The New Sudoku Players' Forum

by **mumdigau** » Wed Nov 22, 2006 4:15 pm

Hi,

in the grid

Code: Select all: 1348 138 348 | 5 2 9 | 46 468 7 458 29 7 | 14 1468 146 | 3 2458 289 458 6 29 | 347 348 47 | 2459 1 289 --------------------+----------------------+-------------------- 2 1378 368 | 149 1469 146 | 168 37 5 146 15 456 | 8 7 3 | 1269 26 1269 9 1378 368 | 2 16 5 | 168 37 4 --------------------+----------------------+-------------------- 368 4 23568 | 137 13 127 | 256 9 2368 368 2358 1 | 349 349 24 | 7 2568 2368 7 239 239 | 6 5 8 | 124 24 123

I can identify a finned X-Wing for digit 9 in columns 7 and 9. I read through the 'Hunting finned X-Wing and other fishy stuff'-thread http://www.setbb.com/sudoku/viewtopic.php?mforum=sudoku&t=750&postdays=0&postorder=asc&start=0&sid=04af3857c652331664b3eeb80d34c315&mforum=sudoku
but didn't manage to find a solution for the puzzle. Can anybody show me how to solve the grid above?

And in general, does a solution depend on how the fin is attached to the X-Wing, and wether there are one or more fin cells?

Many thanks in advance for any help.

mumdigau

by RW » Wed Nov 22, 2006 4:44 pm

Hi mumdigau!

A finned x-wing usually doesn't appear in only two boxes, but is spread over four. A finned x-wing in only two boxes doesn't allow any more eliminations than locked candidates does, so it's basically useless. This puzzle can be solved with a UR in r46c28 and one out of many available short forcing chains. I bet there will be many suggestions of those posted here shortly!

I'll give my solution, that doesn't need the UR:

BUG-lite in r2c29, r3c39 and r9c23: if r9c9=3 => r8c8=8, but if r8c8=8 => r9c9=1
([r8c8]=5=[r8c2]-5-[r5c2]-1-[r5c9]=1=[r9c9])

so r9c9<>3

RW

by **Carcul** » Wed Nov 22, 2006 5:23 pm

Mumdigau wrote:Can anybody show me how to solve the grid above?

Welcome to this forum. Regarding your puzzle:

Code: Select all: *--------------------------------------------------------------* | 1348 138 348 | 5 2 9 | 46 468 7 | | 458 29 7 | 14 1468 146 | 3 2458 289 | | 458 6 29 | 347 348 47 | 2459 1 289 | |---------------------+-------------------+--------------------| | 2 1378 368 | 149 1469 146 | 168 37 5 | | 146 15 456 | 8 7 3 | 1269 26 1269 | | 9 1378 368 | 2 16 5 | 168 37 4 | |---------------------+-------------------+--------------------| | 368 4 23568 | 137 13 127 | 256 9 2368 | | 368 2358 1 | 349 349 24 | 7 2568 2368 | | 7 239 239 | 6 5 8 | 124 24 123 | *--------------------------------------------------------------*

ALS xz-rule combined with a box:
A(3,4,6,8)=[r146c3], B(4,6)=[r1c7], Box 5, x=4, z=6, => r4c7/r6c7<>6 solving the puzzle.

Carcul

by **daj95376** » Wed Nov 22, 2006 7:10 pm

No sign of an X-Wing from my solver. However, the puzzle does have an Achilles Heal.

Code: Select all: r7c7 <> 2 lengthy chain/net using Singles only r7c7 <> 5 lengthy chain/net using Singles only r7c7 = 6 (backdoor single) cracks puzzle

by **mumdigau** » Wed Nov 22, 2006 8:09 pm

Thanks for all replies.

Is the UR solution to be regarded "weaker" than the direct ALS approach?

And it would be fine if someone can post a link with a good ntroduction to / explanation of finned seafood.

@RW
Can you briefly explain your first two sentences? BTW, I'm not aware that (finned) X-Wings require four boxes as prerequisite.
AFAIK this is UR type 3. Are the forcing chains used to determine whether 1 or 8 and where set?
Your BUG solution I'll study tomorrow in detail.

@daj95376
Isn`t r2c9, r3c7, r3c9, r5c7, r5c9 a finned X-Wing?
What is an Achill heal?

mumdigau

@

by RW » Wed Nov 22, 2006 8:35 pm

mumdigau wrote:BTW, I'm not aware that (finned) X-Wings require four boxes as prerequisite.

They don't require four boxes, but if they don't have, then you can ALWAYS make the same elimination using locked candidates:

Code: Select all: . . . # . . X * X ----- . . . . . . X . X ----- / . / / . / / . /

In this example we have a finned X-wing in r36c13, fin in r2c1. This would let us eliminate the candidate from r3c2. However, if the finned x-wing really is a finned x-wing, there must be no other candidates in those two columns => box 7 only has candidates in column 2 => the candidate can be eliminated from the rest of the column by locked candidates. This is why the finned x-wing is useless if it isn't spread out over four boxes.

The UR solution isn't weaker than any other solution IMO, but in this case it doesn't solve the puzzle immediately, only eliminates candidate 3 from r46c2, which reveals a couple of more pairs and brings you here:

Code: Select all: *-----------------------------------------------------------* | 13 13 48 | 5 2 9 | 46 468 7 | | 458 29 7 | 14 1468 146 | 3 2458 289 | | 458 6 29 | 347 348 47 | 2459 1 289 | |-------------------+-------------------+-------------------| | 2 178 368 | 149 1469 146 | 168 37 5 | | 146 15 456 | 8 7 3 | 1269 26 1269 | | 9 178 368 | 2 16 5 | 168 37 4 | |-------------------+-------------------+-------------------| | 368 4 568 | 137 13 127 | 256 9 2368 | | 368 2358 1 | 349 349 24 | 7 2568 2368 | | 7 239 29 | 6 5 8 | 124 24 123 | *-----------------------------------------------------------*

At this point there's several short forcingchains available, one could be:
[r5c9]=1=[r9c9]=3=[r9c2]-3-[r1c2]-1-[r5c2]
(if r5c9<>1 => r9c9=1 => r9c2=3 => r1c2=1 => r5c2<>1)

This tells us that either r5c9=1 or r5c2<>1, in either case r5c2<>1.

RW

by **mumdigau** » Fri Nov 24, 2006 11:11 am

@Carcul

Thanks for pointing to ALS. Haven't used that till now, only APE. The basic ALS technique I think I've understood now (very helpful, but not easy to spot IMO). Question though: For xz rule cells with z candidates to eliminate must see all the z candidates in both A and B. This is not the case here, because r4c7 as well as r6c7 can directly see only one of two z candidates in A. But you mentioned Box 5. Can you shortly explain how it helps to manage the trick?

mumdigau

by **Carcul** » Fri Nov 24, 2006 12:48 pm

Mumdigau wrote:For xz rule cells with z candidates to eliminate must see all the z candidates in both A and B.

Correct.

Mumdigau wrote:This is not the case here, because r4c7 as well as r6c7 can directly see only one of two z candidates in A. But you mentioned Box 5.

Yes, because due to the distribution of z candidates in box 5 (we could say that the z candidates in A and box 5 make up an X-Wing), the z candidates in r4c7/r6c7 can actually see the z's in A.
For example, consider r6c7=6. Then r6c3<>6, but also r6c5<>6 => one of r4c456=6 => r4c3<>6. A similar explanation goes for r4c7.

Carcul

by **ronk** » Fri Nov 24, 2006 1:15 pm

mumdigau wrote:For [edit: the ALS] xz rule cells with z candidates to eliminate must see all the z candidates in both A and B. This is not the case here ...
Can you shortly explain how it helps to manage the trick?

I see it as also using the grouped conjugate link in row 5.

Code: Select all: 1348 138 A348 | 5 2 9 |*46 468 7 458 29 7 | 14 1468 146 | 3 2458 289 458 6 29 | 347 348 47 | 2459 1 289 -------------------+-----------------+------------------ 2 1378 A368 | 149 1469 146 |*1-68 37 5 *146 15 *456 | 8 7 3 |*1269 *26 *1269 9 1378 A368 | 2 16 5 |*1-68 37 4 -------------------+-----------------+------------------ 368 4 23568 | 137 13 127 | 256 9 2368 368 2358 1 | 349 349 24 | 7 2568 2368 7 239 239 | 6 5 8 | 124 24 123

r46c7-6-r1c7-4-{ALS:r1c3=4|6=r46c3}-6-r5c13=6=r5c789-6-r46c7, implying r46c7<>6

by **mumdigau** » Fri Nov 24, 2006 1:37 pm

Carcul wrote:Yes, because due to the distribution of z candidates in box 5 (we could say that the z candidates in A and box 5 make up an X-Wing), the z candidates in r4c7/r6c7 can actually see the z's in A.
For example, consider r6c7=6. Then r6c3<>6, but also r6c5<>6 => one of r4c456=6 => r4c3<>6. A similar explanation goes for r4c7.

Yeah, I got it. Very smart! Thanks.

mumdigau

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