Of course i know, there are many other ways to solve it (just looked at rubylip, says "The move (2,5):=3 would lead to a contradiction.."), but the steps might be interesting for others.
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. . 1 | . 5 . | . . 2
6 . . | 4 . . | . . .
. 9 . | . . . | 3 . .
----------------------
. . 8 | . . 7 | . . 1
. . 5 | 6 . . | 4 . .
2 . . | . . . | 9 . .
----------------------
. . 7 | . . . | . 8 .
. . . | . . 3 | . . 5
4 . . | . 2 . | 6 . .
The first time i got stuck here:
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378 3478 1 3789 5 689 78 4679 2
6 23578 23 4 13789 1289 1578 1579 79
578 9 24 1278 1678 1268 3 14567 467
39 346 8 2359 349 7 25 2356 1
1379 137 5 6 1389 1289 4 237 378
2 13467 346 1358 1348 1458 9 3567 3678
135 12356 7 159 1469 14569 12 8 349
18 1268 269 1789 146789 3 127 12479 5
4 1358 39 15789 2 1589 6 1379 379
(The swordfish in 4 doesnt really help)
Elminate 3 from r3c9:
r9c3=3,r7c9=3
. r8c3=9,r6c3=6,r3c3=4,r1c8=4,r7c9=4
=> r9c3=9
Eliminate 2 from r4c7:
r4c7=2,r7c7=1,r8c7=7,r9c8={39},r9c9={39}
not possible, because r9c3={39}
=> (2 numbers for 3 cells) r4c7=5
Eliminate 2 from r8c3:
r2c3=3->r1c1<>3
r3c3=4,(r1c2<>3),r1c2={78},(r1c7={78}),r1c1<>{78}
=> (no number left for r1c1) r8c3=6
Eliminate 2 from r4c8:
r4c8=2,r5c6=2,r4c4=2,r3c3=4
. r4c2=6,r4c5=4,r6c3=4,r3c3<>4
Eliminate 3 from r4c1:
r4c1=3,r6c3=4
. r4c8=6,r4c2=4.r6c3<>4
=>r4c1=9, r4c5<>9
Elminate 3 from r6c3:
r6c3=3,r4c2=4,(r4c5<>9),r4c5=3,r1c4=3
r2c3=2, r3c3=4 (r1c2<>4), triple {378} in row 1 (c127)->r1c4=9
=>r6c3=4
With this i got here:
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7 4 1 3 5 69 8 69 2
6 58 3 4 89 2 1 59 7
58 9 2 18 7 168 3 56 4
9 6 8 2 4 7 5 3 1
13 7 5 6 139 19 4 2 8
2 13 4 158 138 158 9 7 6
135 13 7 15 6 4 2 8 9
18 2 6 9 18 3 7 4 5
4 58 9 7 2 58 6 1 3
Matter of taste, eg
weak chain in 8: r2c2-r9c2~r9c6-r8c5~r2c5-r2c2 -> r2c2=8
or double forcing chain:
r8c1=1,r3c1=5,r2c2=8
r8c5=1,r2c5=9,r2c2=8
