The following is a partially illustrated solution for the
sudoku.com.au tough puzzle of 01/15/08. I found this particular
tough puzzle to be somewhat vexing. Hopefully, some of the Sudoku tips,tricks and strategies used
in this solution will be helpful. I believe some of the steps that I found are at least interesting.
If this is your first visit to this blog, WELCOME!!
Previous blog pages may be helpful. Links to these pages are found to the right,
under Sudoku Techniques. I may later list specific pages.
The illustrations of steps shown on this page will share this key:
 black line = strong inference performed upon a set (strong link)
 red line = weak inference performed upon a set (weak link)
 black containers define a partioning of a strong set(s)
 candidates crossed out in red = candidates proven false
Please be aware that, for me,
strong and weak need not be mutually exclusive properties.
Given Puzzle  22 givens
Three Unique Possibility exist:
 (3)a7 % column & box (hidden single both in column a and box b8
 (5)e9 % row & box
 (5)g1 % box
Thus UP 25.Note also:
 step 2a Hidden Pair 48 b1,c3
 => b1 & c3 limited to only 4,8
Steps 2b, 3a, 3b, 3c  Locked candidates
 2b Locked 3s gi3 => de3≠3
 (3)e5 % column => UP 26
 3a Locked 2s c79 => c45≠2
 3b Locked 7s a23 => b2≠7
 3c Locked 8s hi5 => df5≠8
A few more easy steps are available, but not required in this solution path and thus not illustrated.
Puzzle Markup
The markup above is my actual markup, appended with some unneccesary steps plus the next step  an extended AUR deduction. There are many steps
available, and practically every cell with two or more candidates marked has a chain passing through it. My notes at this point
in the puzzle included steps such as:
 (3=7)f9(7)e7=(7hidden pair 16)e13=(6)f2 => f2≠3
 It seems I took this step before markup => (3)d2 % box, row
 (6)ab2=(hidden pair 16)a13by AUR(hidden pair 16)e13=(6)f2 => h2≠6
I think a less deep solution than the one that follows is possible. However, for really tough puzzles the following
published steps may be more instructive. Hopefully!
Step 3d  an AUR (Almost Unique Rectangle) chained
Above, find:
 (9=7)b9(7)h9=(714)h8=(hidden pair 14)h45 AUR(naked pair 14)c45=(triple(1=4)59)c245
 => c79≠9
The almost AUR 14 at ch45 is the key. To find that it might be useful, one needs to note:
 Almost triplet X59 at c245, where x is (1=4)
 Almost Hidden Pair 14 at h458. Note puzzle markup indicates this almost hidden pair overlaps strong 7s column h
It is nice to use the Hidden pair qualities and the naked pair qualities of an Almost Unique Rectangle in the same
chain. One needs to think clearly, though. Note that at least one of 14 cannot exist at c45 if only 14 exist at h45.
After making the eliminitions,
 (2)c9 % cell => UP 27
 step 4a Locked 9's b789 => b25≠9 (not illustrated)
Step 4b  A chain found by counting
Above, I picked this chain to hasten the solution process. It is longer than I would prefer. I ignored many chains
such as:
 (9)e7=(98)d8=(85)d6=(5)b6(5=6)b2(6=1)a1(1)a3=(17)e3=(7)e7 => all weak links are sis.
However, seeing this chain let me know that the chain shown above would solve many cells. Actually illustrated above:
 (7)e7=(7)e3(7=4)d3(4=8)c3[(8=1)c7 & (8)i3=(hidden pair 582)i45=(2)i7](12=9)g7
 => e7≠9 => (9)d8 %box => a cascade of hidden and naked singles to UP 42
Counting makes this one possible, for me anyway, to locate quickly. I actually started as follows:
 (18)c7 +2 (1 row, three columns  leaving result column empty for now
 (84)c3 +0
 (47)d3 +0
 (8)i35 +0
 (5)i45 +0
 (2)i457 +0
 (129)g7 +0
 (7)e37 +0
 Note a target exists (9)e7. Place (7)e7, (9)g7 into result column (2) total(0) done.
The resultant BTM (Block Triangular Matrix) can be made into a simple Triangular matrix, however, this is not required.
c7 

1 
8 





c3 
  8  4 
   
d3 
   4 
7    
(8)i 
  i3  
 i5   
(5)i 
   
 i5  i4  
(2)i 
   
 i5  i4  i7 
g7 
9  1   
   2 
(7)e 
e7    
e3    
As AIC, one could directly write:
 [(9=2)*g7(2)i7=(hpair258)i45=(8)i3(8=4)c3(4=7)d3(7)e3=(7)e7]
 =(1)*g7(1=8)c7(8=4)c3(4=7)d3(7)e3=(7)e7
 =>(9)g7=(7)e7 => e7≠9
However, merely counting allows one to not traverse the same chain segments mutliple times. Also, counting need never anticipate
an intitial target.
Step 5  Locked candidate 1 by AUR 26
Above, a visually apparent AUR deduction is available. One could write:
 (pair 16)g56=(xwing(2))ag56AUR(xwing(6))ag56=(pair12)g56
 =>(pair16=pair12)g56 => g78,h45≠1
Perhaps it is less convoluted to think:
 (pair26)a56(pair26)g56=(1)g56
Although this latter representation loses the AIC quality of the deduction, it is more faithful to the typical AUR conclusions.
Regardless of how one chooses to think about it, (1)g5=(1)g6 => g78,h45≠1 reduces the puzzle to naked singles to
the end.
Solution
Notes
 Sets: 2 + (5)1 + 7 + 8 + 4 = 26
 Maximum depth 8 at step 4b
 Rating: .01(3 + 5 + 127 + 255 + 15) = 4.05
Certainly, a less deep solution path is possible. This probably could reduce the rating.
This particular puzzle has many AUR type deductions possible. It also has many chains available that do not use
uniqueness of solution. The solution provided above is primarily meant to illustrate one path available using relatively
simple concepts combined in what I find to be interesting ways.