Solution to Nurikabe #8

 1 2 6 1 3 5 4 1 7 1 2 1 3 1 4 1 3 5 1 1 8 2 7 2

Solution to Nurikabe #9

 1 2 6 1 3 5 4 1 7 1 2 1 3 1 4 1 3 5 1 1 8 2 8 2

Detailed solutions

(The first several deductions are the same for both Nurikabe #8 and #9.)

In the following,

• 'a' denotes a cell which must be white to allow white chain to grow.
• 'B' denotes a cell which must be black to allow black chain to connect.
• 'C' denotes a cell which must be black to provide border for white chain.
• 'D' denotes a cell which must be black to separate two white chains.
• 'e' denotes a cell which must be white to prevent 2-x-2 black square.
• 'F' denotes a cell which cannot be reached by any white chain.

 C 1 C C e 2 D C 6 C 1 C e D 3 a C 5 C e 4 D C 1 C 7 C C 1 C C 2 C C C 1 C 3 C 1 C a 4 D C B C 1 C a D 3 a 5 C 1 C B B C C 1 C D 8 a C e 2 D e 7/8 2

 F C C 1 a a C 2 B 6 D 1 C 3 5 a 4 1 7 1 2 B 1 3 1 4 1 3 5 1 1 C 8 C C 2 C F C C 7/8 C 2

A black group in lower left is almost isolated, so linking black cells can be deduced across the next-to-last row.
 1 a a B 2 6 a 1 3 C 5 4 1 7 1 e 2 1 3 1 4 1 3 C 5 1 a e B 1 C 8 a 2 B D B B 7/8 a a 2

The next diagram shows our first non-trivial deduction. The point y must be white: Otherwise only the 5-chain could prevent a 2-x-2 black square starting above the y and in doing so would be unable to prevent 2-x-2 black square near the z.

The point z must also be white to prevent 2-x-2 black square.
 1 2 6 1 F 3 5 B B C 4 1 a 7 C a C 1 2 y 1 3 1 4 z 1 3 5 1 1 8 2 e 7/8 2

 1 2 6 1 3 5 4 e 1 7 C C 1 2 C 1 C 3 1 4 1 3 5 1 1 8 2 7/8 2

The next diagram demonstrates that the black chain in lower left has no connections to blacks in upper part of board except via a route below the 5-chain. If such a connection route existed it would use the cells shown in green with G, and the 8-chain would not be able to grow to full size without isolating three black cells to its left.
 1 2 6 1 3 c 5 4 1 7 1 2 1 3 1 4 G 1 3 a G 5 a 1 a G G a 1 8 a G 2 G e 7/8 a 2

To break up the 2-x-2 black square at G-x-y-x in lower right, either y or G must be white. These two possibilities lead to our two solutions but first we make deductions common to both cases.

If G is white, the black chain at bottom has no outlet except to the right and the cells B at center right must be black for connection. If y is white, the black chain connects above it (shown in green), e is white (to prevent a 2-x-2 black square) and again the only outlet for the black chain is with the cells B at center right.

Meanwhile, the black chain in upper right must connect via M-M, N-N or P-P. M-M is out because 5-chain wouldn't be able to grow to full size; and we can now see that P-P is out because 7-chain wouldn't be able to grow to full size. So N-N are black.
 1 2 6 1 3 5 4 1 M N P 7 M N P 1 2 B e 1 3 1 B 4 1 3 5 1 e 1 8 G x 2 y x 7/8 2

 1 C 2 6 1 3 5 4 1 e e 7 a e a 1 2 C B 1 a 3 1 4 D 1 3 5 1 1 8 A 2 B 7/8 2

The above diagram brings us up to date. Recall that one of the reddish cells (A or B) in lower right is white. We pursue these cases separately. Note that A can connect only to the upper 8-chain, while B can connect only to the lower 8-chain.

Case A

Assuming A is white we quickly derive solution as shown. Note that the bottom chain must be a 7, not an 8.
 1 2 6 1 3 5 4 1 7 e 1 2 e 1 3 1 4 C 1 3 e a 5 e 1 C C C C C 1 8 A 2 C C C C 7 a a a 2

Case B

Assuming B is white we recall deductions mentioned above. Note that the bottom chain must be an 8, not a 7.

The white cell 'x' must be part of the 8-chain, so the cell to its left must be black for connection.
 1 2 6 1 3 5 4 1 7 e 1 2 e 1 3 1 4 B x 1 3 5 1 a e 1 8 B B 2 B 8 2

In the following note that 'G' (in green) must be black or the two large black groups are unconnected.

 1 2 6 1 3 5 4 1 7 C 1 2 1 C 3 1 4 G C 1 3 a D 5 e 1 a 1 8 2 8 2

 1 2 6 1 3 5 4 1 7 1 2 1 3 1 4 1 3 5 1 D 1 8 a D a 2 B D B 8 a a 2