This puzzle is Copyright © 2007 by James Dow Allen.
 
 

Solution to #1, step-by-step

Although this puzzle is quite difficult, no trial-and-error is required to solve. We will show the solution, step by step. (We show all deductions made, including some that turn out ``useless'' -- we take no further advantage of them until easier deductions arrive in that region.)

Phase 1

Cells between two numbers must be black; white chains with only one direction to grow must grow in that direction ('a'); some black cells are forced by connectivity ('B'); others when a white chain completes ('C').

The green cell marked 'd?' in the lower left (or perhaps the cell just below it) must be white in order to prevent an illegal 2x2 black square. Only the 17-chain is capable of reaching to 'd?'.
aaaa   3       
14 B         3 2
 1       2     
B   4          
 3     1    3  
      B 1      
       17      3
    1    2  7  
               
 6     1 3 C 1 
  4       Ca2 B
a         BC 1 
3        1 aC  
 1 d?       2 4 
B B     aa5 B  

Phase 2

Continuing, the 17-chain must extend at the light green y-y-y-y in order to eventually join its necessary destination near lower left. This isolates a black cell which must extend to 'B'. We also show some cells ('B') near the upper right which must be black because none of the white chains are big enough to reach them. Two of the white cells in the lower right corner can be located ('e').

Note that there are only sixteen cells ('x' & 'a') the 14-chain can possibly reach, so it must reach all but two of them. In any case the cell ('blk') to left of 3 must be black (otherwise the cell to its left would be an isolated black). That black cell ('blk') must extend down and to the right at 'bb'. (Otherwise it would be unconnected.) This reduces the possibilities for the 14-chain to just fifteen cells. Furthermore. the cell ('kk') just above the 4 must be black (it's next to three of the possible 14-chain's cells) and one of the cells adjacent to 'kk' must also be black to avoid isolation.

One of the cells marked 'g?' must be white to break up a 2x2 black. That white can only be part of the 4-chain above it.
    xxblk3  B    
14  axaxbb    3 2
 1 xkkxax 2     
    4 x B     B
 3     1    3  
    g?g?  1      
     B 17      3
    1 y  2  7  
     yy        
 6   y 1 3   1 
  4        a2  
             1 
3        1    !!
 1 d?       2 4e
          5  w!e

(In the lower right of this diagram, there is an OWFU deduction available. Suppose the cell marked "!!" were white. There is no possible reason why the other place ("w!") for the 4-chain to grow couldn't be white instead; the puzzle would then have TWO solutions. Trusting the designer (Yours truly!) to have designed according to the unique solution rule, we know that "!!" must be black; indeed we know that that blackness must be essential: the two black cells connected to "!!" can have no other black connection! We will not, however, take advantage of this "OWFU" deduction.)

Phase 3

Between the 14-chain and 4-chain is a black cell ('C'). To its right a white cell is needed (to keep the 14-chain connected). Its two remaining neighbors include a white and a black to complete the 14-chain. The assignment is forced (to avoid isolating 'C').

Several cells ('C') next to the 14-chain are forced black. The nearby 2-chain must extend down to 'a' to prevent a 2x2 black square. The cell 'V' near 4 in upper left must be black (else the black above it is isolated); this makes the cell ('Y') below it white to prevent a 2x2 black square. The cell 'Z' near the 'Y' must be black to keep blacks connected.

Since the 4-chain is about to complete, only the 3-chain can break up the 2x2 black square near 'ee'; it has only one way to do that.

The cell 'd' in the middle left must be white (otherwise there'd be a 2x2 black square), and is part of the 6-chain. The cell 'B' to left of 6 on the left side must be black (otherwise the 3-chain below would isolate the black cell below the 6), and in fact that black cell will be cut off by the 3-chain below it (whichever way that 3-chain goes), so it must connect to the black chain above it as shown ('F-F-F-F').
       3aa     
14   a  CCC  3 2
 1 BC   C2     
  ee 4C   a     
 3a aV 1    3  
B    Y  1      
Fd    Z17      3
F   1 a  2  7  
F    aa        
B6   a 1 3   1 
  4         2  
             1 
3        1     
 1 d?       2 4 
          5    

Phase 4

Now we come to an interesting long-distance deduction. Recall that the 17-chain must arrive near the 'd' in lower left to prevent a 2x2 black square. This leads to a blocking chain of white cells (shown in green below) which makes it hard for the black chains in the left of the board to connect to those in the right. (Cells shown in darker green are conceptual -- we know the chain connects near 'd' but we don't yet know the precise route.)

Indeed the only way the black chains can connect up is via the bottom row: the cells shown in brown there must be black cells. (The cell marked 'W' can be shown, with some difficulty, to be part of this black chain but we will postpone this deduction until it is easier.) Similarly, since the 3-chain at lower left doesn't leave room for a black connection, the black must connect above via the brown path shown to the left of the green path.
       3       
14           3 2
 1      C2C    
    4   C C    
 3     1 C  3  
   Ca   1      
 d    B17      3
    1  aC2  7  
        B      
 6     1 3   1 
  4   B     2  
             1 
3        1     
 1 d?   W   2 4 
          5    

Phase 5

Several more forces occur as shown. The large black chain in upper left can connect only at the brown cell 'X'. This forces the 6-chain completion, and forces a black connecting cell at the brown cell 'Y'. This will force completion of the 4-chain, and dictate further black connecting cells, shown in brown ('Z').
       3       
14           3 2
 1       2     
    4          
 3     1    3  
        1      
 d X   17      3
 a  1    2  7  
 aBY           
 6 a   1 3   1 
  4a a      2  
    Za       1 
3  Z     1     
 1 d       2 4 
          5    

Phase 6

The 17-chain has only 18 possibilities (shown as 'x'). (It cannot penetrate closer to the 5, or 5-chain can't grow, and it cannot extend to 'B' or the small black chain shown here in blue will be isolated.) The black chain at the bottom will be constricted by the 5-chain, so for it to connect to the right, both cells 'W' must be black for the connection.

The 3-chain includes the green cells 'Y' (else 2x2 black square) and 'd'. This cuts off the bottom black chain so that it can connect above only via the right side, as shown in brown ('X'). The 7-chain must include one of the 'g?' cells (else 2x2 black square); this continues the forced black connecting chain further up, as shown in brown ('Y').
       3       
14           3 2
 1       2     
    4          
 3     1    3  
        1      
       17     Y3
    1    2  7Y 
           g?g? X
 6     1 3Y  1 
a 4     Bd  2  
 Ca   xxx    1 
3   daxx 1    X
 1 aax WW  2 4 
          5    

Phase 7

The 5-chain at bottom completes ('a-a'). The black chains must connect at 'B'.

The 3-chain at the right-side completes (we use our usual symbols: 'd' to prevent 2x2 black, 'a' for only growth direction, 'C' for adjacent black cells).

The 7-chain has only nine possibilities for growth ('x' or 'a'). This leads to more forces, including another force (shown in brown) that is only way our ever-growing black chain can connect to blacks in center and top. That in turn dictates that the nearby 3-chain go to 'g' and stop a 2x2 black square.
       3       
14           3 2
 1       2     
    4        g 
 3     1  x 3 C
        1 xa Ca
       17   ax 3
    1    2 x7 d
          Cxx  
 6     1 3   1 
  4         2  
             1 
3      d 1     
 1   B     2 4 
      aa  5    

Phase 8

Various forces ('d', and less obviously, 'd!') arise to prevent 2x2 black squares. Forced growth (in brown) of our main black chain continues.
       3       
14           3 2
 1       2 dCC 
    4      Ca  
 3     1  dC3  
        1 daC  
       17   ad 3
    1    2d! 7  
            d!  
 6     1 3   1 
  4         2  
             1 
3      d 1     
 1         2 4 
          5    

Remaining deductions are straightforward.

Final Solution to Nurikabe #1

       3       
14           3 2
 1       2     
    4          
 3     1    3  
        1      
       17      3
    1    2  7  
               
 6     1 3   1 
  4         2  
             1 
3        1     
 1         2 4 
          5    

 
 

This puzzle is Copyright © 2007 by James Dow Allen