19 Jan, 2006 Nightmare

[David Bryant]

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]

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]

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: Select all

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: Select all

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: Select all

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: Select all

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]

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.