The New Sudoku Players' Forum

by **mith** » Tue Mar 01, 2022 5:59 pm

Code: Select all: +-------+-------+-------+ | 5 7 . | . . . | 9 . . | | . . . | . . . | . . 8 | | . 1 . | . . . | . . . | +-------+-------+-------+ | . . 1 | 6 8 . | . 4 . | | . . . | . . 2 | 8 . 9 | | . . 2 | . 9 4 | 1 6 . | +-------+-------+-------+ | . . . | . 2 . | . . . | | . 6 . | 9 . 8 | 2 . 4 | | . . . | 4 1 . | 6 . . | +-------+-------+-------+ 57....9..........8.1.........168..4......28.9..2.9416.....2.....6.9.82.4...41.6.. ED=11.9/1.2/1.2 "Loki"

by **999_Springs** » Wed Mar 02, 2022 12:11 am

solved by hand, no computer assistance other than automatically filling in the pencilmarks after the singles at the start. took about an hour - most of it trying to find a chain when there was a hidden single available

singles and a hidden pair to here

Code: Select all: +-------+-------+--------+ | 5 7 . | . . . | 9 . . | | . . . | . . . | . . 8 | | . 1 . | . . . | . . . | +-------+-------+--------+ | . . 1 | 6 8 . | . 4 2 | | . . . | 1 . 2 | 8 . 9 | | . . 2 | . 9 4 | 1 6 . | +-------+-------+--------+ | . . . | . 2 6 | . 89 . | | . 6 . | 9 . 8 | 2 . 4 | | . . . | 4 1 . | 6 89 . | +-------+-------+--------+

trivalue oddagon r7c9=/=1

singles to

Code: Select all: +-------+-------+--------+ | 5 7 . | 8 . 3 | 9 2 1 | | . . . | . . 1 | . . 8 | | . 1 . | . . 9 | . . 6 | +-------+-------+--------+ | . . 1 | 6 8 . | . 4 2 | | . . . | 1 . 2 | 8 . 9 | | . . 2 | . 9 4 | 1 6 . | +-------+-------+--------+ | 1 . . | . 2 6 | . 89 . | | . 6 . | 9 . 8 | 2 1 4 | | . . . | 4 1 . | 6 89 . | +-------+-------+--------+

3,5,7 are only located in r1, so replace them with A,B,C in b9 using eleven's method

Code: Select all: +-------+---------+---------+ | # # . | 8 . BC | 9 2 1 | | . . . | . . 1 | . . 8 | | . 1 . | . . 9 | . . 6 | +-------+---------+---------+ | . . 1 | 6 8 CA | BC 4 2 | | . . . | 1 AC 2 | 8 CB 9 | | . . 2 | B 9 4 | 1 6 A | +-------+---------+---------+ | 1 . . | C 2 6 | A 89 B | | # 6 # | 9 BA 8 | 2 1 4 | | . . . | 4 1 AB | 6 89 C | +-------+---------+---------+

# cells must be ABC

simple colouring on the remaining unsolved cells in b5689 gives either

case I: letters on the left are all true, or
case II: letters on the right are all true

since r1c6=BC=3, this immediately gives A=/=3

suppose case I is true
then r1c18=AC is a naked pair
r4c2=A, r6c2=C hidden singles in row, then no value for r1c2, contradiction

so case II is true, implying that C = 3

Code: Select all: +---------+-------+--------+ | AB AB . | 8 . 3 | 9 2 1 | | . . . | . . 1 | . . 8 | | . 1 . | . . 9 | . . 6 | +---------+-------+--------+ | . . 1 | 6 8 A | 3 4 2 | | . . . | 1 3 2 | 8 B 9 | | . . 2 | B 9 4 | 1 6 A | +---------+-------+--------+ | 1 . . | 3 2 6 | A 89 B | | . 6 . | 9 A 8 | 2 1 4 | | . . . | 4 1 B | 6 89 3 | +---------+-------+--------+

r8c3 is a hidden single for B in column 3, and all singles from here to the end (this step took much longer than it should have done)

Code: Select all: +---------+-------+-------+ | B A 6 | 8 4 3 | 9 2 1 | | 4 9 3 | 2 6 1 | B A 8 | | 2 1 8 | A B 9 | 4 3 6 | +---------+-------+-------+ | 9 B 1 | 6 8 A | 3 4 2 | | 6 4 A | 1 3 2 | 8 B 9 | | 8 3 2 | B 9 4 | 1 6 A | +---------+-------+-------+ | 1 8 4 | 3 2 6 | A 9 B | | 3 6 B | 9 A 8 | 2 1 4 | | A 2 9 | 4 1 B | 6 8 3 | +---------+-------+-------+

B=5 and A=7

Code: Select all: +-------+-------+-------+ | 5 7 6 | 8 4 3 | 9 2 1 | | 4 9 3 | 2 6 1 | 5 7 8 | | 2 1 8 | 7 5 9 | 4 3 6 | +-------+-------+-------+ | 9 5 1 | 6 8 7 | 3 4 2 | | 6 4 7 | 1 3 2 | 8 5 9 | | 8 3 2 | 5 9 4 | 1 6 7 | +-------+-------+-------+ | 1 8 4 | 3 2 6 | 7 9 5 | | 3 6 5 | 9 7 8 | 2 1 4 | | 7 2 9 | 4 1 5 | 6 8 3 | +-------+-------+-------+

yay i solved the hardest sudoku in the world by hand with no pencilmarks

i'm a genius

by **mith** » Wed Mar 02, 2022 12:18 am

Nicely done

I've been refining my solution over the afternoon, and have it down to three non-basic steps. I'll post it later when I have a chance to finish writing it up.

Website · by **denis_berthier** » Wed Mar 02, 2022 7:48 am

.
See my comments on this puzzle here: http://forum.enjoysudoku.com/the-hardest-sudokus-new-thread-t6539-1048.html
As I'm not a genius

I tried to solve this puzzle with SudoRules after eliminating n1r7c9

[For the users of CSP-Rules] This can be done by typing:
(bind ?*simulated-eliminations* = (create$ (nrc-to-label 1 7 9)))
before typing:
(solve "57....9..........8.1.........168..4......28.9..2.9416.....2.....6.9.82.4...41.6.. ED=11.9/1.2/1.2 Loki)
Beware of not keeping the double quotes around Loki, as this would confuse CLIPS[/For the users of CSP-Rules]

Code: Select all: Resolution state after Singles and whips[1]: +----------------------+----------------------+----------------------+ ! 5 7 46 ! 8 46 3 ! 9 2 1 ! ! 23469 2349 3469 ! 257 4567 1 ! 3457 357 8 ! ! 2348 1 348 ! 257 457 9 ! 3457 357 6 ! +----------------------+----------------------+----------------------+ ! 379 359 1 ! 6 8 57 ! 357 4 2 ! ! 3467 345 34567 ! 1 357 2 ! 8 357 9 ! ! 378 358 2 ! 357 9 4 ! 1 6 357 ! +----------------------+----------------------+----------------------+ ! 134789 34589 345789 ! 357 2 6 ! 357 135789 357 ! ! 137 6 357 ! 9 357 8 ! 2 1357 4 ! ! 23789 23589 35789 ! 4 1 57 ! 6 35789 357 ! +----------------------+----------------------+----------------------+ 162 candidates.

hidden-pairs-in-a-column: c8{n8 n9}{r7 r9} ==> r9c8≠7, r9c8≠5, r9c8≠3, r7c8≠7, r7c8≠5, r7c8≠3, r7c8≠1
singles ==> r8c8=1, r7c1=1
finned-x-wing-in-columns: n3{c4 c9}{r6 r7} ==> r7c7≠3
whip[1]: b9n3{r9c9 .} ==> r6c9≠3
naked-triplets-in-a-row: r7{c4 c7 c9}{n3 n7 n5} ==> r7c3≠7, r7c3≠5, r7c3≠3, r7c2≠5, r7c2≠3
z-chain[3]: r1c3{n4 n6} - c1n6{r2 r5} - c1n4{r5 .} ==> r2c2≠4, r3c3≠4, r2c3≠4
t-whip[3]: r8c1{n3 n7} - c3n7{r9 r5} - r5n6{c3 .} ==> r5c1≠3
t-whip[3]: c1n6{r5 r2} - r1c3{n6 n4} - c1n4{r3 .} ==> r5c1≠7
t-whip[3]: r1c3{n6 n4} - c1n4{r3 r5} - r5n6{c1 .} ==> r2c3≠6
t-whip[3]: c5n3{r5 r8} - r8c1{n3 n7} - c3n7{r9 .} ==> r5c5≠7, r5c3≠3
biv-chain[3]: c4n3{r7 r6} - r5c5{n3 n5} - c6n5{r4 r9} ==> r7c4≠5
whip[1]: r7n5{c9 .} ==> r9c9≠5
biv-chain[3]: b6n3{r4c7 r5c8} - r5c5{n3 n5} - r4c6{n5 n7} ==> r4c7≠7
whip[3]: r4n7{c1 c6} - r6n7{c4 c9} - r9n7{c9 .} ==> r8c1≠7
singles ==> r8c1=3, r7c4=3, r5c5=3, r4c7=3, r6c2=3, r6c1=8, r3c3=8, r2c3=3, r3c8=3, r9c9=3
whip[1]: c3n9{r9 .} ==> r7c2≠9, r9c1≠9, r9c2≠9
x-wing-in-columns: n7{c1 c6}{r4 r9} ==> r9c3≠7
biv-chain[2]: c8n5{r2 r5} - r6n5{c9 c4} ==> r2c4≠5
biv-chain[2]: c8n7{r2 r5} - r6n7{c9 c4} ==> r2c4≠7
stte

This path could probably be shortened, but that's irrelevant here.

The point is, a puzzle that was not in T&E(2) - the first known such puzzle, the hardest of the hardest - and not even in gT&E(2) is turned into a relatively easy puzzle (solvable in W3) by the elimination of a single candidate (whichever way this elimination is justified).
As of now, this is probably the largest known effect of a single elimination.

by **shye** » Wed Mar 02, 2022 3:08 pm

first off, congrats on the new 11.9 mith! surprised it is actually solvable

after singles and hidden pair in b9

Code: Select all: .-----------------------.---------------------.-------------------. | 5 7 3468 | 238 346 13 | 9 123 136 | | 23469 2349 3469 | 2357 34567 13579 | 3457 12357 8 | | 234689 1 34689 | 23578 34567 3579 | 3457 2357 3567 | :-----------------------+---------------------+-------------------: | 379 359 1 | 6 8 #357 |#357 4 2 | | 3467 345 34567 | 1 #357 2 | 8 #357 9 | | 378 358 2 |#357 9 4 | 1 6 #357 | :-----------------------+---------------------+-------------------: | 134789 34589 345789 |#357 2 6 |#357 89 1357 | | 137 6 357 | 9 #357 8 | 2 #357+1 4 | | 23789 23589 35789 | 4 1 #357 | 6 89 #357 | '-----------------------'---------------------'-------------------'

trivalue oddagon
+1r8c8

singles, naked triple in r7, an empty rectangle (3r6c4 = 3r7c4 - 3b9p13 = 3b9p9 => -3r6c9) and consequent lc
then:

Code: Select all: .---------------------.---------------.----------------. | 5 7 46 | 8 46 3 | 9 2 1 | | 23469 2349 3469 | 257 4567 1 | 3457 357 8 | | 2348 1 348 | 257 457 9 | 3457 357 6 | :---------------------+---------------+----------------: | 379 359 1 | 6 8 57 |#357 4 2 | | 3467 345 34567 | 1 357 2 | 8 #57-3 9 | | 378 358 2 | 357 9 4 | 1 6 #57 | :---------------------+---------------+----------------: | 1 489 489 | 357 2 6 | 57 89 357 | | 37 6 357 | 9 357 8 | 2 1 4 | | 23789 23589 35789 | 4 1 57 | 6 89 357 | '---------------------'---------------'----------------' .---------------.----------------. .---------------.----------------. | 6 8 B57 |A357 4 2 | | 6 8 C57 |A357 4 2 | | 1 C357 2 | 8 B357 9 | | 1 A357 2 | 8 B357 9 | |A357 9 4 | 1 6 C57 | |B357 9 4 | 1 6 C57 | :---------------+----------------: :---------------+----------------: |B357 2 6 |C57 89 A357 | |A357 2 6 |C57 89 B357 | | 9 A357 8 | 2 1 4 | | 9 C357 8 | 2 1 4 | | 4 1 C57 | 6 89 B357 | | 4 1 B57 | 6 89 A357 | '---------------'----------------' '---------------'----------------'

set 357 in b6 to ABC and label b5689 accrdingly. there are two possible arrangements, but in each case the 57 pair in c6 is always comprised of B and C
-3r5c8, 3 is placed in b6

then, a small chain: 5r8c3 = 5r8c5 - (5=7)r9c6 - 7r4c6 = 7r4c1 - (7=3)r8c1 => -3r8c3
a two-string kite: 3r6c4 = 3r7c4 - 3r8c5 = 3r8c1 => -3r6c1
another chain: 8r6c1 = (8-3)r6c2 = 3r6c4 - 3b8p1 = 3b8p5 - (3=7)r8c1 => -7r6c1
an empty rectangle: 7r6c9 = 7r6c4 - 7b9p39 = 7b9p1 => -7r7c4
a crane: 7r4c1 = 7r4c6 - 7b8p9 = 7b8p5 => -7r8c1

many singles and 9lc in b7
then to finish

Code: Select all: .-----------------.---------------.-------------. | 5 7 46 | 8 46 3 | 9 2 1 | | 2469 249 3 | 2-57 4567 1 | 457 #57 8 | | 24 1 8 | 257 457 9 | 457 3 6 | :-----------------+---------------+-------------: | 79 59 1 | 6 8 57 | 3 4 2 | | 467 45 4567 | 1 3 2 | 8 #57 9 | | 8 3 2 |#57 9 4 | 1 6 #57 | :-----------------+---------------+-------------: | 1 48 49 | 3 2 6 | 57 89 57 | | 3 6 57 | 9 57 8 | 2 1 4 | | 27 258 579 | 4 1 57 | 6 89 3 | '-----------------'---------------'-------------'

remote pair
-57r2c4 stte

by **mith** » Wed Mar 02, 2022 4:19 pm

Alright, here is my three step solution. After basics...

Step 1: trivalue oddagon b5689 => -357r8c8, +1r8c8

Code: Select all: +--------+-------+--------+ | 5 7 . | . . . | 9 . . | | . . . | . . . | . . 8 | | . 1 . | . . . | . . . | +--------+-------+--------+ | . . 1 | 6 8 . | . 4 2 | | . . . | 1 . 2 | 8 . 9 | | . . 2 | B 9 4 | 1 6 A | +--------+-------+--------+ | . . . | C 2 6 | A 89 B | | 37 6 . | 9 . 8 | 2 1 4 | | . . . | 4 1 . | 6 89 C | +--------+-------+--------+

(There are singles we can do here, but they aren't needed for the next step so for the sake of clarity we'll use the above diagram with just the 1 filled.)

Step 2:

I've borrowed 999_Springs' lettering - we can see r6c9 == r7c7 (A), r7c4 == r9c9 (C), and as a consequence of these r6c4 == r7c9 (B).

Note that r6c4, r6c9, and r7c4 form a remote triple. An immediate consequence of this is that the other connected (by row/column) cells in b568 also form remote triples - r5c5/r5c8/r8c5, and r4c6/r4c7/r9c6

(You can see this by looking at the cases as in 999_Springs' post, but more generally - one of the cells in b6 must be C, and whichever it is places A in b5 and B in b8. So that is a remote triple, and if we have two remote triples among 9 cells which we know are three triples from the boxes, the third group must also be a remote triple. This is related to the "parity flow" idea proving the trivalue oddagon - the permutation of letters from b6 to b5 to b8 is ABC for our starting triple, so the permutation for the other two triples must be BCA and CAB; that is, they all have the same parity.)

The important remote triple is r5c5/r5c8/r8c5, and its relationship to r8c1. In chain terms:

Code: Select all: 7r46c1 = r5c13 - r5c58 =(rt)= r8c5 => -7r8c1

Or as a contradiction: if 7 were in r8c1, it would also be in r5c3, and 7 would see all three cells of the remote triple (which we know contains 7). #

After singles:

Code: Select all: +-------+-------+--------+ | 5 7 . | 8 . 3 | 9 2 1 | | . . 3 | . . 1 | . A 8 | | . 1 8 | * . 9 | . 3 6 | +-------+-------+--------+ | . . 1 | 6 8 A | 3 4 2 | | . . . | 1 3 2 | 8 B 9 | | 8 3 2 | B 9 4 | 1 6 A | +-------+-------+--------+ | 1 . . | 3 2 6 | A 89 B | | 3 6 . | 9 A 8 | 2 1 4 | | . . . | 4 1 B | 6 89 3 | +-------+-------+--------+

Step 3: r3c4 is the only place for A in b2/c4 => -2r3c4, stte

----------------

As it turns out, you can use the remote triple idea to prove the trivalue oddagon elimination, as follows:

Step A: r5c8 and r8c5 cannot contain the same digit. If they were the same (say "B"), the other cells in b68 are A/C in some order, and either:
i. r4c7 == r7c4 (A), r6c9 == r9c6 (C), in which case both r6c4 and r4c6 would need to be B
ii. r4c7 == r9c6 (A), r6c9 == r7c4 (C), in which case both r7c7 and r9c9 would need to be B
So r5c5/r5c8/r8c5 is a remote triple, as are r4c6/r4c7/r9c6 and r6c4/r6c9/r7c4.

Step B: If r8c8 were also from 357, the same remote triple argument would apply to b689. However, since r7c7 and r9c9 each cells from different b568 triples (r4c7/r7c4 and r6c9/r9c6 respectively), this implies all four of r4c7/r7c4/r6c9/r9c6 are different. # -357r8c8

(You do still need the TH deduction to make use of step 2; eliminting 7r8c1 on its own isn't sufficent for progress.)

Website · by **denis_berthier** » Mon Mar 14, 2022 7:00 am

.
Now that I have coded this pattern in SudoRules, I can propose a solution where it is integrated with my usual chains, instead of a priori eliminating the target of this pattern.
Notice that, as tridagons rely on 12 CSP-variables (the 12 rc-cells), they are assigned complexity 12. That's why one appears only after a whip[9] in the following path. But the tridagon was available before applying the whip[9].

Code: Select all: Resolution state after Singles and whips[1]: +----------------------+----------------------+----------------------+ ! 5 7 3468 ! 238 346 13 ! 9 123 136 ! ! 23469 2349 3469 ! 2357 34567 13579 ! 3457 12357 8 ! ! 234689 1 34689 ! 23578 34567 3579 ! 3457 2357 3567 ! +----------------------+----------------------+----------------------+ ! 379 359 1 ! 6 8 357 ! 357 4 2 ! ! 3467 345 34567 ! 1 357 2 ! 8 357 9 ! ! 378 358 2 ! 357 9 4 ! 1 6 357 ! +----------------------+----------------------+----------------------+ ! 134789 34589 345789 ! 357 2 6 ! 357 135789 1357 ! ! 137 6 357 ! 9 357 8 ! 2 1357 4 ! ! 23789 23589 35789 ! 4 1 357 ! 6 35789 357 ! +----------------------+----------------------+----------------------+ 205 candidates.

hidden-pairs-in-a-column: c8{n8 n9}{r7 r9} ==> r9c8≠7, r9c8≠5, r9c8≠3, r7c8≠7, r7c8≠5, r7c8≠3, r7c8≠1
whip[9]: b9n1{r7c9 r8c8} - b3n1{r2c8 r1c9} - r1c6{n1 n3} - c4n3{r3 r6} - c5n3{r5 r8} - r8c1{n3 n7} - r6n7{c1 c9} - r9n7{c9 c6} - r4n7{c6 .} ==> r7c9≠3
tridagon for digits 3, 5 and 7 in blocks:
----- b9, cells: r8c8 (target cell), r9c9, r7c7
----- b8, cells: r8c5, r9c6, r7c4
----- b6, cells: r5c8, r6c9, r4c7
----- b5, cells: r5c5, r6c4, r4c6
==> r8c8≠3,5,7
singles ==> r8c8=1, r1c9=1, r1c6=3, r1c8=2, r1c4=8, r2c6=1, r3c6=9, r3c9=6, r7c1=1
finned-x-wing-in-columns: n3{c4 c9}{r6 r7} ==> r7c7≠3
hidden-single-in-a-block ==> r9c9=3
whip[1]: b9n5{r7c9 .} ==> r7c2≠5, r7c3≠5, r7c4≠5
whip[1]: b9n7{r7c9 .} ==> r7c3≠7, r7c4≠7
singles ==> r7c4=3, r5c5=3, r4c7=3
naked-pairs-in-a-row: r6{c4 c9}{n5 n7} ==> r6c2≠5, r6c1≠7
finned-x-wing-in-rows: n7{r4 r9}{c6 c1} ==> r8c1≠7
singles ==> r8c1=3, r6c1=8, r6c2=3, r3c3=8, r2c3=3, r3c8=3
whip[1]: c3n9{r9 .} ==> r7c2≠9, r9c1≠9, r9c2≠9
biv-chain[2]: r6n7{c4 c9} - c8n7{r5 r2} ==> r2c4≠7
biv-chain[2]: r6n5{c4 c9} - c8n5{r5 r2} ==> r2c4≠5
stte

Website · by **denis_berthier** » Fri Oct 21, 2022 11:05 am

.
By assigning Tridagons a higher priority, SudoRules now requires only W3 to solve it.
The tridagon elimination rule brings down Loki complexity from T&E(3) to W3 - the low part of T&E(1)

Code: Select all: hidden-pairs-in-a-column: c8{n8 n9}{r7 r9} ==> r9c8≠7, r9c8≠5, r9c8≠3, r7c8≠7, r7c8≠5, r7c8≠3, r7c8≠1 +----------------------+----------------------+----------------------+ ! 5 7 3468 ! 238 346 13 ! 9 123 136 ! ! 23469 2349 3469 ! 2357 34567 13579 ! 3457 12357 8 ! ! 234689 1 34689 ! 23578 34567 3579 ! 3457 2357 3567 ! +----------------------+----------------------+----------------------+ ! 379 359 1 ! 6 8 357 ! 357 4 2 ! ! 3467 345 34567 ! 1 357 2 ! 8 357 9 ! ! 378 358 2 ! 357 9 4 ! 1 6 357 ! +----------------------+----------------------+----------------------+ ! 134789 34589 345789 ! 357 2 6 ! 357 89 1357 ! ! 137 6 357 ! 9 357 8 ! 2 1357 4 ! ! 23789 23589 35789 ! 4 1 357 ! 6 89 357 ! +----------------------+----------------------+----------------------+

Code: Select all: tridagon for digits 3, 5 and 7 in blocks: b9, with cells: r8c8 (target cell), r9c9, r7c7 b8, with cells: r8c5, r9c6, r7c4 b6, with cells: r5c8, r6c9, r4c7 b5, with cells: r5c5, r6c4, r4c6 ==> r8c8≠3,5,7

Code: Select all: singles ==> r8c8=1, r1c9=1, r1c6=3, r1c8=2, r1c4=8, r2c6=1, r3c6=9, r3c9=6, r7c1=1 finned-x-wing-in-columns: n3{c4 c9}{r6 r7} ==> r7c7≠3 whip[1]: b9n3{r9c9 .} ==> r6c9≠3 naked-triplets-in-a-row: r7{c4 c7 c9}{n3 n5 n7} ==> r7c3≠7, r7c3≠5, r7c3≠3, r7c2≠5, r7c2≠3 z-chain[3]: r1c3{n4 n6} - c1n6{r2 r5} - c1n4{r5 .} ==> r3c3≠4, r2c3≠4, r2c2≠4 t-whip[3]: r8c1{n3 n7} - c3n7{r9 r5} - r5n6{c3 .} ==> r5c1≠3 t-whip[3]: r1c3{n6 n4} - c1n4{r3 r5} - r5n6{c1 .} ==> r2c3≠6 t-whip[3]: r5n6{c1 c3} - r1c3{n6 n4} - c1n4{r3 .} ==> r5c1≠7 t-whip[3]: c5n3{r5 r8} - r8c1{n3 n7} - c3n7{r9 .} ==> r5c3≠3, r5c5≠7 biv-chain[3]: b6n3{r4c7 r5c8} - r5c5{n3 n5} - r4c6{n5 n7} ==> r4c7≠7 biv-chain[3]: b8n3{r7c4 r8c5} - r5c5{n3 n5} - c6n5{r4 r9} ==> r7c4≠5 whip[1]: r7n5{c9 .} ==> r9c9≠5 whip[3]: r4n7{c1 c6} - r6n7{c4 c9} - r9n7{c9 .} ==> r8c1≠7 singles ==> r8c1=3, r7c4=3, r5c5=3, r4c7=3, r6c2=3, r6c1=8, r3c3=8,r2c3=3, r3c8=3, r9c9=3 whip[1]: c3n9{r9 .} ==> r7c2≠9, r9c1≠9, r9c2≠9 x-wing-in-columns: n7{c1 c6}{r4 r9} ==> r9c3≠7 biv-chain[2]: c8n5{r2 r5} - r6n5{c9 c4} ==> r2c4≠5 biv-chain[2]: c8n7{r2 r5} - r6n7{c9 c4} ==> r2c4≠7 stte

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