Alphabetical Sudoku Trivia Quiz

In my explanations, I will use the word square or blank to refer to the smallest squares, each of which holds one letter (or, at the start, number). The word box refers to a 9x9 group of squares; there are nine boxes, each outlined with a heavier line, which form the grid - the term that describes the entire assembly of 81 squares.

The top two rows in this bottom set of boxes already have an H, so in the righthand box the H must be in the bottom row, where this is the only available square. this means that 1 = H.

If you started here, there will not be much more to fill yet, as there are at least two places where each of the five remaining Hs could go. It might be time to move on, and check later on, with more information, if there are more spots where an H is going to be located.

The leftmost column has two blanks, and the two letters to be filled into them are A and B. The top blank is in a row that already has an A, so the A must go in the lower one, left box of the middle row of the grid. This means that 2 = A. Let's fill in more As to reduce the number of blanks in the grid.
Looking at the middle three boxes, there is already an A in the left and right columns, so there must be one in the middle column of the bottom box. Since the lower blank is in a row that already has an A, there must be an A in the upper box.

We now have A in two of the three rows of the bottom set of boxes, so there must be one in the middle row of the left box.

Looking at the right set of boxes, there is an A in each of the left and right columns, so there must be one in the middle column of the top box. There is only one blank, so it goes there.

The last A only has one spot where it can fit. The leftmost box of the top row has a blank in its bottom row, and the A goes there.

You may have noted that the leftmost column of the grid now only has one square without a letter, and you can proceed to identify the letter that replaces the 8. You may also choose to move on to one of the other numbers where only one letter is possible (which is how my explanation will proceed). This decision was based on the fact that proceeding to the 3 will immediately give me a bunch of filled in squares, more than the 8.

The right hand boxes have C in the middle and right columns, so there must be one in the left column of the top box. The only blank is the bottom one, so 3 = C. We can fill in a few more of them: the bottom row of boxes has C in the top and bottom rows, so the middle row must have C in its only blank, the right one.

The middle vertical group of boxes now has C in its left and right columns, so there must be one in the middle column of the top box. Since the upper of the two that are still blank is in a row with C already there, this must be the lower of the two. That is C completely in place.

The box numbered 4 is the only blank in its row once you have identified and filled the three letters that have always been defined (1 = H, 2 = A, 3 = C). This means that 4 = E.

In that top group of boxes, the leftmost must have an E in its middle row; since the middle one is in the same column as another E, the E goes in the right blank in the middle row. In the bottom group of boxes, there are two spots, for a B and an E. The B has to go in the left one (as the right one is in a column that already has a B), and the E in the right one.

By this time, you should have filled in all the instances of A, C and H (which were determined at the start). My explanation assumes you have also filled in a number of B and E squares; if this is not the case, your explanation of how you determined this space will be different.

Looking at the column that includes the number 5, it has two blanks left, for a D and an E. The lower of them is in a row that already has a D, so it must be one of the remaining Es, and the one numbered 5 must be D. So 5 = D. The middle group of boxes now has D in its middle and lower rows, so the center box must have D in its upper row, where there is only one blank for it.

My explanation assumes you have also placed a number of other letters. If you have all the As and Cs in place, and have updated as many entries as possible for E, D and H, things are starting to get clear.

The top row of the lower righthand box has a space for a D (if not already there), which must go in the righthand box, since the left one is in a column that already has a D. This means that the only letter left to go in the box numbered 6 is the I.

The bottom set of boxes now has I in two spots; the last one must be the blank in the bottom row of the left box. The left column of boxes now has I in its left and right columns, so the third one must be in the middle column of the top box. Since the lower of these is in a row that already has an I, it must be the top one. There is one I left, and it goes in the very right spot in the middle row of the grid.

By this time, you should have filled in all the instances of A and C (which were determined at the start) as well as a number of the other letters. My process at this stage has left the box where the number 7 appears having two blanks, which must be an F and an H.

The lower one is in a row that already has an F, so F must go in the top one, which means the 7 = F. The other square is an H, which can be filled in (but is only necessary if you want a complete solution, as is also true for the remaining Fs).

Once you have worked out that 2 = A (one of the two letters whose position was completely determined from the start), the leftmost column only has one letter to be added, so 8 = B. (You may already have worked this out!) Again, you can fill in more of them, but it is not necessary to finish identifying which letter goes with each number.

You may have worked out that the letter in this spot is G before I did, but it is certainly not the place to start, as there is no clear way to decide without filling in some of the empty squares.


In my solution process, I identified the other eight numbers before the 9, so by elimination 9 had to be G. It may well have been possible to reach this conclusion earlier, but that is not the way I approached things. It is important to remember that there is not only one way to produce a logical solution!