Solution to the weekly extreme Str8ts puzzle #39 from www.str8ts.com

1. Weekly Extreme Str8ts Puzzle #39 March 20 – March 26, 2011www.str8ts.com Solution bySlowThinker

2. Start

3. C4=3c, A4=1, E5=7c, G5=8

4. A7=7h, B7=8h, H2=8h, J2=9c

5. H6=7h, J6=8, H7=6c, H1=9

6. G3=7p, C8=8s, C9=7, J1=7s

7. G4=9p, A9=8s, B8=7s Note that F89 cannotbe 6, becauseof F3.

8. Swordfish on 3 The orange swordfish combination in EFJ789 (three sure 3s in three rows appear in the same three columns) removes the 3s in the green fields.

9. Starfish on 6 See variationusingSetti‘srule atthe end Applying the same principle (same number of rows and columns): looking at the five rows BCEFJ, we find that the 6s (all sure candidates in each row) are in five columns (13469) and thus remove the 6s marked green  G8=6s

10. G8=6s, D6=3p, D2=2, D1=1, G1=3 Thereisanotherswordfish on 4 at ABC126.

11. C3=2w, C5=5, J7=3p, F7=2, E7=1 I forgot to remove the 4s from A3 and B3 (due to the swordfish shown on the last slide.) Don‘t know why, apparently I went off on a tangent. Removing the 4s immediately yield A3=3, B3=1, J3=4, J5=1, A2=4, B2=3, B1=4c, …

12. E8=3c, E9=2c, F8=4, F9=3

13. D8=5, D7=4, G7=5, G9=4, A1=2s

14. A6=5p, B6=6, C6=4, B9=9, C1=6

15. F1=5, B1=4, F3=6, E3=5, E4=6

16. B2=3, A2=4, A3=3, B3=1

17. J3=4, J5=1, J4=5, J9=6, H9=5c Done

18. Variation by Walter: Setti‘srule This is the position at slide 9, where I applied the Starfish. Walter came up with an application of Setti‘s rule.

19. Setti‘srule on 6 6 must be in row F Why? Because otherwise you‘d only have four numbers (2..5) left for five open fields. Looking at the number 6: column 2 (orange) cannot contain a 6  according to Setti’s rule there must be (at least) one row without a 6  only row D (green) can be without a 6  row D contains no 6  D678=345  D6=3

20. D6=3S6, D2=2, D1=1, G1=3, G8=6s From here you can continue with the Swordfish on 4 at ABC126, like in the main variation. Alternatively, if you like chains…

21. B6=6y Blue path: B6=5  B1=4  B2=3Orange path: B6=5  A6=4  A2=3  B2=4Paths produce a contradiction, therefore B6 cannot be 5  B6=6y, B9=9

22. B6=6y, B9=9, C1=6s, F3=6h, E3=5, E4=6, J4=5 And continuing from here: J7=3p, J9=6s, F7=2, E7=1, E8=3c, E9=2c, …

23. Glossary Letters appended to steps indicate the last strategy used, just before filling in a field: No letter … number was last candidate in field s … single (last) candidate for that number in compartment c … compartment range check d … stranded (unreachable/impossible) digits removed h … high/low range check across compartments p/t/q … naked pair / naked triple / naked quadruple ph/th/qh … hidden pair / hidden triple / hidden quadruple x … X-wing (2 rows / 2 columns) w … Swordfish (3 rows / 3 columns) j … Jellyfish (4 rows / 4 columns) L … large gap field Sx … Setti’s rule (count the numbers rule) – ‘x’ is the analysed number u … unique rectangle y … Y-Wing or XY-chains

24. Weekly Extreme Str8ts Puzzle #39www.str8ts.com Solution by SlowThinker(with a variation by Walter) Note: there are other (maybe easier) ways to solve this puzzle.