A new nameless game

Question

This is a new game I came up with the night after my birthday. It's so simple that I couldn't sleep just thinking about it. The rules are easy: you have to place all the two-digit combinations (numbers from 00 to 99) in a grid, so that in each row and column, no two numbers have the same tens digit or the same ones digit. It’s like a double 10x10 Sudoku, but without regions, and forming a magic square (without counting the diagonals). Here I share an easy example, and I will share more later. The colors are decoration. Do you have any ideas for a good name for this simple but interesting game? Leave your comments!

Answer 1

PHEW! I know the post says this was an easy example - but the 13 hours and 4 complete restarts it took me to solve says otherwise!

Fantastic idea! Here's the solution:

As for a potential name

There are lots of great solutions in the comments, but my personal favourite I've thought of is DECADOKU!

Although their actual real name is a Graeco-Latin Square... but that doesn't sound quite as catchy :P

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Here's how to solve:

1.

Starting off, creating another grid that will automatically keep track of the numbers (just done with conditional formatting - custom formula =COUNTIF($B$4:$K$13, M4) > 0 for green, and =COUNTIF($B$4:$K$13, M4) > 1 for red to highlight if a number appears twice):

2:

Now starting with the edges as these have the most amount of clues. Labelling what left and right hand digits are required:



We can make some quick deductions. There is only one place for 9- in the top row, leaving the top right cell 6-. In fact the top right cell can only be 62.

This leaves one place for 3- in the right hand column, and in fact this must be 39 as 31 and 34 are taken. This means the bottom right cell is -4, and further up is -1. As 81 is taken, -1 must be 11, and the far right column is complete.

3:

Continuing around the edges, consider the bottom row. There is only one spot for 6-, and this must then be 60. This leaves one spot for -3 and -9, and as 23 is taken the bottom row can be solved.

This means the top left cell can only be 1-, placing 2- in the top row.

4:

Lets look at the remaining -9s, as there are only 3. 49 must be in the 5th row, and 09 in the 4th column. Lets also look at 3-s.

It turns out 35 must be in the 5th row as well, whilst 37 must be in the 4th row, leaving a 36/8 square.

Finally, checking the -3s. 13 must be in one of two cells in the middle of the 6th row, whilst 93 is also in the 5th row.

5:

Now looking at 2-s, and filling in all the candidates there is nothing obvious. However the candidates will be useful later on so we will keep them.

Filling in the -1 candidates also doesn't provide anything useful for now.

However going back to the first column and filling in the candidates, we at least learn 57 must be in this column. One small extra deduction is that the cell in R7C4 cannot be 20 as that leaves 2 cells that must be 27.

But consider the second digit of R6C8. The second digit can only be 3 or 7, but as 13 is in this row, it cannot be 3 so it must be a 7. Also consider R8C4. This has the candidates 67|16 - but 61 and 66 are both taken, as is 71 - so this cell must be 76!

6:

Finally a break through! (The last step alone took about 4 hours)

This means 38 must be in row 8, and 36 in row 3. It also means 26 must be in row 2, whilst 24 must be in row 4.

However in row 4, there is a cell with 09|04 candidates - this now can't end in 4 so it 00 or 90.

Now R8C8 can only end in 3, and as 43 is taken it must be 53 or 63.

This means further up the column, the 49/93 cell can now only be 49 - which means there is only one cell possible for 93! This in turn place 13 just below to the right.

At this point, lets write out the candidates properly:

7:

Note in column 7 we have a pair of cells that both must end with 4 or 8 - so these form a pair and 4,8 can be removed from other candidates.

This is important, as it takes out one of the 38 candidates, placing both 38 and then 36!

This leaves a cell in row 3 that must end with 7, and in fact must be either 57 or 77. However, 57 must be in the first column, so this cell must be 77!

The cell to the right of 36 now cannot end with 5, leaving only one place for -5 in column 8. In the same column, the cell that ends in 7 now must be 97 - which places 00 higher up, which in turn solves the 09|1 which must be 91.

8:

The top row can now be solved as 97 is taken. There is then only one cell possible for 90, and in the same column as 90 we can also place 24, 88, 53 and 41 to complete it. This means the 53/63 bottom right must be 63, which solves the 8th column.

From now on, there is not much need to update the candidates, as deductions can be made simply without them - so ignore them if they're not updated.

The 3rd row can now be completed, as well as the first column.

9:

We can now go row by row to complete the puzzle. The second row can be solved starting with 64, then 87, 45, 70 and then 26.

The 4th row then also solves, including solving the 35/37 pair.

The 5th row also quickly solves, but the 6th, doesn't quite solve, and neither does the 7th.



Instead we need to solve the other rows first, which takes out 52 and allows the rest of the puzzle to be solved! Giving the final solution:

Thanks for the great puzzle!