The following is page one of an illustrated proof for the
sudoku.com.au tough puzzle of 01/30/08. This particular
tough puzzle presented some perplexing challenges. I found many, many potential steps. This proof illustrates those
I deemed either important to the proof, or just plain interesting. Hopefully, some of the Sudoku tips, tricks
and strategies used in this solution will be found helpful.
If this is your first visit to this blog, WELCOME!!
Focus Change
Previous proofs and solutions have been heavy in the proof department. However, I have failed to provide much
insight in how to locate complex chains. Although I have described my preferred puzzle markups, I have not really
spent much time describing the process that takes the puzzle marks and produces deductions. With this proof, I shall
endeavor to describe the process of finding deductions in much greater detail. Unfortunately, the ride might be bumpy.
Some of my deductions come from mentally jumping from one area to another. The process is definately not straightforward.
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
 boxes highlight some groups
Please be aware that, for me,
strong and weak need not be mutually exclusive properties.
Given Puzzle  22 givens
Two Unique Possibilities exist:
 (4)e8 % column & box (hidden single both in column a and Box b8 <=> Box 1)
 (6)g2 % column & box
Thus UP 24.
Steps 2a & 2b  Locked Candidates
Illustrated above are the only two easy steps this puzzle allows:
 2a Locked 5s bc8 => i8≠5
 2b Locked 9s i13 => i78≠9
At this point, I printed out the puzzle and marked it up, highlighting hidden strengths in my typical fashion:
 Circle bilocals
 Underline the singular part of
 (a)cellxrowt=(a)cellsxyzrowtboxr
 ditto for columns
 V obvious chain snippet strengths derived from underlines
MarkUp
This puzzle has only five bivalues, and relatively few bilocals. No obvious bivaluebilocal Alternating Inference
Chains (AIC), (also forbidding chains) that yield an elimination exist. Given this dearth of immediate
fruit, I note that candidate 5 is particularly strong. Box h8 (also called Box 3) is heavily populated with strengths. I then
mentally catalogue items not listed well on the markup.
Of particular note, to me anyway, were the following Almost Hidden Pairs (AHP):
 (47)cgi1
 (16)ade7
 (29)efi3
 (49)i135
 (25)bc8c9
 (15)b138
 (58)f156
 (37)gi8g9
 (49)a45c5
Some of these form some links, such as
 (1)b3=(hp15)b18(2)b8=(2)b89
 (4=7)g1=(hp47)ci1(9)i1=(9)i3
 (hp25)bc8=(hp27)gi8
After the puzzle markup, I explore some chain snippets. I try to do this rapidly, and not too deeply. This allows most
of the snippets to remain in short term memory. Two short snippets lead to a relatively shallow elimination, shown below.
Step 2c  Snippets linked through a Kraken cell
Illustrated above is my thought process:
 (hp16)ad7=(1)e7
 Circled 6's ad7 and underlined 1 a7,f8 highlit this snippet
 (9)h7=(94)h9=(43)g9=(3)df9
 Circled 9s h79, circled 4s gh9, underlined 3s g9,f8 highlit this snipper
 (9)f8=(13)f8
 With very little thought, the underlined 1 and underlined 3 at f8 expose this partition as potentially
useful
The chain could write itself
 (9)f8=[(hp16)ad7=(1)e7(1=3)f8(3)df9=(34)g9=(49)h9=(9)h7] =>
 strong inference set (sis):[(9)f8,(9)h7,(16)d7]
 => d7≠9
Of course, one could reverse engineer the snippets to produce the sis:[(9)f8,(9)h7,(6)d7] which is slightly stronger. I do
this reverse engineering in my head to determine if further eliminations may exist within a pattern. However, during the
initial search, since no target has been determined, the more general snippets are often better. The more specific snippets
will fall out if they are required anyway.
By chance, perhaps, the focus on the Kraken cell (139)f8 caused me to muse about the Kraken Cell (139)f3. I had already
noted, mentally, the following snippets:
Step 2d prelude 1  markup induced snippet A
Illustrated above is the snippet:
 (2)g789=(21)g5=(147)g1=(hp47)ci1(9)i1=(9)i3
 Included in the illustration are two of the many possible weak links emanating from the snippet
 (1)g1(1)b1
 (9)i3(9)f3
 I also mused about stuff like
 (92)i3=(2)e3
 but found little immediate traction with most of those
My puzzle markup has some holes. I have found no good way to represent a number of recurring themes and still be able
to read the marks easily. One such theme not represented in the markup is:
 multicandidate hinges such as (3)f89=(32)d9=(2)de7
The next snippet pair shows how this can be a useful piece of information:
Step 2d prelude 2  more markup induced snippets B
Above, I actually have two snippets:
 (2)de7=(23)d9=(3)f89
 (2)de7=(2)d9(2)c9=(hp25)bc8
Also illustrated are two of the many available weak links emanating:
 (hp25)bc8(1)b8
 (3)f89(3)f3
Now one may be able to deduce a potential chain. Note (1=39)f3 and (1)b3=(1)b18 and (1)b3(1)f3. This means that
I can add both those sis
 (1)b138
 (139)f3
To the already considered snippets to link them.
Step 2d Combining short snippets
There can be no mistake about it. This comination of snippets does not graph well. However, the combination is
easily checked by AIC. I will show the AIC's as starting from either snippet. With both, however, one must recognize
multiple weak links emanating from the snippets:
 (2)g789=(2[1)g5=(147)g1=(hp47)ci1(9)i1=(9)i3](9)*f3&(1)**b1=
 [(3)d9=(3)f89(3=1)*f3(1)b3=(1)**b8(hp25)bc8=(2)b9](2)d9=(2)de7
 (2)de7=(2)d9[(3)d9=(3)f89 & (2)c9=(hp25)bc8](1)b8*&(3)**f3=
 [(47)g1=(hp47)ci1(9)i1=(9)i3(9=1)**f3(1)b3=*(1)b1](1)g1=(12)g5=(2)g789
Either way, (2)g789=(2)de7 => hi7≠2.
I wish I could say that I was able to immediately see the link between these snippets. I did not. Instead, even after
mentally noting these snippets (and many others), I did not see the elimination. Instead, I suspected that something was going on regarding
kraken cell (139)f3. Thus, the search began with looking at the Kraken cell (139)f3. Mentally, I built a group of matrix counts from that cell.
Below, find the applicable links that I used, and the order in which I considered them. The matrix below does not
prove anything, but is merely a reflection of what the link counts mean.
f3 
** 
1 
3 
9 








9i 
   i3 
i1    
   
47r1 
   
i1  cg1   
   
1g 
   
 g1  g5  
   
1b 
 b3   
 b1?   b8 
   
3B2 
  f89  
   
d9    
25B1 
   
   bc8 
 (2)c9   
2B2 
   
   
d9  d9  de7  
2g 
   
  g5  
   g789 
Above, I have counted myself into a 12x9 matrix. Thus, my count is +3. However, it can be easily fixed.
Always, I know that all two member only columns are strictly ok, regardless of row order. To produce a valid deduction,
I need to move columns and rows to force the following:
 (1)g1 needs to be in a top position. This is because I am using its weak link in two dimensions. Another way to look
at the same idea is that clearly (1)b1(47)cg1 is not a native weak link. Nor is it a derived weak link in evidence. It may
in fact be false.
 If (2)d9 becomes a top column item, then the matrix count is reduced by 1.
 If (2)de7, (2)g789 are placed in the result column, the matrix count is reduced by 2.
Below, find the Mixed Block Matrix (MBM) that I could build from the link counts.
2g 
g789 
g5 









1g 
 g5  g1  
   
  
4r1 
  g1  c1 
i1    
  
7r1 
  g1  c1 
i1    
  
9i 
   
i1  i3   
  
2B2 
de7    
  d9  
  
3B2 
   
  d9  f89 
  
f3 
   
 9   3 
1   
1b 
  b1  
   
b3  b8  
5B1 
   
   
 b8  c8 
2B1 
   
  c9  
 b8  c8 
Note the use of the symmetric pm of almost hidden pair 47. Typically, one can dodge one of these. With this step, however,
since there are almost hidden pairs used in each snippet, one can dodge using a symmetric pm with the matrix on only one side.
Of course, one can write a Block Triangular Matrix to do the job. In fact, in might look exactly the same. In eiher case,
the overall difficulty is no greater than a simple triangular matrix if one thinks in useful groups. In other words, treating
the hidden pair 47 as a single Boolean that must exist in at least one of gi1.
Since I am densely (puns are a low form of humour) explaining these steps, it seems perhaps that I should end this page here. Not much has been accomplished
yet to unlock this puzzle. However, some groundwork has been laid. The first step on the next page is one of my favorites.
Eventually, I will get around to making
page two.