This puzzle is Copyright © 2007 by James Dow Allen

Solution to Nurikabe #5


 
 
      6      
 5        1  
  3      3  3
2         4  
             
 2 4        1
     9 4     
        1    
  4      1   
    1      3 
4     6      
   3         
       2    2

 
 

Detailed solutions

In the following,

Such deductions which are non-obvious are shown in green with an exclamation mark.

In the following diagram one of two blue "e?" cells must be white to avoid 2-x-2 black square.
 
 
 aa   6   CF 
 5DB     C1Ce
B!D3     a3DB3
2        D4  
            C
 2D4 a! a   C1
     9D4C e! C
       C1C e! 
  4 C   C1C  
   C1C   C 3 
4   C 6      
   3     e?e?C 
       2  F 2

 
 

The 6-chain at bottom is almost located by the following diagram.
      6     C
 5        1  
  3      3  3
2B        4Ce
aC       CaC 
C2 4D D eCa 1
 a   9 4 C C 
        1 C C
  4    e 1 aC
    1 CCC C3C
4    C6aa  CF
   3e CD  e e
       2   C2

 
 

The 9-chain must grow to the green "a!": otherwise nearby 4- and 3-chains cannot both fit.
      6      
 5  B!     1  
  3      3  3
2      F  4  
     a!B      
 2 4        1
C C  9 4     
BC   aB 1    
  4  aa  1   
    1      3 
4     6  e   
   3    C    
  F  F 2 F  2

 
 

The green "B!" cell at left must be black: otherwise the 2-sized black chain above it would be isolated.
      6      
 5        1  
  3a     3  3
2  D a    4  
   a!         
 2 4        1
   aD9 4     
  B!     1    
  4      1   
    1      3 
4     6      
   3         
       2    2

 
 

In the following diagram we see three deductions which each require a bit of look-ahead (trial-and-error).

Furthermore note that either all the cells "x" or all the cells "y" must be white. In either case, the black chain at celter left will have no route to the center of the board except via the lower left: the cells there shown in dark blue must therefore be black.
      6      
 5        1  
  3 x    3  3
2 y B!     4  
  x          
 2 4        1
     9 4     
   yB!   1    
 a!4x     1   
    1      3 
4     6      
   3         
       2    2

 
 
      6      
 5        1  
  3   C  3  3
2         4  
             
 2 4        1
     9 4     
        1    
B 4      1   
B a 1      3 
4     6      
a  3         
       2    2

 
 

From here I don't know how to proceed except by trial and error. Suppose the 3-chain extends to the right at "R" (red). Several forces ("B") follow to keep black chains connected, and then there is no room for the 6-chain at the top to grow to its full size.
      6 C    
 5   BeCa 1  
  3 RC B 3  3
2     a D 4  
       Ba    
 2 4        1
     9 4     
        1    
  4      1   
    1      3 
4     6      
   3         
       2    2

Therefore "R" above is not white: it is black (R!). Several forces then follow at once:
      6      
 5        1  
  3 R!    3  3
2 a       4  
  D          
 2 4        1
     9 4     
   a    1    
  4D     1   
    1      3 
4 aD  6      
  D3         
       2    2

 
 

There is only one way (shown as "B!" in green) for the chain at left to connect to the right. Following that there is only one way (shown as "B!" in brown) for that chain to connect to the center, leaving room for 6-chain to grow.
      6  B!   
 5     B!B! 1  
  3      3  3
2         4  
             
 2 4        1
     9 4     
        1    
  4      1   
    1      3 
4     6   a  
 B! 3 a   B!   
   B!B! B!2a   2

 
 

The 9-chain must expand to the green cell "e!": otherwise we would end up with a 2-x-2 black square somewhere near, or to the right of, the "e!".
     a6ae    
 5        1  
  3      3  3
2     e!   4  
             
 2 4        1
     9 4     
        1    
  4      1   
    1      3 
4     6      
   3         
aa     2    2

 
 

Completion follows soon after the following deductions:
      6      
 5   eB!   1  
  3  C a 3  3
2         4  
       Ba    
 2 4        1
     9 4     
        1    
  4      1   
    1      3 
4     6      
   3         
       2    2

 
 

This puzzle is Copyright © 2007 by James Dow Allen