Back to the blackboard.
P.O. wrote:...the one step loop of yzfwsf is very impressive to me.
Same to me !
yzfwsf wrote:(8=9)r9c9 - n9r5(c9=c6) - n9r1(c6=c3) - n3r1(c3=c9) - r5c9(3=7) - r5c15(7=4*1) - r1c1(1=5) -r1c5(5=6) - r9c5(6=8) - 8r9c9,loop=> r1c369<>5, r369c1<>1,r3c5<>6,r36c9<>9,r5c36<>7,r5c6<>4,r9c136<>8
LC+HP=>stte
In order to catch yzfwsf's solution, I have written it my way:
It's a memory chain, that I'd write comprehensively as a net:
- Code: Select all
- - - - - - - - - - - - (9)r1c9
/ ||
(8=9)r9c9 - r5c9 = r5c6 - r1c6 = (9-3)r1c3 = r1c9 - (3=7)r5c9 - (7=41)r5c15 - (1=5)r1c1 - (5=6)r1c5 - (6=8)r9c5
\ || \ ||
- - - - - - - - - - - - - - - - - - - - - - - -(9)r5c9 - - - - - - - - - - - - - - - - - -(4)r9c5
Without memories, it could be written as a kraken (cell r9c5):
- Code: Select all
(4)r9c5 - - - - - - - - - - (4=7)r5c5
|| \
(6)r9c5 - (6=5)r1c5 - (5=17)r15c1 - (7)r5c9 = [(9*=*3)r5c9 - r1c9 = (3-9)r1c3 = XW(9)r15\c69] - (9=8)r9c9
||
(8)r9c5
-------------
=> (8)r9c5 == (8)r9c9
Here is the Triangular Matrix of this move (TM 10x10) with yzfwsy's 15 eliminations shown in the bottom line:
- Code: Select all
8r9c9 9r9c9
9r5c9 9r5c6
9r1c9 9r1c6 9r1c3
3r1c3 3r1c9
9r5c9 3r5c9 7r5c9
7r5c5 4r5c5
7r5c1 1r5c1
1r1c1 5r1c1
5r1c5 6r1c5
8r9c5 4r9c5 6r9c5
---------------------------------------------------------------------------------
-8r9c136 -9r36c9 -7r5c36 -4r6c5 -1r369c1 -5r1c369 -6r3c5
This matrix is not far from a symmetric pigeonhole matrix. The double presence of 9r5c9 in the second column prevents it. Nevertheless, column 2 and columns 6 to 10 can be brought as the first column, hence the eliminations at their bottom. It not possible for columns 3,4,5 without placing 9r5c9 as a top entry of a column, which would make an invalid TM.
So, I agree with yzfwsf's list of eliminations. My question to you is: how did you build your list of eliminations, which is not so obvious to derive from your "chain" writing (and maybe not much more from the nets writing) ?
Anyhow, kudos for your finding