jovi_al01 wrote:some digit relabeling with [57] is possible (cells in the middle stack marked with A and B)...
I hadn't coded eleven replacement for 2 digits in 2 cells but I added this to CSP-Rules [minor update on GitHub]. Here is what it gives, starting from the same resolution state as you, i.e.:
- Code: Select all
+----------------+----------------+----------------+
! 38 38 9 ! 57 6 57 ! 4 1 2 !
! 2 7 1 ! 49 38 49 ! 38 6 5 !
! 5 6 4 ! 2 38 1 ! 389 389 7 !
+----------------+----------------+----------------+
! 7 1 6 ! 3 2 8 ! 5 4 9 !
! 9 358 2 ! 1 4 57 ! 378 378 6 !
! 348 3458 358 ! 69 57 69 ! 2 378 1 !
+----------------+----------------+----------------+
! 1346 9 357 ! 8 157 46 ! 67 2 34 !
! 3468 3458 3578 ! 4567 9 2 ! 1 57 34 !
! 146 2 57 ! 4567 157 3 ! 679 579 8 !
+----------------+----------------+----------------+
- Code: Select all
(solve-sukaku-grid-by-eleven-replacement2
5 7
5 6
6 5
+----------------+----------------+----------------+
! 38 38 9 ! 57 6 57 ! 4 1 2 !
! 2 7 1 ! 49 38 49 ! 38 6 5 !
! 5 6 4 ! 2 38 1 ! 389 389 7 !
+----------------+----------------+----------------+
! 7 1 6 ! 3 2 8 ! 5 4 9 !
! 9 358 2 ! 1 4 57 ! 378 378 6 !
! 348 3458 358 ! 69 57 69 ! 2 378 1 !
+----------------+----------------+----------------+
! 1346 9 357 ! 8 157 46 ! 67 2 34 !
! 3468 3458 3578 ! 4567 9 2 ! 1 57 34 !
! 146 2 57 ! 4567 157 3 ! 679 579 8 !
+----------------+----------------+----------------+)
AFTER APPLYING ELEVEN''S REPLACEMENT METHOD to 2 digits 5 and 7 in 2 cells r5c6 and r6c5,
the resolution state is:
+-------------------+-------------------+-------------------+
! 38 38 9 ! 57 6 57 ! 4 1 2 !
! 2 57 1 ! 49 38 49 ! 38 6 57 !
! 57 6 4 ! 2 38 1 ! 389 389 57 !
+-------------------+-------------------+-------------------+
! 57 1 6 ! 3 2 8 ! 57 4 9 !
! 9 3578 2 ! 1 4 5 ! 3578 3578 6 !
! 348 34578 3578 ! 69 7 69 ! 2 3578 1 !
+-------------------+-------------------+-------------------+
! 1346 9 357 ! 8 157 46 ! 657 2 34 !
! 3468 34578 3578 ! 4576 9 2 ! 1 57 34 !
! 146 2 57 ! 4576 157 3 ! 6579 579 8 !
+-------------------+-------------------+-------------------+
THIS IS THE PUZZLE THAT WILL NOW BE SOLVED.
RELEVANT DIGIT REPLACEMENTS WILL BE NECESSARY AT THE END, based on the original givens.
After Singles and whips[1], we have a few easy chains (possibly reducible to only one):
- Code: Select all
biv-chain[3]: r4c7{n5 n7} - r7n7{c7 c3} - r9c3{n7 n5} ==> r9c7≠5
biv-chain[3]: r7c6{n6 n4} - r9n4{c4 c1} - b7n1{r9c1 r7c1} ==> r7c1≠6
biv-chain[3]: c1n6{r8 r9} - r9n4{c1 c4} - b8n7{r9c4 r8c4} ==> r8c4≠6
hidden-single-in-a-row ==> r8c1=6
biv-chain[3]: r1c2{n3 n8} - c1n8{r1 r6} - b4n4{r6c1 r6c2} ==> r6c2≠3
biv-chain[3]: r8c8{n5 n7} - b8n7{r8c4 r9c4} - r9c3{n7 n5} ==> r9c8≠5, r8c2≠5, r8c3≠5
stte
Without using the assumption of uniqueness and without replacement, the solution was in Z4, so there is some gain. Of course, replacement has own complexity cost.
P.S. I understand that what you're doing after replacement is different. For me, it was more a question of trying eleven replacement in 2 cells.
