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Weekly Extreme Str8ts Puzzle#29, January 9- January 15, 2011www.str8ts.com Solution by SlowThinker
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Possiblecandidates
B3=7c, C6=4c, G3=3c, E6=6c
E5=5s(DE3 compartment)
C2=7h(C5..9=1..5)
G9=7h(G1..5), G8=8c Removingstrandedcandidates
J6=8c, H6=9s H6=89h, becauseof H123=1..4  J6=8c, H6=9s Removingothercandidates: compartment, high/low, strandeddigits
D2=8sh FGHJ2=1..5  D2:-4; D2: -9: strandeddigit(ABCD2) D2=8s
D9=2s, B7=2s Also:	F4:-9 (strandeddigit)DEFG4and	D4:-9, because 6 isonly in DF4  9 strandeddigit
C5=5s
Removingcandidates EF7/HJ7 followsfrom ABC7=1..4 andstrandeddigits A7:-4pB4:-9 (strandeddigitwithregardto A4)
JellyfishAFJHon 6  D45:-6
D3=6s F4:-8 (strandeddigitbecauseofsingle 6 in DEFG4)  DEFG=1..7  ABJ4:-4
X-WingFH57 B5:-8
B1:-49 B1:-49 either ,[object Object]
B5=49 isa„biggap pair“Big gap pair (byErwin): If a cell has only two candidates and their difference is at least a str8t's length, you can delete these from all other cells in this str8t.
A8:-1 (chaining) Either: A7=1  A8:-1 / Or: A7=3  orange chainleadsto D8=1  A8:-1 (Side note: thisstepis optional forsolvingthe puzzle)
Possibility: Link chains: H8=7 Type of links digitposition compartment weak link a !b,but !a  ? Wecan find thefollowing link chainwithonlyoneweak link: H5,H7,H8,J7,J8.Startingwith H5 (hastwo strong links (blue, alternative digitposition) andthedigit 7 (hasmost links): H5=7  H7=8, H8:-7  J7=7, J8=7  contradiction  H5:-7  H8=7
Possibility: Linked alternative pairs(Ideaby Siegbert / Text by John) Restrict your search to "necessary" candidates. Search for a column which has a pair of digits, ab, which can occur in alternative positions (that is, in more than one place). In this puzzle, 67 in column 4 can occur either in [A4J4] or in [E4F4]. (*) Search for another column which has a pair of digits, bc, where one digit is the same as one digit from Step 2, which can occur in alternative positions. In this puzzle, 78 in column 7 has one digit the same, and the pair can occur either in [E7F7] or in [H7J7]. Search for a row which has one of the digits a, b or c in both columns. In this puzzle, the 7 in J4 is common to both. Because of the common digit identified in Step 4, the two "alternative positions pairs" are linked and if the pair in one column is put in one of its alternative positions, the pair in the other column might be forced to go into one of its alternative positions which produces a contradiction. So...  Search for a candidate x which, if used for the solution of its cell, "forces" the position of the pair ab in the first column and also, at the same time, forces the position of the pair bc in the other column. If this produces a contradiction, then x cannot be the solution for its cell and x must be removed. In this puzzle, if H5 = 6, it forces a contradiction, so 6 must be removed as a candidate in H5. (*) SlowThinker: I‘m not sureitreallyworkshere: 76 canbe in EF4 and D4, thusno pair?
Linkedcompartments: B4=8(solutionpathchosen) Type of links digitposition compartment weak link a !b,but !a  ? Checking J4 we find that for J4=2356 in combination with DEFG4/A4 B4:-5 For J4=7 J7:-7  E7=7  E4:-2  DEFG4=2..6  B4:-5 Therefore: B4 = 8
B4=8, B1=5s … and A4:-6c
A5=6s, A6=2s
G5=1s, G1=2s, F2=2s
Link chains: J7=4 Type of links digitposition compartment weak link a !b,but !a  ? BecauseofH5=78, H7 and H8 must not be78 atthe same time (aligned pair exclusion.) But:J7=7 forces H7=8c and J8:-7,H8=7s  J7:-7  J7=4
J7=4L, H7=5c, E7=7s, F7=8c
H9=6s, H8=7s, H5=8s, J9=5s
J8=6s, F9=4s, J2=3s
F8=5g, F4=6g F5=39 is a biggap pair: neither 3 nor 9 canappearanywhereelse in F4..9
A2=5s, G2=4s, H2=1s, G4=5s
H1=3s, F1=1s
J5=2s, J4=7s, A4=9s
A7=3c, A1=7s, A8=4s
D1=4s, D4=3s, D8=1s
C7=1s, C9=3s, B9=1s, B8=3c
E4=4s, D5=7c, E5=3s
F5=9s, B5=4s Done

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Str8ts: Solution to Weekly Extreme Str8ts #29

  • 1. Weekly Extreme Str8ts Puzzle#29, January 9- January 15, 2011www.str8ts.com Solution by SlowThinker
  • 8. J6=8c, H6=9s H6=89h, becauseof H123=1..4  J6=8c, H6=9s Removingothercandidates: compartment, high/low, strandeddigits
  • 9. D2=8sh FGHJ2=1..5  D2:-4; D2: -9: strandeddigit(ABCD2) D2=8s
  • 10. D9=2s, B7=2s Also: F4:-9 (strandeddigit)DEFG4and D4:-9, because 6 isonly in DF4  9 strandeddigit
  • 11. C5=5s
  • 12. Removingcandidates EF7/HJ7 followsfrom ABC7=1..4 andstrandeddigits A7:-4pB4:-9 (strandeddigitwithregardto A4)
  • 14. D3=6s F4:-8 (strandeddigitbecauseofsingle 6 in DEFG4)  DEFG=1..7  ABJ4:-4
  • 16.
  • 17. B5=49 isa„biggap pair“Big gap pair (byErwin): If a cell has only two candidates and their difference is at least a str8t's length, you can delete these from all other cells in this str8t.
  • 18. A8:-1 (chaining) Either: A7=1  A8:-1 / Or: A7=3  orange chainleadsto D8=1  A8:-1 (Side note: thisstepis optional forsolvingthe puzzle)
  • 19. Possibility: Link chains: H8=7 Type of links digitposition compartment weak link a !b,but !a  ? Wecan find thefollowing link chainwithonlyoneweak link: H5,H7,H8,J7,J8.Startingwith H5 (hastwo strong links (blue, alternative digitposition) andthedigit 7 (hasmost links): H5=7  H7=8, H8:-7  J7=7, J8=7  contradiction  H5:-7  H8=7
  • 20. Possibility: Linked alternative pairs(Ideaby Siegbert / Text by John) Restrict your search to "necessary" candidates. Search for a column which has a pair of digits, ab, which can occur in alternative positions (that is, in more than one place). In this puzzle, 67 in column 4 can occur either in [A4J4] or in [E4F4]. (*) Search for another column which has a pair of digits, bc, where one digit is the same as one digit from Step 2, which can occur in alternative positions. In this puzzle, 78 in column 7 has one digit the same, and the pair can occur either in [E7F7] or in [H7J7]. Search for a row which has one of the digits a, b or c in both columns. In this puzzle, the 7 in J4 is common to both. Because of the common digit identified in Step 4, the two "alternative positions pairs" are linked and if the pair in one column is put in one of its alternative positions, the pair in the other column might be forced to go into one of its alternative positions which produces a contradiction. So... Search for a candidate x which, if used for the solution of its cell, "forces" the position of the pair ab in the first column and also, at the same time, forces the position of the pair bc in the other column. If this produces a contradiction, then x cannot be the solution for its cell and x must be removed. In this puzzle, if H5 = 6, it forces a contradiction, so 6 must be removed as a candidate in H5. (*) SlowThinker: I‘m not sureitreallyworkshere: 76 canbe in EF4 and D4, thusno pair?
  • 21. Linkedcompartments: B4=8(solutionpathchosen) Type of links digitposition compartment weak link a !b,but !a  ? Checking J4 we find that for J4=2356 in combination with DEFG4/A4 B4:-5 For J4=7 J7:-7  E7=7  E4:-2  DEFG4=2..6  B4:-5 Therefore: B4 = 8
  • 22. B4=8, B1=5s … and A4:-6c
  • 25. Link chains: J7=4 Type of links digitposition compartment weak link a !b,but !a  ? BecauseofH5=78, H7 and H8 must not be78 atthe same time (aligned pair exclusion.) But:J7=7 forces H7=8c and J8:-7,H8=7s  J7:-7  J7=4
  • 29. F8=5g, F4=6g F5=39 is a biggap pair: neither 3 nor 9 canappearanywhereelse in F4..9
  • 38. Weekly Extreme Str8ts Puzzle #29www.str8ts.com Solution by SlowThinker Note: There are other (maybe easier) ways to solve this puzzle. Whether or not you find the steps logical or just to be “guessing” (linked chains, linked compartments) is up to you. I simply have not found a “more logical” way.