## Solution to Ocean's #11/18

Advanced methods and approaches for solving Sudoku puzzles

### Solution to Ocean's #11/18

Here is a solution to Ocean's puzzle #11/18:

Code: Select all
`178: Ocean #11/18 (9.4)   1 2 3   4 5 6   7 8 9 +-------+-------+-------+a| . . 1 | . . . | . . 2 |b| . 3 . | . 4 . | . . . |c| 5 . . | 6 . . | 3 . . | +-------+-------+-------+d| . . 4 | . . 7 | . . 1 |e| . . . | . 8 . | . 3 . |f| 9 . . | 2 . . | 6 . . | +-------+-------+-------+g| . . . | . . 8 | . . 9 |h| . 7 . | . 2 . | . 4 . |k| 2 . . | 1 . . | 8 . . | +-------+-------+-------+`

1. k2=4 or k6=4, d4=4, f9=4, c2=4 > ag2!=4

2. e7=4 or a7=4, c2=4, k6=4, e4=4 > e69!=4

3. e4=5, (e9=7, e7=4, f3=7, f6=4, f5=3, d4=9, row5!=9 contradiction >) e4!=5

4. c9=7, (e9=5, f9=4, b9=8, a4=8, e4=4, c58=19, c6=2, a7=4, c2=4, c3=8, ab1=67, a2=9, b3=2, e1=1, hk9=6, g1=4, k6=4, df2=8, d3=3, f56=13, g2=1, h7=1, gh4=3, d45=5, h1=8

4a. f3=7 or f3=5, k2=5, k8=7 > f8!=7, f8=8, f2=5, k2=6, k9=3, h9=6, e6=6, box4!=6 contradiction >) c9!=7

At this point I wanted to force the 6 boxes withs alternating 4 into one of the two positions. But I ended with a rather lengthy solution step 5 including 12 substeps - some of these quite difficult - to prove at last that c9!=8, the only other merit of this approach being the relatively low number of contradictions (6) of the total solution. At least I was "rewarded" with a nice non-uniqueness pattern in substep 5l with 6 cells:

5. c9=8, (a7=4, c2=4, g1=4, k6=4, e4=4, f9=4, de2=2

5a. d8=8 or f8=8, f3=7, d1=3 > d1!=8

5b. d1=3 or f3=3, d1=6, h1=3, h3=8, h7=1, e6=6, e7=9, f8=7, e9=5, h9=6, b9=7, b7=5, b1=8, b4=9, d5=9, g7=7, k8=5, g2=1, g3=5, g4=3, d4=5 > d45!=3, d1=3

5c. d5=6, h6=6 or d2=6, e6=6, e7=9, d8=8, d7=2, g8=2, e2=2, e1=1, g2=1, h7=1, f56=13, ef3=7, (k2=5, a2=9, c3=2, b6=2, b8=1, b4=9, a4=8, g4=7 or k2=9, f2=5, f8=7, c5=7, g4=7 >) g4=7, g7=5, b7=7, c5=7, f5=1, f6=3, bc6=12, a1=7, (h4=3, g5=6 or a4=3, a2=8, f2=5, f8=7, k8=6, g5=6 >) g5=6 > k5!=6

5d. g5=6, (e6=6, e7=9, d2=6, e2=2, d8=8, d7=2, g8=2, e1=1, g2=1, h7=1, ef3=7, (k2=5, a2=9, c3=2, b6=2, b8=1, b4=9, a4=8, g4=7 or k2=9, f2=5, f8=7, c5=7, g4=7 >) g4=7, g7=5, g3=3, b7=7, a1=7, c5=7, f5=1, f6=3, bc6=12, a8=6, k8=7, f8=5, f2=8, k2=5, a2=9, a6=5, h6=9, k5=3, a5=nil contradiction >) g5!=6, h6=6, d5=6

5e. h1=1, h3=8, h4=9, d4=5 or e1=1, f56=13 > f56!=5, f56=13

5f. a2=9 or k2=9, h4=9, e6=9 > a46!=9

5g. e7=2, d2=2, d8=8 or e2=2, e1=1, e3=6, g3=1, k2=6, g8=6, g7=2, d8=2 > d7!=2, d47=59, de7=9

5h. e7=2, g8=2 or e2=2, e1=1, e3=6, g3=1, k2=6, g8=6, b9=6, h1=8, b1=7,g7=2, e7=7 > e7=27, g8=26, d7=9, d4=5, e6=9, ab6=5, bc8=1

5i. h9=3, h4=9 or h9=5, h7=1, e9=7, e7=2, g7=7, g4=3, h4=9 > h4=9

5j. e1=1 or h1=1, h7=5, b7=7, b4=8, b1=6 > e1!=6, ab1=6

5k. h1=1, e1=7, h7=5, b7=7, b4=8, b1=6, b9=5, e9=nil contradiction > h1!=1, h1=8, e1=1, g2=1, h7=1, ab1=7

5l. c5=7, b4=8, a4=3, a6=5 or c8=7, b7=5, a6=5 > a6=5, f6=3, f5=1, b179=567, b4=8, a2=8, d2=2, f2=5, e2=6, e3=7, f8=7, (already non-uniqueness pattern at bc368= 129) e9=5, b7=5, e7=2, g7=7, k8=5, k5=7, c5=9, g8=2, a8=6, row1!=9 contradiction >) c9!=8, c9=4, a1=4, k2=4, g4=4, f6=4, e7=4, e4=9, ab4=78, d78=29, d12=8, hk3=9

After this, the remaining steps are relatively easy:

6. e9=5 or e9=7, f3=7, f5=3, d4=5 > e6!=5

7. e1=7, e9=5 or b1=7, b4=8, f9=8, c8=7, e9=7, f8=5, f2=1 > e1!=1, e3!=7, gh1=1, f9!=5

8. f3=3 or d1=3, h4=3, k9=3, g3=3 > hk3!=3

9. d5=6 or e6=6, f5=1, f3=3, g23=56, k3=9, h9=6, g5=7, k5=6, k9=3, h4=3, d5=3 > g5!=6, d5!=5

10. g2=6 or g2=5, f2=1, e6=1, d5=6 > d2!=6

11. a8=6 or a2=6, c2=9, g2=5, f2=1, e6=1, d5=6, d2=8, d1=3, g3=3, g5=7, ab7=7, c3=7, c8=8 > a8!=8

12. c8=7, a4=7, b4=8, box3!=8 contradiction > c8!=7, c3=7, e1=7, e9=5

13. h7=1 or h1=1, h3=8, h6=9, h7=5, h4=3, k9=3, b1=8, c8=8, f8=7, g7=7, g8=2, h7=1 > h7=1, g1=1

14. c8=8 or c2=8, d2=5, d4=3, d5=6, e6=1, c5=1 > c8!=1, b8=1

15. g2=6 or g2=5, f3=5, g3=3, g8=6, b9=6, a2=6 > e2!=6, g3!=6, fg3=35

16. h1=3 or g3=3, f5=3, a6=3, h4=3 > h69!=3, k9=3 etc.

To avoid the lengthy solution step 5., I tried other eliminations by contradiction (EBC), which works, but seems to be a whole bunch of EBC, and I didn't work it out.

Greetings, Maria
maria45

Posts: 54
Joined: 23 October 2005