Fourth page for January 30, 2008

The following is page four of an illustrated proof for the tough puzzle of 01/30/08. Pages three, two, and one finely detail a manner for locating eliminations. This page will continue in similar fashion.

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.

Previous Page change notification

For those who may be interested, I have altered the previous page to show a slight alternative to step 3f This addition may make my thoughts after step 3c easier to understand. Hopefully, it serves to illustrate that taking the time to find alternative elimination paths is often fruitful, as puzzle weaknesses tend to overlap.

Mark-Up From Page 3

Since I am trying to detail how I find eliminations, I thought perhaps I needed to return to the previous mark-up. Typically, as I make the new marks, I try to think about how the new stronger strong inference sets (sis) effect what I alrady know about the puzzle.

Puzzle Mark-up revisited

Below, find a list of marks and thoughts as eliminations were made.

  • Step 3b => a5,c56≠8
    • (8)c9=(8)c12 => underline (8)c9.
    • Note (8)b456=(ht489)a45c5-(ht479)a45c5=(7)c6
  • Step 3c => e25<>1 => not much to note on the mark-up, but.... make a mental note to return to the resultant stengths later.
  • Step 3d Prelude => note (1)f6=(7)c1
  • Steps 3d & 3e Note (1)f6=(8)f1 => f6≠8 => Hidden pair 58 f25 => d5≠1369, d1≠19
    • (8)f1=(8)f5 => circle (8)f5, promote underline at (8)f1 to circle.
    • (1)d5=(1)gh5 => underline (1)d5
      • Not noted in the mark-up, a potential hinge (1)d12=(1)ef3
      • This hinge, when combined with the knowledge of the almost jellyfish, caused me to mentally note:
      • Almost coloring: (1)a2=[(1): g5=g1-h2=d2]
      • I spent some time thinking about candidate (1) here, and the old underlines:
        • (7)a2,(12)b8, with the circled (5)b8 became very interesting
        • Add mental note: (2)d7=(6)c6, (some detail of this will be provided later)
      All of this is a precursor to the manner in which I found 3f and also step 4, which with a little added depth could easily solve more cells.
    • (3) no updates
    • (6)f9=(6)f6 => circle (6)f6, keeping underline & promote underline at (6)f9 to a circle.
    • (6)d5=(6)f6 => add circles to both (6)d5 & (6)f6. Note
      • (6)d5=(6-1)f6=(7)c1-(7=26)c6 => (6)d5=(2)c6
      • (6)d5=(6-1)f6=(8-5)f1=(5)f5 => d5≠5 - is this useful?
    • (9)f3=(9)f89 => underline (9)f3.
    • (9)d1=(9)i1 => circle (9)d1, upgrade underline at (9)i1 to a circle.
    • note ALS (1)b1=(pair58)bf1. My proof does not use this, but one can eventually eliminate (58)c1 using this concept.
  • Steps 3f & 3g => d5≠1 => (1)e6=(1)f6, circle both (1)ef6, => h6≠1
    • (1)g5=(1)h5, add circles to both (1)gh5
    • (1)h2=(1)h5 => add circle to (1)h5, promote underline at (1)h2 to a circle
    • (1)d7=(1)d12 => underline (1)d7
    • Note (1)a2 eliminated by (12)column d, (1)Box9, (1=2)hi2 ALS. => look into (15)overlap h5,d2
    • Note (58)c1 eliminated by above plus ALS(1=58)bf1 => look into whether to execute this step also
    • Note (1)f6=(1)e6-(1=2)e7-(2)d7=(2-5)d2=(5)c2-(5)c8=(5-2)b8=(2)b56-(2)c6=(6)d5 => f6≠6 - again, decide whether to use this possible step
    • Note new bivalue (27)h6 => AUR 27 hi26 with (7)h2 underline => i6≠7 (again,not used)
  • step 3h chosen to solve a cell: h9≠2 => (4)h9% cell
    • (2)h2=(2)h56 => underline (2)h2, AUR27 => i6≠27 - this looks not very powerful
    • (4)a4=(4)g4 => circle (4)g4, promote (4)a4 underline to circle. Muse briefly about almost coloring.
    • (4)f1=(4)g45 => underline (4)g1.
Now I am a bit confused as what to execute, as a number of eliminations are available to me. After some careful consideration, I estimate that eliminating (6)d7 is sufficient to break the puzzle if eliminating (6)f6 breaks the puzzle, At this point, I decide to use the less deep elimination, and if that gets me stuck, to return and use the deeper elimination.

Thinking as I update the puzzle marks is generally very helpful. It lets me know what is really new. There is no reason that I can fathom to treat the puzzle as a new puzzle after each elimination. The main problem that I have with this puzzle at this point, and in fact with many puzzles at similar points, is that there are too many possible paths to take. The path illustrated below is one of many. The choice is fundamentally arbitrary.

Step 4 - a simple chain using 3f prelude

Below, find a straightforward chain, given (6)d5=(2)c6 as proven in the prelude to step 3f.

AIC using derived sis

Above, find:

  • (6)c5=using 3f(2)c6-(2)b56=(2-5)b8=(5)b1-(5)c2=(5)d2-(2)d2=(2)d7 =>
  • (6)c5=(5)d2 => c5≠5
  • (6)c5=(2)d7 => d7≠6
  • (6)a7 %row => UP 31
If I were to not use the derived inference above, I also thought as follows:
  1. (2)d7=(2)d2-(2=7)i2-(7)a2=(7)a45
  2. (2)d7=(2-5)d2=(5)c2-(5)b1=(5-2)b8=(2)b56
  3. (2=7=6)c6
  4. => (2)d7=(6)c6
  5. (2)d7=(6)c6-(6)f6=(6)d5
  6. => (2)d7=(6)d5 => d7≠6
Usually, I am not certain after I solve a puzzle what I saw first.

In both cases, (6)a7 % row. The (5) elimination from d5 is an added bonus of the illustrated version. I knew it anyway immediately after establishing the Hidden Pair 58 in column f.

The elimination graphed and the musings noted are an accurate reflection of how I work difficult puzzles. Remembering sis that have been proven is not only helpful, but silly to deny. Note that part of step 3f led to the conclusion above.

After (6)a7,d5≠5 puzzle marks are again updated as follows:

  • (1)d7=(1)e7 => f8≠1, circle both (1)de7.
  • (1)a8=(1)b8 => circle both (1)ab8
  • (1)a8=(1)a23 => underline (1)a8
  • (1)f6=(1)f3 => circle both f36
  • (6)d9=(6)f9 => circle both df9
  • (6)b3=(6)c3 => circle both bc3
  • (5)f5=(5)h5 => circle (5)f5, upgrade underline at (5)h5 to circle.
    • (5-1)h5 new possible long chain node
  • (5)f5=(5)d4 => add circles to both of these - note (triple circled 5 at f5
  • (5)d4=(5)d12 => underline (5)d4 => Note Almost Hidden triple 125 conjugate with ALS 3695 in column d.
This group of updates were only mildly interesting by themselves. However, one of the steps considered earlier on this page seems like a great step to execute now, especially becuase of the added circled 5 at h5. One could write the next three steps together, but I am not sure that is the clearest way to proceed.

Steps 5a&b - Locked candidates and a very pretty ALS/AHS overlap

Below, find the locked candidates elimination of f8≠1. The other elimination shown, a2≠1, I found very pretty to think about in the manner illustrated. There are, of course, other ways to find the same exclusion using the exact same sets.

Fabled Hidden Pair Naked Triple Overlap

Above, step 5b is shown exactly the way I saw it:

  • (1)h2=(1)g1-(1)d1=(Hidden Pair 12)d27&(naked triple 127)dhi2
  • More classically, without the &:
    • (1)h2=(1)g1-(1)d1=(Hidden Pair 12)d27-(35)d2=(naked triple 127)dhi2
    • Also, (1)h2=(pair27)hi2-(2)d2=(2-1)d7=[(1)Xwing d12,gh2]
  • Regardless of view points, (1)h2=(1)d2 => a2≠1
Significant updates to the mark-up occur here, noting (1)a3=(1)a8 and (1)d2=(1)h2. At this point, the overlap between candidates 1 & 5 at h5,d2 is hard to miss. Below find how I parce this relationship.

Step 5c Prelude - An Almost Proving Loop


  • (8)b1*=[(1)h2=(1-5)d2=(5)c2-(5=1)*b1-(1)g1=(1)h2]
  • => (1)h2=(8)b1.

    Almost Proving Loop

    The continuation of the chain above practically writes itself:

    Step 5c

    Below, find:

    • (1)h2=(8)b1-(8=5)f1-(5)f5=(5)h5
    • => h5 ≠1

    Proving Loop linked up

    After making the elimination, (1)h2 % column and (1)g5 % row, box gets the puzzle to UP 31. Lots of updates can be made to the mark-ups. The most significant one, for me, was recognizing the new bilocal: (1)d1=(1)d7. This led to an easy step.

    Step 6 - Standard bilocal AIC

    Standard fare AIC

    Above, find;

    • (1)d1=(1-2)d7=(2)e7-(2)e3=(2-9)i3=(9)i1 => d1≠9
    • => (9)i1% row => UP 32
    The puzzle now can be cleaned up easily using standard fare AIC. The last page of this proof will illustrate one manner.


    Although one may prefer a better organized approach to puzzle analysis, I find that quickly brainstorming progressive chain snippets is quite helpful, and fairly fast. I do not want to bog down too long on one narrow analytic channel.

Indicate which comments you would like to be able to see

Sorry about the long delay between pages. I cannot promise that the next page will be any faster! If only life were as logical as Sudoku!
06/Mar/08 10:18 PM
Hi Steve,
"If only life were as logical as Sudoku!"
Something is not good for you (your business, family...)? Hope, I'm wrong...
07/Mar/08 2:02 AM
Thanks for asking ttt!

All is well!!
- I only meant that life is often not well served by purely logical analysis.
07/Mar/08 2:31 AM
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