Okay, so here we’re solving the following grid. Each square has a number from 1 to 7. No digit appears more than once in each row or column. The digits must obey the inequalities and if there is a circled number, the two digits either side must differ by that number.
1) The only two digits that can be 6 apart are the 1 and the 7.
2) Consider the fifth row: in which squares could a 4 appear? It can’t be in the first, fourth, or fifth column, because there aren’t any digits between 1 and 7 that are 4 or 5 away from 4. The 4 also can’t be in the second or third columns, because there aren’t four or five digits greater than 4. Finally, the 4 can’t be in the seventh column, because the only digits 3 away from 4 are 1 and 7, and those are both already spoken for in the sixth row. Therefore, the 4 in the fifth row must be in the sixth column.
3) Consider the sixth column. The three squares linked by circled 2s must either be all even or all odd. They can’t be even, though, because we just placed the 4 elsewhere in that column, so they’re all odd. That puts either the 3 or the 5 in the middle of that run, the 2 and 6 surrounding the circled 4 at the bottom, and either the 1 or the 7 at the very top.
4) Now consider the circled 5 between the fifth and sixth rows. Since the 1 and 7 in the sixth row are already spoken for, the square below the circled 5 must be either 2 or 6.
5) Now ask where we can put the 5 in the fifth row. It can’t go in the second or third columns (because there aren’t enough larger digits), nor in the fourth column (because that puts a 1 above it, and 1 isn’t greater than anything), nor in the seventh column (because the 2 in the sixth row is already spoken for). That puts the 5 in the first column and a 1 above it, which decides our original 1-7 question around the circled 6.
6) Similar reasoning places the 6 in the fifth row.
7) And then there’s only one place for the 7 in that row, and that decides several previously open questions.
8) Now there’s only one place for a 1 in that row: it can’t be in the second column (there’s already a 1 in that column), nor in the third (1 isn’t greater than anything else), so it must go in the fourth column, with a 5 above it.
9) We only have the 2 and 3 left in the fifth row, and the greater-than sign tells us how to fill them in. Then, there must be a 4 above the 3 because that’s the only digit between 3 and 5.
10) That allows us to finish off the sixth row as well, and to make some additional notes nearby.
11) There’s now only one place for a 2 in the fourth row. We can also now refine our understanding of the chain of circled 2s in the sixth column: the bottom square of the chain can only be 3 or 7, so the middle one must be a 5 and the top one also either 3 or 7. That leaves only 1 for the top square in that column.
12) There’s now only one way to fill in the 1 and 2 in the third column.
13) The 2 in the second row can’t be in the first column, because the greater-than sign would force the next square over to be a 1, and there’s already a 1 in that row. So that 2 can only go in the fourth column, and the final 2 (in the third row) must therefore go in the first column.
14) Similarly, the 1 in the third row can’t go in the fifth column (the circled 1 would put a 2 next to it, but all of the 2s are already placed), so it goes in the seventh column, and the final 1 goes in the fifth column of the bottom row.
15) That circled 1 in the third row can only be linking the 3 and 4, and those can only be placed in one order. That only leaves the 6 and 7 in the fourth column.
16) The bottom-left corner square must be one of 3, 4, or 6. It can’t be 6 (because that’s spoken for in that row) or 4 (because the circled 2 would force the next square to be either 2 or 6, and both are already taken), so it must be the 3. That then also places the 4 and 5 in that row.
17) The upper two squares in the first column now must be either 4 or 6. Less obviously, that’s also true for the second column: those are the only remaining digits in that column that differ by two. The greater-than sign in the second row forces the precise placements.
18) From here on out, everything is forced as we avoid putting more than one of any digit in any row or column.
Now we have the two grids, it is straightforward to identify the unblackened squares in the first grid that are in the same positions as unshaded squares in the second grid. Once you take the letter in each such square and advance it in the alphabet (wrapping around from Z to A if necessary) by the digit in the corresponding square, you get QUIS CUSTODIET.
Thanks again to Pavel Curtis for this puzzle. If you like this kind of material why not buy his book of Adalogical AEnigmas.
I set a puzzle here every two weeks on a Monday. Send me your email if you want me to alert you each time I post a new one. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
My puzzle book Can You Solve My Problems? is just out in the US. It is already out in the UK with a slightly different subtitle. I’m also the co-author of the children’s book Football School: Where Football Explains The World, which was a runner-up in the Blue Peter Book Awards 2017.
