Nurikabe: The Super-Kabe Attacks!

Question

This puzzle is another Nurikabe, but with a new twist for me: it's a gigantic Super-Kabe! It's also, in my opinion, quite tricky and has some interesting deductions. I hope you enjoy!

Rules of a Nurikabe (copied from my previous puzzle):

This is a Nurikabe puzzle. The goal is to paint some cells black so that the resulting grid satisfies the rules of Nurikabe:

• Numbered cells are white. (Think of them as "islands.")
• White cells are divided into regions, all of which contain exactly one number. The number indicates how many white cells there are in that region.
• Regions of white cells cannot be adjacent to one another, but they can touch at a corner.
• Black cells must all be orthogonally connected. (Think of them as "oceans.")
• There are no groups of black "ocean" cells that form a 2×2 square anywhere in the grid.

Now, the puzzle:

And as always, here is a handy puzz.link solver in case you would like one. The image should also be MS Paint compatible, if you click through to the larger version.

Answer 1

COMPLETED GRID

REASONING

First we start with basic connectivity and counting deductions, to get to:

Reaching around:

Looking in the lower right corner, the 2 on the right edge must go down to prevent a 2x2. Just above it, the 2x2 R6-7C19-20 must have an unshaded square, but nothing can reach it except the 2 in R5C19. Then the unshaded square in R3C17 can only be reached by the 5 in R7C17, resolving most of the right side.

Moving left, the unshaded square R6C6 can only be reached by the 4 in R6C9. Then the 2x2 block R4-5C4-5 is only reachable by the 3 in R3C5, forcing R4C5 to be unshaded. Now looking at the 5 in R4C10, it can have at most 1 square going to the left. Moreover there is a 2x3 block, R6-R7C10-12 that can only be reached by the 5, and in particular R6C11 must be in its area. The grid thus far:

Moving to the left side:

There is another 2x3 area which cannot be covered except by the 8 in the lower left, namely R9-10C6-8, so we must have R10-C67 unshaded in the 8s area. We cannot isolate any squares underneath this area, so we must in fact have the whole row R10C1-7. Looking at the 3 in R9C10, we see that at least one of its cells must go right, as otherwise R9-10C11-12 would be a shaded 2x2. In particular, this forces the 8 area to continue into R10C8, so that R9-10C8-9 will not be entirely shaded. Now R8-9C1-2 must have an unshaded square, which must include at least R8C2.

Moving up for a moment, we see that the 6 in R2C3 cannot go left, since it would block the 2 in the upper left corner, so the 2 in R5C1 must go up. This forces the 4 in R8C3 to continue up into R7C2, completing the lower left. Finally, the 6 cannot cover the R2-3C1-2 2x2, so the 2 in the upper left corner must come down. The grid thus far:

Lower middle:

The 2x2 R9-10C12-13 can only be covered by the 3 in R9C10, which must extend right to R9C12. This then completes the 4 in R9C16, forcing it to extend to R10C14, and also completes the 5 in R4C10, since R7C11 must be unshaded. Now, the 2x2 R5-613-14 can only be reached by the 6 in R1C13, which must extend down, and easy logic finishes this section. The grid thus far:

Finishing up:

Since the 6 in R2C3 cannot use R2C4, the 2x2 square R1-2C8-9 can only be reached by the 3 in R3C8, and the rest is just fill-in.