This last observation partially answers your third question as well: the argument I gave above clearly generalizes to show that every $n imes n$ puzzle of this kind will require at least $n$ clues to force a unique solution: with $n-1$ clues, they have to be, without loss of generalization, all along the diagonal, and hence if there is any solution at all, there will always be another. Indeed, I didn't have to provide two solutions at all to make the point that the $4$ clues as indicated cannot force a unique solution, since given that all the clues are on the diagonal, then if it has any solution at all, then it has a mirror solution as well.

Note that the second solution is the first solution mirrored along the diagonal with the clues (which itself can be achieved through a single rotation together with a vertical or horizontal mirroring, i.e. So, if we can find wo solutions to this puzzle, then we know that $4$ clues can never force a unique solution. Without loss of generalization, we can therefore say that the clues are $1A$, $B2$, $C3$, and $D4$, and also without loss of generalization we can assume the clues are placed as follows (remember that with the clues in all different rows and columns, we can keep swapping any two rows and columns to end up in this configuration): So, if you could do it with only $4$ clues, you definitely want all clues to be of all different numbers and all different colors. Likewise, if you have two clues with the same number (or color), then there will be two numbers (or colors) that are not used any of the clues, and hence you can swap those numbers (colors) for any solution to obtain another solution. So, if you could do it with only $4$ clues, you definitely want all clues to be in different rows and columns. If you have $4$ clues, and two of the clues are in the same row (or column), then there will be two rows (or two columns) that have no clues, and those rows (columns) can be swapped for any solution to obtain another solution. You showed that there is indeed a way to force a unique solution with $5$ clues, and below is a proof that you cannot force a unique solution with less than $5$ clues. $5$ clues is the minimum to force a unique solution on a $5 imes 5$ board. This enables us to figure out where '4A' and '5A' go, and also '2B' and '2E'.

It is easy to deduce that '2A' must go in the top right corner (all other rows and columns have either a '2' or an 'A'). The large numbers in bold are the starting numbers. How would the answers to the above questions change if we played on a different size of grid?Д®dit: this is a board in which 5 determines a unique solution. What is the minimum number of squares which must be filled in order to determine a unique solution? I have done it with 5, but might it be done with fewer?ДЄre there other solutions which cannot be generated through the transformations described above? If so, how many? I have several questions relating to colour sudokus: interchanging any two colours with each other, or interchanging any two numbers.One can then conduct various transformations of this solution: One simply cycles through the integers in one direction (first row is 12345, second row is 23451, third row is 34512 etc) and cycles through the colours in the opposite direction (first row is ABCDE, second row is EABCD, third row is DEABC). It is easy to create a solved colour-sudoku. there is one and only one of each integer-colour combination (eg one blue 3).each row and column contains one and only one of each colour (eg red, blue, yellow, green, black).each row and column contains one and only one of each integer 1-5.

I have designed a new type of sudoku-like puzzle, done on a 5*5 grid, with the following rules:

Easy color sudoku