As a compensation for the bad sample last time i want to show, how a pretty hard sudoko can be solved. Its nr 149 from the outlaws in the sudoko gallery.

Of course i know that there are many ways to solve it (rubylip's says, a 3 in r2c5 would lead to a contradiction), but i think, the steps might be interesting for others.

- Code: Select all
`. . 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 r9c3:

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=8,r3c1=5,r2c2=8

r8c5=8,r2c5=9,r2c2=8