Solution to the weekly extreme Str8ts puzzle #45 from www.str8ts.com

1. Weekly Extreme Str8ts Puzzle #45 May 1 – May 7, 2011www.str8ts.com Solution bySlowThinker & Walter

2. Start

3. Caution! There is severe testing/guessing ahead. You cannot solve the puzzle without it, which does take some of the fun away for some people. If you don’t like to test/guess, solve the puzzle with E8=7 added. This puzzle will still be very challenging, but can be solved without further guessing. Have fun! 7 Solution for E8=7 isshownatthe end.

4. G8=6c, H9=4c, B9=2, C9=1

5. C7=2c, D8=2d, A2=8d, B6=3h

6. J9=8h, E9=6

7. Setti‘srule on 5 Checking the number 5 we find that there must be a 5 in every column. Therefore there must be a 5 in every row  G2=5

8. G2=5S5, C2=6, B1=6s There is a hidden pair 56 in the green area  other candidates can be removed.

9. X-Wing on 5 The blue X-Wing on 5, removes the 5s in the green fields.

10. Unique rectangle & Setti on 6 Setti‘s rule on 6: every column has a 6  every row must have a 6  F8=479 Blue rectangle: F56 cannot be 89  9s can be removed in F56. This works because the only 8s are in F56 too, e.g. one cannot apply this to row E.

11. An unorthodox chain 3 4 G1=3 G1=4 Looking at the lower left, examining possible values for G1: G1=3  H1:-3, G3=4  J3:-4 G1=4  HJ1=23. Together with HJ4 we can apply the unique rectangle rule, which forces J4=4  H4=3  J3:-4, H1:-3 In both cases, 3 gets removed from H1 and 4 gets removed from J3  we can remove both candidates.

12. Anothercombination If B5 (green) is either 8 or 9, then we have the naked pair 89 together with G5 (blue) that removes 89 from the orange fields. What happens if B5=4? B5=4  B4=8 (only other 8 in row)  C4=79  C5=8 With C5=8 we get G5=9 and again 89 is removed from the orange fields.  We can remove 89 from EFH5and 9 from C5 in either case. 4

13. Currentstatus: 24 fieldsfilled in I was stuck at this point…

14. Variation by Walter: Setti‘srule on 8 Column 7 doesn‘t have an 8, column 3 and the rows C and F may have no 8. Either row C or row F doesn‘t have an 8, or both have no 8, in which case column 3 must not have an 8 either. Testing row F: If there’s an 8 in row F  no 8 in row C. If there’s no 8 in row F  F3=2  no 8 in column 3 (range!)  two columns with no 8  no 8 in row C.  There’s never an 8 in row C  C45=45!

16. E3=8s (only one left in row E)

17. F5..9=2..7  F23: 1289 (3,4 are removed, blue marks)

18. F3=2

19. F3=2 and E3=8 is illegal (range!)

20. C45 cannot be 789

22. More combinationswithSetti Ideaby Walter Lookingat F5..9 again: Ifthereisno 8 in F59, F5..9=2..7. Ifthereis an 8 in F59  E3=8 (becauseofSetti‘srule on 8)  J3=7  J8=9  F8=47  F5..9=2..8.In bothcases, F5..9 contains a 4!  F8=47 (9 is out ofrange)

23. Setti‘srule on 4 Wenowknowthateveryrowcontains a 4  everycolumn must contain a 4 aswell  H1=2

24. H1=2S4, H4=3 F2 cannot be 3, because F2=3  F3=2  F5..9 contains an 8  E3=8 (Setti) which contradicts F3=2 (range!) [idea by Walter]

25. Testing F3=3 Ideaby Walter If F3=3  F2=2c, F6=8h  E3=8S8, E2=139, F5=67  E5=3s  no 2 left in row E  F3 cannotbe3. FromthereitfollowsthatF2=19, F3=28  EJ3 : -28 (large gap pair F3), F6!=8 (S8), J2=23p  J3=7s!

26. J3=7s, J8=9

27. E3=9u The blue fields form a Swordfish on 8 and on 9, the one on 9 is disturbed by E3. But if E3!= 9  E46=89 and we do not get a unique result, because the 8s and 9s in BEG456 could be swapped  in order to get a unique solution E3=9

28. Variation by Walter Testing F5=2 F5=2  F7=3s, F8=4c  B8=5, A7=5  no cell left for 6 in column 7. We therefore know that F5!=2 and F5..9=34567. Setti`srule on 7 Now 7s are confirmed everywhere except in column 7 and row A. Both have a 7  D7=1. No 7 in row A/column 7  H7=19, H5=57  X-Wing on 1 at AE58  neither E7 nor A7 can be 1 D7=1. So, in both cases there is no 3 in row D  according to Setti`s rule no 3 in column 3 G3=4, F3=8, E3=9. From there the puzzle solves easily.

29. E3=9u, F3=8c, G3=4c, F2=9c, G1=3 Continuingfrombefore…

30. H2=1, J1=4, J4=2, J2=3, E2=2

31. X-Wings on 3 and 1 The blue (horizontal) X-Wing on 3 (EF57) removesthe 3 from D7  D7=1 The orange (vertical) X-Wing on 1 makes 1 a surecandidate in A4..9  7 isremovedfrom A..9, A4..9=123456

32. D7=1x Setti‘s rule on 7: row A doesn‘t contain a 7  at least one column doesn‘t contain a 7 either  H7=9 and eight columns contain a 7  row F must contain a 7  A5=2s

33. H7=9S7, A5=2s, A6=4, A4=6, A7=5

34. A8=1, B7=4, B8=5, D4=7, D5=6

35. J5=5, J6=6, H5=7, H6=5, E7=3, F7=6

36. F6=7, F5=3, F8=4, E5=1, E8=7

37. E6=8, E4=4, G6=9, G5=8, B5=9, B4=8 Done

38. Variation: E8=7 7 In the following, just the most important points are shown, when trying to solve this variation of the puzzle.

39. Setti‘srule on 5 and X-Wing Setti‘s rule: there‘s a 5 in every column, therefore also in each row  G2=5 This then gives an X-Wing on 5 (marked blue,) and a hidden pair J56=56

40. Setti‘srule on 6 There‘s a 6 in each column, therefore a 6 in each row  F8=4. Next important steps are the resulting singles D4=7, A4=6, J3=7 and F6=7.

41. Large gap pair & Swordfish F3=28 is a large gap pair, which removes 2 and 8 from E3. With that we have a Swordfish on 8 that results in C4=5.After that the puzzle solves itself.

42. 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

43. Weekly Extreme Str8ts Puzzle #45www.str8ts.com Solution by SlowThinker & Walterwith a variation by Jan Note: there are other (maybe easier) ways to solve this puzzle.