How many of you have had a chance to look at the puzzle from yesterday? You will notice that I am trying to solve the grid without using any kind of pencilmarks. I think this is a valuable exercise for your logical thinking and you will only find that it gets even easier once you have mastered this and add pencilmarks too. I’m interested to hear your views on this, throw a comment if you feel strongly one way or the other.
Back to the sudoku puzzle then. I left the grid in a position where there were several more values that you could find by using the same sudoku technique I showed you in the last post.
Here is the grid with those values added in.
The 2 in the first box in 3,1 means that the second box must have its own 2 in either 1,4 or 1,6. This means that the third box must have 2 in row 2 but we already have 2 in columns 7 and 9, so 2 must be in 2,8.
In the same way, the 7 in the bottom-left box can only be in 8,1 when you look at how 4,3 affects the possibilities in box 1 (remember we number boxes 1,2,3 across then 4,5,6 etc).
The other 2 in this diagram is a bit harder to see immediately. The 2 in box 9 constrains the 2 in box to be in column 4. This is turn forces the 2 in box 2 to be in column 3. Then we just do the crossing of all the areas that these possibilities affect and also the 2’s in boxes 4 and 6.
Here is the next group of values that you can now find.
I have highlighted the central column in green to show a new sudoku technuque. Here we know that we must have 4 or 9 in these two cells as these are the only values left for this house of 9 cells. The 9 already in 5,1 constrains our 9 to be in row 6 and so the 4 must go in row 5. Once we have these, the 9 in 4,7 is the next obvious value. It is often true that you should look for how your latest entry affects all the other possibilities for that value.
I have colored the remaining 4’s in red, with the color getting deeper as you progress. As an exercise for this post, see if you can work through these and check how I filled them in next.
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