Luke451 wrote:Not to beg the last question, but why would you not want to use pencil marks?

Keep in mind that this is a Times puzzle. I think many people don't actually like to mark candidates on newspaper puzzles, so Richard's pondering makes sense to me.

I'm no expert in pencilmarkless solving (

RW is the boss in that department here), but still I can offer an approach here:

After Richard's position where all trivial hidden singles are resolved:

- Code: Select all
` 4 6 . | 9 . 7 | . 8 5`

8 7 . | 6 . . | 4 2 9

. . 9 | 4 8 . | . . .

-------+-------+-------

3 . . | 7 4 . | 9 6 1

. . 7 | . 9 . | 8 . .

9 4 # | . - # | . . 3

-------+-------+-------

. . . | 8 7 9 | . . .

1 8 . | - * 4 | . 9 .

7 9 4 | 5 * 1 | . 3 8

Focusing on the digit 6, you can see r89c5 (*) are the only cells which can have 6 in box 8. That means r6c5 can't be 6. And then if we look closely at r6, we'll see r6c36 (#) are the only cells possible to have 6 & 8 on that row. Therefore together they must have {6,8}. This is what we call a "hidden pair".

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` 4 6 . | 9 * 7 | . 8 5`

8 7 . | 6 * - | 4 2 9

. . 9 | 4 8 - | . . .

-------+-------+-------

3 . . | 7 4 . | 9 6 1

. @ 7 | . 9 . | 8 . .

9 4 . | # - . | . . 3

-------+-------+-------

. . . | 8 7 9 | . . .

1 8 . | . . 4 | . 9 .

7 9 4 | 5 . 1 | . 3 8

Now let's focus on the digit 1. Again you can see r12c5 (*) are the only cells possible to have 1 in box 2. That means r6c5 can't be 1. And then if we look close on r6 (again), we'll see r6c4 (#) is the only cell possible to have 1 on that row (remember we have concluded r6c36 must be {6,8} in the previous step). This is a "hidden single".

Also, we can immediately deduce that r5c2 (@) is the only cell possible to have 1 on r5 now (another "hidden single").

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` 4 6 . | 9 . 7 | @ 8 5`

8 7 . | 6 . . | 4 2 9

. . 9 | 4 8 . | * * *

-------+-------+-------

3 . . | 7 4 . | 9 6 1

. 1 7 | . 9 . | 8 . .

9 4 . | 1 . . | . . 3

-------+-------+-------

. . . | 8 7 9 | . . .

1 8 . | . . 4 | . 9 .

7 9 4 | 5 . 1 | . 3 8

And then we need to focus on r3. You can see r3c789 (*) are the only 3 cells which can possibly have the 3 digits {1,6,7} on that row. Therefore they must collectively have those 3 digits. This is what we call a "hidden triple".

And then immediately we can conclude that only the cell r1c7 (@) can have the digit 3 in box 3 (another "hidden single").

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` 4 6 # | 9 - 7 | 3 8 5`

8 7 . | 6 . . | 4 2 9

. . 9 | 4 8 . | . . .

-------+-------+-------

3 . . | 7 4 . | 9 6 1

. 1 7 | . 9 . | 8 . .

9 4 . | 1 * . | * . 3

-------+-------+-------

. . . | 8 7 9 | . . .

1 8 . | . . 4 | . 9 .

7 9 4 | 5 * 1 | * 3 8

Here comes the critical step. Obviously the 2 on r9 must be held in one of r9c57 (*). Also the 2 on r6 must be held in one of r6c57 (*) (remember r6c36 must have {6,8} from the first step above). This is a classical pattern which is called "x-wing" by most sudoku players. The immediate conclusion is that the 2 on c5 and the 2 on c7 must be locked on r6 & r9, one way or the other.

Hence it can be concluded that r1c5 can't be 2. Which leaves r1c3 (#) as the only possible cell to have 2 on r1 (another "hidden single").

I believe the rest can be solved relatively easily by Richard and other pencilmarkless solvers.

Afterthoughts: personally I think puzzles with x-wings (or simple turbot fishes) are about the boundary of pencilmarkless solving and I frankly don't see any easy way to apply more advanced techniques such as xy-wings, ALSs and URs without using pencilmarks (

RW will probably disagree here). The key for pencilmarkless solving is to have a good memory (e.g. here you have to keep remembering r6c36 must contain {6,8} and nothing else all the way through).