Solving a colour sudoku

discrete mathematicspermutationspuzzlerecreational-mathematics

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

each row and column contains one and only one of each integer 1-5
each row and column contains one and only one of each colour (eg red, blue, yellow, green, black)
there is one and only one of each integer-colour combination (eg one blue 3)

It is easy to create a solved colour-sudoku. 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).

One can then conduct various transformations of this solution:

interchanging any two colours with each other, or interchanging any two numbers
swapping any two rows or columns
rotating the board

I have several questions relating to colour sudokus:

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?
Are there other solutions which cannot be generated through the transformations described above? If so, how many?
How would the answers to the above questions change if we played on a different size of grid?

Edit: this is a board in which 5 determines a unique solution. The large numbers in bold are the starting numbers.

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'). This enables us to figure out where '4A' and '5A' go, and also '2B' and '2E'. It's pretty straightforward after that.

Best Answer

$5$ clues is the minimum to force a unique solution on a $5 \times 5$ board. 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.

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. So, if you could do it with only $4$ clues, you definitely want all clues to be in different rows and columns.

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 of all different numbers and all different colors.

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):

\begin{array}{|c|c|c|c|c|} \hline A1&.&.&.&.\\ \hline .&B2&.&.&.\\ \hline .&.&C3&.&.\\ \hline .&.&.&D4&.\\ \hline .&.&.&.&.\\ \hline \end{array}

So, if we can find wo solutions to this puzzle, then we know that $4$ clues can never force a unique solution. And indeed there are two solutions:

\begin{array}{|c|c|c|c|c|} \hline A1&E3&D2&B5&C4\\ \hline E4&B2&A5&C1&D3\\ \hline D5&A4&C3&E2&B1\\ \hline B3&C5&E1&D4&A2\\ \hline C2&D1&B4&A3&E5\\ \hline \end{array}

\begin{array}{|c|c|c|c|c|} \hline A1&E4&D5&B3&C2\\ \hline E3&B2&A4&C5&D1\\ \hline D2&A5&C3&E1&B4\\ \hline B5&C1&E2&D4&A3\\ \hline C4&D3&B1&A2&E5\\ \hline \end{array}

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. swapping of rows or columns). 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.

This last observation partially answers your third question as well: the argument I gave above clearly generalizes to show that every $n \times n$ 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.

OK, but can you always force a unique solution with exactly $n$ clues? That is still an open question ... we know it works for $n=5$, but frankly I doubt you can do it for $n>5$.

As far as your second question goes, I found four valid boards that cannot be obtained from one another through swapping colors, numbers, row, columns, or doing any rotation or mirroring:

\begin{array}{|c|c|c|c|c|} \hline A1&B3&C4&D5&E2\\ \hline D4&A2&B5&E3&C1\\ \hline E5&D1&A3&C2&B4\\ \hline B2&C5&E1&A4&D3\\ \hline C3&E4&D2&B1&A5\\ \hline \end{array}

\begin{array}{|c|c|c|c|c|} \hline A1&B3&C4&D5&E2\\ \hline D3&A2&E5&C1&B4\\ \hline E4&C5&A3&B2&D1\\ \hline B5&E1&D2&A4&C3\\ \hline C2&D4&B1&E3&A5\\ \hline \end{array}

\begin{array}{|c|c|c|c|c|} \hline A1&B3&C2&D5&E4\\ \hline C4&A2&E5&B1&D3\\ \hline B5&D4&A3&E2&C1\\ \hline E3&C5&D1&A4&B2\\ \hline D2&E1&B4&C3&A5\\ \hline \end{array}

\begin{array}{|c|c|c|c|c|} \hline A1&B3&C2&D5&E4\\ \hline C5&A2&D4&E3&B1\\ \hline B4&E5&A3&C1&D2\\ \hline E2&D1&B5&A4&C3\\ \hline D3&C4&E1&B2&A5\\ \hline \end{array}

I am pretty sure that all other valid boards can be transformed into one of these $4$ through swapping colors, numbers, row, columns, or doing any rotation or mirroring. For example, the two earlier boards can be seen to be of the third type by putting all the $A$'s along he diagonal in order, followed by a diagonal mirroring. So, I am pretty sure the answer to your second question is $4$.

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[Math] Does a solved sudoku game always have same sum? Is this sum unique to solved game

Apropos your comment,

I'm guessing... that there is not an arithmetic way to check for set uniqueness?

there actually is a way, using only arithmetic, to check whether a sequence of nine numbers each between $1$ and $9$ contains all unique elements: for each number $n$ in the sequence, replace it with $2^{n-1}$, then add them all up and check that the sum is $511$. For example, the row $[5 | 3 | 4 | 6 | 7 | 8 | 9 | 1 | 2]$ is valid because $$2^{5-1} + 2^{3-1} + 2^{4-1} + 2^{6-1} + 2^{7-1} + 2^{8-1} + 2^{9-1} + 2^{1-1} + 2^{2-1} \\ = 16 + 4 + 8 + 32 + 64 + 128 + 256 + 1 + 2 \\ = 511,$$ but $[5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5]$ is not because $$2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} \\ = 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16 \\ = 144 \ne 511.$$

It should be clear that any permutation of the same set of numbers should give the same sum, so any valid solution should have the sum $511$. What is not immediately obvious is that no other sequence of nine numbers can give this sum. This is true because the Hamming weight of $511$ is $9$. The sum of $N$ numbers that are powers of $2$ can have Hamming weight at most $N$, and if there are duplicates among them, the weight of the sum is strictly less than $N$. So the only way to get a Hamming weight of $9$ by adding nine powers of $2$ is if they are all distinct.

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