Help with finned X-Wing

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Help with finned X-Wing

Postby 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
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Postby 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

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Postby 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.

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Postby 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
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Postby 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

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Postby 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.

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Postby 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?

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Postby 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.

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Postby 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
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Postby 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.

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