Slog thru Tough January 30, 2008 Page 1

The following is page one of an illustrated proof for the 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.

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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 mark-ups, 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

Puzzle as given

Two Unique Possibilities exist:

  1. (4)e8 % column & box (hidden single both in column a and Box b8 <=> Box 1)
  2. (6)g2 % column & box
Thus UP 24.

Steps 2a & 2b - Locked Candidates

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


Puzzle Mark-up

This puzzle has only five bivalues, and relatively few bilocals. No obvious bivalue-bilocal 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 mark-up.

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 mark-up, 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

Snippet Links

Illustrated above is my thought process:

  • (hp16)ad7=(1)e7
    • Circled 6's ad7 and underlined 1 a7,f8 highlit this snippet
  • (9)h7=(9-4)h9=(4-3)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=(3-4)g9=(4-9)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 - mark-up induced snippet A

Snippet Links prelude 1 for step 2d

Illustrated above is the snippet:

  • (2)g789=(2-1)g5=(1-47)g1=(hp47)ci1-(9)i1=(9)i3
  • Included in the illustration are two of the many possible weak links emanating from the snippet
    1. (1)g1-(1)b1
    2. (9)i3-(9)f3
  • I also mused about stuff like
    • (9-2)i3=(2)e3
    • but found little immediate traction with most of those
My puzzle mark-up 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 mark-up is:
  • multi-candidate hinges such as (3)f89=(3-2)d9=(2)de7
The next snippet pair shows how this can be a useful piece of information:

Step 2d prelude 2 - more mark-up induced snippets B

Snippet Links prelude 2 for step 2d

Above, I actually have two snippets:

  1. (2)de7=(2-3)d9=(3)f89
  2. (2)de7=(2)d9-(2)c9=(hp25)bc8
Also illustrated are two of the many available weak links emanating:
  1. (hp25)bc8-(1)b8
  2. (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. (1)b138
  2. (139)f3
To the already considered snippets to link them.

Step 2d Combining short snippets

Step 2d

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=(1-47)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=(1-2)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 i1cg1
1g g1g5
1b b3 b1?b8
3B2 f89 d9
25B1 bc8 (2)c9
2B2 d9d9de7
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. (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.
  2. If (2)d9 becomes a top column item, then the matrix count is reduced by 1.
  3. 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 g5g1
4r1 g1c1 i1
7r1 g1c1 i1
9i i1i3
2B2 de7 d9
3B2 d9f89
f3 93 1
1b b1 b3b8
5B1 b8c8
2B1 c9 b8c8

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.

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