This puzzle is Copyright © 2007 by James Dow Allen

Solution to Nurikabe #11


 
 
 4    1      
       1     
3           1
   2    5    
          14  
2            
  3          
          1  
3  1 5 5    2
    1        
             
  14    15   4 
             

 
 

Detailed solution

In the following,

Such deductions which are non-obvious are shown in green with an exclamation mark. In the following "e!" is the cell in a potential 2x2 black square that cannot otherwise be reached from any white chain. Only the 14-chain can prevent a 2x2 black square in the upper right corner: to get there it must go through "a!".
 4  e!C1CB    
      C1C   C
3      C  a!C1
   2    5 a C
          14  
2            
  3       C  
   CBa   C1C 
3 C1C5D5  C 2
   C1C       
    C        
  14    15   4 
             

Whichever way the 14-chain breaks up the 2-x-2 black square in the upper right corner, a small black group near the upper right will be isolated and have no way to connect except via the cells "B!" and "F!". Another consequence of the 14-chain's journey to the corner is that it cannot reach the cell "F" (brown) in upper center without exceeding 14 cells. So "F" is black and the cells next to "F" must be white to break up 2-x-2 black squares.

It isn't hard to see that the 14- and 15-chains at the bottom of the board cannot reach anywhere except the cells marked "x", along with at most one of the cells marked "?". This is a total of 33 cells (including the "14" and "15" marks themselves) which must accomodate the 14- and 15-chains and the border between them. The border will contain a minimum of two of the "x" cells, so all but two (33 - 14 - 15 - 2) of the "x" cells will be thusly consumed.
C4aa  1      
BCCCCB 1     
3  eFe      1
   2    5  B! 
          14 F!
2           F!
  3         F!
          1  
3  1 5 5 x  2
 x? 1 ? xxx  
xxx? ?xxxx   
xx14xxxx15xx 4 
xxxxxxxxxx   

Reasoning from the extreme restrictions just described on the large chains at the bottom, we can place most of the border of the 15-chain.
 4    1      
       1     
3eC     C   1
  C2C aa5C   
B  C  CCC 14e 
2          e 
 C!3          
          1  
3  1 5 5 x  2
Dx? 1 ?D!xxx  
xxx? ?xxxxB!  
aa14xxD!x15xXXD!4 
aaaaxD!xxxx   

Just one or two of the "x" cells near the "15" will be black. Using the OWFU ("Only Way for Uniqueness") inference we could eliminate several possibilities. For example the cell marked "XX" cannot be black: if that led to a valid Nurikabe, the same configuration with the cell just above "XX" black instead would be valid and solution would not be unique. Such reasoning would lead to shortcuts in this problem, but I won't use them: Instead I use the standard process which not only finds the solution but proves its uniqueness.
 
 

The green "B!" cell at left cannot be white: There would then be an isolated black cell underneath it.
 4    1      
       1     
3           1
   2    5    
 F  F     14  
2            
B! 3          
          1  
3  1 5 5 x  2
 B? 1 ? xxx  
aaa? ?xxxx   
  14aa x15xx 4 
    a xxxx   

 
 
 4    1      
       1     
3           1
 e 2    5    
  F F     14  
2aCe e       
  3a         
 B        1  
3  1 5 5 x  2
    1 ? xxx  
     ?xxxx   
  14   x15xx 4 
      xxxx   

The black chain on the left side cannot connect to the right except by using one of the two cells marked in green with "B!" In fact, because of the 5-chain's needs, if either "B!" cell is black, so is the other one.
 4    1      
       1     
3     B!     1
C  2    5    
     B!    14  
2a  C        
  3 Ca       
a C       1  
3a 1 5 5 x  2
  B 1 B xax  
   a Bxxax   
  14   x15ax 4 
      xaaxD!  

The "e!" inference in the center of the board may not be obvious, but if that 5-chain grows any other way, its neighbor 5-chain won't be able to break up the 2-x-2 black square near "e!". The 14-chain would be able to reach far enough, except that that would starve the 5-chain below it.
 4    1      
       1     
3           1
   2 a  5    
          14  
2     e!      
  3          
          1  
3  1 5 5D!x  2
    1   x x  
      ex x   
  14   x15 x 4 
      x  x   

The final 5-chain is almost determined by "e!". As shown, this will constrain the 14-chain: It will occupy all the "x" cells in the upper right and one of the "?" cells.
 4    1  xxxx
       1 xx  
3        ?  1
   2 a  5    
         ?14  
2      Ce! x  
  3   C    x 
      Ca  1  
3  1 5 5 x  2
    1   x x  
       x x   
  14   x15 x 4 
      x  x   

The black chain on the upper right side of the board has only one way ("B!") to connect, shown in green. (This leaves the fifteen cells of the 15-chain completely determined.)
 4    1  aaaa
       1 aaB 
3        ?  1
   2 a  5    
         ?14  
2      C Ca  
  3   C aC a 
      Ca  1 B!
3  1 5 5 B! B!2
    1   x x  
       x x   
  14   x15 x 4 
      x  x   

Some simple deductions follow, after which there is only one way, shown in green as "B!" to link the black chains.
 4    1      
       1     
3           1
   2 a  5    
         B!14  
2            
  3          
        e 1  
3  1 5 5    2
    1      Ca
           BC
  14    15   4 
             

 
 

This puzzle is Copyright © 2007 by James Dow Allen