SteveG48 wrote:Still figuring yours out!
Good luck with that!
Actually it shouldn't be hard at all for you, except maybe for those possibly confusing spread end points on the right hand side (marked with '@'). I chose to go with them instead of monster split nodes to keep the chain simpler. Either way, the overall logic should be read as:
(1r8c1 & 4r2c1 & 8r9c3) == (4r5c2 & 4r9c7 & 6r2c1) => -4 r59c1,r9c3, -6 r8c1
But yikes! You need all those eliminations!
Of course! Our motto: more is more!
Actually I got a bit frustrated when I found a bunch of eliminations easily but none gave much of anything by themselves or even combined. That horror combo finally worked, and I was lucky to even notice it. It was kind of fun, though, since the chain turned out simpler than I'd feared. Figuring out the matrix was the hardest part. I was pretty sure I wouldn't be able to write a valid matrix for that piece of logic, but it actually might be. (That said, I trust only Cenoman's judgment with those).
Here's a matrix for your solution too, if you're interested. It seemed like another non-trivial practice case, and it was. I hope I got it right.
- Code: Select all
3r1c3 2r1c3
2r1c7 7r1c7
7r3c8 1r3c8
3r8c2 ................. 6r8c2
7r8c7 ..... 6r8c7 2r8c7
7r9c7 ........... 2r9c7 4r9c7
4r9c2 5r9c2
4r9c1 5r9c1 7r9c1
|3r7c8 ........... 1r7c8 ............................. 7r7c8
|3r7c6 ............................................... 7r7c6
&3r9c8 ========================================= 7r9c8 =====
------------------------------------------------------------
-3r79c3
Here's one for eleven's logic too, which was also interesting (a bit harder with a 3-SIS at both ends):
- Code: Select all
4r7c3 3r7c3 8r7c3
8r7c9 1r7c9
1r3c9 6r3c9
6r3c2 6r8c2
3r8c2 ................. 6r8c2
6r3c2 7r3c2
7r3c8 1r3c8
|3r7c8 ................................... 1r7c8 7r7c8
|3r7c6 ......................................... 7r7c6
&6r7c7 ======================= 6r8c7 =================
------------------------------------------------------
-3r7c3
-4r7c7
As an AIC:
4r7c3 = 36b7p35|81r7c39 - 16r3c92 = 76r38c2 - 7,1r37c8|6r8c7 = 73r7c86,6r7c7 => -3r7c3, -4r7c7
That's structurally closest to mine, except mine has a 4-SIS at both ends:
- Code: Select all
&1r8c1 1r8c9 ===========
&4r2c1 ===== 6r2c1 =====
&8r9c3 ........... 8r7c3
8r7c9 8r8c9
2r8c9 2r2c9
2r8c9 ................. 2r2c9
6r2c9 6r2c1
6r2c9 ................. 6r2c1
6r7c1 6r7c7
4r7c7 4r9c7
&4r5c2 ......................................... 4r9c2
&4r9c7 =================================== 4r7c7 =====
&6r2c1 ============================= 6r2c9 ===========
------------------------------------------------------
-4r59c1
-4r9c3
-6r8c1
Phil's was normal (2-SIS at both ends), thank goodness:
Phil 12x12 BTM (or 10x10 TM with folded ALS): Show - Code: Select all
6r2c9 2r2c9
2r1c7 7r1c7
7r1c2 3r1c2
2r8c9 ........... 8r8c9 1r8c9
8r7c9 1r7c9
1r7c8 7r7c8 3r7c8
7r7c6 3r7c6
3r8c2 ................. 3r7c3 3r9c3
8r9c3 8r9c5
2r9c5 2r9c7
4r9c7 4r7c7
6r7c1 6r7c7
------------------------------------------------------------------------
-6r2c1
And lastly, rjamil's clean moves as three small matrices:
rjamil 4x4 PM + 4x4 SPM + 3x3 PM: Show - Code: Select all
4x4 PM
7r8c6 3r8c6
3r8c2 6r8c2
6r3c2 7r3c2
7r1c7 7r3c8
------------------------
-7r8c7
As an AIC: (7=3)r8c6 - (3=6)r8c2 - (6=7)r3c2 - r3c8 = (7)r1c7 => -7 r8c7
- Code: Select all
4x4 SPM
6r8c2 3r8c2
3r1c2 7r1c2
7r1c7 2r1c7
6r8c7 2r8c7
------------------------
-6r8c1 -2r9c7
As an AIC: (6=3)r8c2 - (3=7)r1c2 - (7=2)r1c7 - (2=6)r8c7 - loop => -2 r9c7, -6 r8c1
- Code: Select all
3x3 PM
1r8c1 7r8c1
7r9c1 7r9c7
1r7c9 7r7c9
------------------
-1r7c1
-1r8c9
As an AIC: (1=7)r8c1 - r9c1 = r9c7 - (7=1)r7c9 => -1 r7c1,r8c9