These puzzles are Copyright © 2007 by James Dow Allen

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,


 
 
   C1C    C  
 e2DC 6  C1Ce
  D3a C 5 C  
 e4D C1C    7
C     C      
1C C2     C  
C C1C   3C1C 
a4DC     BC1C
aD3a   5 C1CB
 B C      C1C
  D8a      C 
 e2D        e
     7/8      2

 
 
FCC 1  aa    
C 2  B6 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 
FCC  7/8     C2

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   6a  1  
   3   C5    
  4   1     7
             
1 e 2        
   1    3 1  
 4         1 
  3 C  5  1  
a e B      1 
C  8 a       
  2 BDB      
   B 7/8aa    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    5B B 
C 4   1 a   7
 CaC         
1   2y       
   1    3 1  
 4   z     1 
  3    5  1  
           1 
   8         
  2          
    e7/8      2

 
 
    1        
  2   6   1  
   3    5    
  4 e 1     7
    CC       
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  aG5a 1  
     aGGa  1 
   8  aG     
  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  MNP7
         MNP 
1   2      Be
   1    3 1 B
 4         1 
  3    5  1  
         e 1 
   8      Gx 
  2       yx 
     7/8      2

 
 
    1       C
  2   6   1  
   3    5    
  4   1  e e7
        a  ea
1   2   CB   
   1   a3 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  ea5e 1  
     CCCCC 1 
   8      A  
  2    CCCC  
     7  aaa 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  Bx     1 
  3    5  1  
        ae 1 
   8     BB  
  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  aD5e 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  aDa    
  2    BDB   
     8  aa  2

 
 
 
 

These puzzles are Copyright © 2007 by James Dow Allen