- Code: Select all
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- Code: Select all
Resolution state after Singles and whips[1]:
+----------------------+----------------------+----------------------+
! 68 7 158 ! 2 4 158 ! 3568 358 9 !
! 24689 15689 12458 ! 13589 1358 1589 ! 7 258 56 !
! 289 3 258 ! 589 6 7 ! 1 258 4 !
+----------------------+----------------------+----------------------+
! 2468 168 9 ! 1568 1258 3 ! 2568 4578 1567 !
! 7 168 1238 ! 15689 1258 4 ! 235689 3589 1356 !
! 5 168 12348 ! 7 128 12689 ! 23689 3489 136 !
+----------------------+----------------------+----------------------+
! 1389 4 6 ! 1358 13578 158 ! 359 3579 2 !
! 13 2 7 ! 1345 9 15 ! 345 6 8 !
! 389 589 58 ! 3468 2378 268 ! 349 1 37 !
+----------------------+----------------------+----------------------+
203 candidates.
Starting from here, there is no 1-step solution with whip of reasonable length, but there is an easy, simplest-first solution:
hidden-pairs-in-a-column: c6{n2 n6}{r6 r9} ==> r9c6≠8, r6c6≠9, r6c6≠8, r6c6≠1
singles ==> r5c4=9, r2c6=9, r3c1=9, r9c2=9, r9c3=5, r2c2=5, r2c9=6, r1c1=6
whip[1]: c2n1{r6 .} ==> r5c3≠1, r6c3≠1
whip[1]: c2n8{r6 .} ==> r4c1≠8, r5c3≠8, r6c3≠8
whip[1]: c3n8{r3 .} ==> r2c1≠8
whip[1]: c9n5{r5 .} ==> r4c7≠5, r4c8≠5, r5c7≠5, r5c8≠5
;;; Resolution state RS2. This is where the other solutions start from:
hidden-triplets-in-a-column: c8{n4 n7 n9}{r6 r4 r7} ==> r7c8≠5, r7c8≠3, r6c8≠8, r6c8≠3, r4c8≠8
whip[1]: b9n5{r8c7 .} ==> r1c7≠5
biv-chain[3]: r5c8{n8 n3} - r5c3{n3 n2} - r3n2{c3 c8} ==> r3c8≠8
biv-chain[3]: r2c8{n8 n2} - r3c8{n2 n5} - r3c4{n5 n8} ==> r2c4≠8, r2c5≠8
naked-pairs-in-a-block: b2{r2c4 r2c5}{n1 n3} ==> r1c6≠1
hidden-single-in-a-row ==> r1c3=1
whip[1]: c6n1{r8 .} ==> r7c4≠1, r7c5≠1, r8c4≠1
biv-chain[3]: r2n8{c3 c8} - r5c8{n8 n3} - r5c3{n3 n2} ==> r2c3≠2
biv-chain[4]: r8c1{n3 n1} - r8c6{n1 n5} - r1n5{c6 c8} - b3n3{r1c8 r1c7} ==> r8c7≠3
biv-chain[3]: r9c1{n8 n3} - r8n3{c1 c4} - b8n4{r8c4 r9c4} ==> r9c4≠8
biv-chain[3]: r8n3{c1 c4} - r8n4{c4 c7} - r9c7{n4 n3} ==> r9c1≠3
naked-single ==> r9c1=8
biv-chain[4]: r8c7{n5 n4} - r9c7{n4 n3} - r1n3{c7 c8} - r1n5{c8 c6} ==> r8c6≠5
singles ==> r8c6=1, r8c1=3, r7c1=1
biv-chain[4]: c3n3{r6 r5} - r5c8{n3 n8} - r2c8{n8 n2} - b1n2{r2c1 r3c3} ==> r6c3≠2
biv-chain[4]: r4n7{c9 c8} - b6n4{r4c8 r6c8} - r6c3{n4 n3} - r6c9{n3 n1} ==> r4c9≠1
biv-chain[4]: r6c3{n3 n4} - r6c8{n4 n9} - r7c8{n9 n7} - r9c9{n7 n3} ==> r6c9≠3
naked-single ==> r6c9=1
naked-triplets-in-a-row: r6{c2 c5 c6}{n6 n8 n2} ==> r6c7≠8, r6c7≠6, r6c7≠2
whip[1]: r6n2{c6 .} ==> r4c5≠2, r5c5≠2
hidden-pairs-in-a-block: b6{n2 n6}{r4c7 r5c7} ==> r5c7≠8, r5c7≠3, r4c7≠8
stte
Instead of changing the starting point as the other solutions and considering that the first hidden-pairs eliminates a lot of candidates (60), I tried the fewer-steps method. Bingo at the first try: a mere bivalue-chain[6] is enough as a second step:
hidden-pairs-in-a-column: c6{n2 n6}{r6 r9} ==> r6c6≠9, r9c6≠8, r6c6≠8, r6c6≠1singles ==> r5c4=9, r2c6=9, r3c1=9, r9c2=9, r9c3=5, r2c2=5, r2c9=6, r1c1=6
whip[1]: c2n1{r6 .} ==> r5c3≠1, r6c3≠1
whip[1]: c2n8{r6 .} ==> r4c1≠8, r5c3≠8, r6c3≠8
whip[1]: c3n8{r3 .} ==> r2c1≠8
whip[1]: c9n5{r5 .} ==> r4c7≠5, r4c8≠5, r5c7≠5, r5c8≠5
;;; RS2
biv-chain[6]: r5c8{n3 n8} - r2c8{n8 n2} - c1n2{r2 r4} - r4n4{c1 c8} - b6n7{r4c8 r4c9} - r9c9{n7 n3} ==> r6c9≠3, r5c9≠3, r7c8≠3stte