## Times Superfiendish 2451 (Friday 27th Feb 2009)

Post the puzzle or solving technique that's causing you trouble and someone will help

### Times Superfiendish 2451 (Friday 27th Feb 2009)

The following puzzle was set last Friday in the Times.
Code: Select all
`4 6 .  9 . 7  . . 5. 7 .  6 . .  . 2 9. . .  . 8 .  . . .3 . .  . . .  . 6 1 . . 7  . 9 .  8 . . 9 4 .  . . .  . . 3. . .  . 7 .  . . . 1 8 .  . . 4  . 9 .7 . .  5 . 1  . 3 8`

Code: Select all
`4 6 .  9 . 7  . 8 58 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`

This solves readily by noting that 2 cannot be a candidate for r1c5 thus forcing 2 into r3c6.
Beacuse:

Code: Select all
`r6: x x x   x  25  x  257 57 xr9: x x x   x  26  x  26  x  x `

implies that 2 must appear in either r6 or r9 of c5.
However I'm dissatisfied with my solution because I had to resort to PM.

My questions are:
1) is this the simplest way to solve this puzzle
2) if it is what technique am I employing?
3) is there a method for spotting this situation without having to use PM?

Richard
richardm

Posts: 53
Joined: 27 December 2006

This may be the position you have reached:
Code: Select all
` *-----------------------------------------------------------* | 4     6     12    | 9     12    7     | 3     8     5     | | 8     7     135   | 6     135   35    | 4     2     9     | | 25    235   9     | 4     8     235   | 167   17    67    | |-------------------+-------------------+-------------------| | 3     25    258   | 7     4     258   | 9     6     1     | | 256   1     7     | 23    9     2356  | 8     45    24    | | 9     4     68    | 1    *25    68    |*257   57    3     | |-------------------+-------------------+-------------------| | 256   235   2356  | 8     7     9     | 1256  145   246   | | 1     8     256   | 23    236   4     | 2567  9     267   | | 7     9     4     | 5    *26    1     |*26    3     8     | *-----------------------------------------------------------*`
richardm wrote: implies that 2 must appear in either r6 or r9 of c5.

You might have said, "implies that (2) must appear in either r6 or r9 of c5 and c7."

My questions are:
1) is this the simplest way to solve this puzzle
2) if it is what technique am I employing?
3) is there a method for spotting this situation without having to use PM?

You’ve found an x-wing, and it’s as good a way to go at this point as any. Any (2) in c5 or c7 outside the marked cells can be eliminated.

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

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

Ah, but I didn't get to that point. I don't have 1 in r5c2 or in r6c4, or 3 in r1c7
And so I still have 2 as a candidate for r6c3.

So how did you eliminate these last three and the 2 candidate from r6c3?

Richard
richardm

Posts: 53
Joined: 27 December 2006

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".

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`

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").

Code: Select all
` 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").

Code: Select all
` 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).
udosuk

Posts: 2698
Joined: 17 July 2005

richardm wrote:Ah, but I didn't get to that point. I don't have 1 in r5c2 or in r6c4, or 3 in r1c7
And so I still have 2 as a candidate for r6c3.

So how did you eliminate these last three and the 2 candidate from r6c3?

You had isolated the naked triple, but it seems you hadn't pulled the trigger on it . You were here, then:
Code: Select all
` *--------------------------------------------------------------------* | 4      6      123    | 9      123    7      | 13     8      5      | | 8      7      135    | 6      135    35     | 4      2      9      | | 25     1235   9      | 4      8      235    | 1367   17     67     | |----------------------+----------------------+----------------------| | 3      25     258    | 7      4      258    | 9      6      1      | | 256    125    7      | 123    9      2356   | 8      45     24     | | 9      4      12568  | 12    *25     2568   |*257   *57     3      | |----------------------+----------------------+----------------------| | 256    235    2356   | 8      7      9      | 1256   145    246    | | 1      8      256    | 23     236    4      | 2567   9      267    | | 7      9      4      | 5      26     1      | 26     3      8      | *--------------------------------------------------------------------*`
There's three candidates for three cells in r6. This means you can eliminate 2, 5, or 7 in any of the unmarked cells in r6. That will place the other candidates you mentioned when you clean up the singles, and get you to the x-wing.

I gather you do use pencil marks, but would rather not. I'd say udosuk has you well covered from that angle.

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

udosuk wrote: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).

I think that UR's could be kept in mind. Reasoning :
1. only certain of the unfilled cells will be suitably lined up for UR consideration
2. the solver can select a set of 4 such cells, note mentally the candidates for one of the cells, continue clockwise (as a discipline) to a second cell. At that stage he has a possible UR and continues clockwise or stops.
3. he can repeat this for other possible configurations
4. he doesn't need to be exhaustive in the approach
5. he can do this at any stage of the puzzle
6. he can limit his UR investigation to the basic type ab ab ab abc or to ab ab abc abc
7. in this way on a certain random basis he has some chance of a close encounter and introduces a bit of variety into his solving
aran

Posts: 334
Joined: 02 March 2007

Thank you, yes that's it - the hidden pair that I missed. The rest follows easily. For some reason I didn't do my analysis correctly - probably beacuse I was watching Monty Python at the same time.

I should have got this without PM. Generally I have no problem with 2,3 or 4 fish using no PM. But I can rarely spot an xy or xyz wing. I suppose as you say, a good memory for cells with a pair of candidates is what's needed.

Richard
richardm

Posts: 53
Joined: 27 December 2006

richardm wrote:For some reason I didn't do my analysis correctly - probably beacuse I was watching Monty Python at the same time.

"Is this the room for an argument?"

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

5 minutes or a full half-hour?
richardm

Posts: 53
Joined: 27 December 2006