- R6C7 and R6C8 form a <59> Twin
- R2C8 is also a g<59>, therefore R2C7 cannot be either <5> or <9> and therefore must be 3;
- AN1:7,6; AN2:4; CN7:1,9,2,3; CN8:2,7,g1,g8,3; CN9:7,6,1,3,g4,g9; XN7:g7,6; XN1:6,7; XN7:1,7; CN1:7;
- DN5:5; DN8:5; DN7:5; XN9:g5→XN3:5→BN1:5; XN3:8; XN1:8; XN4:8; XN5:8; XN2:8; XN9:8; DN3:8; DN2:8; DN5:8; DN7:8; DN4:8;
- AN5:7
- XN1:g1→XN3:1; XN2:1; (XN1:g1) →AN8:1
- DN8:g7,g2 (twin) →1,3; DN9:3,4; DR7:6,2; DN4:6,7; DN6:7; DN5:6,1,3,9; DN6:6; DR4:2; [DN8:<27> (twin) →potential Unsolvable Rectangle with DR1C4/DR1C5]→DR1C5:3; XN9:3; XN7:3; DN3:3; [DN8:<27>(twin)+DN2:<27>(twin) →potential Unsolvable Loop with XR7C9/XR8C9]+R7C9/R8C9:g2→NOT g7→XR9C9:7; XN3:7; XN2:7; XN8:7; XN6:7;
- BN6:8,g5; BN9:5;
- XN9:g1; [DN7:g1,g9(twin)→ potential Unsolvable Rectangle with XR7C8/R7C9; but XN9:g1→ XR7C8/R7C9 is NOT g9→ XC8:g9]; [XC8:g9+XN3:g9=XC7:g9]; [XC7:g9+XN9:g5=XR8C7/R9C7:<59>(twin)] →DN4:9,4,1; DN6:9,4,g1,g2; DN9:9,2; DN6:2,1; DN3:4,9; DN2:4,9; DN1:5,9; DN3:1,5,7; DN2:7,2; DN8:7,2; DN1:2,4,1; DN7:1,9; XR5C9:6 (nb g9 in XN3); XN3:6→BC3:6,7;
- XR5:g2; XC3:2; XN2:2; XC4:3,6,9; XN4:6;
- (TO FINISH OFF GRID B:) BN8:8,2,9; BN7:9; BN9:9; BN3:9; BN5:g2→BN4:2,5; BN6:5; BN9:8,6,7,4,3; BN7:3,4; BN8:4; BN3:3,2,1; BN6:1,6,3,7; BR4:1,3; BC5:3,7,4; BR1:2,1; BR2:1,6; BR6:7,9,4; BC2:4,9; BR7:1,3; BR3:3,4,6; BR5:6,2
- XN1:3,9,4,1; (TO FINISH OFF GRID A:) AR1:1,6,2; AC7:9,3; AN3:3,4,6; AR2:4,8; AN1:8; AN8:9; AR9:5,4; AN7:9,g3→AC4:3; AN8:5,8,2; AN2:5,2; AN7:8; AR8C2:3; AR4:8; AC8:9,2; AC2:2,8; AC6:8,3; AR6:3,7,6; AN6:6,1; AR5:1,5; AN7:1,7; AR4:7,5,2;
- XN4:4; XN6:4,2; XN2:3,4; XN8:3; XR8C1/R8C5=<45>(twin); CR2C4/R2C7=<45>(twin) →CR2C3:2; CN2:2; CN3:g2; CN3:g2+g9=(twin) →4,5; XN8:5,4,2; XN5:2,1,5,9; XN6:1,9; XN4:1; (n.b.: apparent unsolvable rectangle in XN4 and XN7, but- because this is a Samurai- box XN7 is also CN3, so the solution to this rectangle can come from grid C, hence this rectangle is not actually unsolvable)
- (TO FINISH OFF GRID C, AND THE REST:) CN2:5; CR1:4,6,9; CN1:9,3,8; CN2:3,4,1; CN8:1,8; CC4:1; CR4:1,6; CC1:2,3; CN6:2,5; CN4:5,8,4; CC6:5,3,g4→CR7:4,9; CN3:9,2; CR5:9,4; CN5:4,8; CN6:8,3,9; XN4:2,9.
- XC5:7,4; XN5:6,9,3,1;
- CN8:5; CN1:3; CN3:g3→XN9:3→DN7:3; DN1:g4→XN8:4;
- BN4:2; BN9:2; BN7:2,3; XN2:3;
- AN4:1,2; AN1:2;
- BN3:g7,g3→5,8; BN7:g1→BN1:1→BN2:1; BN6:1;
- DN1(R3C2):7; DN4:7; DN6:4,2,g6,g7,g8; DC2:4; DN5:3; DN8:3;
- XN7:g1→XN1:g1→AN3:1;
- XN9:g9→XN7:9; XN1:g9→AN6:9; CN7:g9→CN9:9; CN7:g6=<69>(twin) →<27>(twin) →CR8C4:1; CN9:1;
- XN3:g1; XN6:1; XN9:1; XN7:1; XC7:7; XN6:g4→XN3:4; BN1:4;
- XN3:g6; XN9:g6; DN7:6; DN4:9,1; DN5:9,4; DR5:8,7; DN8:4;
- XN1:g6→XN3:6,5;
- CN6:3; CN9:g3; CN3:3; XN4:3; XN1:3,9,6; AN3:3,9;
- CN3:g6→CN6:6;
- DN7:<45>(twin),1; DC1:2; DN8:2,1,9,7; DN5:1,8; DN2:1,3,7,6; DN3:7,3,1,5; DN9:7; DR9:9; DN6:9,6; DN3:9; DN7:9; DN1:9,6; DN3:6,2; DN9:2,6; DR2:8; [n.b. no potential Unsolvable Rectangle between the <25>s in DN1 and DN2, because DN1 is a link-nonet and so the correct order of the numbers can be forced from elsewhere in the puzzle: this is unique to Samurai and other inter-locking grid Sudokus]; XR2:1,7; XC2:8; XN6:g8,<48>(twin) →XN9:8; XN3:8; XN6:5,9; DC1:5,4; DC3:4,8; BN4:8; BN7:7,9; [X-Wings on <9> in BC6 and BC9 (affecting rows 4 and 8)→BR4C2 is not <9>, therefore 3]; BN4:9,6; BN1:9,7,6,8,3; BN2:6,7; BC4:6,3,5; BR1:8,3; BN3:7; BN9:7,1; BC8:8,5; BN8:6,8,7,9,4; BC6:3,5; BC9:9,4; BN6:4,3; BN5:3,4,9; BN2:5; BN9:5,8,9; XR1:1,4; XN2:1,g2,g8 (twin) →potential Unsolvable Rectangle with XN8, therefore XR8C6:5; XN7:5; XR9:4,2; DN2:2,5; DR1:5;
- AC8:6; AN1:6; AN7:6; AN3:4; AR1:4,7; AR8:4,1;
- XC2:5,2; XN8:2,8; XN2:8,2; AN9:2,8; AN3:2,5; AC7:8,6; AR9:1,9,7,3; AN4:7; AC3:5; AR8:5,3,8; AN1:5,8; AC6:5,8; AN6:5,4; AN4:4,5,6; AN5:6,8; AN2:8,9,5,3; AN8:9,7; AN5:7,3; XC3:2,5,7; XN5:2,5,8; XN6:8,4; XN4:4,7; XN7:6,8; CC9:8,2; CR8:8,3; CR9:7,4; CR7:4,3,7; CR4:4;1; CN2:4; CC3:4,9; CR1:9,1,8; CC6:8,2; CN5:1,7,6; CN2:5,6,2; CC1:8,9,6,2,7; CC3:2,6,7; CN6:7,2,4; CC2:6,2,7,9.
- Semi-colons separate each step in the logic.
- Numbers in bold are the numbers that can actually be filled in for sure.
- “R1”- means “look in Row 1”; similarly for C1 (column 1) and N1 (nonet 1).
- “11[2] (C2)” means “refer to the clue 11[2] in column 2”; if no row, column or nonet reference is given that is because it should either be obvious, or it follows on from the last number to be filled in.
- “g4” means ‘ghost 4’ and would refer to a ghost number for which a note can be made.
- <12> or <357> means either that a cell could be contain one of the numbers between the brackets; or it can mean that a certain group of clearly named cells contain those numbers in some order.
- T10 means that certain cells (specified in the text) add up to a Total of 10.
Nick came second in the first-ever National Championships for Sudoku last year. And no, it wasn't the Times (whatever they claim); it was the Independent who hosted the event; there were 2000 entries... and one guy beat me. Soundly. But that was then, this is now, and there's a book just out that explains how to do those Sudokus fast...
Teach Yourself Advanced Sudoku and Kakuro
(click here for step-by-step solutions)
This book was released on May 26th. Unlike every other book produced for the Sudoku market it is designed to show you how to do those tough puzzles, and how to do them fast. This book is far more advanced than any other Sudoku book, and even introduces techniques that are rarely needed for puzzles printed in newspapers. All techniques are thoroughly explained, so you won't just be told what to do, you will understand why the techniques work, and you will have practice puzzles to try each technique out as you learn it.
If you have already bought the book then you will know that this website uniquely has a step-by-step guide to solving each puzzle. I don't see much point in having just the answers when what you want to know is why you couldn't unlock the puzzle in the first place. See below for step-by-step answers.
I will shortly be writing up some more techniques or variations on techniques that I have discovered since writing the book.
STEP BY STEP SOLUTIONS TO THE PUZZLES
Here are step-by-step solutions to the Puzzles in Teach Yourself Advanced Sudoku and Kakuro. It seems to me that simply having answers at the back of the book is largely pointless since they don’t show you how to come to the solution. I hope this fills that gap, and helps you improve your ability with Sudoku. As far as I know, no one has tried to do this on any scale before, so I have no idea whether the coding I use below will make sense. I hope it does, because it took ages to type up! Apart from the most advanced bits of thinking which I write out in full, I have simply tried to say which group is the next one to look at, and what can be found there. For example, N3:5 means that if you look at nonet 3 you should be able to place a 5 in it. Or C4:g2 means that if you look at C4 you should be able to make a note of a ghost <2> there.
There are more detailed notes at the top of the Killer
Sudoku section, not least because that was more complicated to notate. Let me
know if it either does or doesn’t make sense please!
Incidentally, there are many ways to come to a solution. What follows are merely the routes I happened to spot when I last did the puzzles. And I think I had done several of them differently before. If you are looking to see what you’ve missed then don’t worry if you found things in a different order to me. Just make sure that you can spot each step in the solution: the first one that you missed is probably what’s been holding you back!
Choose your chapter:
Chapter 3 Advanced Number Work
Puzzle 1, Puzzle 2, Puzzle 3, Puzzle 4, Puzzle 5, Puzzle 6, Puzzle 7, Puzzle 8
N2:6; N9:6; R9:4,8,5; N8:7,5,1; N1:4; N9:5,1; N3:5; N2:9,1,8; C6:9,2; N4:9; C3:1; N4:1; R8C3:6; N1:6; N4:6,1; R1C2:3; R1:4; C3:3,2; N1:7; N7:7,8,2,9; R3:2,8; R2C8:7; N6:7; R5C5:4; R5:1,3,2; C8:4,8; N5:8,3,5; R4:1,2; N9:8,9; N3:9,3; C6:3,4,5.
C5:7; R9:3,8,6,2; N1:3; N3:3; N8:5,3,8,6; N5:6; C6:9; C2:9; R1:6; N4:6; C3:2; C4:3; N6:3,8; C3:8,1; R5C5:4; R1:4,1; N5:1,2; N6:2,7,4,9; R7:4,2,7,1; N3:1,7,6,5; N9:7,6,9; R6:7,8,5; N1:5,1,4; N2:4,7,2; R4:8,5; C1:9,1.
N5:8; N6:4; N1:4; N2:4; N8:4,2; N7:1; N2:1; N3:1,9; C5:3; N2:3; N8:8; R7:8; C6:7; N3:7; N1:7; N4:7; C2:3,8,6; R2:2,5; N1:5; C4:9,6; N2:6; R6:5,3,2; N7:8,5,6; N4:6,8,9; N5:9,2; R9C6:9; N9:9,3,7,5,6; N8:6,5; N6:5,2,7,3,6; N3:2,6.
N2:3; N8:6,9; N7:5; C9:8; R8C9:4; R3C1:7; R3:6,4,5,1; N3:5; R2:7,9,8,3; N2:8; N7:6; C2:1; N1:4; R5C4:1; N8:7; C4:2; N4:2; N7:2,3,1; N9:3,9,1; N8:2; N9:2,1,7,6; N3:7,9; C1:3; N6:3,5,9,6; N5:9,7,3; C2:7,4; N5:4,5; N6:4,8; N4:8,5.
N3:5,1; R2:3,2. N9:6. N7:2,5. N8:7,5. N5:4. R1C4:1. N1:1,7,4,2. N3:4,9. N4:1. C1:8. N4:9,5,2. C3:6. N7:4. N8:4. R7:9. R5:2. N6:2. R4:5. C7:3,4,7,8. N5:8,6. N2:6,9,8. N8:9,8. N7:9,7. N9:8. N3:8,7. N6:7,4,6. N4:6,7.
N5:8,6,2,7. N6:7,4. R5:3. R4:3. C7:2. N3:5. N2:5. C4:2. N9:8. C8:3,9,1. N6:5,9. N4:9,5,6. R9:3. C4:1. R3:7,3. R3C2:8. N4:8,1. C2:5,6,2. N2:2,3,8,9. R1:8,9,1. N9:1. N3:9,4. N1:4,1. R9:1,6,9. C5:9,4. N8:1,7. N7:4,6. N9:7,6.
All The 8s (in N2, N4 and N9). N1:7,9,5. R8:5. N3:9. C1:9. N6:9. N4:3. C2:6. N4:1,4. R6:6,2. N5:2,4. R4:6,4. N6:5,7. N9:7. N1:4. R1:2. N2:2. R2:3,6. R1:6,1. N1:1. N2:5,4,3. N8:4,6,9,1. N7:1,4. N9:4,1,3,9,2. N7:3,2. N8:3,2. N3:1,4.
N1:58. N4:6,3. N5:4. N2:6. N9:6,5. N4:4. C9:4,2. N9:8,7. R8C6:2. R8C2:1. N4:1,7. N6:1,9. C9:9. N2:8. C1:4. R9C2:3. C2:2. R5:5,7,2. N5:2,9. N8:9. N7:9,8,7. C4:2. C7:2,6,4,7. C8:4,6. R3:3,1. R2:1,3,7. R1:7,3,9. N8:3,7,8.
Chapter 3 Advanced Number Work
Roulette Practice Puzzle, Easy X-Wings Practice Puzzle, Twins Practice Puzzle, Triplets Practice Puzzle;
End of chapter: Puzzle 1, Puzzle 2, Puzzle 3, Puzzle 4, Puzzle 5, Puzzle 6,
Puzzle 7, Puzzle 8
This particular puzzle can be solved in all sorts of ways, and the whole point of Roulette notation is simply to have the right note in the right place at the right time... therefore there are many ways to attack this puzzle. However, I’m sorry to say that it does include a Twin, so sorry if you’ve been stuck with it!
N3:9; N6:5,g1; R4:7,g1; N1:5; N2:g5; N3:5 (in R3C8); N9:5; N8:5; N2:5; N1:g7; N7:7; N7:g1; N4:6; N7:g6; N8:g8; R3C2:3; R3C1:7; R7:9; R4:9; N4:g8 so N7:g8; N7:g1 and g8 make a Twin, so N7:6; N9:6; C7:4; N6:7; N3:7,8; R3:1; N5:1; C7:1,3; N9:g1 so N7:1,8; R7:4,1,3; N9:3; N6:3,2; N3:2,1,6; N1:2,4; N2:4; N7:4,3,2; N8:4; N4:4; C4:2; R5C5:8; N4:8,3,2; N8:8,9; N2:8,6,3; N5:2,3,6,9.
N5:2; N9:2; N7:6; N8:1; N7:7; N3:g6 so N6:g6; N4:g6 (making Easy
X-Wings) so N5:6; N2:6; N3:6; R1C2:1; N3:1; R3:g3,8; N7:8; C2:7; N7:g1,g7,3; N7:g5; R3C3:9; N2:9,5,3; N5:5; R5C6:9; R5:3,7; N7:7,1; N4:1; N6:1,7; N5:7; R6C4:3; N4:3,6; N6:6; C5:3; N9:3; N3:3; R8:4,9; C8:5,8; N9:8; N6:9; N3:9,5,8,2; N1:2,5; N4:9,4; N5:4,8; N8:8,9,4; N7:4,5
N3:g3,g5 (making a Twin); C8:8; N4:7; N6:7; N1:g3,g6 (Twin); (therefore also N1:g8,g5,g1); C1:g9,g2 (Twin),4; N9:g4,g9 (Twin) so g7,g9 (Twin); N8:7; N7:g3,g5 (Twin); R7:g6,g8 (Twin); N8C4:2; N7:2,9; N9:9,4; N8:4,1; N2:4; N6:4; N8:6; N1:6,3; N3:3,5; N1:5; N7:3,5; N2:3; N5:3; R5:9,1,2; N7:1,8; N2:2,9,6,8; N3:9; N8:8,6; N5:1,6; N6:6,2,1,9; N4:2,5,9,8; N5:8,5; N1:8,1; N3:6,1,7,2,8; N9:7,1.
R7C1:3; N1:4; N7:g6 so N8:6; N7:g2,g6,g8 (Triplet) so g1,g9 (Twin); R7C7:5; N3:g9 so N2:9; R2C7:2; N2:g2; C1: <2> can only be in
R5C1 or R9C1 and also <8> can only be in those two cells, therefore
<28> Twin, therefore C1:6;
N7:6; R2:7,5; N8:g7; N9:7; R7:4; N5:4; R3C6:3; N2:5,2; N1:g2,g8
(Twin),1,7; N4:7; N5:7,6,9; N8:7,1,3; N7:1,9; N5:1; R6C1:5; R3C7:1; N3:7,5; C7:3,8; N7:8,2; N1:2,8; N4:2; N5:2,3; N4:3.9,8; N6:9,8,6,5,1; N3:9,3; N9:2,4,1,3.
N1:9,g4. N8:6,9,g7,g1. N7:9. N2:3,5,g4,g1. R1:3,g5,g4,g1. N5:4,g7,g1. C7:7,6,g3. Isolated Twins: R7C2 and R7C4 could both only be <17>, therefore R7C7=3. R7C1:2. N4:2. R4C4:5. N7:3. R9:2,g1,g5. N7:7. N8:7,1. N6:2,g3,g5 (making a Twin),9,6. N3:6,7,8. N9:8,4,5,1. N3:5,1,4. N5:2,3. N6:3,5. R4C6:8, C6:9,6. R5C2:1. N5:1,7. N4:6,7. N1:7,8,4,1,5. N2:1,4. N4:8,5. N7:1,5.
N3:6,8. N1:g4. N7:4,6. R9C2:2. N7:g7,g8. N4:8. N8:9. N9:4. N1:g4,g5 (making a Twin), g1. N3:1,7,2. C8:1,5. n6:5. R3:7. N1:2. Easy X-Wings: g6 in both N8 and in N2 gives us N5:6. N5:g7,g4 (making a Twin). R5:1. N4:2,3,9. N6:3,9,2. R6:9. R4C5:5. C5:8,2. R3:5,4. R1:4,5,3. R4:3,1. N7:8,7. R7:2,1. R9:1,7,3. N9:2,8. N5:4,7,2,8. N8:7,6. N2:6,1. N1:1,9,3.
N6:3. R9:2,g3. N8:3. N5:3. N1:g3. N3:3,g2. N9:2, N4:g2. N1:2. N3:2. N1:4. N7:g4,g1 (making a Twin),7. N9:g7,g1 (making a Twin), g4. N6:4. N5:4. C1:3. N7:5,g6,g8 (making a Twin). C2:3,5. R5:1,6. C1:9,8. N6:9. C9:g8,6. N1:6,5. C4:5,9,8,7. N6:8,7. N9:7,1,6. N7:6,8,1,4. N8:7,8. N9:8,4. N4:8,2. N5:2,7,9,6. N3:8,9,1,5. N2:8,1,6,5.
N3:g3,g8 (making a Twin), so g6,g4 (making a Twin). N5:2,3,g4,1,g9 (making a Twin with the g4),7. N6:7,6,4,9. N5:4,9. N4:9,6,8,5. N1:9. C3:1. N2:1,g5,g8 (because of the g8 in N3!) (making a Twin), g6. N9:1,7,2. Here’s a particularly hard moment: R1C1=2 (note both the g6 and the g3 in R1!). N7:2. N8:2,7,g8. N2:8,5,7,3,4,6. N1:7,8. R3:4,6. N3:6. N7:6,4,3. N1:4,3,5. N8:6,8,5,9,4. N9:6,3,9,8,5,4. N3:3,8.
N3:1. N7:6,g3. N9:1. R8C8:5. R7:6. R3C6:8. R3:2,9. C6:3,1,4,5. N3:5. C2:7,2. N1:3 (note the g3 in N7),8. R2:8,6. N9:8,9. R4C7:2. C7:7,3. R7:8,3. N8:3,7,2,4. N7:4,9,5. C1:5,1. N5:1,4,8,9. N6:9,3,5,6,8. N3:6,2. N2:9,7. N4:9,8,6,4,3.
N2:6. N3:3,7. N1:7,5,8. R3C1:4. N3:g1,g2 (making a Twin),4,8. R3:5. N8:5,7. N9:5. N1:g3,g9 (making a Twin). N7:3,8,2. This is a hard moment: look how the numbers 5 and 6 affect cells in row 5: this leaves R5C2 and R5C8 as an Isolated Twin in row 5. Therefore R5:3. after which it all comes out easily: N2:3,4. N1:3,9. N8:g4. N5:4. C5:9. R4:3. C7:9. N4:9,7. N5:1,7,8. N2:8,9. N8:9. C9:9,7,2,8,1. N9:2,4. N3:2,1. N6:1,8,6,5. N4:5,6,2,1. N7:1,6,4. N8:1,4,6.
N3:2. N4:1. N5:1. N6:2. N9:1. N3:1. N7:2. N8:g6. N5:6. R4:2. N8:2,6,5,8. N5:8,7. R4:5. N9:6.
This is the tough bit: look at N6, C9 and R9: both N6 and R9
are only missing <347>. Now C9 has three cells that intersect with either
N6 or R9, so those three cells have
to contain a <3>, a <4> and a <7> in some order. This gives
us a ghost triplet on <347> and means that R2C9 cannot be a <3>,
which means that the <3> in row 2 has to be in R2C2. Therefore R2C2=3. And from here it comes out easily:
N7:3. R9:7,4. N6:7. N7:7,5. R7C7:8. N3:8. N1:8. N7:8,9. N4:9,4. N1:9,7. R5:4,5. R4:4,3. R6:9,3. C8:4,9,3. C9:9,5. R3:4,5. N2:9,4,3. R1:4,2,5.
N8:3. N5:3. N7:2. C2:8. N1:g1. N2:g4,g8 (making a Twin), so 6,7. N5:g6,g9 (making a Twin). C6:7. R1C9:5. R1:3,1. N1:5,7. C2:5,9. N3:6,1. R6: notice how <582> wipes out R6C1, R6C3 and R6C9 which makes a triplet in those three numbers over the other available cells; so that leaves a ghost <1> in R6 (in nonet 4), giving us: N6:1. N5:1. C5:2,8,4. N8:4. N7:8. R9C3:6. N4:6. N5:6,9. R5:9. R4:2,4,7. R5:7,5. R4:8,9,5. N9:5,7,4. N7:7,5,4,1. N4:1,7. N3:2,4,7,8. N2:8,4.
Twins Practice Puzzle 1, Twins Practice Puzzle 2, XY Wings, Magnetism, Distant Twins, Unsolvable Rectangles 1, Unsolvable Rectangles 2, Unsolvable Puzzle;
End of chapter: Puzzle 1, Puzzle 2, Puzzle 3, Puzzle 4, Puzzle 5, Puzzle 6,
Puzzle 7, Puzzle 8, Puzzle 9, Puzzle 10, Puzzle 11, Puzzle 12, Puzzle 13, Puzzle
14, Puzzle 15.
N1:6; N7:6; N6:6; N5:6; N8:6; N7:4,9,3; N1:9,2; N2:9; N5:7,g1,g5 (making a Twin); therefore R4C9:2; N3:g9,g3 (making a Twin); R2C4:<48> and R4C4: <48> making a Twin; therefore R8C4: 3; N9:3; N3:3,9; N6:9,g1,g4 (making a Twin),5; N5:3,4 (because of the g4 in N6, the 4 in N8 and the <48> Twin in C4, forcing R4C4 to be the 4); C4:8; R4:8; C1:4,2; R8:5; N3:5; R3:4; R2C7:4; N2:4,1; N1:1,8; N4:1,5; N7:5; N6:4,1; R5:1,2; N8:2,1; N9:1,2,7,8; N7:8,7; N8:8,7; N5:8,5; N2:5,7; N3:2,7,8.
N7: 9,g1; N4:1; N5:8; N9:2,g8; N6:8; N1:9; N7:g1,g8 (making a Twin),g5,g3 (making a Twin), so R5C3:4; R5:5,3; C7:5,1; R1C4 = <34>; R3C4 = <34> (making a Cell-Twin); therefore R1C6:2; R1:4,3; R3C4:3; N1:2,6,5,4; N3:3,9,4; N2:6; N8:6; C2:3; C7:3,9; R4C8:7; C1:7,5; N5:5; N8:5,7; N7:5,3; C4:7,1; R9:7,1; N2:1,9; N5:9,2,4; N4:2,6; N6:6,4; N8:3,4; N9:3,8,5,1; N7:1,8.
The first cell is the <12> in R4C1; the buddy cells are the <24> in R4C4 and the <14> in R6C3. Depending on what the first cell is, one of them must be <4> and therefore R6C4 cannot be a <4> and must be 6 after which the puzzle comes out easily.
N8: 8; N9: 2; N2:4; N8:3; R9:6; N5:6; C5:8; N4:8; R5:7; N5:2; N3:g7,g8
(making a Twin), 3; N9:3;
Now it’s possible to use magnetism on <4>:
C9: there are two options for
<4>: R6C9+ and R7C9−
N9: there are two options for
<4>: R7C9− and R9C7+
R9: there are two options for
<4>: R9C7+ and R9C2−
C2: there are two options for
<4>: R9C2− and R6C2+
Now, in row 6 there are two cells
marked +, therefore they must be wrong, and the − cells
must be correct: therefore R7C9:4;
R9C2:4.
Now magnetism on <9>: (otherwise you can use
Quads/Triplets which are harder to spot!):
R2: there are two options for
<9>: R2C8+ and R2C2−
N1: there are two options for
<9>: R2C2− and R1C1+
C1: there are two options for
<9>: R1C1+ and R7C1−
R7: there are two options for
<9>: R7C1− and R7C8+
Now, in column 8 there are two
cells marked +, so they must be wrong, and the − cells
are all <9>s. Therefore: R2C2:9;
R7C1:9.
Now it finishes off easily: N1:8,3,2; N3:9,8,7; N4:3,2,4; N6:2,4; C1:7; N2:7; N8:7,5; N7:7,6,5,1; N4:1,5; N9:1,5,6,9; N6:9,1,5; N3:5,1; N1:1,5; N2:5,1; N5:5,1,7,9.
N3:2; N6:2,9; N8:7; R8:6; C8:7; R8C8:8; R8:5,1; N7:2; R9C5:8; N9:5; R2C7:8; R2:9; N2:9; N4:9; R6C3:8; N7:8;
Notice all the <34>s that are missing... therefore
Distant Twins on <34>:
In C8: R1C8+ and R7C8−;
Therefore, in N9: R9C9+
Therefore, in R9: R9C1−
Therefore, in N7: R7C3+
Therefore, in C3: R3C3−
And as a result, R1C1 cannot be a <3>. Therefore, N1:3; and all the other − cells
are also 3; and all the other +
cells are 4; N3:3,1,7; N6:3,6; R1:8; R6C4:6; R7C4:2; N5:2,3,5; N4:5,1; N6:5,8,7; N1:1,4,6; N2:1,4,6,8; N8:1,6; N5:4,8,7; N4:6,7,4
Unsolvable Rectangles Puzzle 1
N2:1,4; N4:4,6; N7:4; N8:3; R6C3:1; N7:1; R6:6; N5:9,1,g5,g7 (Twin); N4:7; R4:1; N6:6,2; N9:1; C1:7; N1:2; C6:7,6; N8:6; R9:7;
Potential Unsolvable Rectangle:
N1:3; N4:3,8; N7:3,7; N9:7,2,4,9; R7:2,9:
[n.b. at this point there is another potential Unsolvable
Rectangle involving ghost <58> in cells R8C3, R8C4, R9C3, R9C4. However,
it is not strictly necessary]
N3:7,4,8,2; N2:8,9,2,5; N1:5,8; N3:5,9; N6:5,9; N8:8,9,5; N7:5,8
Unsolvable Rectangles Puzzle 2
N6:9; C7:1,4; N5:5,1,4,6; N2:4; N8:5,8,2; N1:4; N5:2,7,3; N2:3,8; C7:8,2; N1:8; N4:7,2; N6:3;
Potential Unsolvable Rectangle:
In column 4, N8, the ghost <39> forms a Twin and the
base pair (R7C4 and R8C4)
In column 9, N9, the ghost <9> (in R7C9 or R8C9) means
that there cannot be a <3> in either of those cells, therefore N9: 3 (in R9C9);
N9:8,6; N6:8,1; C3:3,6,7; N2:7,9; N1:9,6,2 (because R1C2 cannot be a
<5> or a <1>); N3:2,1,5,7;
N9:7; N7:7,2; N1:1,5; N4:1,3; N7:3,5; N8:3,9; N9:9,5.
Unsolvable Puzzle practice puzzle
N6:9,2; N2:5; N4:g7; N7:7,2; N1:9,4; C1:5; N3:5,8; N9:5,8; N7:5,8,4; R9:9; N5:9; C5:8; N4:8,5; C3:6; N2:6; R1:3; N1:1; N4:1; R4:1; C9:2,1; N3:1,4; R3:1,8; N8:8; N5:g4 so N6:4
Now, all cells have exactly two candidates except for R5C2
which must therefore now be <3> otherwise (if it were either a <6>
or a <7>) then the puzzle would be unsolvable. Therefore R5C2:3 and the puzzle naturally solves
itself from there.
N2:1. N8:6. C6:2. C4:4. N7:8. N1:8. N7:4,g1. R1C9:9. R3C9:5. N1:5,9. N2:9. N4:5,g3,g6 (making a Twin). N7:6. N3:g3,g4,6. N9:6,5. N6:5. N7:g1,g2 (making a Twin). R8:g3,g7 (making a Twin). N2:g3,g7. Now note that in Column 5 there is an Isolated Twin on <37>; in N5 there is a g3 g7 which form a Twin (in R5C4 and R5C6), so in N4:3,6. N6:6. R5:2. N9:2. C7:9,1. R5:1,8. N6:7. N9:7,1,8,9. N8:1,9,2. N5:8,4. R4:7. N1:7. N4:g1.
Potential Unsolvable Rectangle: R8C2 and R8C3= g<12>. R6C2 and R6C3 contain g1, therefore cannot contain a g2 else the rectangle formed would be unsolvable. Therefore R6C1=2. N1:2,4,3. N3:4,3. N2:7,3. N5:7,3. N8: 7,3. N7:2,1. N4:4,1.
N1:2. N5:1. N8:1. N5:g9. N6:9. N2:g9. N1:9. N7:9. R2C2:5. R9C2:4. N4:5. N7:5. R9C1:1. N9:1. N3:g6,g4 (making a Twin). N3:1. N1:1. R9:5,3. N8:3. R7:4. R1:3,5,8. N9:8. N3:5,7. N9:7,6. N3:6,4. N6:3,7. N4:7. R4C5:8. R6:6. Magnetism on <3>: shows that R3C6 cannot be <3> and therefore R3C6=8. N1:8,3. N4:3,6,2. N7:2,6,8. N6:2,4. R6:3,9. N2:9,3,7. N5:2,7,4. N8:2,4.
N6:6. N4:7,2. R8:6. N9:2,3. R8:4. R3C6:2. N2:8. N1:8,g3. N7:g3,g4,g5 (making a Triplet), so N8:3. N5:3. N8:7,6. N2:6,9. N8:9,1. N9:1. N5:1. N4:6. N1:6. Magnetism on <5> shows that R4C8 cannot be <5>, therefore R4C8=4. R2:4,5. C4:5,9. N5:4,2. R4:5,9. N6:9,2,g1,g8 (making a Twin),7. R3:4,5. N4:4,1,3,8. N6:1,8. N9:9,4. N3:7,1,4. N1:5,3. N7:3,5,4.
N3:8,4. N8:4. N5:3,5,g8. N4:8,4,7. R5:1,3. N6:4,6,7. N5:6. N4:2,9. R1C3:5. N3:6. R8C8:3. N7:g1. N9:1. C8:7. N7:g3 (which twins with g1). N1:g1,g3 (making a Twin). N1:4. Cell-Twins: R8C1 and R8C6 are both <29>. R8:8,5. N3:5. N9:2. R9:5. N2:5. R2:1,3. N7:1,3. N3:9. N1:3. Magnetism on <2> shows that R3C2 cannot be <2> therefore R3C2=7. N1:2,9. N7:2,9. N8:2. N2:2,1,7,9,6. N8:6,9,7,8. N9:9,8. N5:8,7,1.
R5:2,9. R5C8:5. N9:2,1. C8:2,6,8. R7C9:8. N3:8,6. N2:2. N1:2. N7:6. C5:6,5. Distant Twins on <34>: R4C5 (+), R4C7 (−), R5C9 (+), R2C9 (−) therefore no <3> or <4> in R2C5, therefore R2C5=7. C5:3,4. N6:3,4. C9:4. R8:7,8. N1:7,1. N4:7,9,g5,g3 (making a Twin),4. N5:4. N7:4,8. N1:4,5,3. N2:3,4,6,1. N5:1,9. N8:9,3. N3:1,3. N7:3,5. N4:3,5. N9:5,7. N8:6,7.
N2:3. N9:3. N4:3. N7:5,8,1. N2:5,1,g7,g4 (making a Twin). R3:1. R3C3:8. N5:5. R9:g4, so R9C2 must be 7 to avoid an Unsolvable Rectangle with N2 twin. N7:6. N8:g2,g7 (making a Twin). R8:1. N2:7,4. N8:4,6. C4:1. N4:1. N7:1,5. N5:g4,g9 (making a Twin). C1:g4. Easy X-Wings on g4s in N4 and N5 allows us to see that R5C8=4. N9:4. R5C3=9 (because there is a <49> Twin in N5 and a g4 in corresponding cells in N4, so there cannot be a g9 in those cells, therefore in R5C3). C3:6. R2C2:2. R2:9,7. R3:7,2. C8:7,3,6. C7:6,2. C6:2,6. N4:6,8,2,4. N5:4,9. N6:9,8. C9:6,2. N8:2,7.
N2:9,6. N7:6. N1:6. N3:5,7. C5:6,3,4. N6:6. R8:5,9,8. N7:g9. So, N4:9. C1:g7 (which twins with N7’s g9), so N7:8. N4:3. R4C4:8. R4:g4,g1 (making a Twin), so R6C8=2. R6:8,7,4. N4:1,4. N6:8,3. N9:3. N8:8. N1:8,2,4,7. N3:4. Now it’s a potentially unsolvable puzzle! All the cells have only 2 candidates except for R7C4. If that cell is not a <2> then the puzzle actually will be unsolvable (there will be two potential solutions) therefore R7C4=2 and all the rest follows.
N2:2. N1:6. N3:g9. So, N1:9,8. N4:8,9. N7:5,2,g3. N1:3. N2:8. N8:9. N9:9,2,8. R7C8:4. N8:4. N3:9,2. C7:8. C9:4. R8:6,1. C1:7. N4:4. N1:4,1. N3:4,3. R5C7=1 (because R8C7 and R8C9 have a base Twin <57>, and R5C9 is also <57> which means the final cell in the rectangle (R5C7) cannot be either a <5> or a <7> else there would be an Unsolvable Rectangle). C7:7,5. C9:7,5. N3:1. N2:1,4,5,7. N8:1. N5:7,5,3,6. N6:6,5,2. N5:2,1. N4:1,2,3. N8:7,6,3. N7:3,7.
N5:1,2,4. N8:6. N1:6. N7:g2. N1:2. N7:g4. N1:4. N4:4. C2:g5. C3:5,1. N4:5. R5:g3,g7 (making a Twin). C5: g3,g7 (making a Twin). This is a tough one to see: notice that whichever of <3> or <7> goes in R5C9, it will wipe out all three of the cells in column 9 of Nonet 3 and must therefore be the same number as goes in R3C7, since N3 still needs a <3> and a <7>. Therefore R3C7 is <37>, and we can therefore place N3:1. R3:5. C5:5,9. N8:9,3,7 N2:3. R9C7:6. N3:g5, so R9:5. R6C7:9. N9:9. N6:6,g4,g8 (making a Twin). N4:8,9,7,3. N5:3,7. N1:8,9,5. N3:5,3,7. N2:8,7. N1:7,3. C7:7,1. N7:1,2,4. N9:3,4,2,8. N6:4,8,3,7.
R5C5:5. N4:6. N7:6. N9:6. N3:6. N2:6. R5:4. N9:4. N5:4. N1:4,8. N2:8. C5:8,9. N3:1. N1:1. N9:1. R4C6=1 (because there’s a g7,g1 Isolated Twin in Column 4, and there’s a g7,g1 Twin in Nonet 8, which threatens to make an Unsolvable Rectangle if R4C6 were a <1> or <7>, therefore it cannot be and must therefore be a <3>). R4C3:5. N6:5. R4:1,7,8. N9:8,9,3,2. N6:2,3. N5:7. N8:7,1,2,9,5. N7:5,3. N1:5. N3:5,3,2. N2:3,2,5. N1:2,3,9. N7:9,2. N4:3,2,9,7.
N2:7. N3:3,6. N2:6. N5:6,2. C4:8. C9:9. N4:7. N6:7,3,1,8,5,4. C7:7. N8:7. C4:9. N4:3. C9:4. N1:4. C6:3,5,9. C5:9. N4:5,4. R7C1:1. N8:1. N2:1,4,8. N5:1,4. N8:4. R2:5,1. R3:1. C2:9. N4:2. Magnetism on <8>: R8C7 (+), R1C7 (−), R1C2 (+), R3C3 (−), R8C3 (+) therefore there are two +s in R8, so all the +s are wrong and the −s are all 8s. Therefore finish off N3 and N1, then C1; N7:9,5. N8:5,3. N7:3,8. And finish off N9.
N7:2. N1:1. N3:1,g8. So, N3:8. N4:5,6. N1:g9,g7 (making a Twin). N4:g7. N5:g6,g7 (making a Twin),2. Magnetism on <3>: make R1C6 a +, and you will find two −s in R8, which means all the +s are ‘correct’ and therefore 3s (i.e. R4C3, R1C6, R3C9 and R6C7). C3:4. N4:g4 (making a Twin along with the g7). R6:9. N5:4. N2:4,5,7,8 (notice the g8 in N1 for that),2. N1:7,9. N3:5. N6:5,4. N5:7,6. C4:8,1. N7:1. N3:4,9,2,6. N6:9,7. N9:7,6,4. N7:6. C6:6,5. C7:5,9. R9:9. N8:9,3. N7:7,3,8. And finish off.
N3:6. N4:6. N2:5. N5:4,g2,g3 (making a Twin). N2:3,g3,g9 (making a Twin). R3C6:4. N1:g2,g7 (making a Twin). N2:7,1. C4:3,7,4. N9:5. C6:2. N6:7. R6:3. N6:9. There’s now a potential Unsolvable Rectangle that needs to be avoided: R1C2 and R1C3 form a Twin on <27>, and R4C2 and R4C3 share a g7, so there cannot be <2> in either of those two cells. This now allows us to use Magnetism on <2>: R5C2 (+), R8C2 (−), R8C1 (+), R6C1 (−); therefore R6C3 cannot be <2> and therefore R6C3=5. N7:5. N1:5,1,4. N7:4 N9:4. N3:4. R9:9. N7:9. C3:2 (remember R4C3 cannot be a <2>!) C3:7. N1:7. R4C2=8 (because there is a <23> Twin in Nonet 5 and R5C2 is also <23> so R4C2- which would make up the fourth cell in a rectangle- must not be either a <2> or a <3>, hence an <8>. N6:8,2,1. N3:2,1. N9:8,1. N7:1,8,2. N4:2,3. N5:3,2. N8:1,9,8. N2:9,8. N3:9,8.
N5:g1,g6 (making a Twin),5,g8,g9 (making another Twin!). N6:9. N3:1. N8:2,g1. N5:1,6. N1:g2. N4:2. N2:2,6,g3. N3:3,6,8. N2:8,4,7,3. N5:6. N8:6. N1:2. C9:8. R8:9 R9:8,9,1. N8:7,9. R8:1,3. N7:8. N4:1. N5:9,8. Magnetism on <7>: R9C3 (+), R9C8 (−), R7C7 (+), R3C7 (−), R3C3 (+), therefore C3 has two +s, therefore the −s are the correct ones and are all 7s. (R9C8=7, R3C7=7). N1:7,5,4. N3:5,4. C3:7,5. C2:7,4. C8:5,3. N6:7,4. Finish N4.
N5:8,5. N4:3,g1,g4 (making a Twin), g7,g9 (making a Twin). N9:3. N7:5. R7C3:9. R7:3,1. R3C3:1. R3:2,5. N9:g2,g8 (making a Twin), g4,g6 (making a Twin). N7:4 (note the g4 in N4),8,7. N9:8,2. N8:2,6. N2:6. N1:8,2,3,5,7. N6:6. N9:6,4. N3:5. R2C7:4. C7:7. R6:4. C5:7. Magnetism on <9>: R1C9 (+), R4C9 (−), R5C8 (+), R5C5 (−), therefore R1C5 cannot be <9>, therefore R1C5=4. N5:4,7,9. N6:9,1. N3:9,1. N4:9,7,4,1. C6:9,1. C4:1,7,9.
Magnetic Attraction, X-Wings, Swordfish, Inconsistent Loop, Consistent Loop;
End of chapter: Puzzle 1, Puzzle 2, Puzzle 3, Puzzle 4, Puzzle 5, Puzzle 6,
Puzzle 7, Puzzle 8
Magnetic Attraction Practice Puzzle
Just to let you know: the Magnetic Attraction example (i.e. the one with the 9s on page 93, rather than the actual Practice Puzzle on page 94) is not a real puzzle, and does not have a valid solution. It was just a very good example of what I was trying to explain!
The real puzzle on page 94:
N5:5,6; N3:3; R2C6:9; N8:9; N4:8,3; N5:8; C5:3; N8:3; N2:3; C4:8; N2:8; N7:g5 so R2:5,1; N7:1,g6 so N9:6; C3:6;
Magnetic Attraction on <1> (n.b. the g45 Twin in N6)
R6C5+ and R1C5−
(in column 5), and R6C1− (in row 6), and R4C1+ (in nonet
4)...also...
R4C9a and R1C9b
(in column 9).
Therefore, − is opposite b (in row 1) and + is
opposite a (in row 4), and therefore a is the same as −,
and b is the same as +.
Make this alteration and therefore R4C8 is <27>; R9C8 is also <27>
giving a Cell-Twin in column 8; and therefore R5:7; now the rest falls out easily:
N1:7,4,6,9,2; N2:2,1; N3:4,1,9; N6:4,5,9,1,2,7; N9:1,2,9,7,5,4; N8:4,2; N7:6,2,5; N4:2,1; N5:1,4,2.
N5:1,7,6,3; C6:2,3,9,8; R1C7:1; R6C7:9; N8:g3,g6
(Twin); N4:6; N6:6; N9:g6
Now there’s a potential Unsolvable Rectangle with <36> Twin in N8 and the
g6 in N9, which means that there cannot be a <3> in R7C8 or in R8C8 and
therefore R7C7:3;
N8:3,6,5; N9:6; C7:8; R2C4:7; N1:7; N8:7; R1:4,5; N1:4; C1:9; N4:9; R5:3; C8:4
Put in pencil marks to find the X-Wing in columns 1 and 8 on the number
<2>; therefore there is no <2> in R6C3 or in R7C9; then the X-Wing
in columns 1 and 8 on the number <8> means that there is no <8> in
R7C9 and that R7C4:9!
N9:9; N8:8; R9:2; N1:2; C2:1,8; C1:2,1,8; N1:8; N3:8,5; and after that it all finishes off easily.
The swordfish is found on the number <3> in rows 1, 4
and 7; this means that the number <3> cannot appear in the following
cells: R3C4, R3C7, R3C8, R5C7 and R6C8. Therefore R3C7:1, after which it all comes out very easily:
N1:1; N6:1; R5C5:2; R5:3,7,9; N6:9; R1:4,7,8,5,3; N3:8,6; R3:5,4,3,9,2; C4:3,1; N7:1,8,7,9; R9:9,2,5; N8:4,2,7; N7:2,5; N4:5,4,2,7; N5:1,6,5; N6:6,2,3; N9:3,5.
Inconsistent Loop Practice Puzzle
The chain can actually start in R1C1, in the top left-hand corner. If R1C1 is <2>, then R1C8 is <8>, then R9C8 is <2>, then R8C7 is <3>, then R8C1 is <2> therefore R1C1 cannot be <2>! Therefore R1C1:8, and the final squares fall.
Consistent Loop Practice Puzzle
Start the loop at R9C5... therefore...
Either: R9C5 is <6>, so R9C8 is <2>, so R2C8 is <5>, so R2C6
is <6>, so R8C6 is <4>, so R8C4 is <9>, so R9C5 is <6>
(where we started)
Or: R9C5 is <9>, so R9C8 is <4>, so R2C8 is <6>, so R2C6 is
<5>, so R8C6 is <2>, so R8C4 is <6>, so R9C5 is <9>
(where we started)
Therefore, in row 9, either R9C5 or R9C8 is going to be <6> therefore R9C2
cannot be a <6> and therefore is <29>.
Now we can do XY Wings: R9C2 is <29>, and R1C2 is <69> whilst R9C8
is <26>, therefore either R9C2 or R9C8 must be <6> and therefore
R1C8 cannot be a <6> and must be 4!
And the rest falls out easily.
N3:6,8. N7:6. N4:6. N2:6,3. N5:6. R3:9,7. N4:g8. N1:8,3. N5:8. N7:g4,g3 (making a Twin),2,9. N4:8. R8:9. N9:g3,g7 (making a Twin),1. C7:g4. R7C9:2. C7:2. R5:2. N5:3. R6:5,1. XY-Wings: R9C9 is either <3> or <7>; each choice would make either R9C6 or R5C9 be <4>, therefore R5C6 cannot be <4> and must be 1. N4:1,2,5. N1:2,1,5. N2:5,7. N3:7,1,4. N6:5,4,3. N9:3,7. N7:3,4. N9:4,5. N8:4,7,5,9. N5: Finish off.
C2:7. N4:9. N5:9. C6:8,7. C4:1,4. R5:7. N9:7. N7:7. N2:g2,g8 (making a Twin), g3,g6 (making
a Twin). N5:g2,g5 (making a Twin). N8:g6,g2 (making a Twin), g4,g1 (making a
Twin). R4:8. N7:8. C1:g2. R1:g2,g9 (making a Twin). R8C1:6. C6:6,2. R7C8=2 (because N3 has a Twin <29> and
N9 has a g9 that corresponds, so those cells cannot contain a g2 as well,
leaving R7C8 to contain the <2>). R5:2.
C3:g6.
Another potentially Unsolvable Rectangle: there’s a <36> Twin in Nonet 2, and a g6 in Nonet 1 which prevents a g3 being in those same cells, therefore N1:3,2. R1:6 (in R1C3). N2:6,3,2,8. N3:8,6. C1:5. N7:3. N9:3,9. N6:6. R5:5. N4:6,4. N7:4,1. N8:4,1. N9:4,1. N1:1,5. N3:5,4,1,9,2. N6:1,2,4,5. N5:5,2.
N7:8. N9:8. C8:5. C4:8. R1:3. Magnetism on <7>: R4C1 (+), R4C4 (−), R6C6 (+), R9C6 (−), therefore R9C1 cannot contain a <7>, therefore N7:g7.
Swordfish on <4> (or Magnetic Attraction on <4>) concentrating on the possible <4>s in rows 1, 5 and 9, and how they affect columns 3, 4 and 7. As a result there cannot be a <4> in R7C4, which is also affected by that g7 in N7 and therefore R7C4=2.
N9:2,3,1,4. N3:3. N2:3. N8:4. N2:4. N1:4. R5:4. N6:4. N4:2. R4:7. N8:7. C4:1,5. R9:9,6. R7:6,9,7. N7:5,1. C1:1. N5:1. R5:2,6. N6:1,9. C7:9,7. R3:1,6,2,8. N2:5. N3:5,2. N1:2,9,6,7. N4:7,6. N5:6,5,9. N8:8,9.
N3:1. N6:7. R9:9. N4:9. N8:1. N3:g9. N2:9. C4:7. C5:8. N4:8. N7:8. N9:8. R9C7:3. N6:3. N7:3,1. N1:1. N4:1. N8:3. R7:7. R9:7,5. R6C2:6. C2:7.
Swordfish on <5>: look at where <5> can go in columns 3, 5 and 8, and how that only affects rows 3, 4 and 7. Therefore <5> cannot go in any other cell in those rows (i.e. there can be no <5> in any of the following cells: R3C1, R3C6, R4C6 and R7C9).
Magnetism on <5>: R5C1 (+), R4C3 (−), R4C8 (+), R7C8 (−), R8C9 (+), R8C6 (−), R5C6 (+), therefore there are two +s in R5, therefore the −s must be the ‘correct’ cells for the 5s. Therefore R4C3=5, R7C8=5, and R8C6=5.
R7:9,6. R8:5. R6:4,2. R4:2,4,6. R8:4,6. N5:5,6. N2:2,6,5,4. N3:4,9,6,2. N1:6,5,4,2. N4:4.
N1:7,6,g9. N7:4. N4:4,7,2,8,3. N7:8. C2:9,2. C3:3. R7:8,1,3. N3:1,g6. R3:6,3. C9:6. Swordfish on <5>: look at where <5> can go in rows 1, 6 and 9, and how that only affects columns 4, 6 and 7. Therefore <5> cannot go in any other cell in those rows (i.e. there can be no <5> in any of the following cells: R4C4 or R4C6). N5:g5 (in row 6), therefore R6C7=6. N5:6. N6:g5,g2 (making a Twin),8,9. R5:1,3. C5:8,5. N9:5. C7:2. N3:5,9. N6:5,2. N9:9,2. R1:9,5. N1:9,2. N5:8,7,5,4. N2:4,2. N8:7,3,2,1. N7:1,3.
R7:2. N8:6. N9:2. N4:6. N6:6. N9:6,3. N6:3,7. N4:2,7g8. N7:g8. N9:8. N3:7. C7:9. N3:8,4. N2:2. N1:2. R3C1:8. N7:8. N2:8. N5:8. N4:8. R2:3. Cell-Twin: R5C5=<59> and R6C4=<59>, therefore N5:g3,g4 (making a Twin).
X-Wings on <9> in Rows 2 and 5 (affecting columns 3 and 5), therefore cell R8C5 can only be <15> which twins with cell R8C8 (also only <15>), leaving R8C6 to be <49>.
Swordfish on <5> in columns 2, 6 and 9 (affecting rows 1, 4 and 7), forcing removing the possibility <5> from the following cells: R1C3, R1C4, R4C8. Also we can now see that R1C3=1.
Magnetism on <5>: R5C5 (+), R6C4 (−), R6C8 (+), R8C8 (−), R8C5 (+) therefore there are two +s in Column 5, so they must be the ‘wrong’ ones, and all the −s are actually 5s: R6C4=5, and R8C8=5.
N6:5,1,9. N4:5,9. N7:9,4. N8:5,7,9,1,4. N9:1,7. N5:4,9,3. N2:1,3,5,9. N3:1,5. N1:4,5,9.
N5:2,4. N6:4. N9:4. N2:4. C5:9,7,1. N8:5,1,g2,g8 (making a Twin). N2:8,g5,g6 (making a Twin),9. N5:9,g6,g3 (making a Twin). N1:8,g2,g3 (making a Twin). N7:8. N8:8,2. R8:g3,g6 (making a Twin).
Potential Unsolvable Rectangle to avoid! R7C7 and R9C7 form a <27> Twin, R9C3 is also <27>, therefore R7C3 cannot be either <2> or <7> and is therefore <16>.
N3:g3,g9 (making a Twin). C3: Cell-Triplet: R1C3=<15>; R5C3=<56> and R7C3=<16>. Therefore R6C3 cannot be <1> or <5> or <6> and is therefore <37>.
Magnetism on <5>: R5C7 (+), R5C3 (−), R1C3 (+), R1C7 (−), therefore R4C7 cannot be a <5>, so must be a <16>. Since R6C9=<16> there is a Cell-Twin in Nonet 6. Therefore R4C8=<57>, R5C7=<59>, R6C8=<79>. R5:g6, which means that R4C2=<13>. R6C2=<139>. R4:g5,g7 (making an Isolated Twin in R4C1 and R4C8). N4:g1 (in column 2), therefore R7C2=<26>.
Potentially Unsolvable Puzzle to avoid! If R6C2=<19> then the puzzle would actually be unsolvable, therefore R6C2=3 and the puzzle can be finished off.
N1:7. N3:9. N6:6,7,g4,g2 (making a Twin). C8:9. N9:7. N8:7. N5:7,6. N8:6. N7:6,g1. N9:1,8,4,5. N3:1,4,g2,g5 (making a Twin). N2:4. N1:9,g5. C2:g3 (making a Twin along with the g5). N5:g4. R3:g2 (in nonet 2).
Swordfish on <2> (or Magnetic Attraction): involving R3, R6 and R7 (and affecting C2, C4 and C6). Therefore R4C6 is not a <2> therefore R4C6=8. R8C6=3. N7:3,2. N8:8. R6:2,9. C6:9,5. N1:5,3. N3:5,2. N1:2,8. N2:2,8. N8:2,5. N4:2,1,8. N7:1,8. N6:2,4. N5:4,1,5,3. N2:3,1.
Puzzle 1
R1:0; C11:5; R3:7; R4:B; N14:B; N12:E; N8:E; N5:E; N6:E; N1:E; N3:E; N3:5; R3C12:9; N7:9; R5:7,1,0,A,B; N15:1; C11:0,F; N5:F,B,A; R7:8,5,7,3; N8:3,C,6,A; N6:5,2,4,F; C5:9,8; N11:8; R10:1; R12C14:2; R2C14:A; R6C9:A; C9:0,2,C,7,4; N13:4; N9:4;
R10:B; N11:B; N7:B,4; N11:4,7; C5:7,2; R12:6,A; N1:6; N9:9; R11C1:0; R11:C,3,1; N2:1,2,3,4,F,A; N3:A,D; R3:4,C; R1:C; R1C4:8; R1C2:1; C1:1,8,D,F; R1:D; C16:5; N16:6,8; N15:8,E,3,9; R14:F,2; N14:D,A,0; N10:0,F; C14:8,D; R16:F,A,C; N12:C,B; N16:B,2,7,9; N7:E,8; R13:0,5,2; N1:5; C16:2,9; N1:9,3; N4:3,1,7,D; N13:9,3,A; N9:3,2,A; N11:A,3.
Puzzle 2
N16:1,g5,g7; R16:4,gD,g9,gC,gB,gE,0,F; N16:F,0; R13:g4,g9,E,6,1; C16:8,0; C15:7; N13:C; R15:5,7; N16:7; R15C12:0; R15:E; N13:0,6,E,4,9; C2:E; R1:gA,gB (n.b. <AB> twin in C3); C3:0,1,7,6; N11:8; C3:5,8; N11:5; N12:5; R12:2; R6C4:gAB→(n.b, <AB> twin IN C4); C4:9; R11C2:3; R5C7:F; R5:9,E,D,8; C9:3,0,1; N3:g1,gE,C,3,9; N13:9,D; R9:D; N12:D; N13:gB,gD; R14:5,3; R1:8,D,9; C6:2; R3C6:4; R12C6:7; R12:4,3; N12:2,E,9,B; C16:3; N8:3,2,4,6,F,5; R16C13:C; N12:C,0; N4:0; R2C13:4; N1:4; N9:4,B,F; N11:4; C5:F; N10:3; N2:3; C5:1,C; N10:8; R11:7; N11:C,A; N13:A,B; R1:A,B; C4:A; N1:5; N4:5,8; R4:8; C4:D; N7:D; R4:D,7,F; N15:F,B; N7:F,B,A; C6:A,8,B; N5:8,B,5,9,6; N6:0,1,7,C; N2:C,F,2,1; N3:1,E; N4:E,B; N16:E,B; N1:D.
The central grid is grid X; the top-left grid is grid A; top-right is grid B; bottom-left is grid C; bottom-right is grid D.
N1 to N9 are the nonets within each grid. This means that AN9 is the same as XN1; BN7 is the same as XN3; CN3 is the same as XN7; and DN1 is the same as XN9: those are the link-nonets. I normally use the ‘X’ version to refer to the link-nonets except where the thinking is very much confined to a particular grid, in which case the specific grid is referred to.
Bold means new numbers that can be entered in to the grid
Small ‘g’ before such a number means it’s a ‘ghost’ and therefore straddles two cells
New trains of thought are represented by a new bullet point.
Most entries naturally led on to the next entry, but where entries are grouped together inside square brackets it shows a few pieces of thinking being brought together to make a conclusion.
‘→’ means ‘therefore’.
Puzzle 1
Puzzle 2:
Puzzle 1, Puzzle 2, Puzzle 3, Puzzle 4, Puzzle 5, Puzzle 6, Puzzle 7, Puzzle 8, Puzzle 9, Puzzle 10, Puzzle 11, Puzzle 12, Puzzle 13, Puzzle 14, Puzzle 15
R6C6:1
(intersection); complete 4[2]; R7C6:2
(intersection and note that 41[8] has no <4>); complete 7[3]; R5C7:1 (intersection and note that 43[8] has
no <2>); complete 3[2];
R2C1:1 (intersection with 43[8]
which has no <2>); complete 3[2]; 4[3](C5):1,3; 10[4](C6):4 (in
R2C6 which cannot be <1>, <2> or <3>), also 3; R1C3:7 (intersection); R3C3:9
(intersection); complete 24[3] and 16[2]; R3C6:2 (7[3](R3) needs a <2>, and this is the only place it can
go); complete 10[4] and 7[3]; 3[2](C5):2,1;
11[4](C8):3,5; 16[5](R4) must be
<12346>, therefore:6; R4C4:4 (intersection); complete 16[5], 7[2]
and 13[2];
16[2](R4):7,9; 43[8](C7) needs
<8> and <9> which cannot go in R1C7 or R8C7, therefore in R2C7 and
R7C7 as Cell-Twins; now 43[8]:7,5;
R8C4:6 (by elimination); complete
15[2] then 43[8](C7) then 13[2](C4) then 28[7](R1) and 43[8](R2);
12[2](C8):7,5; R6C2:5 (intersection and by elimination);
complete 37[8](C2) then 28[7](R8); 20[5](R5) requires a total of 4, i.e. <13>,
therefore complete 10[5]; 25[4](C1) requires a total of 15, i.e. <96>,
therefore complete 25[4] and finish off.
3[2] (C6):g12 (Twin); 23[3] (C1): g986 (Triplet) therefore
11[2] (R3) is either <92> or <83> or <65> (all Twins); but 7[3] (C2) g124 (Triplet) therefore
11[2] (R3) is <92> (Twin);
38[8] (R1) is missing a <7> therefore 16[2] (C3):7,9;
39[6] is <987654>; 10<4> (C2) is <1234>, therefore R8C2 must
be 4.
R6C2 = 3 (otherwise the 20[3] is
unsolvable); R7C2 = 2 (because 43[8]
(in R5) is missing a <2>); therefore R5C2 = 1 (10[4]);
24[3] (C8):9 (needs a <9> but
not in R7 because of the 36[8] and not in R6 because of the 10[3]), 8 (can’t go in R6 because of 10[3]); 7;
23[3] (C1) 9,8,6; 20[3] (C6): 8;
36[8] (R7): g13 (Twin) in R7C6 and R7C7 (since they cannot go anywhere else);
R7C4:5 (by elimination); therefore
14[2] (C4):9; R7C5 cannot be
<4> else there’d be another <9> in R8, therefore: 7; 13[2] (C5):6;
38[8] (R7):4; 39[6] (R8):8 (in R8C7, since it cannot go anywhere
else), 5 (can only go in R8C6), 7.
34[5] (C3):9,6; 17[5] (C6): has to
be <12356>, therefore 6,3,2,1;
36[8] (R7):3; 10[3]:1; 17[4] (C7):5;
43[8] (R5):7 (not in 16[5]),4; R4C4:4 (only possibility for both clues);
21[4] (R4):2; 16[5] (C5) must be
<12346>: therefore g6 (in R1 and R2 but not R3 else R3C4 would be a <1> which is not possible)
34[5] (C4):9,8,6; 23[3] (C1):8,6; 7[2] (R3):1; 16[5] (C5):6,3;
Potential Unsolvable Rectangle: 6[3] (in C7) must be <123>; if R3C7 is
<3> then R1 and R2 would be <12> which would create an Unsolvable
Rectangle... therefore R3C7 is not
<3> and therefore R2C7:3;
R1C6 is <12>; R1C7 is <12> (Cell-Twin), therefore R1C2:4; 7[3] (C2):1; 3[2] (C6):1,2; 6[3]
(C7):1,2 and finish off.
R2C1:7, therefore
complete 9[2] and 16[2]; complete 3[2]; complete 4[2] (R5); therefore 5[2] (R6)
is not <14>, therefore it is <25> and complete it.
R3C2:4 (note 10[4]); R5C8:5; R2C6:5 (15[5] meets 14[2]) therefore complete 14[2];
34[5] (R1):8 (in C4 as not possible
anywhere else); R1C8:4 (34[6] meets
15[5]); 17[2] (C6) is <98> but
R6C6 is <8> (as 14[4] cannot have a <9>) therefore complete 17[2]
and 29[4]:8 (not in C7), 7.
34[5] (R1):7,6; 37[8] (C7):9; complete 16[2] (R8) and 16[2] (C8);
complete 11[2] (R7);
37[8]:5,4; (note the 14[4] (R6) must
be <1238>); complete 8[2] (R4); complete 4[2] (R3); complete 15[5] (C8)
and 37[8] (C7); complete 10[3] (C5):2,1;
15[5] (R2):3;
14[4] (C4):2,1; 10[4] (R3):3; 11[2] (C3):8; R6C5:3
(intersection); 14[4] (R6):2; R4C2:7 (not <6> or <5> because
of effect on R4C1); therefore R4C1:5;
15[5] (C1):2 (not in 33[5]),4; 33[5] (R8) is <98754>
therefore 37[8] (C2):6,5; 33[5]:8,7,9; complete 29[5].
R3C3:9
(intersection 16[2] and 23[3]); complete 16[2];
R6C4:4 (intersection 34[5] and
5[2]); complete 5[2];
R6C6:1 (intersection 4[2] and 42[8];
complete 4[2];
16[5] (C5):3; complete 11[2];
17[5] (C8) must be either <12347> or <12356>, therefore:3; therefore 10[2] (R5):7;
16[2] (R6):7,9; 16[2] (R1) meets
14[2] (C5), therefore R1C5:9;
complete 16[2] and 14[2]; then 15[2] (C4); then 23[3] (C3);
R2C1:9 (because of 16[2] (C1) –v-
38[8] (R2)); complete 16[2] and 13[2];
R1C7:8 (not <6> because of
9[2]); complete 15[2] and 9[2];
R6C1:4 (where 34[5] and 5[2] meet);
compete 5[2];
3[2] (C3):1,2; 17[5] (C8):1,2,4; 16[5] (C5):2,4,6; R7C7:5 (forced,
because 42[8] is missing a <3>); R8C7:6; R4C7:4; 42[8]:2,1; complete 10[2] (R3); 34[5] (C4):6,7,9; R8C1:8 (where 34[5] meets 38[8]); complete 38[8]; 42[8]:8,9; 43[8] (C2):9 (because not in R5C2
because of 10[2] and 34[5]); 34[5]:6,7;
10[2]:4; 38[8] (R2):4,3; complete.
R1C1:1 (3[2]
meets 4[2]); complete both; and 7[2];
R6C1:7 (9[2] meets 16[2]); complete
both and 3[2];
26[6] (C2) needs a total of 8, therefore <26>, therefore 15[2] (R3):6,9; 13[2] (C3):4; 26[6] (C2):2; 10[4]
(R4):1,3; complete 12[2] and 16[2];
R1C11:2 (11[2] meets 3[2]); complete
both and 6[2];
R3C9:3 (11[2] meets 4[2]); complete
both; R4C10:3 (38[6] meets 10[4]);
38[6]:7,6 and complete 3[2] (C11);
R2C4:6 (15[2] meets 16[5]); complete
15[2] then 11[2]; R2C8:4 (because
16[7] (R1 is <97>, therefore 13[2] (C8) is either <94> or
<76> but cannot be <76> therefore <94>!); complete 13[2] then
16[2];
16[5] (R2):2 (in R2C7 because the
41[7] doesn’t have a <1> or <3>);1,3; complete 9[2];
28[7] (C5) is not <9> or
<8>, therefore 23[3](R3):9,6;
R6C5:5 (intersection); 22[3]:9,8; complete 11[2]; 10[4](R4):2,4; complete 6[2](C8); complete
10[2](R5); complete 41[7](C7); complete 28[7](C5); R7C8:6 (intersection); 23[6](C8):9
(not R8 because of 14[4] being
<1238>),8; complete 15[2](R9);
R8C9:2 (cannot be <1> or
<3> because of 9[2] and 28[7]); complete 9[2]; 28[7]:3,1; complete 6[2]; R8C10:3
(not possible for <1>);
complete 14[4]; 10[4](C11):4
(intersection with 13[2]), 3
(intersection with 20[3]),2;
complete 13[2]; 20[3](R11):9,8;
complete 10[2] and 10[3]; 11[4](C1) is <1235> therefore R9C1:5 (intersection); complete 13[2];
R8C4 is <12>; if R8C1 is <1> or <2> then R8C4 is the other
one and R8C2 would be <5> which is not possible; therefore R8C1 cannot be
<1> or <2> therefore (11[4]):3;
R10C1:2 (else 19[3] is not
possible); 19[3]:8,9; finish 11[4],
10[2], 10[3], 28[4], 13[4], 6[3], 9[2];
R10C5:2 (else 19[3] is not
possible); finish 10[3]; R10C7 cannot be <9> else 21[3] is not possible,
therefore 8 and complete puzzle.
R7C9:3
(intersection); complete 4[2] then 20[3] then 9[2] then 3[2](R10); R9C10:6 (R9C10 is either <8> or
<6> but if <8> then R11C10 is <2> which is not possible);
complete bottom right corner; R1C1:7
(intersection); complete 8[2] then 16[2](R2) then 24[3]; R3C3:5 (not <9> because of 10[3]);
complete 22[3] then 25[4] then 11[2], 10[3] and 5[2]; R2C4:7 (intersection); complete 8[2], 8[2] and 16[2]; R6C2:3 (intersection); complete 4[2]; R8C1:6 (intersection); R9C1:5 (intersection); complete 14[2] then
23[3] then 13[2]; R6C4:4
(intersection); complete 8[3]; R7C5:7
(intersection); complete 15[2]; R7C2:5
(intersection); R7C6:2 (intersection);
complete 22[6]; R10C1:4
(intersection); complete 21[6]; R10C2:7
(intersection- note that 33[5] must be <45789>); complete 38[6];
16[2](C3):7,9; complete 24[4];
complete 14[2] then 33[6] then 9[2] then 7[3]; R8C7:9 (Outie); complete 16[2] and 26[4] then 14[2] then 15[2];
41[7](C6):8; R6C7:6 (intersection); complete 7[2], 34[5]
and 10[3]; R3C6:5 (intersection);
complete 41[7] then 8[2], 19[4], 8[3] and 6[2]; R1C11:5 (intersection); R2C9:5
(intersection); complete 14[2]; R2C10:3
(intersection- note 38[6] does not include <1>, <2> or <4>);
R3C10:5 (intersection); complete
6[2] then 15[5](R2) then 9[2](C8) then 29[4]; R6C11:3 (else R6C10 doesn’t work); complete 9[2] then 38[6](C10); R4C11:4 (intersection) and finish.
R1C9:9
(intersection); complete 17[2] and 16[2]; 16[2](R4):7,9; R3C2:8 (else cannot
fill 14[2] and 8[2]); complete 14[2] and 8[2];
R9C2:5 (intersection); R9C1:7 (intersection); 29[4]:9,8;
R10C1, R10C2 and R11C1 are all
<12> therefore R11C2 cannot be, else you have an Unsolvable Rectangle,
therefore R11C2:4;
R8C2 is <1> or <2>; if it is <1> then 29[4](C3) becomes
impossible, therefore:2; complete
bottom left corner and 7[2]; 29[4]:8,7
(note the intersection with 28[7]);
R8C7:4 (if it were <1> or
<”> then 10[2] forces <9> or <8> into 28[7] which is
impossible); complete 10[2];
R7C9:5 (intersection); R8C6:2; complete 6[2] and 7[3] and 3[2];
R9C10:5 (else 12[4] unsolvable);
R5C2:6 (can’t be bigger and if
smaller then 12[3](C1) cannot be solved); complete 10[2](R5) and 12[3](C1);
21[4](R6): R6C2 and R6C4 total 12, but R6C4 cannot be <9> (because of
11[3]), therefore R6C2:5; complete
21[4](R6) and 36[6](C2);
33[5](R2):<9> and <8> can only go in R2C9 and R2C11, therefore
R1C11 is <3> or <4>, therefore R1C10 is <1> or <2>;
R4C10 is <1> or <2>, therefore Cell-Twins, therefore R3C10 is not
<1> or <2> therefore R3C9 cannot be <9> therefore 8! R2C9:9; R2C11:8; R1C11:4; R1C10:1; 3[2](R4):1,2;
11[3](R3):3;
R3C7:6 (intersection); 33[5](R2):7; 22[6](C10):5,4; complete 10[2] and 9[3]; R6C9:7 (intersection); complete 15[4]; complete 28[7]:1,3; complete 12[3](C8); complete
13[2](R5); complete 11[3](C4) and 10[2](R5);
R2C7:3 (intersection); complete
33[5] and 7[2] and 3[2] and 17[5]; R3C6: <8> or <9>: if <8>
then R4C6 is <4> which is not possible, therefore 9; complete 12[2], 9[3], 23[3]; R2C5 is not <9> therefore 4; complete 33[5], 22[5] and 10[2];
29[4](C9):9,8; complete 9[2]; R9C11
is an Outie, therefore 8; complete
corner; complete 29[5](R9);22[3[ is left with g69 Twin; 16[2]:g79 Twin;
therefore Easy X-Wings on <9> therefore 15[2] cannot be <69>,
therefore it is <78>; Easy X-Wings on <7> and <9> means that
13[2] is not <94> nor <76> therefore must be <85>; both top
and bottom rows (i.e. bother 32[5] and 30[5]) must have a <9> and an
<8> and a <7> now, therefore the extra 2 for the 32[5] comes from
having <6> and <2> rather than <5> and <1>. Now
complete the lines.
R9C10:1 (intersection); complete 4[2] and 8[2]; 3[2](R4):1,2; complete 8[2] and 7[3]; R3C2:1 (intersection), complete 4[2] and 7[2]; 7[3](C2):4,2; complete 12[2] and 11[2]; 3[2](C7):2,1; complete 5[2]; R2C6:5; R2C4:6 (intersection); complete 28[7], 10[2]; R4C4:9 (intersection); complete 16[2], 23[3] ad 13[3]; R4C6:1 (intersection); complete 15[5]; R5C7:5 (intersection); R4C7:2 (intersection); complete 6[3] and 8[3]; 4[2](C11):1,3; R6C9:5 (intersection); 29[5](C9):9,8,7; complete 10[3], 9[2]; 15[5](R6):2,4; R5C8:7 (intersection); complete 16[4], and 33[5]; R1C10:5 (intersection); complete 6[2]; R2C10:6 (intersection); complete 9[3] and 41[7] and 11[2] and 13[4] and 27[4]; R5C3:4 (intersection),g12; 3[2](R8):g12 therefore Cell Twin with R5C2 and R8C2, therefore R10C2:4; R9C3:5 (intersection); complete 7[2] and 7[3] and 3[2](R8) and 7[3](R5) and 5[2](C1); R9C2:7 (intersection); R6C2:6 (intersection); R11C2:5 (intersection); complete 14[2] and 20[4] and 29[5] and 28[7](C2); R7C3:8 (intersection); R7C6:4 (intersection); R7C4:2 (intersection); complete 18[5]; R8C4:8 (intersection); complete 26[4] and 34[5] and 15[3] and 23[3]; R9C6:6 (intersection); complete 9[3] and 6[3] and 3[2] and 34[5] and the rest.
R1C8:6
(intersection); complete 7[2] and 8[2]; R2C11:4 (intersection); R11C11:7
(intersection); complete 16[2] and 8[2] and 8[2]; R1C1:9 (intersection); complete 17[2] and 16[2] and 13[2]; R5C3:6 (intersection); R5C2:8 (intersection); complete 23[3] and
14[2] and 9[2];
21[6](C3):1 (can only go in R2C3);
25[6](R2) needs <235> therefore R2C4 is 5 (intersection) and R2C5 is 3
(intersection); complete 25[6] and 5[2]; R1C4:9 (intersection); complete 19[3]; R4C5:8 (intersection); complete 27[4]; 28[4](C4):8,6; complete 21[3]; R4C6:4
(intersection); R3C7:4
(intersection); complete 13[2] and 23[6] and 28[4]; R6C7:2 (intersection); 10[4](C7)1,3;
complete 29[7] and 21[6] (note the intersection with 41[7]); R7C1:5 (intersection); R8C2:5 (intersection); complete 6[2]; R9C2:7 (intersection including 9[2] and 11[4]); complete 9[2] and 11[4];
10[4](C11):1 (not R4C11); complete 8[2]; 10[4](C11):3,2; complete 10[2] and 29[5];
R5C9:2 (25[6] needs a <2> and
R5C9 is only place it can go)
R5C8:8,6 (this 25[6] needs a
<6> and <8> and <6> cannot go in 12[2]); complete 12[2];
R8C6:7 (intersection); R6C9:7 (intersection);
both R7C2 and R7C6 can only be <89> therefore Cell-Twin, therefore R7C4
and R7C5 are <67> Cell-Twins, therefore 13[2](C4) is <67>,
therefore 7,6 and R7C5:6;
R8C10:6 (intersection); complete
15[2]; 22[3](R7) cannot be <976> therefore it is <985>, therefore
R7C9:5 (intersection); complete
22[3] and 13[2]; R8C7:4
(intersection); R10C7:8
(intersection); R10C8:5
(intersection); 38[6](R10):6,3;
complete 15[2]; 28[4](C7):9,7;
complete 17[3]; 22[6](C9):1,4;
complete 21[3] and 17[4] and 28[7]; R6C5 cannot be <9> therefore 8 and work out the rest from here.
R10C3:5
(intersection); complete 13[2]; R10C4: 2
(intersection- cannot be <3>); complete 11[2] and 21[3]; R10C1:3 (intersection); complete 11[4]; R9C1:9 (intersection); complete 19[3] and
14[2] and 15[2];
30[7](C2) is missing <9> and <6>, therefore R5C2:7 (intersection); R5C1:5 (intersection); complete 6[2];
R1C2:4 (intersection); complete
13[2] and 12[2]); R3C2:2
(intersection); complete 7[3] and 11[2] and 14[2]; 6[3](C5):2,3,1; complete 3[2]; 7[3](C6):1,4; complete 22[6]; R3C4:4 (intersection); complete 13[4]; R4C4:8; complete 28[4] and 29[4] and 9[2];
R4C7:3 (intersection); complete
12[2] then 11[2]; 11[4](C7):2,1;
30[7](C2): g23; 24[4](C3):g69; therefore 12[3](R6) needs a total of 11,
therefore <29>; complete 30[7] and 24[4];
R8C4:9 (intersection); complete
16[2] and 13[2]; R7C5:5 (intersection);
complete 19[5]; R3C10:7
(intersection); complete 16[2]; R4C11:3
(intersection); complete 4[2] and 19[3]; R7C10:5 (intersection); R1C10 cannot be <9> or <8> or
<5> therefore 22[3](R1) must be <976>; therefore R1C10:6, R1C9:7; complete 22[3] and 14[2] then 8[2] and 17[4] and 28[7];
R9C10:2 (intersection); complete
11[2] and 16[2]; 37[6](R10) lacks a total of 8, and does not contain <1>,
therefore 6[3]:1,3; complete 5[2]
and 10[2]; therefore 37[6](R10) is missing <125> therefore R10C7:4, R10C8 is <98> and R8C8 is also
<98> making a Cell-Twin, therefore R7C8:7 (intersection) and R9C8:4;
R7C9 is <98> and R8C9 is <98> making a Cell-Twin, therefore
28[4](C9) needs a total of 11, which is either <56> or <47>,
therefore R6C9 is either <6> or <7> and since it is not <6>
because of the 10[3](R6) is must be 7;
complete 10[3], 10[2], 29[4], 17[2](R8), 28[4](C8);
37[6](R10):8; complete 28[4](C9);
15[5](R5); 14[3](C6); 19[3] and 29[4]; R9C6:2 (else 17[3] is not possible) and finish off.
R1C1:1 (intersection);
complete 4[2] and 6[2]; R2C4:6
(intersection); complete 7[2] and 7[2]; 17[2](C3) is <98> therefore R3C1
is <12>; R4C2: is <12> making a Cell-Twin, therefore R5C2:4; complete 6[2] and 3[2](R4); R3C2:1; complete 10[2], 17[2] and 8[3](C1);
R2C2:7 (intersection); complete
33[5] and 22[6]; 25[4](R6) needs <89>; R6C4:8 (intersection); complete 25[4]; R9C3:5 (intersection); in the corner: 7[3](C1):4,2,1; complete 6[3](C2); R8C3:8
(intersection); complete 29[4](C3); R3C5:4
(intersection); R4C5:7 (intersection);
complete 34[5] and 11[2]; R7C4:2
(intersection); complete 14[4]; R4C6:2
(can’t be <1>); complete 10[3], 5[2], 9[3];
R9C9:3 (Outie); 4[2](R10):g13;
3[2](R11):g12; therefore COUPLES: 1+2 (total 3) and 1+3 (total 4), therefore
13[3](C11); must be <931> and 11[3] must be <812>, so complete
them;
R7C9:5 (intersection); 11[4](C9):1,2; R7C5:4 (intersection); R7C6:6
(intersection); R8C6:4
(intersection); 13[4](C6): g12; R9C5:5
(intersection); R8C5:3; 15[5](C5):
g12; 34[5](C7):4; R10C5 and R10C6
both <12> therefore Cell-Twin; R10C8 cannot be <3>, therefore 6; complete 7[2] and 8[2], and 16[5] (3) and 10[2] and 8[2] and 15[5] and
16[5] and 13[4] and 15[3] and 34[5](C7) and 29[7] and 41[7];
25[5](C7) needs a <9> therefore in R1C7:9; complete 14[2] and 9[2]; R2C7:6 (intersection); complete 25[5] and 8[2] and 23[4]; 16[5](R2):2 (note the 38[6] in C10 cannot have a
<1> or <2> or <4>); complete 8[2] and 14[2]; 16[5](R2):1,3; complete 4[2] and 8[2]; R4C10:6; complete 7[2]; 38[6]:9,7 and finish off.
R1C1:1
(intersection); complete 4[2] and 3[2]; R2C2:8 (intersection); complete 19[3] and 11[2]; R5C2:6 (intersection); complete 15[2];
R11C1:3 (intersection); complete
11[2] and 4[2]; R10C2:5
(intersection); complete 22[3] and 13[2]; R10C4:4 (intersection); complete 13[2] and 16[2]; R1C10: 7 (intersection); complete 16[2] and
10[2]; R4C10: 6 (intersection);
complete 15[2];
14[3](R2) is either <761> or <851> and cannot be <761>
therefore <851>; therefore R2C10:5
(intersection); complete 14[3] and 12[2]; R10C8:4 (intersection); complete 6[2] and 8[2]; R6C9:4 (intersection);
R11C11 is <5> or <6> or <8> or <9> (because of the
14[2]) but if <8> or <9> then there’s a problem with the 11[2]
intersecting with 34[5]; therefore R11C11 is <5> or <6>, therefore
11[2] is <56>, therefore R10C11:6
(intersection); complete 11[2], then 14[2]; R10C10:7 (intersection); complete 17[3] and 8[2]; R8C9:6 (intersection); R8C8:7 (intersection); complete 15[2], then
34[5](C9), then 34[5](R10);
R9C7:8 (intersection); complete
25[4]; R3C10:1 (Outie); complete
8[3](R6); R7C10:4 (can’t be
<2>); complete 27[4], then 24[4], then 10[2];
28[7](R8) needs a <4> which can only go in R8C6 (if you try out the
effects of putting it in R8C5 on 23[3], 10[2] and 15[5]); therefore R8C6:4 and complete 10[2], then 23[3];
R10C5:2 (intersection); complete
23[4], then 15[5]; 28[7]:3,1;
complete 8[2]; 17[5]:7,2; 11[4](R7):3,1; 29[7](C2):2; complete 15[3] then 26[4] then 12[2] then 29[7];
15[4](R3) needs a total of 10, either <28> or <37>; if R3C8 is
<2> or <3> then it is impossible to fill R5C8, but it cannot be
<8> since R3C7 cannot be <2> (note the 38[6]); therefore R3C8:7; complete 15[4];
R5C7:6 (intersection)l R1C6:7 (if it were <9> then impossible
to fill R1C7); complete 16[2]; R1C7:5;
complete 13[3]; R2C7:8 (cannot be
<7> because of R2C5); complete 22[3]; 38[6](C7):9 (because R4C7 cannot be <9>),7; complete 15[2] then 19[3]; R3C5:7 (intersection); complete 22[4]; 3[2](R4):1,2; complete 11[3]; 16[5]:1,3
and finish off.
R1C2:6 (intersection);
R3C2:8 (intersection); complete
23[3] then 9[2] then 7[2] then 7[2](C3) then 9[3](R1); R1C6:1 (intersection); complete 3[2], then
4[2]; 16[2](C5):9,7; complete
14[2](R3); R2C4:4 (intersection);
complete 41[7]; 15[5](C4):3,2;
complete6[2] then 5[2]; 11[4](R5):5,2;
complete 5[2] then 11[2]; R7C2:2
(intersection); complete 3[2] and 5[2]; R9C2:4 (intersection); complete 6[2]; R10C2:6 (intersection); complete 9[3] then 3[2]; R8C1:6 (intersection); complete 12[4] then
11[3] then 41[7]; R3C7:5
(intersection); complete 6[2] then 7[2]; R5C7:6; complete 7[2]; R6C7:9
(intersection); complete 38[6]; 23[3]:6,8;
complete 13[3] then 7[2]; R10C10:2
(intersection); complete 3[2] then 6[2] then 6[3];
R2C9:4 (intersection); complete
13[2] then 7[3] (note the intersection with 41[7]); R8C4 is not <8> else there’d be a bad
knock-on effect, therefore 7;
complete 34[5] then 9[2] then 7[2] then 38[6] then 8[2] then 7[2];
R9C7:8 (if it were <9> there’d
be a problem); complete 15[2] and 17[2]; 41[7](R10) needs <8> and
<5>, therefore 13[2](C9) is <58>; R11C9 cannot be <5> else
clash in 31[5](C8), therefore R11C9:8;
complete 17[3] then 13[2] then 41[7];
16[2](R8):9,7; complete 31[5],
11[2], 11[4], 10[2] and 16[2]; R4C11 cannot be <9> because R1C11 cannot
be <2> or <3>, therefore 34[5](R4):9; complete 17[2] and 15[2]; 41[7]:6; complete 22[3] and 34[5] and finish off.
R9C12:5
(intersection); R12C11:6
(intersection), therefore R11C11 is <8> or <9>, therefore R11C12 is
<5> or <6>, therefore 6;
complete 14[2] and 23[3];
R13C12:4 (intersection); complete
28[4] (because there is only one way round for <7> and <8> in
12[2], 11[2] and 28[7];
R12C7:7 (intersection);
complete11[2] then 5[2] then 3[2](R11); 28[7]:1,2,4; complete 11[3] then 13[4]; R10C7:4 (intersection); R8C2:4
(intersection); complete 5[2] and 4[2]; 27[4](R9):9,7,8; complete 8[2] and 9[2]; R11C2:6; complete34[5] then 14[2] then 9[2]; R12C3:2 (see what happens if you try the other ‘possibilities’!);
complete 6[2] then 9[2], 8[2], 18[5], 13[2], 12[2], 21[3], 9[2](R9), 7[3](C7),
9[2], 11[2], 34[5] (R10), 16[2];
R9C11:8; complete 29[4]; R8C11:9 (Outie); complete 23[3] then 13[3]
then 7[2](R10); R8C5:4
(intersection); complete 5[2] then 11[3], 13[2]; R7C3:7 (intersection); complete 26[4] then 9[2], 5[2]; R6C3:2 (intersection); complete 10[3]l R6C1:8 (intersection); complete 19[3], then
11[3] and 6[2]; R5C2:6
(intersection, and it can’t be <3> because of 12[2]); complete 12[4] and
12[2]; R4C7:3 (intersection);
complete 9[3] then 8[2]; R4C6:6
(intersection);
15[4](R7) needs <23>, therefore R6C9:6
(intersection); complete 8[2] then 15[4], 15[2] and 16[3]; R5C11:3 (intersection); complete 12[2] then
12[2], 4[2](C13), 4[2](R6) and 10[4];
R4C8:8 (massive Outie in top right
corner); complete 15[3], 8[2] and 9[2]; R2C10:6 (intersection); complete 7[2], 6[2], 13[2]; R2C13:9 (intersection); complete 11[2] then
7[2] and 18[5]; 33[5](R4):7; R3C2:3 (intersection); complete 5[2];
7[3](C3) needs <1> and <4>; 38[6] needs <7> and <8>,
therefore R1C3:1 else 13[4](R1) is
not possible; complete 7[3]; R2C4:9
(only place that the <9> can go for 38[7]); complete 11[2]; R1C2:7; complete 38[6] then 13[4] and 10[2];
38[7] is missing a total of 7, therefore missing either <25> or
<16>; since <5> cannot go in any of those cells in 38[7] (try it!)
it is missing <25> and therefore R2C5:3 (intersection), and finish off.
R2C6:7
(intersection); complete 16[2]; R1C5:2
(intersection); complete 3[2]; R1C3:7
(intersection); complete 26[4]; R6C3:6
(intersection); R5C1:7
(intersection); complete 15[2] and 23[3]; R7C11:5 (intersection); complete 6[2]; R5C6:4 (intersection); R4C6:6
(intersection); R4C8:2
(intersection); complete 9[3] then 3[2] and 10[2];
29[4](R1): R1C9 and R1C10 are <57> Twin, therefore R1C8:9, and R1C11:8; complete 26[4](C8); R3C10:7
(intersection); complete 16[2] then 29[4](R1) and 8[2];
note R2C10 and R2C11 have a total of 7; R4C10 and R4C11 have a total of 10;
total for all four cells is 17; R2C11 and R4C11 together have a total of 7,
therefore R2C10 and R4C10 have a total of 10; therefore R5C10, R6C10 and R7C10
have a total of 6, and are therefore <123>; therefore R7C10:3 (intersection) and R6C10:2 and R5C10:1; complete 9[3] and 10[2]l R2C10:4 (intersection) and complete top right section;
38[6](R7):9,8l 41[7](C9):7; 38[6]:g98(Twin), g63 (Twin),
therefore R10C11:5; complete 6[2];
R7C2 is <12> and R8C2 is <12> making a Cell-Twin; therefore R9C2
cannot be <2> therefore R8C1 cannot be <4>, therefore is <1>
or <2>; therefore R10C1 and R11C1 make a <34> Cell-Twin; R10C1 is
<3> or <4>, therefore 26[4](R10) needs a <9> which must be in
R10C3:9; complete 11[2];
11[4](C4):2 (can only go in R8C4);
complete 3[2] then 5[2] and 11[3] and 34[5]; R6C7 canno be <3>, therefore
1(for 11[3]); complete 11[3] and
4[2], 8[2] and 38[6]; R10C8 is <6> or <3> therefore 3; so R10C6:3; complete 7[2];
17[5] needs a <5>, therefore R9C8:5;
therefore 41[7]:5 complete 6[2],
then 17[5], 11[2], 10[4], 41[7] and 13[2];
R10C4:5 (else 26[4] is unsolvable);
complete 11[4]; R10C1:4 (else 26[4]
unsolvable); complete 26[4]; R11C1:3;
complete 12[4]; 35[7](C2) missing a total of 10, therefore missing <37>,
therefore 35[7]:4; complete 6[2]
then 10[4] and 3[2](R8), 3[2](R7) and 35[7] and 22[6](R5);
22[6](R2):2; complete 5[2]; 22[6]:3; complete 20[5] then 6[2] and 22[6] and
finish off.
How the code works:
Choose a Puzzle to look at:
Puzzle 1, Puzzle 2, Puzzle 3, Puzzle 4, Puzzle 5, Puzzle 6,
Puzzle 7, Puzzle 8,
Puzzle 9, Puzzle 10, Puzzle 11, Puzzle 12, Puzzle 13, Puzzle 14, Puzzle 15
KILLER PUZZLES
N6: Double Innie (R4C7 and R5C7) has total of 16, therefore
contains <97>, therefore R4C7=7;
R5C7=9; fill in 8[2]. Note g8.
N4: Double Innie (R5C3 and R6C3) has total of 15, therefore
either <96> or <87>. R5C3
cannot be <9> therefore R6C3 cannot be <6>. R6C3 cannot be
<9> (it’s a 9[2]) or <8> (g8 in N6), therefore it’s a 7 and R5C3 is 8. Complete 9[2].
N4: g1 and therefore 5[2] = <41>
N7: R9C3=9
(Innie), complete 16[2].
N3: R1C7=6
(Innie), complete 11[2].
C6: 7 (in R5C6);
C6: 14[2]=<86>
N7: 11[2]=<56>, because of C3; therefore N7:
8[2]=<71>, therefore 10[2]=<82> and 7[2]=<34>
N4: g1, g4, therefore R7: 8[2]: 1, 7.
R1C1= <3> or <9> (by elimination) therefore,
because of 22[3], 9.
R1: 10[4]=<1234>, therefore R1C8 and R1C9=<78>,
therefore R2C8 and R2C9=<14> or <23> but not <5>.
N3: 10[2] can’t contain a <5>, therefore
9[2]=<54>, therefore R1C8 is not <7>, so 8., so R1C9 is 7, complete
11[2], 9[2] (C9); g19 in 10[2];
N6: 6[2]=<15>, therefore 1,5; complete 10[2] (C3); 8[2]=<26>, therefore
15[3]=<348>, therefore R6C8=4
N9: 7,4 (in R7C9
because it cannot be in 21[3] with the <7>), 5 (because cannot be in R7C8), g6, therefore 21[3]=<768>;
complete 15[3] (R6)
R7C8=9 (by
elimination); so 15[3]: g3;
N8: g4 (because not in 14[2] or R7 or 11[4]), therefore 7[2]
(R2): 4,3; so N9: 3,1, so 11[4] (C9): 1,2; complete 10[2] (C1); so N9: 8,6; 14[3] (C8)= <149>, therefore
R8C4= <5> or <6>, but cannot be <5>, therefore 6; complete 11[2] (N7) and 14[2] (N8);
so R7: 5,3; C6: 4,9; N8:9; 23[4] (R6/7):6; R6:5,1; N4: complete 11[2] and 5[2]; and it all finishes easily from
there...
N9: 4[2]=<13>, so 3[2] (C5): 1,2; R8C3=4; so 7[2]
(N7): g12;
R4C8: 2 (Outie);
R6C2: 3 (Outie); Double Innie in
rows 7, 8 and 9: R7C8+R7C9= Total of 6; NOT <15>, therefore <24>,
therefore R7C8:4; R7C9:2; complete 6[2];
Double Innie in N9: R7C7+R8C7= total of 17, therefore
<98>; therefore 18[3] = <567>. Therefore 13[2] (R9) = <49>,
therefore 14[2] (N8)=<86>; Easy X-Wings on <8> in N8 and N9,
therefore R9C1=8 and R9C6=3. So 19[3]=<379>, therefore
R8C7=9; R7C7=8; complete 13[2], 19[3], 14[2] (N8);
N7:3; so
24[4]=<3795>, therefore R8C2=5;
therefore R8C1=6;
N2: 15[2]=<96>, therefore N5: 17[2]:9,8; so complete N8: 13[2], and C5:6;
N2: 13[2]=<58>, therefore N5: 12[2]:5,7; also C5:3; complete 7[2]
R2C4:7, so
complete 11[3]; C4: g34; therefore 2;
complete 10[2] then 13[2] (C5);
C2:8; C1: 9[2]
cannot be <81> or <72> or <63> therefore <54>: 5,4; N3:4,2; complete 13[3]; N1:7;
R7:7,9; N6:8; N3:8; R1:5,6,9; N4:7,4; so 16[3]:1;
complete 3[2] (R2); N7:1,2; N5:4,3,1,2; R6:9,6; N4:2,6; 15[3] (C9)
needs an extra 7 in total = <61>:1,6;
so N2:6,9; N3:9,3,5; N9:1,3,6; N6:3,1,7,8,5; N9:5,7.
N2: Double Outie, Total 2, therefore both outies = 1 (R3C6 and R4C9); complete 8[2], and
note in 6[3]: g23 (Twin);
N7: Double Outie, total 15, therefore either <96> or
<87>, but since R6C1 is part of 7[2], it must be a 6, therefore R7C4=9, and
complete the 7[2];
28[7] (in N6 and N9)= <1234567>, therefore R8C8=1; complete 3[2] (C5); N3: 1; N5:1;
15[2] (R6)=<87>, therefore 15[2] (R4)=<69>,
therefore 9,6 (notice the 16[2] in
C5);
8[2] (R4)= <35>; 14[2] (C6) cannot be <95>,
therefore <86>, which means 15[2] (R6): 8,7; so, N5:8,2 in
11[3]; 7[2] is <34> therefore R4C4=5,
and complete 8[2];
7[2] (C6) must be <34> so C6: 5; N2:5; C5:g34, so R7C5
must be <3> or <4> therefore 12[3] must be <345>, therefore
R7C7:4; complete 12[3], and N5: (the
7[2]):3,4; 14[2] (C7)= <59> so
fill in C7:6,7; 28[7] (in N6 and
N9): 3,5,4,2; N6: 2,9;
R6: g19 (both part of 37[7]), therefore N1:1 (in R2C3), so complete 7[2]; R6:1,9;
13[3] (R3) needs a total of 8 more, therefore <26>:6,2; so C7:2,3;
N3: g45 (making a Twin), 6, g89 (making a Twin); C8:7;
N8:7,4; 19[3] (C9):8; R7:6,7; so complete 20[3]; R3:8,5;
so N8:8,6; N9:6,2; N2:2,8;
15[3] (C1) involves a <4>- either in R3 or R5,
therefore the other two cells have a total of 11; R4C1= <2> or <8>,
therefore <29> or <38>; <38> is not possible, so it must be
<29>, therefore R4C1=2; R3C1=9; so 15[3]:4; R5:7; R4:8; N3:9,8; R3:4; so 37[7]:3; and finishes off.
R1C9 (Innie in R1): 9;
R9C1 (Innie in R9):7; All cages in
rows 6, 7, 8 and 9 total 173, therefore there’s a Double Innie (R6C8 and R6C9)
with a total of 7; R6: 3[2] = <12>, therefore the Double Innie (R6C8 and
R6C9)= <34>; 4[2] (C9) = <13> therefore R6C9:4; R6C8:3; 9[2] (C9):5;
N4: Double Innie has a total of 6; R6C1 must be 5, so R4C3=1, and complete 8[2]; R6C4=6,
complete 7[2]; 4[2] (in C9): 1,3;
N7:1; N1:1; complete 8[2]; 15[2] (in R6) cannot be <96> so must be
<78>, therefore 7,8; R6:9; complete 11[2];
N4:12[2] is <93>; C1: 12[2] (in N1) is <84>; C1:2,6; N7: 9[2] is <45>; N7: 8,3,9; 14[3] (in R9) is 5; N4: g46; C2: g25, therefore R2C3=6; C3:9,3; so complete 12[2] in R3;
R9: g3; you can’t put the <2> in 13[3] (try if you
like!) so it must go in 11[2], making 11[2] <29>, therefore 9,2;
13[2] (R7) = <58>, therefore 5,8; therefore N7: 5,4;
12[3] (in R1) = <345>, therefore 5,4; N5: 5; therefore
complete 6[2];
12[3] (in R5) = <138>, therefore 8; N5: 2; R6:1; and it comes out from here.
R4C1 (Innie from rows 1, 2, 3 and 4): 5; complete 6[2]; R5C9 (Innie from R5): 5; complete 9[2];
14[2] (C1) = <86>, 16[2] (C1) = <97>, therefore
the three cells in C1 of N1 are <234>, therefore 7[2] cannot be
<34>, therefore it must be <25>, therefore 2,5;
N7: 12[2] is not <57> therefore N7:5, so 12[2] must be <84>; so
complete 14[2] (C1), so 10[2] (R7) must be <82>, so 12[2] (N7):8,4; 6[2] (N4):4,2; N1:8 (since not in
10[2]); thus complete 12[2]; C1:3;
complete 10[2]; N8: 15[2] is <96>, so 16[2] (C1): 9,7;
11[2] (R9) is <38> therefore N9:6; R7C7 is <1> or <5> therefore 12[3] is <156>
therefore 6[2] (N9) is <24>, so 4,2;
N7:2; N3:g4, so 6[2] = <24>,
therefore: 2,4;
N8: 16[3] must include a <7>, therefore not a
<1> or <3>, therefore R8C6 = 5
and 12[3] (N9): 5,1; N7:1,3; 11[4] (in R8 and R9): 1; N8:3;
19[5] (mainly in R6) = <12367> by default, therefore
R6C7=7; so 15[2] (N6) is <96>
therefore 6,9; C7: 8,3; N2:3; R2: 8 (because not in
10[2]), so N9:8,3; N6:3; complete 9[2] (C9); therefore 10[2]
(C9) is <19>, therefore place the 9
(because of the 16[2] in N2) and the 1;
C9: 7; N9:9;
N1:9,1,6; N3C8:5; 9[2] (R3) must be <18>
therefore R3C4:4; N3:7 complete 20[4] (Rows 1 and 2); R5:3; N5:4; complete 18[4]; N6:2;
R6:5 (because not in 19[5]),3,9; N4:6,7; C6:9; and comes
out.
R3C4 (Innie in N2):9;
R7C6 (Innie in N8):4; R6C4 (Innie on
C4):2; R4C6 (Innie on C6):7;
5[2] (C4) = <14>; N1 (Innie):6; N9 (Innie):8; N1:
<2> cannot go in 20[3], therefore in 15[3], therefore [7] must go in
20[3];
N9: <9> must go in 21[3], <7> must go in 21[3]
because <731> in 11[3] would make the 5[2] impossible to complete,
therefore 21[3] is <975>
C9: g7 goes in either R5C9 or R6C9 therefore 21[4] (C7)
needs an extra 9 in total, which must be <63> because it cannot be
<45>, <81> or <72>, therefore 7[2] (C7) must be <52>
11[3] is either <632> or <641>, but R9C7 cannot
be <6> or <3> or <2>, therefore it must be <641> and
5[2] (N9) is therefore <32>
N7: <6> is not in 9[2] (because of the 4[2] in C1,
which must be <13>), and the <6> cannot be in R9 or in C3,
therefore there’s a ghost <6> in R7C2 or R8C2;
R7C3 is either <1> or <2> (not a <4> or a
<3> because of the 5[2]), therefore R6C3 is either <3> or
<4>; R6C1, R6C2, R6C3 add up to a total of 14 (triple Outie from N7)
therefore R6C1 and R6C2 add up to either T10 or T11, depending on R6C3. If T11
then the 23[4] in nonets 4 and 7 splits into 11[2] in nonet 4 and 12[2] in
nonet 7; but there’s already a g6 in what would be the 12[2] which makes the
12[2] impossible; therefore the 23[4] must be a 10[2] in nonet 4, and a 13[2]
in nonet 7; and R6C3 must be a 4;
complete 5[2];
27[4] (in columns 3 and 4) requires an extra total of 12,
therefore <75>, so place the 7,5;
and N6:7; so 7[2] (in C7):5,2; N6: g5 in 24[4] therefore 24[4] is
<8754>, therefore there’s a g19 in R4N6; 5[2] (in C4): 1,4; N4: g1, therefore also g9 to make
the total of 10 required; 23[4] must be <1967> (note the g7); N1:7; N7:4; complete 9[2]; N9:5;
N6:5,4,8;
20[4] (in N3 and N6) is <9173>, therefore g73; 14[2]
(N3) is <86> therefore 6,8;
N3:2; N9:6; C7:7; N9:9; N7:7,6; R6:1,9; N9:g14
(making a Twin).
N3: 16[4] needs <149>, but to avoid an Unsolvable
Rectangle with the g14 in N9 the <9> needs to be in either of R1C7 or
R1C9; however that means that C7 can only have its 9 in R1C7, therefore place it there; 15[3] (N1) is therefore
<924> and 20[3] is <758>, therefore: 5,8; 13[2] (C4) cannot be <58> therefore must be <76>,
so 7,6; C4: 5,8,3; N8:7,6; complete
15[3], then 5[2] (C9); N8:1,9; and
it all finishes from here.
R5C5 (Innie on R5):8;
so 22[5] must be <985>; C5: 6[3] must be <123>, therefore C5: 17[3]
is <467> therefore 14[2] (N2) is <95> therefore N8:5 and complete 11[2];
N2: R3C4:8 so
complete 10[2];
N8 (C4) is <467>; R8C4 and R9C4 must include the
<7> else 16[3] becomes impossible to solve therefore R7C4 is either
<4> or <6> and is part of the 15[3]; C4 (R5C4 and R6C4) = g13,
therefore R6C4 is either <1> or <3> and is also part of 15[3]; Look
at how those two results for 15[3] affect the third cell in it (R6C3): here are
possible combinations:
R7C4 4 4 6 6
R6C4 1 3 1 3
R6C3 10! 8 8 6
In the first case 10 is impossible, in the fourth case 6
repeats the <6> in R7C4, therefore R6C3 can only be an 8, so place
that in the grid.
Therefore 20[3] (R6) is <974>, so complete 22[3] (C5)
R1: the 16[3] plus the 15[2] give a total of 31[5], so
R1C4+R1C5+R1C6+R1C7 have a total of 14, therefore R1C4 cannot be <9> so
it must be 5 and complete 14[2];
R3C6 is either <1> or <2> or <3>; R4C6 is
either <4> or <7>; both are part of the 16[3]. Here are the
possible combinations for those results and how they affect R4C7:
R3C6 1 1 2 2 3 3
R4C6 4 7 4 7 4 7
R4C7 x 8 x 7x 9x 6
The results are either impossible (where marked with an x).
Only possible results leave R4C6 as a 7
(so place that in), making R5C6:4;
15[3] (R4) must now be <456> therefore R4C7 cannot be <6> and must
be 8; so complete the 16[3];
therefore R1C7 is 1;
R1C5+R1C6 have a total of 8, therefore <62>, therefore
R1C5:6, R1C6:2; and complete 6[3]; 15[2] (R1) is <78>; N3:9; 16[3] (N1) is <943>, so 14[2]
is <86>: 8,6; R4C8 is
<1> or <3> therefore N3:3
and complete 8[2]; 7[2] (C8) cannot be <61> so must be <34>: 4,3; N2:4,7; R4:1; N9:1 (since not in 26[4]); so 6[2] (C9)
must be <24>, so N3:2,6; N6:4; 20[3] (R6):9,7; 15[2] (N3):7,8; C7:2; N6:6,5; C8:5,8; 13[3] (R9)
must now be <931> therefore 9,3;
17[2] (C6): 9,8; N7:8,5; R9:7,6,4,2; complete 16[3] (in N8 and N7); C9:4,2; N8:6; so complete
15[3]; C4:3; R6:3,2; complete 11[2]; N7:2;
so 14[4]:1; N8:1,3; N7:3,7; N9: 6,7; R3:5,2; R2:7,1; R5:7,9,1; R1:3,4,9; R4:5,4,6.
N7: 16[2] is <97>; Double Innie (Total 8); 3[2] (C1): R7C1
is <1> or <2>, therefore R9C3 is <7> or <6>, but cannot
be <7> because of 16[2], therefore 6
and R7C1=2; complete 3[2] (C1) and
3[2] (C4)
N7: 11[2] must be <83> and 10[3] is <145>
N4: Innie:4, so
rest of 16[3] has a total of 12, therefore C4: Innie:5; 28[4] is <9865>: 9,8;
17[2] (C6) is <98> therefore N5: g98, but <8> cannot be in 16[3]
(else it would be <844>) so R4C4=8;
16[3]=<943> and C4:g467; R1C5 and R2C5 have a total of 5, therefore g32.
14[2] (N9) is <86>: 8,6; Double Innie (in N3) has a total of 15, therefore <96>
or <87>; but R3C9 = <6> or <7> (because of the 8[2]) but
cannot be <6> therefore it’s 7
and R1C7=8; complete 8[2] (C9);
3[2]: 1,2; 7[2] (C9):4,3; C9:9,5; complete 11[2]; 14[4] (in R1 and R2): 1; N6 now has an Innie:6;
so 14[2] is <95>; N6:2 and
g78;
N9: 8[2] is <35> therefore 3,5; R9:3,4; N9:9,7; N6:9,5; 14[4] (in rows 1 and 2):2,3;
R1: 10[2] must be <46>; R1:5,7;
14[2] (C1) = <86> therefore R4C1 = 6;
14[2]:8; N6:8,7; 14[2] (R6) is <95>, therefore: 5,9; N7:5,4,1; therefore
8[2] (C2) is <17>, so 7,1; N3:1,3,6,4; N2:6,4,5; N1:2,8,3,9; N2:8,9; N7:9,7,8,3; N1:4,6; N4:7,3,2; R4:2,4; N8:4,7,6; N5 finish
off.
N1: Innie: 6;
complete 9[2]; N1: 4[2] is <13> therefore 7[2] is <25>; C1: 11[2]
is <74>;
N9: Innie: 3;
complete 9[2];
R5 (mainly!): 21[5]+9[3]+23[5] makes 53[13], which means
that R4C2+R4C8+R6C2+R6C8 have a total of 8!They cannot all be <2>s, and
the <3> in N4 means that they can’t be two 1s and two 3s; so the four
numbers are either <1223> or <1124> with duplicate numbers
obviously not being in the same row or column or nonet. Note 5[2] in row 6: it
is either <23> or <14>, therefore R6C2 and R6C8 cannot be
<12> or <13> or <24> since that would make the 5[2]
impossible, so must be either <14> or <23>; that means that the numbers
<1234> are already spoken for on Row 6 (in columns 2, 6, 7 and 8);
therefore 11[2] (C1): 4,7; and 16[2]
(C5):7,9; so R6C4 is <5> or
<8>, therefore 13[2] is <58>
N8: Double Innie has a total of 10, therefore R7C4 cannot be
<5> so it must be 8; R9C4=2; 13[2]:5 and R6C3=8; 18[3] (R9)
= <972>; N7:3;
Columns 1, 2 and 3 have an Outie in R4C4, which must be 7; so complete 13[2];
C1 needs a <1>, <8> and <9>; <8> and
<9> cannot both be part of 18[4], therefore R5C1=9 (since it cannot be <8>) and g18 in N7; complete 18[4] and
note g52 in N7.
N5: g6 and so 9[3] = <621>;
Nonets 2, 5 and 8 have a Double Innie: R1C6+R6C6 have a
total of 5, therefore there’s a double innie in C6, where R2C6+R5C6 must have a
total of 15, which is either <96> or <87>; since R5C6 can only be a
6, you place that in and make R2C6:9;
9[3] (N5):2,1;
C4:9; N9:9; 18[3] (N8): must be <963> so 3,6; thus 10[3] is <145>; so 5[2] (R6) is not <14> and
must be <23>; so R6C6:3; and
complete 5[2]; N4:2,1; R6:4; and R4C8:1 (to make up the total of 8 required by that Row 5, four-cell
Outie noted near the start!); N4:5,4;
N5:8,4; complete 9[2] and 15[2]; N2:2; complete 7[2] (C1); 8[2] (N9):7,1; R7:2,5; complete 15[3] (C3); N9: 6[2] must be <24>, so 2,4; N8: 4,1; N7:1,8; N9:8; N3:2; 13[2] (C9) must be <85> so 8,5; N6:9,8,3,7; C9:9,1; so 19[4]:7; C8:5,6; complete
11[3] (R1); N9:6; N3:5,4; N2:1,8,6,3,4; and finish.
Rows 8 and 9: Double Outie, Total 8; R7C2 must be 7 and therefore R7C3 is 1; complete 3[2]; 17[4] (R9) must be
<4562>, therefore R9C4=2; so
R7C7=2 and R7C1=3;
Rows 1 and 2: Double Outie, Total 12. Therefore row 2 of N3
has a Total of 6, since the 8[2] and the 10[3] must add up to 18; total of 6
must be <123>, so place the 2 because
of the 4[2] in C8; now whichever way round the <1> and the <3> go,
the Total of 12 must be made up of <57>, so we have a g57 in R3.
Therefore 6[2] (R3) is <24>; R3C9 must therefore be 6 (since it cannot be any of 1,2,3,4 or 5), so 23[3] (in rows 3 and
4):6, and g98.
R3:3,1 so
complete 10[2] (C1), 8[2] (C2), 24[3] (N7).
Innie in N4:3;
Row 1 in N3 is <984> therefore R1C6=1;
N1:<2> is either in R1C1 or R1C2 but cannot be in R1C1
because of the 11[2] so R1C2:2 and
complete 10[2]; N4:8; 23[3] (in rows
3 and 4): 8,9; N1: g6, so 11[2] is
<65>, therefore N7:6,5,4;
20[4] (in Nonets 1 and 2) must therefore be <7463> therefore g63 in N2:3,6; N1:6,5; N2:5 and g97;
therefore N1:7,4;
C4: R7C4+R8C4 have a total of 12, therefore <57>, so 7,5; complete 14[2]; N2:9,7; 12[3] (N8) = <714>,
therefore: 4,1; 4[2] (N9):1,3; R8:3,6,5; complete 12[2]; R9:9,8;
N8:6; complete 15[3]; C1:7,2; N5:7; N2:4,2; C6:2,5; so 22[4]:8; R4C8=4; N9:4,8; complete 14[2], then 13[2]; N3:9,4,8,7,5,3,1; N6:1,5,2,3; N4:1,4 and
finish off.
Double Innie in N4: Total of 17, therefore g98; Triple Outie
in Rows 7, 8 and 9: total of 21, therefore R6C4=4;
Innie in N5:7;
Double Innie (N6): total of 17, therefore g98, therefore
13[2]=<76>; N5: g98; N3: 17[2]:g<98>, so Cell-Twins (R2C8 and R4C8)
making R7C8 a 6.
19[3] (in columns1 and 2)= <973> or <874>,
giving us a g7;
Double Outie in N7: R7C4+R9C4 make a total of 8; Double
Outie in N1: R1C4+R2C4 have a total of 11; therefore in C4, R3C4 and R8C4 must
add up to 11 as well. This means that in C4 there are three different pairs of
cells that add up to 11, and between them they must share the pairs:
<29>, <38> and <56>; therefore the T8 (R7C4 and R9C4) must be
<17>, and therefore, because of the g7 in row 7 of N7, we can place the 7 and the 1.
Thus, 27[5] (in columns 3 and 4) must be either <14985>
or <14976> but because R7C3 cannot be any of 9, 7 or 6, it must be
<14985>, so R7C3 must be 5;
R1C3+R2C3= Total of 6, therefore <24>; C3: Cell-Twin:
R6C3 and R8C3 share <89>, therefore 9[2] (C3)=<36>; C3:7,1; N1: 13[3]=<157>; N4: g1, so
N1:1,5; N4: 9[2]=<54>;
10[3]=<172> and N7:7.
If R4C8=<8> then 19[3] becomes unsolvable; therefore
R4C8=9 and R4C7=8; 19[3]=<946>, so: 6,4;
R3:g8; R1C4+R2C4=T11 but not <29> or <38> or <47> so must be
<56>; N2:g7, so 21[5]=<87123>, allowing us N2:4,9; therefore 29[5] (in columns 6 and 7) is <98723>; row 1
of N3 is g157, therefore N2:5,6;
R8C7+R9C7=T5, therefore <14>: 1,4; making 19[4] (in columns 6 and 7) <8641>; therefore 8[2]
(C6) is <53>; therefore C6:1,2;
N9:8,9; N3:8,9; C7:5; R4C4:2; complete 11[2]; N5:8; N2:8; C4:3; N9:g3;N7:3; complete 19[3]; N4:8; complete 27[5]; N8:9,5,4; N9:5,3; 5[2] (C8) is <23>, therefore N9:2,7; N3:7,1; N2:7; N1:4,2; N2:2,3; N3:2,3; R1: 9,8; R2:3,6; N7:g2,
therefore 8,2,4,6; N8:6,8; R4C1=1; N5:1,6; N4:6,3; N6:6,7; N4:7,2; N6:2,3; and finish off.
N1: Double Innie has total of 17, so <98>; 16[2] (C1)
= <97>;
Columns 8 and 9 have a Double Innie with a total of 5.
Notice also the 6[2] in C8. If the 6[2] were <42> then the T5 would be
impossible to fill, therefore the 6[2] is <51> and the T5 (Double Innie)
is <32>.
N9: 22[4] = <9832>; 11[3] must be <146>; N9:5,7;
N6: g7 (in C7); N3: g7 in 30[5]; g98 in 30[5], therefore
30[5] is either <98751> or <98742> but R3C8 is either <1> or
<5>, therefore 30[5] is <98742>.
C4: g4, therefore one of <9> or <2> has to be in
the column 9 part of 30[5], and so 11[2] (C9) cannot be <92>; it cannot
be <65> or <74> either, so must be <83>; therefore 27[4] (in
columns 8 and 9) must be <9765>, so we can note 5 and g69; 6[2] (C8):5,1;
C8: g48.
Columns 1 and 2 have a Double Outie with a total of 7, so
19[4] splits into 12[2] and 7[2]; the 12[2] part cannot be <93> or
<84> (because of the g89) therefore it is <75>
Now for the tricky bit... label the following cells, to make
it easier to talk about!
Cell a = R3C3
Cell b = R3C4
Cell c = R4C3
Cell d = R4C4 (notice
that a+b+c+d make up the 22[4])
Cell e = R3C7
Cell f = R6C7
Cell g = R6C6 sorry,
I know that ‘g’ is also being used for “ghost” but it shouldn’t cause a problem
Cell h = R7C6
Double Innie in N5, involving cells d and g has a total of
12.
Double Innie in rows 1, 2, and 3 involve cells b and e and
has a total of 5, so either b=4 and e=1 or b=2 and e=3;
Therefore 12[3] (C7) is either <147> or <327>
but either way involves a <7>
So cell f= <2> or <4>: if e=1, f=2 and if e=3,
f=4. Notice that this makes f the ‘opposite’ value to b: if f is 4, b is 2 and
if f is 2, b is 4;
Now, if cell a is <8>, then you might think that the
22[4] could be any of <8761>, <8752>, <8743>, <8653>,
<8941> or <8932>. However, none of a, b, c or d can be a <1>
so we can already narrow down the options. Also none of them can be a <7>
(since if d is a 7 then g is a 5 which is impossible). So we only need to
consider <8653> and <8932>.
<8653> is impossible because cell b has to be either a
<2> or a <4>.
<8932>: If cell a is <8>, and the 22[4] contains
a <9> as well, then it would have to be in cell d, which would make cell
g be <3>. Cell b would be the <2>, making cell e <3> and cell
f <4>; cells f and g are part of the 19[4] and those values would make
cell h be a <7> which is impossible. Therefore cell a cannot be
<8>, so it must be 9. That
makes R3C2:8.
Since the 22[4] does not include a <7> the options for
it are now: <9652> and <9643>, which either way involves a
<6>. Since cell b is not a <6> and cell d cannot be a
<6>(since that would make cell g a <6> as well!), cell c must be 6.
This means that the 19[4] in N1 is not <7561> so it
must be <7543>, which means that there is a g8 in the 15[3] in N4, which
means the 15[3] is <852> allowing us to place the 5,8,2.
Cell d is either a <3> or a <5> according to the
two options. If d=<3>, then g=<9>, but there’s an Easy X-Wings on
<9> in the middle stack: there’s a g9 in C1 matching a g9 in C8, which
means that cell g cannot be a <9>. So cell d is not a <3> and must
be a 5, and b=2, e=3, f=4, g=7, and h=3.
Now it all falls out: 16[2] (C1):7,9; N6:7,2,9,6; 12[2] (C6) cannot be <93> therefore R4C5=9 and 12[3] (C5) is <921>: 2,1; N5: 9[2]=<63>, so: 6,3; R6:1; complete 10[2]; R3C6=<16>; R2C7=<16>, therefore 16[3] (in columns 6 and 7) is <169> and R2C6 is 9; R3:4,7; N3:9,2; N1:2; N4:2; R8C3 cannot be <7> because of the 18[3], therefore 1; thus C3:7; 18[3]:9,8; N8:7; 16[3] (R9):5,4; N8:1,2,6; R7:1,4; N4:4,3; and it all comes out from here...
17[2] (C4) = <98>, therefore R2C4 = <5> or <6> and R2C5 corresponds by being either <9> or <8>. Double Outie for Columns 1, 2, 3 and 4: R2C5+R9C5=Total of 10. Since R2C5 = <9> or <8>, R9C5 = <1> or <2>; however, can’t be <1> because row 9 has a triple Innie (R9C1, R9C4 and R9C5) which have to add up to 16, and two of those cells (R9C4 and R9C5) are part of a 15[5] which must contain <12345>, which means that if <1> were in R9C5 then the maximum that those two cells (R9C4 and R9C5) could have is 6, which isn’t enough for the triple Innie total of 16!
Therefore R9C5 = 2,
R2C5 = 8, and complete the 14[2]
(R2)
R7C4 = <8> or <9>; 14[2] in N8 involves either
<8> or <9>, giving us a fuzzy cell-twin; therefore R7C6 = 7 and 24[3] has a g89. 11[4] (C7) =
<1235>, therefore C7 needs a <4>, <6> and <7>. 14[2]
(R1C7) cannot be either a <4> or a <7>, therefore 6, and complete the 14[2].
C5: R1C5+R8C5 make a total of 11; cannot be <92> or
<83>; cannot be <74> since R8C5 cannot be <4> or <7>,
therefore <56>; R1C5 is not <6>, therefore 5. R8C5 is 6, complete
14[2], 24[3], 17[2] (C4).
10[2] (C4) = <73>; N2:7; C5:3 and g914;
7[2](N5) = <52>, giving us a g5 in N8; N5:6; complete 7[2]; R9C6=4;
N8:1 (not in R9 for reasons given at
the top), 5; N7: 14[2] is <86>;
N7:9; 15[5]: g34; N7:7; complete 18[3]; N7: 5 (cannot be in 10[2]), 1; complete 10[2].
R8: g25, therefore R7C8=6;
R7:4; complete 11[2]; C7: 4,7; N3:7; R1: 14[4] = <7421>, therefore N3:1 and therefore R1C9 is not <3> so is 9; complete the 15[3]; N3:2
(not in 12[2]), 3; complete 12[2];
R8:5,2; N9:1,3; C9:6,8; N2:9,3; N1:9; R2C3 cannot be either <5> or <3> because of the
15[3], so must be 2; complete 15[3];
N2:2; R2: g53; complete 16[3]; R3: 6; complete 11[2]; R2:5,3; C1:8; N7:8,6; N4:6; so 15[4] (C3) has g13; N7: 3,4; C1:1,4; R1: 4,7; C8:4,2; C6:2,5 and finish it off.
N9: Double Innie has a total of 14, which is <86> or
<95>; if 15[2] = <96> then that T14 would be impossible to
complete; therefore 15[2] is <87> and T14 is <95>; 6[2] = <42>
and 10[3] is <631>; therefore 7[2] (R9) = <25>.
Double Innie for Columns 6, 7, 8 and 9 has a total of 17,
which = <98>
Double Innie for
columns 1, 2, 3 and 4 has a total of 15, which is <96> or <87>
R2C4+R3C4 = Total of 8, which is either <71> or <62>.
If R9C4 is <5> then R4C4+R5C4 also
has a total of 8, which would be the other one of <71> or <62>; but
that would use up both the <6> and the <7> for C4, making the
Double innie with total of 15 impossible to complete. Therefore R9C4 does not
equal <5> and must be 2;
complete 7[2].
N8: <7> is part of 27[4], as is <9> and
<8>, therefore 27[4] = <9873>; the <3> cannot be in R9, (due
to g3 in N9) therefore in R8. But not in R8C4 since that is part of the double
innie with a total of 15, therefore R8C5=3;
N7 has a double innie with a total of 5, which means that N4
has a double outie (R4C4 and R5C4) which have a total of 11. Since that cannot
be <92> and the <9> for C4 cannot go in the T8 or the 9[2] then it
must go in the T15 Double Innie, which is therefore made up of <96>.
<6> is not part of the 27[4] in N8, therefore R1C4 must be 6 and R8C4 must be 9. Which means we can place the 9 for column 6, the 8 for
column 6 and therefore in 27[4] the 3 and
7; R9: g49; N7: g8, therefore 27[4]
(N7)= <9846>, note g6; Double Innie in N7 has a total of 5, which cannot
be <41> so must be <32>; therefore 8[2] is <71>: place 7,1; N9: 9,5; R8:1; 21[4]:6; R7: g45; therefore 9[2] (C4) =
<45>, which means that R4C4 and R5C4 must = <83>, therefore fill: 8,3; complete 7[2]; R2C4 and R3C4 =
<17>; 28[4] (N2): g58; N2: 3
(by elimination of possible places it could go); complete 10[2]; N2: 2,4; complete 11[2]; C6:5,2; complete 10[2]; C4:5,4; N3 has a double innie now with a
total of 13, which cannot be <94> or <76> therefore must be
<85>; 13[2] (R3) = <76>; N2: 7,1;
N3: 5[2]:3,2; R3:9; complete 20[4]; 6[2] (R2) =
<15>; N2: 5,8; N8:4; N6: g4 (not in the 14[2]), therefore
21[5]=<85431>; 14[2] (C9) = <95>; N3:5,8; N9:8,7; N6:7,2,6; C9:3; N9:2,4,6,1; C7:4; N6:3,1; C8:9,4; N3:6,1; C3:6,2,8; 21[4] (R4)= <2568>, therefore N6:5,9; C5:6,9,1; R6: 3 (because cannot go again in 23[5])
and then it all finishes off.
N4: 16[2] = <97>
Let’s get the big bit out of the way first...
Cell a = R6C1
Cell b = R6C2
Cell c = R6C3
Cell d = R7C1
Cell e = R7C2
Cell f = R7C3
Triple Innie (from N1 and N4) tells us that cells a+b+c have
a total of 12, which leaves d+e+f with a total of 9.
Cell e cannot be <1> or <3> (because cell b is part of the 10[2] and cannot be either <7> or <9>) and cell e cannot be <5> because it is part of the 10[2]. The 5[2], which involves cells a and d, is either <14> or <23>, and the 6[2], which involves cells c and f, is either <24> or <15>.
Try the options:
If d=1, a=4; then, if
e=2, f=6 (which is too big for the 6[2])
or if
e=4, f=4 (which is the same)
or if
e=6, f=2 and c=4 (which is the same as a)
or if
e=7, f=1 (same as d)
Therefore d cannot be <1>
If d=4, a=1; then, if
e=2, f=3 (impossible in a 6[2])
or if
e=4, it’s the same as d!
or if
e=6, then it’s too big for the T9 that d+e+f make.
So d cannot be 4, therefore the 5[2] is <23>. Allowing
us to solve the 3[2] in row 1: 2,1;
now cell e cannot be <2> either.
If d=2 and a=3, then: if
e=4, b=6, c=3 (impossible on 6[2])
or if
e=6, b=4, c=5, f=1. This is fine.
If d=3, a=2, then: if e=4, b=6, c=4, f=2. This is also fine.
No other possibilities for cell e.
Notice that in the only two combinations that work involve the 10[2] (cells ba dn e) being <64>.
20[3] (C1) cannot be <983> (because of the 5[2]) or <974> (because of the 16[2]), but can be either <965> or <875>; either way, the <9> or the <7> will have to be in R3C1 (because of the 16[2]) and so there is a g5 in N4.
If <965> then b=4, e=6, d=2, f=1, but then c=5 which is now impossible. Therefore it must be <875>; so R3C1 is 7, and 20[3] gives us a g58 (making a twin); now cell c cannot be <5>, so must be 4, making f=2, d=3, a=2, e= 4 and b=6.
Now N1:4;
9[2]=<63>; C1: g69; N7: double innie with a total of 13, which is not
<94> or <67> therefore it’s <85>, giving N7 g71; N4:7,9,1,3;
N1:g58,9;
R8C3 is either <5> or <8>, therefore 22[3] is
<985>, giving a ghost 9 in N8; R9:9;
N7:6;
More naming of cells...
Cell p = R3C7
Cell q = R3C8
Cell r = R3C9
Cell s = R4C7
Cell t = R4C8 on
the board these cells are arranged: p
q r
Cell u = R4C9 s t u
Cell q is either <5>, <6> or <8> (because
of 14[2]); p cannot be either <1> or <2>;
If p=3 and q=5 then r=7 (not allowed); if p=3 and q=6, then
r=6 (can’t both be 6); if p=3 and q=8 then r=4 (not possible in an 8[2]);
therefore p cannot be <3> either.
If p=5, q cannot be 5; so if q=6, r=4 (impossible on 8[2]);
if q=8, then N1 (R3C2) would be unsolvable. So p cannot be 5.
Cell p cannot be <6> (because of 12[2])
Therefore p can only be <4> or <8>, and the
12[2] is <48>.
Possibilities:
If p=4, and q=5, then r=6, s=8, t=9, u=2: fine.
If p=4, and q=6, then r=5, s=8, t=8 (not possible)
If p=4, and q=8, then r=3, s=8, u=5 (which makes R4C1
unsolvable)
If p=8, and q=5, then R3C2 is unsolvable
If p=8, and q=6, then r=1, s=4, t=8, u=7: fine
If p=8, and q=8, impossible, because both numbers the same.
So the possibilities are: 4
5 6 or 8 6 1
8
9 2 4
8 7
Therefore either s or t is an <8>, so we can solve N4:8,5; 11[2] (R5) is a <65>;
Either r is a <1> or u is a <2>, therefore the
12[3] (C9) cannot be <912>, so -in N9- the <9> must be part of the
30[6]; therefore the 12[3] is either <741> or <732> (not
<831> because of the g8 in either s or t), but either way contains a g7,
therefore u cannot be <7> and must be 2, allowing us to give the values: p=4, q=5, r=6, s=8, t=9.
Thus 12[3] must be <741>, so place the 4; and note g71; 11[2] (R5): 5,6; N1:5,8; N3:g12 (making a Twin); N3: Double Innie has a total of 9, and is not <18> therefore must be <27>: 2,7; N3:1; N2: g123, therefore 18[3] is not <873> or <972>, so must be <765>, giving us g56; N1: 6,3; N3:3,9,8; R1:4; N2: 14[3]=<923>, therefore 9,3; 10[2]: 2,8; N2:1,7; 20[5] (C4) needs an extra total of 8, which must be <26> or <35>, therefore R4C4=6; and R5C4 = 2; R5:3,1; N5 has an Innie:8; C4:3; R7C8=7; so C9:7,1; N6:1,3; R7:8; therefore complete 22[3]; C3:8; C9:8,5; complete 9[2]; C8:2; so 30[6]:9; N9:3,6; R9: 12[2]:1; N7: 1,7; R9:2; N8: 10[2]=<64>:4,6; N2:6,5; R6:5,9; R4:4,7; N8:7,5.
Thanks to Rick from Maine for spotting the first typo in these step-by-steps!
And to be frank, if you made it throught all that, then you're probably one of the finest sudoku players in the world. Hope you enjoyed the book!