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There are no other Step 1-5 clues, so we will move on to Step 6.
There are 3 circumstances that establish the potential for a Step 6 exercise: - Look for
**just 2 unsolved cells**in a**box**that contain the**same option**where these 2 cells are**not in the same row or column**. - Look for
**just 2 unsolved****cells**in a**column**that contain the**same option**where these 2 cells are**not in the same box**. - Look for
**just 2 unsolved cells**in a**row**that contain the**same option**where these 2 cells are**not in the same box**.
In column 3 we find just 2 unsolved cells that contain #1 as an option … C3R1 & C3R9. These cells are not in the same box, thereby qualifying as a candidate for a Step 6 exercise. These cells are highlighted below in Example #31.4 below:
One of these two cells must be a 1. We will consider them as “starter cells” which “drive” the exercise. Here is the theory. We will perform two exercises. First, we will assume C3R1 is the 1, and see which other cells cannot be a 1. Then we will assume C3R9 is the 1, and see which other cells cannot be a 1. What happens if we find a cell or cells that cannot be a 1 regardless of which cell in column 3 is a 1? Quite simply it means that that cell cannot be a 1 and you can eliminate the 1 from that cell or those cells. We will mark C3R1 with a “Y” and mark C3R9 with a lower case “y” to keep track of the exercise as per Example #31.4 above. Starting with C3R1 we see that if it is a 1 (Y for yes), then C4R1 & C6R1 are not a 1. We will mark these two cells with a “N” indicating they cannot be a 1 if C3R1 is a 1 (see Example #31.5 below). The only other cell in box 2 that now could be a 1 is C4R2 so we will mark it with a Y. If C4R2 is a 1, then C4R4 & C4R5 are not a 1. The only other cell in box 5 that could be a 1 is C6R5, so we will mark it with a Y. Then C7R5=N. Then C9R4=Y. Then C9R9=N. Now we will switch to C3R9 and assume it is the 1. Then C7R9 & C9R9 cannot be a 1 and we mark them with a “n”. The only other cell in box 9 that can be a 1 is C7R8 and we will mark it with a “y”. Then C7R5 is a “n”. Then C9R4=y. Then C4R4=n. Now take a close look at Example #31.5 below. Cells C4R4, C7R5 & C9R9 have a “N,n” designation. We have proven that these 3 cells are not a 1 regardless of which starter cell in column 1 is a 1, so the 1 can now be eliminated from those cells as an option. Conversely, cell C9R4 has a “Y,y” designation, meaning it is a 1 regardless of which starter cell is a 1. Therefore, C9R4=1.
You can now clearly see that C4R4=7. Then C6R3=7. C2R2=7. C3R7=7. C7R8=7. C9R6=7. C1R8=1 (only cell in row 8 that has a 1 as an option). C3R1=1. C4R2=1. C6R5=1. C7R9=1. That retires the 1’s and 7’s. C8R8=6. C1R7=6. C6R6=6. C7R5=6. C3R4=6. C5R2=6. C9R3=6. C5R1=5. C9R1=4. C6R1=3. C4R1=8. C4R3=4. C4R5=2. C4R6=3. C5R4=8. C5R6=4. C5R8=3. C3R5=5. C8R5=8. C2R5=4. From this point the puzzle is easily solved, giving you the final version Example #31.7 below:
May this Sudoku puzzle add to your future Sudoku success & may the gentle winds of Sudoku be at your back.
Dan LeKander, Wellesley Island
[See Jessy Kahn’s Book Review, “3 Advanced Sudoku Techniques…” by Dan LeKander, June issue of TI Life.] ## CommentsThere are currently no comments, be the first to post one. |