.
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Resolution state after Singles and whips[1]:
+----------------------+----------------------+----------------------+
! 5 12379 1379 ! 178 12379 4 ! 6 3789 1379 !
! 139 13479 8 ! 6 1379 3579 ! 3579 3479 2 !
! 1239 123479 6 ! 178 12379 23579 ! 35789 34789 13479 !
+----------------------+----------------------+----------------------+
! 139 139 5 ! 2 4679 679 ! 379 134679 8 !
! 2389 6 349 ! 5 4789 1 ! 2379 3479 3479 !
! 7 1289 149 ! 3 4689 69 ! 29 1469 5 !
+----------------------+----------------------+----------------------+
! 6 3579 379 ! 17 1237 8 ! 4 23579 379 !
! 4 3578 37 ! 9 2367 2367 ! 1 23578 367 !
! 1389 13789 2 ! 4 5 367 ! 3789 3789 3679 !
+----------------------+----------------------+----------------------+
There is an easy solution with swordfish and bivalue-chains[3}
Apart from allowing hidden Subsets, there is no 1-step solution with chains of reasonable length. (There's only 1 W1-anti-backdoor: n8r9c7.)
As for 2-step solutions (with no undeclared Subset steps), there are lots of them. Here is the simplest:
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biv-chain-rn[2]: r8n5{c8 c2} - r8n8{c2 c8} ==> r8c8 ≠ 2, r8c8 ≠ 3, r8c8 ≠ 7
singles and whips[1]
z-chain[4]: r5n9{c3 c5} - c5n7{r5 r7} - r7n3{c5 c9} - r8n3{c9 .} ==> r5c3 ≠ 3
stte
The first eliminations can also be done by:
biv-chain-cn[2]: c8n2{r8 r7} - c8n5{r7 r8} ==> r8c8 ≠ 8, r8c8 ≠ 3, r8c8 ≠ 7
Notice that the first version is also hidden pairs in a row and the second hidden pairs in a column - so that
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z-chain[4]: r5n9{c3 c5} - c5n7{r5 r7} - r7n3{c5 c9} - r8n3{c9 .} ==> r5c3 ≠ 3
is a 1-step solution modulo Pairs