dailysudoku.com

[lda697]

Hey, Lee here. This is my first message. For today's 11/21 su doku, I was able to identify all the 2s and some 3s. I highlighted 4 Xs in red where I know it has to be a "3".I also identified some 9s and one 6, so that my puzzle looks like this:

93X X5X 827
XX8 9X2 53X
X2X X8X XXX

XXX 7XX XX2
XXX 2X5 XXX
3X2 XX1 XXX

XXX X2X X9X
273 X69 4XX
694 X1X 2X3

I was wondering if somebody out there could identify another number or two and explain how they figured it out. Sincerly Confused!

[Guest]

Hey, Lee here. This is my first message. For today's 11/21 su doku, I was able to identify all the 2s and some 3s. I highlighted 4 Xs in red where I know it has to be a "3".I also identified some 9s and one 6, so that my puzzle looks like this:

93X X5X 827
XX8 9X2 53X
X2X X8X XXX

XXX 7XX XX2
XXX 2X5 XXX
3X2 XX1 XXX

XXX X2X X9X
273 X69 4XX
694 X1X 2X3

I was wondering if somebody out there could identify another number or two and explain how they figured it out. Sincerly Confused!


Hi Lee,

Your lower red X's in the lower middle quadrant are incorrect as a 3 since there is a 3 in that row.

I am not much further along than you, but have identified a couple of 7's. Mine looks like this:

93X X5X 827
XX8 972 53X
X2X X8X XXX

XXX 7XX XX2
XXX 2X5 XXX
3X2 XX1 XXX

XXX X2X X9X
273 XX9 4XX
694 X17 1X3

The 7 in the Lower Middle quadrant had to be in Column 6 because Column 4 has a 7 and Row 8 has a 7 thus eliminating all squares in Column 4 and 5 in this quadrant. Therefore the 7 in the Upper Middle quadrant has to be in Column 5 and there is only one opening there at Column 5 Row 2.

After looking at the positioning of the 3 and 4 in both the Lower Left and Lower Right quadrant, I determined that the 3 and 4 in the Lower Middle quadrant had to be along row 7, thus eliminating the posibility of a 7 being at Column 6 Row 7 and leaving only Column 6 Row 9 for the 7.

Boy does that sound confusing. I'm sure there is an easier way to explain all that.

Curtis[/b]

[David Bryant]

Hi, Lee!

You already found the "6" in r8c5. Good! Now look at r2c5. It has to be a "7" because there must be a "7" in column 5, and "7" can't fit anywhere else in that column (notice the "7" already in the middle center 3x3 box, and that column 5 is full in the bottom center 3x3 box).

With the "7" filled in you can see that the three digits in r4c5, r5c5, and r6c5 must be {3, 4, 9 }, in some order. Combining those with the {1, 2, 5, 7} already appearing in the middle center 3x3 box you can infer that the pair {6, 8} must appear in r4c6 and r6c4, the diagonally opposite corners of that middle center 3x3 box. This doesn't help you place another number right now, but it will be useful information pretty soon.

To get the next numbers, concentrate on row 7. You already have 2/3 of the lower left 3x3 box filled in, so you know that {1, 5, 8} must fit in r7c1, r7c2, & r7c3, in some order. Now look at the "3" and "4" appearing in both row 8 and row 9. Since all of these are outside of the bottom center 3x3 box you can see that the pair {3,4} must occupy the two cells r7c4 & r7c6. And now if you count what's left over in row 7 you'll see that the pair {6, 7} must occupy r7c7 & r7c9 in some order ... but since there's a "7" at r1c9 you can put the "7" in r7c7 and the "6" at r7c9.

I hope that's enough of a hint! dcb

[Guest]