# Nurikolor (Level 3)

Previous Level: Nurikolor (Level 2)
Next Level: Nurikolor (Level 4)

Level 3 has arrived, and the grid gets larger, now being an 8x8. The rules still apply.

• There are colored numbers on the grid, which indicate the number of tiles the group of its color holds.
• There are tiles with 1 color, which indicate the color of the tile.
• There are tiles with 2 or more colors, which indicate intersections of colors. All intersections are shown, and these are the only intersections.
• Grey tiles are not part of any group; they just serve as barriers.
• The goal is to have every non-grey tile covered by a type of color.
• 2 by 2 non-grey squares of the same color are illegal.
• In future levels, there will be multiple numbers of the same color. Their groups must never intersect or be orthogonally adjacent to each other.
• NEW: There will be colored lines in certain places. The same-color group may not cross through the colored lines, although they must border the line.

Colorblind version:

--- --- c10 xxx -b6 --- -p5 ---
--- -g- --- --- -cb --- --- -bp
--- --- --- -cg xxx --- --- -lp
--- --- xxx --- --- --- --- xxx
xxx -rg --- --- g11 xxx --- ---
--- -r8 --- xxx -o- -oy -ly ---
--- -o7 --- --- --- --- -l7 ---
--- --- --- --- xxx --- --- -y9

r6c3 red, up
r4c6 lime, left

r = red, o = orange, y = yellow, l = lime, g = green, c = cyan, b = blue, p = purple, xxx = gray


Request: Is it possible to create a program that generates puzzles like these?

• Re the request: The answer depends on which part of generating a puzzle you're referring to - generating the puzzle instance or generating the image? As a programmer myself, I think the former is possible, though the result will be likely worse than hand-crafted ones. For the latter, maybe you could use something mentioned here? Oct 12 '20 at 1:43
• @Bubbler: The image that creates the puzzle, of course. The rules stay there, but the puzzle board has to be generated. Oct 12 '20 at 1:51

The completed grid:

Reasoning:

Start in the upper right corner. The square left of the lime/purple square must be lime, otherwise the lime/purple would be lime-isolated. We also know the square right of the lime line is lime, since lime cannot reach the other side. This forces two more lime squares between the grays to connect up with the lime 7, finishing the region.

Continuing:

The square left of the blue/purple square must be blue, otherwise the shared square would be cut off from the blue 6. The square to its left must then also be blue, forcing the square left of the 5 to be purple to avoid a blue 2x2, and the upper right corner to be purple to join the purples up. Finally, the remaining square has to be blue, being the only adjacent color able to expand. The grid thus far:

The bottom right corner:

The orange 7 must connect to the orange/yellow square through the gray squares at bottom, since the green 11 blocks the only other potential path. Moving to yellow, the only color that can get to the squares above the 9 is yellow (lime is finished). We currently have 6 yellow squares, so three of the four remaining unshaded squares adjacent to yellow must be yellow. One of the ones adjacent to the orange/yellow square must be orange, so the other two on the bottom row must be yellow. The grid thus far:

Lower left corner:

Note neither green nor cyan cannot penetrate into the lower left corner, since it is blocked by the red 8, the red line, and the orange passing through the gap between gray squares. Thus the remaining 8 unshaded squares there must be either red or orange. We already have four squares shaded orange, plus we know another square adjacent to the orange/yellow square is orange, so only two of these are shaded orange, thus the remaining six are red. This means the square below the red line must be red, since there can be no additional red squares outside this corner. Connectivity then forces the rest of the corner. The grid thus far:

Finishing up:

The only remaining colors are cyan and green. The cell northwest of the 11 must be cyan to avoid a 2x2 block of green, and this square is the only cyan access to this nook. We also easily color the square southwest of the 10 cyan for connectivity. We must also color the square south of the 10 cyan, for if it were green, we would have a green island surrounded by cyan as we go around to hook up with the 10. We now have the cell two south of the 10 must be green to avoid a cyan 2x2. To finish up, we note that the 2x2 region above the red/green square must have a cyan block, which must come around through the top left corner. This easily forces the rest of the grid.

Note:

In the original version, the solution is not unique. The two squares left of and below the orange/yellow shared square could switch colors and still be a solution. The OP has updated the puzzle to avoid this ambiguity, consistent with the solution as presented.

• Ah, I've noticed. Added orange to square to unique. Oct 12 '20 at 2:09
• Amazing. My god, I love it when people add logic. Good job :) claps hands Oct 12 '20 at 2:40