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#132 documented

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#132 documented

  1. 1. After initial routine eliminations the solver stops here:C3=2 (the only candidate in sequence with C2, and in any case the only 2 in AJ3),eliminating 2 from C7 and C9.D4=7 (D35 must contain 7 and D4 is the only candidate), eliminating the other 7s inrow D and column 4.1 can be eliminated from A1 (not in sequence with B1). DG1 must contain 34 sothose numbers can be eliminated from AB1. 3 can be eliminated from A78 (A15must contain a 3). 1 can be eliminated from A8 (not in sequence with B8). 2 can beeliminated from A7 (not in sequence with A8). 4 can be eliminated from B8 (not insequence with A8). 38 can be eliminated from B4 (neither is in sequence with A4).1 can be eliminated from C57 (stranded without a 2 in the compartment). 89 can beeliminated from C57 (stranded without a 7 in the compartment). 45 can beeliminated from C9 (C57 must contain 45).F13 must contain a 5, eliminating 5s from the rest of row F. F57 must contain 78,eliminating those numbers from F9. 89 can be eliminated from CF9 (strandedwithout a 7 in the compartment). CF9 must contain 34 so those numbers can beeliminated from HJ9. 1 can be eliminated from J9 (not in sequence with H9).J59 must contain a 5, so that number can be eliminated from J3. 6 can be eliminatedfrom J2 (not in sequence with J3). 89 can be eliminated from J59 (inconsistent withJ23). HJ29 is an x-wing, eliminating the other 7s in rows H and J. 89 can beeliminated from H6 (not in sequence with J6).9 can be eliminated from E4 (not in sequence with D4), and from H9 (not insequence with J9). 89 can be eliminated from B6 (not in sequence with C6). 4 canbe eliminated from E6 (not in sequence with F6). 6 can be eliminated from F6 (not insequence with E6). 8 can be eliminated from E8 (not in sequence with D8). Thestranded 6s can be eliminated from DE8 (no 5 in the compartment). 6 can beeliminated from G4 (not in sequence with H4). 9 can be eliminated from G5 (if itwere 9, G35 would have to be [789], but the only 7 was in G5).If E9 were 6, CF9 would have to contain 45, but the only 4 and 5 would both be in D9so that 6 can be eliminated from E9. If D9 were 6, CF9 would have to contain a 5,
  2. 2. but the only 5 was in D9, so that 6 can be eliminated from D9. D9 cannot be 3. If itwere, E9 would be 4 and there would be on 2 or 5 in the compartment. 3 cansimilarly be eliminated from E9. If D9 were 1, CF9 would be [1234], but the only 2and 4 would be in the same cell, so 1 can be eliminated from D9. Similarly, 1 can beeliminated from E9.A15 must be [12345], [23456] or [34567]. The only 2 is in A1 and the only 7 is in A5.If A1 were not 2, AB1 would be a naked pair 56, A2 would be 3, A4 would be 56 andA5 would be 7. In that case, G5 would not be 7, eliminating the stranded 89s fromG35 and thus from H4 also. AB4 would have to contain a 5, eliminating 5 from G4and 6 from H4, leaving a naked pair 34 in GH4 which would eliminate the 4 in B4,leaving B4=[56] which would be a unique rectangle with AB1 and A4. Hence A1=2and B1=1, eliminating the other 1s and the 9s from row B, 7 from A5, 12 from A78and from DG1, and 3 from B8.H3=1 (the only 1 in AJ3), eliminating the other 1s and the 9s in row H along with thestranded 2 in J9. 7 can be eliminated from G7 (not in sequence with G8).G5 must be 7. If it were not, the stranded 89s would be eliminated from G35, and theremaining naked quad 3456 in G1 plus G35 would solve G8=7, G7=8, J8=6, B8=8,A8=9 and A7=8, which would be inconsistent with G7=8. G5=7 eliminates 7 from F5and G8, 8 from G7 and 34 from G3 and G4. 3 can thus be eliminated from H4 (not insequence with G4). 68 can be eliminated from G4 (not in sequence with H4) and 59can be eliminated from G3 (not in sequence with G4). The resultant naked triple 689in column 3 eliminates 68 from the rest of the column. G4 must be 9 else G4=5 andG3=6, and when the other 56s are eliminated from row G there would be three cellscontaining only [34]. Thus G4=9, H4=8, G3=8, J3=9, D3=6, J2=8, H2=7, H9=6, E4=6and B4=4. 6 can be eliminated from H6, leaving a naked pair 34 in HJ6 whicheliminates those numbers from BC6.As H9=6, 6 is eliminated from CF9, leaving a naked pair 13 in row C which eliminates3 from C57. 5 can be eliminated from D9, else CF9 would be [2345] but the only 2and 4 would be in the same compartment.GJ8 must contain a 4, so 4 can be eliminated from DE8 and 6 from D7. DE89 is anx-wing eliminating 2 from D7 and E7 and solving J7=2, which eliminates 7 from J9solving J9=5. Whether F5=6 or F5=9 (the only two possibilities), each of F6 and F7is [78]. 3 can be eliminated from D1 (D79 must contain 3). 6 can be eliminated fromG7.EF67 is an x-wing eliminating 8 from A7 and B7 and thus 9 from A8. The nakedquad 3567 in row B solves B8=8, A8=7, A7=6, E7=9 (the only 9 in AJ7), E6=8, F6=7,F7=8, F5=9 (with 9 eliminated from E5 it is the only 9 in AJ5), B7=7 (the only 7 inAJ7), D7=1 (the only 1 in AJ7), D9=2, D8=3, E9=4, E1=3, E8=2 and E5=1.The naked pair 35 in row A solves A2=1, C2=3, C9=1, F9=3, F3=5, F1=6 and B3=3.C5=4 else BC56 would be a unique rectangle.The rest is routine leading to the solution:

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