where do i go from here, my third hard sudoku i have tried

- - 8 5 2 - 7 9 1

- - 2 - 9 - 5 6 8

9 - - - - 8 2 3 4

2 3 9 8 - - - 7 6

- - 7 2 6 - 9 8 3

- - - 7 3 9 - 5 2

- 9 - 3 - - 6 2 7

- 2 3 - - - 8 1 9

1 - 6 9 - 2 3 4 5

12 posts
• Page **1** of **1**

where do i go from here, my third hard sudoku i have tried

- - 8 5 2 - 7 9 1

- - 2 - 9 - 5 6 8

9 - - - - 8 2 3 4

2 3 9 8 - - - 7 6

- - 7 2 6 - 9 8 3

- - - 7 3 9 - 5 2

- 9 - 3 - - 6 2 7

- 2 3 - - - 8 1 9

1 - 6 9 - 2 3 4 5

- - 8 5 2 - 7 9 1

- - 2 - 9 - 5 6 8

9 - - - - 8 2 3 4

2 3 9 8 - - - 7 6

- - 7 2 6 - 9 8 3

- - - 7 3 9 - 5 2

- 9 - 3 - - 6 2 7

- 2 3 - - - 8 1 9

1 - 6 9 - 2 3 4 5

- sfddoc
**Posts:**17**Joined:**14 August 2005

Sure. Column 6 needs to have 1, 3, 4, 5, 6, 7 put in it. Cells 4, 5 and 7 in the column cannot be 3, 6 or 7, so the other three must be, and they must be a, 4 and 5. That only leaves one place for the 4 in box 2. I'm afraid I called it a pair and a triplet, didn't I? I meant two triplets, and I guess this is what caused the confusion. Sorry, my fault.

- PaulIQ164
**Posts:**533**Joined:**16 July 2005

i do appreciate all your help, but you are talking above me as a newcomer. I too am a steve f - can you explain to me what is a naked pair and then take me through your logic with how that helps. also pauliq164 what is a triplet and how does that help me. thanx so much, iam eager to learn. i do have all my pencil marks inplace but i don not see how to proceed.

- sfddoc
**Posts:**17**Joined:**14 August 2005

Right. If you look at the sixth column - the one you already have 8, 9 and 2 in. You still need to put in 1, 3, 4, 5, 6 and 7. Now, the central 3x3 box on the grid already has 3, 6 and 7 in it, so the two squares in row 6 that are in this middle box can't be 3, 6 or 7 (meaning they can still be 1, 4 or 5). Now, if you look at the seventh row down, you'll see that row already has 1 3, a 6 and a 7 in it too. So again, the cell on the seventh row and the sixth column cannot be 3, 6 or 7. So, out of the six cells left to fill in row 6, there are three that cannot be 3, 6, or 7; and three that can be. So we must have 3, 6 and 7 in these cells.

Now look at box 2 (the 3x3 box in the top-center). On the face of it, there are two places where the 4 can go (hopefully you can see this from your pencilmarks). But, one of them is one of the three cells we've already identified as between themselves containing the 3, 6 and 7. So this one can't take the 4. That means the other cell must be the 4. Luckily, after this the rest of the puzzle is fairly simple.

Now look at box 2 (the 3x3 box in the top-center). On the face of it, there are two places where the 4 can go (hopefully you can see this from your pencilmarks). But, one of them is one of the three cells we've already identified as between themselves containing the 3, 6 and 7. So this one can't take the 4. That means the other cell must be the 4. Luckily, after this the rest of the puzzle is fairly simple.

- PaulIQ164
**Posts:**533**Joined:**16 July 2005

I'll give it a shot.. (Newbie here too, but learning)

The terms used refer to your pencil marks.

In R6C3 and R6C7, you should have only a 1 and 4 in both cells. That means that one of the cells must be a 1 and the other must be a 4. Consequently, none of the other cells in the row can be either. So you can eliminate the 1s and 4s from the other cells in the row.

By the same reasoning, cells R4C6, R5C6, and R7C6 are the only cells in the column that can contain 1, 4, and 5. R1C6, R2C6 and R8C6 can only contain 3, 6, and 7. When you eliminate the 4s from R1C6 and R2C6, you are left with only one cell in box 2 with a possibility of 4. That's R2C4.

The terms used refer to your pencil marks.

In R6C3 and R6C7, you should have only a 1 and 4 in both cells. That means that one of the cells must be a 1 and the other must be a 4. Consequently, none of the other cells in the row can be either. So you can eliminate the 1s and 4s from the other cells in the row.

By the same reasoning, cells R4C6, R5C6, and R7C6 are the only cells in the column that can contain 1, 4, and 5. R1C6, R2C6 and R8C6 can only contain 3, 6, and 7. When you eliminate the 4s from R1C6 and R2C6, you are left with only one cell in box 2 with a possibility of 4. That's R2C4.

- RickM
**Posts:**22**Joined:**18 August 2005

Yes, if you have three cells in a column (or row or box) that only have three different possibilities between them, then those three cells must contain those three numbers in some order. That's a triplet. Then, you can eliminate those three numbers as possibilities from everywhere else in the column.

- PaulIQ164
**Posts:**533**Joined:**16 July 2005

one more question, can a soduku puzzle have more than one solution, ie

325 487 916

186 293 574

749 516 823

217 349 658

958 67- 34-

463 8-- 79-

-31` 7-4 289

-92 13- 467

-74 9-- 135

i believe whtether row 6 column 5 or 6 is a 5, you can have a viable solution, yes?? thanx

325 487 916

186 293 574

749 516 823

217 349 658

958 67- 34-

463 8-- 79-

-31` 7-4 289

-92 13- 467

-74 9-- 135

i believe whtether row 6 column 5 or 6 is a 5, you can have a viable solution, yes?? thanx

- sfddoc
**Posts:**17**Joined:**14 August 2005

12 posts
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