e.g suppose we have:

$\begin{array}{|ccc|ccc|ccc|} \hline 1&2&3&4&5&6&7&8&9\\ 4&5&6&7&8&9&1&2&3\\ 7&8&9&1&2&3&4&5&6\\ \hline 2&3&4&5&6&7&8&9&1\\ 5&6&7&8&9&1&2&3&4\\ 8&9&1&2&3&4&5&6&7\\ \hline {\color{red}3}&4&{\color{red}5}&6&7 &8&9&1&2\\ 6&7&8&9&1&2&3&4&5\\ {\color{red}9}&1&{\color{red}2}&3&4&5&6&7&8\\ \hline \end{array}$

and we replace it with

$\begin{array}{|ccc|ccc|ccc|} \hline 1&2&3&4&5&6&7&8&9\\ 4&5&6&7&8&9&1&2&3\\ 7&8&9&1&2&3&4&5&6\\ \hline 2&3&4&5&6&7&8&9&1\\ 5&6&7&8&9&1&2&3&4\\ 8&9&1&2&3&4&5&6&7\\ \hline {\color{blue}4}&4&{\color{blue}4}&6&7 &8&9&1&2\\ 6&7&8&9&1&2&3&4&5\\ {\color{blue}8}&1&{\color{blue}3}&3&4&5&6&7&8\\ \hline \end{array}$

Note, the sums must be the same but values need not be distinct.

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imagine each digit is a 5, then all summation to 45 are met and we clearly have repition, all that's necessary is a pattern with an average of 5 to pull this off.

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