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Advanced Uniqueness opportunities for the 8/4/06 Nightmare

 
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Myth Jellies
Hooked
Hooked


Joined: 04 Apr 2006
Posts: 42

PostPosted: Sat Aug 05, 2006 5:57 am    Post subject: Advanced Uniqueness opportunities for the 8/4/06 Nightmare Reply with quote

Code:
 *-----------*
 |.4.|5.1|.8.|
 |..2|.6.|9..|
 |...|9.7|...|
 |---+---+---|
 |.17|...|69.|
 |28.|...|.51|
 |.3.|...|.4.|
 |---+---+---|
 |...|789|...|
 |...|...|...|
 |..1|2.6|8..|
 *-----------*

basic methods take you to here...
Code:
 *---------------------------------------------------------------------*
 |B3679    4     B369    | 5      23     1      | 237    8      2367   |
 | 1357   d57     2      | 348    6      348    | 9     -137    3457   |
 |A13568   56    A3568   | 9      234    7      | 145    1236   23456  |
 |-----------------------+----------------------+----------------------|
 | 45      1      7      | 348    345   C23458  | 6      9     C238    |
 | 2       8      49     | 6      79     34     | 37     5      1      |
 |B569     3     B569    | 1      79    C258    | 27     4     C278    |
 |-----------------------+----------------------+----------------------|
 | 3456    256    3456   | 7      8      9      | 145    1236   23456  |
 |A678    D2679  A68     | 34     1      345    | 45   d'267   D2679   |
 | 3457   D579    1      | 2      45     6      | 8    d'37    D34579  |
 *---------------------------------------------------------------------*

...and pickings are kind of slim while you search for that
(4=3)r5c6 - (3=5&4)r4c15 xyz-wing to kill the 4's in r4c46.

In the meantime, there are a few advanced uniqueness rectangle tricks that you can take advantage of. Three of them have extra candidates in all four cells. Uppercase letters mark out the four rectangles.

A) 68-UR in r38c13. Two strong 8-links form a corner at r3c1, and the opposite corner contains just the base candidates, 68. Therefore r1c3 <> 6.

B) 69-UR in r16c13. Two strong 9-links form a corner at r1c1. There is also a strong 6-link that does not include that corner, therefore r1c1 <> 6.

C) 28-UR in r46c69. Two strong 2-links form a corner at r4c6. Two strong 8-links also form a corner at r6c9. For the same reason as in B, we know that r4c6 <> 8, and r6c9 <> 2.

D) 79-UR in r89c29. Here the 9's are locked into the UR, so if the deadly pattern is to be avoided, it will have to be the 7's that move out. If the move occurs out of r89c2, then the only other spot in c2 is r2c2. If the move occurs out of r89c9, then the only other spots in box 9 are r89c8. Since one or both of these alternate landing places has to contain a 7, any cell which sees both of them cannot contain a 7. Therefore r2c8 <> 7.


geek Just for grins and giggles...

If you'd like to express these uniqueness tricks as AICs, you need to note that A, B, and C use a special URCorner avoidance weak link that goes as follows: if you have two strong links forming a corner for one of your base candidates, then you have two URC weak links that exist between the other base candidate in that corner and each of the base candidates in the opposite corner. The URC weak link is represented by "-URC-", and I've emphasized the strong UR corner links with a "= (N&N)[cell1,cell2]" structure. Thus if you'd like to express these tricks as AIC's you could write something like the following....

A: (8)r3c1 = (8&8)[r3c3,r8c1] - (8=6)r8c3 -URC- (6=1358)r3c1 => r3c1 <> 6
B: (9)r1c1 = (9&9)[r1c3,r6c1] - (6)r6c1 = (6)r6c3 -URC- (6=379)r1c1 => r1c1 <> 6
C1: (2)r4c6 = (2&2-8&8)[r4c9,r6c6] = (8)r6c9 -URC- (8=2345)r4c6 => r4c6 <> 8
C2: (8)r6c9 = (8&8-2&2)[r4c9,r6c6] = (2)r4c6 -URC- (2=78)r6c9 => r6c9 <> 2
D: (7)r2c2 = (7&9)r89c2 -UR- (7&9)r89c9 = (7)r89c8 => r2c8 <> 7
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Chuck B
New kid on the Grid
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Joined: 09 Jul 2006
Posts: 4

PostPosted: Sat Aug 05, 2006 8:03 pm    Post subject: Reply with quote

Also, there's an XYZ wing at r8c6: r9c5, r5c6 (sorry 'bout the unconventional notation) which kills the 4 at r4c5. If you happen to find this one first, it reduces the original XYZ wing to an XY wing.

IAE, after both wings the result is that r5c6 = 4 and, except for one instance of coloring, it's almost a sprint to the finish from that point.

-cb
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Myth Jellies
Hooked
Hooked


Joined: 04 Apr 2006
Posts: 42

PostPosted: Sun Aug 06, 2006 12:25 am    Post subject: Reply with quote

Don't worry about the notation. Mine is not standard either.

I don't see how r8c6, r9c5, and r5c6 legitimately kill a 4 in r4c5. In order to be killed, the target candidate needs to see all of the 4's in the xyz-wing cells, right? r4c5 cannot see the 4 in r8c6. I think this xyz-wing should only be able to kill candidates in box 8.
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Chuck B
New kid on the Grid
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Joined: 09 Jul 2006
Posts: 4

PostPosted: Sun Aug 06, 2006 1:27 pm    Post subject: Reply with quote

Myth Jellies wrote:
Don't worry about the notation. Mine is not standard either.

I don't see how r8c6, r9c5, and r5c6 legitimately kill a 4 in r4c5. In order to be killed, the target candidate needs to see all of the 4's in the xyz-wing cells, right? r4c5 cannot see the 4 in r8c6. I think this xyz-wing should only be able to kill candidates in box 8.


Duh!!!! Binary thinking about a trivalent cell - brain cramp on my part? Wink

My bad, MJ - you are quite right! I just made a lucky guess...

-cb

"The difference between ignorance and stupidity? One is curable!"
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Chuck B
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Joined: 09 Jul 2006
Posts: 4

PostPosted: Mon Aug 07, 2006 8:37 pm    Post subject: Plan 'B' Reply with quote

Let me try to recover from my XYZ wing SNAFU. Rolling Eyes Embarassed Here's an implication attack through MJ's UR 'B' to show that r4c5 = 4 only if the puzzle is invalid.

First use MJ's XYZ wing to clear 4 from r4c46, then suppose that r4c5 = 4:
r4c5?=4 -> r4c1=5, exposing the bottom of 'B'; r6c3 is the corner of interest.

r4c5?=4 -> r3c5=3, via the chain r5c6->r5c7->r6c7->r1c7->r1c5, clearing '3' from r3c3's candidate list and exposing r1c3 (UR B's upper right corner) along the way.

The right side of UR 'B' forces the final chain:
r16c3={6,9} -> r8c3=8 -> r3c3=5 -> (r2c2=7, r3c2=6) -> r1c13={9} ?! (2 cells, 1 candidate)

Thus r4c5<>4 -> r5c6=4, etc., as claimed earlier. Not very pretty, but I think the logic is correct this time.

Thanks for your indulgence!

-cb

update: it's not really necessary to do the XYZ wing first, but then you need to do the resulting XY wing after 4 is eliminated from r4c5.
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