This document summarizes the step-by-step solution to a Sudoku puzzle published in Weekly Extreme Str8ts Puzzle #40 from March 27 to April 2, 2011. The summary identifies strategies used such as removing candidate numbers, naked triples, X-Wings, hidden pairs, and more to systematically solve the puzzle cell by cell until completion. A glossary defines strategies referenced in the solution steps.

1. Weekly Extreme Str8ts Puzzle #40 March 27 – April 2, 2011www.str8ts.com Solution bySlowThinker

2. Start

3. Intermezzo: removingcandidates If you have a row with no empty black fields, like row D, then the compartments always start/end at the ends of the possible row range, here at 1 and at 8  D12 cannot be 12 (because of D1)  D12=78

4. Intermezzo: removingcandidates For a similar reasonA4567 cannot startat 1, because itwould leave A9with no candidates. 1 and 2 are already stranded digits in G6789 and can be removed. DEFG9 (orange) cannot include 1 or 2. There are two ways to look at it: (1) If AB9=12 it is obvious, but even if AB9=34, 1 and 2 are stranded digits in DEFG9.(2) If DEFG9 starts at 1 or 2, then there is no possible range left in AB9.

5. H1=2h, H2=1c Can you spot thelarge gap field inrow C? Note that F9 is a large gap field: either 3 is in F9 or 3 is not part of DEFG9 at all  3 can be removed from D9 and E9.Also: 9 is a stranded digit in DEFG9 and can be removed.

6. Naked Tripple To get this far on your own, all you have to do is painstakingly apply the basic rules over and over again and remove the candidates one by one. From here on the puzzle is rather easy. The blue fields form a naked triple on 678  E3=5t Optional:There is an X-Wing on 7 at DE12 that removes the 7s marked green.

7. E3=5t, G3=4, F3=3, E9=4, E7=3

8. F9=7c, F8=8c, F1=6, E2=6s, E1=7s

9. D1=8, D2=7, B2=5 Note on C1234: you reach this state either by finding the hidden pair 67 in C34 or by applying the large gap field rule for C1 and C2.

10. J4=8p, J9=3, J8=4c, J1=1

11. Intermezzo C7 is either 4 or 5. Both solutions yield B2=8, because even if C7=5, B2=4 would be a stranded digit. // There is a unique rectangle at BC34, yielding B4=9. I don’t apply this rule though.

12. B2=8h, G7=8s, D4=3s, F4=1s, G4=2s

13. H5=8s, J5=9c, J4=6c, C4=7, B4=9

14. C3=6, B3=7, B5=6, A5=7

15. A7=6d, A1=3p, A2=2p, F2=4, F6=2

16. E6=1, E8=2, A9=1, B9=2, C1=9t

17. C6=3s, C8=5, C7=4, D7=5, D9=6

18. D8=1, D6=4, D5=2, B8=3, G9=5

19. A6=5, A4=4, H4=5, G6=6c

20. G8=7, H8=6, H6=9, H7=7 Done

21. 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

22. Weekly Extreme Str8ts Puzzle #40www.str8ts.com Solution by SlowThinker Note: there are other (maybe easier) ways to solve this puzzle.