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C1R5 & C2R5 have a monopoly in box 4. These 2 cells also have a monopoly in row 5; therefore, no other cell in row 5 can have a 3 or 8 as an option. Now you can clearly see that the only cell in box 5 that can be a 3 is C5R6. C5R6=3. Now the only cell in box 5 that can be an 8 is C4R6. C4R6=8. Now look again at row 6. The 7 in box 6 prevents C7R6, C8R6 & C9R6 from being a 7; therefore, the only remaining unsolved cell in row 6 that can be a 7 is C3R6. C3R6=7. Then C1R2=7, C8R1=7 & C9R9=7. Now your grid should look like Example #35.3 below:
We are still in the Puzzle Preparation phase, but incorporating Step 1: Pairs, Triplets & Quads. Let’s look for other Pairs or Triplets. The only choice in box 6 for options 45 are C9R4 & C9R5, which is another obvious Pair. Then, the only possible options for C7R1, C8R1 & C9R1 are 129. The only possible options for C1R7, C1R8 & C1R9 are 169. Mark those 8 cells with their only possible options and your grid should now look like Example #35.4 below:
Now you note that the only cell in column 9 that can be a 3 is C9R2. C9R2=3. Then C2R1=3. C2R5=8 & C1R5=3. Now your grid should look like Example #35.5 below:
From here you can see that the only 2 options for C1R3 & C1R4 are 58. Since there is an 8 in row 4, C1R3 must be the 8 and C1R4=5. Now C9R4=4 & C9R5=5. Now C2R4 must be a 2. Now the only 2 choices for option in C2R2 & C2R3 are 19. Since there is a 9 in C4R2, C2R2=1 & C2R3=9. Also, fill in the obvious Pair in column 3, box 4 and the Triplet in column 3, box 1. Now your grid should look like Example #35.6 below:
We can now see that C6R3=1. C7R1=9. The only options for the 3 unsolved cells in column 9 are 289. Since there already an 8 & 9 in row 3, C9R3=2. Then C9R6=9 & C9R8=8. Then C6R9=8. C6R7=3. C8R9=3. C7R2=8. C5R1=8. C7R3=6. Now the only cell in row 9 that can be a 9 is C1R9. C1R9=9. C1R7=1 & C1R8=6. C4R7=7. Then C4R9=6 & C7R9=1. C7R6=2 & C8R6=1. The only options for C7R7 & C7R8 are 45. Since there is already a 4 in row 7, C7R7=5 & C7R8=4. Now your grid should look like Example #35.7 below:
Now look at row 4. The only possible options for the 4 unsolved cells are 1679. There are already a 679 in column 4; therefore, C4R4=1. C3R4=9 & C3R5=1. Then C5R8=1. Now the only cell in row 4 that can be a 7 is C6R4. C6R4=7. C5R4=6. Now your grid should look like Example #35.8 below:
Look at box 8. The only cell that can be a 5 is C4R8. C4R8=5. Now the only options for C4R1 & C4R5 are 24. Since there is already a 2 in row 1, C4R1=4 & C4R5=2. The only cell in column 5 that can be a 5 is C5R2. C5R2=5. Now the only options for C5R5 & C5R6 are 49. Since there is already a 4 in column 5, C5R5=9 & C6R5=4. Then, C5R7=2, C6R2=2 & C6R1=6. C6R8=9. C8R7=9. C8R8=2. C3R1=5. C3R3=4. C3R2=6. C8R2=4 & C8R3=5, giving us the final puzzle as Example #35.9 below:
Well, what just happened here? May the gentle winds of Sudoku be at your back.
By Dan LeKander, Wellesley Island
[See Jessy Kahn’s Book Review, “3 Advanced Sudoku Techniques…” by Dan LeKander, June issue of TI Life.]
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