7
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Rules of Nurikabe: (copied from here)

  • 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.
  • 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.

Special rule:

  • Every row and every column must have exactly half of the cells shaded (and the remaining half unshaded). For a 8x8 puzzle, exactly 4 out of 8 cells must be shaded on each row/column.

Now, solve the following puzzle.

CSV:

,,,,,1,,1
,,7,,,,,
,,,,,,,1
,,,,,,,
,,,?,,,8,
,,,,,,,
,,,,,,,
,,,1,,,,
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0
5
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Step 1:

step 1
Some quick shading around the 1s. We can also confirm R3C6 as unshaded, because otherwise that would create a 2x2 of shaded cells.

Step 2:

step 2
Only one more cell in R2 can be shaded. If it is either of the left two cells, then the shaded cell in R1C5 won't be able to connect to the rest - not enough shaded cells are left for R1 to allow it to move left far enough. Therefore neither of those two cells are shaded. With similar logic, the bottom two cells of C2 must be unshaded.

Step 3:

step 3
Two more cells in R1 must be unshaded. For shaded-cell connectivity, they must be the leftmost two.

Step 4:

enter image description here
If the C8 ocean connects to the rest through R6C7, then the unshaded cells below the 8 will not have an island - they can't connect to the 8, and if they connect to the ? then there won't be room for the 8.

Wherein we take a short detour into Case Bashing Land

impossible
Only one more of R2 can be shaded, and this must be how the R1 cells connect down. If it is through R2C5, then after shading R2C5 and R3C5 (for connectivity), the rest of the cells must be unshaded. If so, R8C5 must use the rest of C6 to connect - except that would be too many shaded cells. Therefore R2C5 is unshaded

Step 5:

step 5
For connectivity, both R2C4 and R3C4 must be shaded. That's all the shaded cells for C4, so the rest are unshaded.

Step 6:

step 6
For connectivity, R2C3 must be shaded. Also the cells around R7C4 must be shaded because that's the only remaining way to connect. Therefore the remaining C3 cells are unshaded

Step 7:

step 7
Shading in R3C2 (for connectivity) forces the 7. Then that forces more shaded cells in R3 for connectivity again.

Step 8:

step 8
If R7C2 is shaded, then a 2x2 is unavoidable. Therefore it is unshaded.

Step 9:

step 9
Now the shape of the shaded cell flow is forced. Some cells must be unshaded; these are part of the ?

Solution:

step 10
The rest of the deductions are trivial: just keep the shaded cells moving up for connectivity, and obey the half-shaded rule

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Here's the solution, was a nice puzzle !

Gradual Deduction.

Step $1$ :-

Start putting the black squares around the 1's . Since they have to be connected, the red square should also be black. Also I am colouring the confirmed unshaded squares to be pink.

Step $2$ :-

Observe the square on R3C6 is pink, if it would have been black then there would have been a $2 * 2$ square. Also in the last column, there are already $2$ pink squares, so there should be $2$ other pink squares. Keeping in mind that the black cells should always be connected, and to make an $8$ in the $7$th column, all the $4$ squares will be pink. the last $2$ squares in both columns $6$ and $8$ , should also be pink. This makes the other squares in the column to be black and stay connected.

Step $3$ :-

Now, all the squares remaining in column $5$ , has to be pink. The first $2$ squares in row $8$ , has to be black. Also in row $7$ , the $1$st and the $3$rd square will be black and the $2$ will be pink, in order to avoid any $2 * 2$ squares. In the $1$st row, in order to make the black squares stay connected, the first $2$ squares will be pink, and the last $2$ will be black, and we get this.

Step $4$ :-

The final step now lies in connecting the black squares in the right path. Notice that to connect the $2$ ends, you must follow the $2$ brown paths. After going to the brown ones, all the other squares in column $4$ will be pink. The $3$rd row already has $3$ pink squares, so in order to add another one, there should be $2$ black squares there. So we follow the green path next. This completes the box which has $7$ pink squares, and next it is easy to follow the red path and connect the black squares.

Hence we have our solution :-

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