9
$\begingroup$

This puzzle's theme is inspired by one from Grandmaster Puzzles' "The Art of Puzzles", but the puzzle itself is original.

Rules: (Nurikabe section shamelessly stolen from an earlier puzzle by @jafe)

  • Numbered cells are unshaded.
  • Unshaded cells are divided into regions, all of which contain exactly one number. The number indicates how many unshaded cells there are in that region.
  • SPECIAL RULE: the regions will form a tetromino set, with rotation and reflection allowed.
  • Regions of unshaded cells cannot be (orthogonally) adjacent to one another, but they may touch at a corner.
  • All shaded cells must be connected.
  • There are no groups of shaded cells that form a 2 × 2 square anywhere in the grid.

the puzzle

I've included all available tetrominoes as a reference. In addition to the picture, here is a Penpa link (no automatic answer checking) and a CSV:

,,,,,,,
,,,4,,,4,
,,,,,,,
,,,,,,,
,,4,,,,4,
,,,4,,,,

Answers are expected to include a description of their solve path. This is not a very large puzzle, and you can probably explain the logical steps in a only few progress pictures, but please refrain from posting an answer with only the final solution.

Special thanks to @Stiv for playtesting the puzzle!

$\endgroup$

1 Answer 1

6
$\begingroup$

First of all, let's give some things names. Call the five 4s, in "reading order", NW, NE, SW, SE, S. We'll also use these names for the regions containing them, and for their tetromino shapes. I'll call the tetromino that's often called "S" "Z" instead to avoid the name-clash with the S region.

Now

the region-adjacency restriction forces the S region to be either T or L. Then one of the regions must be I, and in fact it can only be the NW region, with two possible horizontal positions for the I. We can shade some squares either because they are adjacent to, and can't be part of, known regions, or because they need to be shaded to let other shaded squares connect out, or because they aren't reachable from the 4s.

+---------------+
|# # # # : : : :|
|:     4 : : 4 :|
|# # # # : : : :|
|: : : : : : : :|
|: : 4 #   # 4 :|
|: # # 4   : : :|
+---------------+

Now

the SW region has to extend leftward from its 4 in order to put an unshaded cell in the SW-most 2x2 block, and then we can fill in some more shaded cells to let the ones on the bottom row connect out. After this the remaining cell in the centre-left 2x2 block must be unshaded, which forces the SW region to be an O:

+---------------+
|# # # # : : : :|
|:     4 : : 4 :|
|# # # # : : : :|
|#     # : : : :|
|#   4 #   # 4 :|
|# # # 4   : : :|
+---------------+

Now, is it possible

that the isolated shaded cell near the SE connects out downwards rather than upwards? If so, we must have this:

+---------------+
|# # # # : : : :|
|:     4 : : 4 :|
|# # # # : : : :|
|#     # : : : #|
|#   4 #   # 4 #|
|# # # 4   # # #|
+---------------+

but this is impossible because now

we don't have enough space for both the S and SE regions. So no, that cell must connect upwards:

+---------------+
|# # # # : : : :|
|:     4 : : 4 :|
|# # # # : : : :|
|#     # : # : :|
|#   4 #   # 4 :|
|# # # 4   : : :|
+---------------+

At this point, consider

the cell in the SE corner. If this is shaded then it must connect out somehow. If it connects upwards then there is no room for the SE region. If it connects leftwards, we have this:

+---------------+
|# # # # : : : :|
|:     4 : : 4 :|
|# # # # : : : :|
|#     # : # : :|
|#   4 #   # 4 :|
|# # # 4   # # #|
+---------------+

after which

the only tetrominoes there's space for in the SE are O (already used) and Z:

+---------------+
|# # # # : : : :|
|:     4 : : 4 #|
|# # # # : : #  |
|#     # : #    |
|#   4 #   # 4 #|
|# # # 4   # # #|
+---------------+

This would force

the S region to be an L, and the shaded cell near the NE corner to connect upward:

+---------------+
|# # # # : # # #|
|:     4 : : 4 #|
|# # # # # : #  |
|#     #   #    |
|#   4 #   # 4 #|
|# # # 4   # # #|
+---------------+

but now

there is no room for the NE tetromino. So the cell in the SE corner isn't shaded after all. What is the SE region, then? If it includes the cell immediately below its 4 then (1) the S region is an L and (2) the SE region can only be L or O, both already used. So that cell is shaded; it must connect out leftward, so in fact the S region is an L and we have this:

+---------------+
|# # # # : : : :|
|:     4 : : 4 :|
|# # # # # : : :|
|#     #   # : :|
|#   4 #   # 4  |
|# # # 4   # #  |
+---------------+

Now let's see whether the SE region

can be a Z. It would have to look like this

+---------------+
|# # # # : : : :|
|:     4 : : 4 :|
|# # # # # : # #|
|#     #   #   #|
|#   4 #   # 4  |
|# # # 4   # #  |
+---------------+

and then

the NE region is forced, and then so is the NW region, giving

+---------------+
|# # # # #      |
|#     4   # 4 #|
|# # # # # # # #|
|#     #   #   #|
|#   4 #   # 4  |
|# # # 4   # #  |
+---------------+

which is in fact legal. So this is our solution, but we haven't yet proved it's the only one. The other possibility would be

a T in the SE region:

+---------------+
|# # # # : : : :|
|:     4 : : 4 :|
|# # # # # : : #|
|#     #   # #  |
|#   4 #   # 4  |
|# # # 4   # #  |
+---------------+

and this is impossible because

now the cell in the NE corner can't be covered by, but must be disconnected by, the Z-shaped NE region.

So we're done.

$\endgroup$
2
  • 1
    $\begingroup$ There shouldn't be a need to bifurcate that deeply, but the logic holds as far as I see $\endgroup$
    – bobble
    Jan 16, 2023 at 21:55
  • 1
    $\begingroup$ I haven't tried hard to optimize the solution. I am extremely unsurprised that more efficient solution-paths are possible :-). $\endgroup$
    – Gareth McCaughan
    Jan 16, 2023 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.