*The puzzle above is from the WPF Sudoku GP, Round 4 2016 #13. To solve it, each row, column and marked 3×3 box must contain the digits 1 to 9 once each, and the solution must have, in the marked rows, maximal odd substrings summing to the given totals in the correct order.*

I’ve always enjoyed a good puzzle, though I’ve never considered myself especially good at them. That said, I’ve done my fair share of sudoku, including several books where puzzles were labelled as very difficult; yet, fairly often identifying triples, or the odd X-wing, XY-wing or unique rectangle would be sufficient to blow through the puzzles. At least, I thought that clearing some “very difficult” puzzles in 8 or 10 minutes was blowing through them; I found the World Puzzle Federation’s Sudoku GP, where some of these had benchmark times along the lines of 3-4 minutes!

Furthermore, in addition to your standard “classic” puzzles which feature the standard rules (i.e. each row, column and marked 3×3 box must contain the digits from 1 to 9 once each), there are a couple of interesting variants that require understanding new rulesets and figuring out how to deal with them. After a while, as mentioned the classic puzzles largely tend to be solved with standard techniques, and the challenge there is dispatching them with extreme speed; the variants, on the other hand, tend to maintain some degree of freshness (at least as far as I’m concerned).

The above puzzle was an example of this; several of the rows are marked with additional clues, and for these rows, the largest odd substrings in the row must sum to the values indicated by the clues. For example, if a row was 178356924, the clue associated with it would be 8, 8, 9. Note that this is also generated by several other valid sequences, such as 532174689 or 174283569.

For the above puzzle, the 9, 5, 3, 8 clue is a good starting point, and indicates that we have 9, 5, 3 and (17 or 71) as substrings of the solution to that row in some order; given the position of 6, we can conclude that the 9 and the 3 must occur before the 6 (though we don’t know). We thus have 9 in the bottom left hand corner, an even number (either 2 or 4), and then a 3. Now, reading the first column clue 1, 12, 3, 9; we have 1, (57 or 75), 3 and 9 as substrings. The 9 is at the end, and 3 cannot occur before the 4 as there’s no space to fit it. Thus 3 occurs after the 4, but it has to be separated from the 9 by an even number. More simply, the second column clue 10, 9, 6 indicates that our 5 has to be preceded by an even number and then a 1. There’s only one spot left for a 7 in this lower box, so we can insert it.

We thus reach this position:

Moving on, we can consider the middle column. 3 can only be satisfied by a singular 3, so we know that the 3 in the bottom row **cannot** be in the middle column. It thus must be in the sixth column, with 1 and 7 taking up the eighth and ninth columns (in some order).

Now consider the rightmost column. It is 3, 10, 12; 3 can only be satisfied by a singular 3, so the odd substrings must be 3, (19 or 91) and (57 or 75). Given the position of the 2 (fourth row), it must be the case that the column has 1 and 9 in the fifth and sixth rows, an even number in the seventh, and 5 and 7 in the eighth and ninth. We thus clearly have 7 in the corner, and 5 above it.

Finally, we can actually decide on the ordering of 1 and 9 as well; consider the hint for the middle row, which ends with 8; a 9 would automatically exceed this. Thus the middle row must contain the 1 and the sixth row the 9. The hint gives us for free that we have a 7 to the left of the 1.

There are more interesting deductions we can make as well. Consider the seventh column. The clue is 16, 9 and this can be satisfied in two ways: {1357} in some order and then a singleton 9, or {79} and {135} in some order. However, we can prove that the latter construction is impossible; the middle row is an even number, and also we have already used the 1 and the 5 in the bottom right hand box so we can’t place those numbers there. Thus, the first four numbers must be {1357} in some order and there is a 9 somewhere in the bottom right box.

The clue in the first row requires a 5 at the end. Given the 10, 10, 5 clue and the 4 already placed in the sixth column, the only odd substring in the last box must be the 5. We thus must have the 5 in the first row, and now only the 3 can fit in the fourth row. We also gather that the last two columns of the first row must be even, and we have enough clues to place these (using standard Sudoku rules).

Finally, we can place the 4 in the middle right box (now that the 3 is there, there’s only one space left where it can fit). Similarly, we can place the 5 (knowing that it can’t be in the seventh column any longer).

From this point on, solving the rest of the puzzle should be reasonably straightforward. The next step would probably involve placing 1, 9, 6, 7 and 3 in the first row (in that order – using the column hints and the knowledge that we have two tens)!

For these contests, I’d say the challenge lies largely in figuring out how the ruleset can be used to help solve the puzzle (and, well, speed; you’re thrown 13-15 puzzles to solve in the space of 90 minutes and I can only usually manage about seven or eight, and for that matter the ones I do finish are the ones on the easier end of the scale!).