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David Bryant
Gold Member

Joined: 20 Jan 2006
Posts: 86

 Posted: Sat Jan 21, 2006 8:51 pm    Post subject: 19 Jan, 2006 Nightmare Hello! I'm new to this forum ... this is my first post. I really liked this puzzle, Ruud. It had me scratching my head for quite a while. When I finally reduced the possibilities for the digit "6" to just 19 cells (I had to use coloring and a swordfish to get there) the solution became obvious -- there was only one possible way to fit the nine "6"s in those 19 cells. So the final solution was obtained all at one blow, more or less. Brilliant composition! dcb
lac
Hooked

Joined: 02 Jan 2006
Posts: 43
Location: Göteborg, Sweden

 Posted: Sat Jan 21, 2006 11:01 pm    Post subject: Jan 19th nightmare Where's the swordfish? I solved it using the technique I explain here: http://www.sudocue.net/forum/viewtopic.php?t=49 in this forum. But guess I found a different way. I still cannot see the swordfish.... confused, Laura
David Bryant
Gold Member

Joined: 20 Jan 2006
Posts: 86

Posted: Sun Jan 22, 2006 12:53 am    Post subject: Here's how I did it ...

Hi, Lac!

I don't think you're confused ... if you solved this puzzle, you have quite a head on your shoulders, no matter how you did it! No doubt there are many roads to the one solution.

After I had filled in 19 cells, with 27 left to go, my copy of the puzzle looked like this. (None of the moves so far were very hard ... the trickiest thing I had to do so far was to see the "hidden triple" at r1c8, r2c7, & r2c8.)
 Code: 456  368  3468   9    1    7     2    58   346 456   1   4689   3   46    2    578   578  469  2   369    7    8    5   46    36     1  3469  8   69   1569   7   46   456   146    3    2  3    2    56    1    9   456   467   76    8  7    4    16    2    3    8    16     9    5  1    5    46   46    8    3     9     2    7  9    7    38   56    2    1   3568    4   36 46   38     2   456   7    9   3568   568   1

At this point I could start to trace a binary chain in the cells containing a possible "6", from r9c1 to r7c3 to r7c4. Combining this with the two spots for a "6" in row 6 I was able to eliminate the "6" at r9c7. And combining the same short binary chain with the two spots for a "6" in column 8 I was able to eliminate the "6" at r5c3, leaving "5" as the sole candidate in that cell. That allowed me to place another "5" at r4c6, so now my matrix looked like this.
 Code: 456  368  3468   9    1    7     2    58   346 456   1   4689   3   46    2    578   578  469  2   369    7    8    5   46    36     1  3469  8   69    169   7   46    5    146    3    2  3    2     5    1    9   46    467   76    8  7    4    16    2    3    8    16     9    5  1    5    46   46    8    3     9     2    7  9    7    38   56    2    1   3568    4   36 46   38     2   456   7    9    358   568   1

The next thing I noticed was the group of {4, 6} pairs in box 2 & box 5. I traced a short double-implication chain (which I think is what you're talking about in your "novel"):

r3c6 = 6 ==> r3c7 = 3
r3c6 = 4 ==> r5c6 = 6 ==> r4c5 = 4 ==> {1, 6} pair in r4c7 & r6c7 ==> r3c7 = 3

So r3c7 = 3, and this forces r8c9 = 3 (unique in box 9). We can also make some simple moves involving "3", "5", and "8". The matrix looks like this now.
 Code: 46    8     3    9    1    7     2     5   46  5    1    469   3   46    2    578   578  469  2   69     7    8    5   46     3     1   469  8   69    169   7   46    5    146    3    2  3    2     5    1    9   46    467   76    8  7    4    16*   2    3    8    16*    9    5  1    5    46*  46*   8    3     9     2    7  9    7     8   56*   2    1    56*    4    3 46    3     2   456   7    9    58    68    1

The swordfish appears in rows 6, 7, & 8, columns 3, 4, & 7 -- we can use it to eliminate "6" at r2c3, r4c3, r9c4, r4c7, & r5c7.
 Code: 46    8     3    9    1    7     2     5   46  5    1    49    3   46    2    78    78   469  2   69     7    8    5   46     3     1   469  8   69    19    7   46    5    14     3    2  3    2     5    1    9   46    47    76    8  7    4    16    2    3    8    16     9    5  1    5    46   46    8    3     9     2    7  9    7     8   56    2    1    56     4    3 46    3     2   45    7    9    58    68    1

And now the beautiful binary chain of "6"s is apparent -- it leads all around the board from r1c9 through r1c1 to r9c1 to r9c8, etc. If you use simple blue/green coloring you will discover a contradiction in r3c2 & r3c6, and that lets you set the values in 19 cells all at once!

Anyway, that's how I solved this one, lac. I'm curious -- how was your approach different from mine? dcb
Ruud
Site Owner

Joined: 30 Dec 2005
Posts: 601

 Posted: Sun Jan 22, 2006 1:00 am    Post subject: I will not interrupt your discussions, but I would like to welcome you to the forum, David. If you want to know how Laura solved this nightmare, follow the link in her post and prepare for a long read. Many more nightmares are coming. Ruud._________________“If the human brain were so simple that we could understand it, we would be so simple that we couldn't.” - Emerson M Pugh
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