Yesterday I set you the following puzzle by Pavel Curtis, channelling Ada Lovelace. (Here’s a printable pdf.)
My dear Mr Bellos,
I should like to begin by expressing my deep gratitude for the honour you do me in presenting my little logical conundra to the readership of so august a journal as The Guardian. Truly, I am in your debt.
I am cognizant that your publication space is at a premium, so I shall, much against my accustomed nature, endeavour to be brief. In this first grid, I desire you and your readers to blacken some of the squares, in such a way as to leave all of the unblackened squares connected to one another, horizontally and/or vertically. Wherever there is a circled number, it specifies exactly how many of the squares touching it must be blackened.
Within this second, perhaps somewhat less genial grid, I pray that you shall enter a digit, ’twixt 1 and 7, into each and every square, refraining from placing the same digit into more than one square in any row or column. Where there exists a less-than symbol on the line separating two squares, please ensure that your entered digits obey the given inequality. Wherever a circled number has been placed, your digits must differ by precisely that amount.
When you have completed both grids, you may move on to finding the final answer to my challenge. Please begin by identifying the unblackened squares in the first grid that are in the same positions as unshaded squares in the second grid. 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. Reading the resulting letters, left-to-right and top-to-bottom, will reveal an appropriate phrase in a foreign language. Good luck! – ADA
Many people found the puzzle difficult - although there were a few happy solvers!
The solution is QUIS CUSTODIET
The words come from the Latin phrase Quis custodiet ipsos custodes? or Who will guard the guardians? “When one is creating a puzzle for your periodical, how can one be expected to resist such a temptation?” said Pavel. Nice!
Pavel will take you through the solution here step by step. Even though you might have found it tricky, when broken down hopefully it all makes sense, and you can marvel at the beauty of such a carefully crafted puzzle. The explanation for the first grid is here. So as not to have too large a post, the explanation for the second grid is in a separate post.
1) Consider the ‘3’ in the lower left corner of the grid. Since it is surrounded by three black cells, at least one of them must be *above* the three, adjacent to the ‘1’ clue. Since that *is* a ‘1’ clue up there, we can conclude that both squares below the ‘3’ (labelled ‘N’ and ‘O’) must be black. Symmetrically, since we know that one of the squares between the ‘3’ and the ‘1’ (labelled ‘G’ and ‘Q’) must be black, that’s the one black square allowed around the ‘1’, so the two squares above it (labelled ‘O’ and ‘C’) must both be white. This leaves us here:
2) Consider the ‘D’ square in the bottom row. If it were white, then the two ‘1’ clues below it would imply that both the ‘I’ and ‘E’ squares would be black, and that would prevent the white ‘D’ square from being connected to the other white squares. So that ‘D’ must be black, and the ‘I’ and ‘E’ both white:
3) That white ‘I’ square must be connected to all of the other white squares, so that implies that the adjacent ‘N’, ‘B’, and ‘O’ squares must also be white:
4)Consider the now-known-white ‘E’ square in the middle of the bottom row. It must be connected to the other white squares, and its path of connection must, at some point, cross the boundary between the fifth and sixth rows of the grid. It can’t cross on the left or right side of the ‘3’ clue in the middle (since that crossing would comprise two white squares, and that’s too many for a ‘3’ clue). Similarly, it can’t cross along the right edge of the grid, since that would violate the ‘1’ clue there, so it must make its crossing in the sixth column:
5) That leaves only two squares left to be blackened to satisfy the adjacent ‘2’ clue, and then only one path for the white ‘E’ square to reach our new crossing:
6) Continuing with this path of connection, all of these white squares in the lower right must somehow connect across the boundary between the fifth and sixth columns. As before, that connection can’t be either above or below the top-most ‘3’ clue. If the connection was in the fourth row, below the ‘2’ clue, then the ‘O’ and ‘P’ above that ‘2’ would have to be black, and that would imply that the ‘K’ and ‘R’ above that would have to be white, and that’s too many white squares for that top-most ‘3’ clue. So the crossing must be in the third row:
7) That forces resolution of both adjacent ‘2’ clues:
8) And that leaves only one path for our lower-right white squares to reach the crossing point and then to escape from there:
9) And that resolves both ‘1’ clues in the lower-right corner:
10) Consider the ‘1’ clue in the upper-right corner. If the ‘C’ were white, then the ‘N’ would be black, and the white ‘C’ would need a connection path out to the other white squares. As we’ve seen, though, it can’t cross above the ‘3’ clue, so it would be stuck. Thus, the ‘C’ is black and the ‘N’ is white:
11) Continuing in that corner, if the ‘T’ were white, the ‘N’ to its left would be black, keeping the ‘T’ from connecting out to the other white squares. Therefore, that ‘T’ is black, the ‘N’ is white, and the ‘T’ in the middle of the top row is also white (to allow its neighboring white ‘N’ to connect out):
12) The ‘2’ clue in the middle of the grid has, for some time now, had two white squares around it, so the other two (the ‘F’ and the ‘N’) must be black. That also resolves the ‘1’ clue just below it, forcing that ‘Q’ and ‘N’ to be white:
13) That, in turn, resolves our original ‘1’ and ‘3’ clues, forcing the ‘G’ to be black, and also the other ‘3’ down there, forcing the ‘I’ to be black as well:
14) Now consider the upper-left ‘3’ clue, and suppose that the ‘E’ square up-left from it were black. We know that one of the ‘L’ and ‘S’ above it must also be black (because of the top-most ‘1’ clue), so either way, that would complete a black wall cutting off the white squares on the left from those on the right. Therefore that ‘E’ must be white, and that resolves that ‘3’ clue:
15) There is now only one path left for the right-hand white squares to meet up with their left-hand brethren, completing the grid.
You can follow the explanation for the second grid here.
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.
