Necessary conditions for a Sudoku puzzle to have no repetitions

Question

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!

Answer 1

No. For instance, this "sudoku" fulfills your conditions, but has some repetitions: $$ \begin{array}{|ccc|ccc|ccc|} \hline 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ \hline 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ \hline 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ 5&5&5&5&5&5&5&5&5\\ \hline \end{array} $$

Answer 2

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.

Answer 3

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.