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Myth Jellies Hooked
Joined: 04 Apr 2006 Posts: 42

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


Code:  **
.4.5.1.8.
..2.6.9..
...9.7...
++
.17...69.
28.....51
.3.....4.
++
...789...
.........
..12.68..
** 
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 xyzwing 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) 68UR in r38c13. Two strong 8links form a corner at r3c1, and the opposite corner contains just the base candidates, 68. Therefore r1c3 <> 6.
B) 69UR in r16c13. Two strong 9links form a corner at r1c1. There is also a strong 6link that does not include that corner, therefore r1c1 <> 6.
C) 28UR in r46c69. Two strong 2links form a corner at r4c6. Two strong 8links also form a corner at r6c9. For the same reason as in B, we know that r4c6 <> 8, and r6c9 <> 2.
D) 79UR 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.
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&28&8)[r4c9,r6c6] = (8)r6c9 URC (8=2345)r4c6 => r4c6 <> 8
C2: (8)r6c9 = (8&82&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
Joined: 09 Jul 2006 Posts: 4

Posted: Sat Aug 05, 2006 8:03 pm Post subject: 


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
Joined: 04 Apr 2006 Posts: 42

Posted: Sun Aug 06, 2006 12:25 am Post subject: 


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 xyzwing cells, right? r4c5 cannot see the 4 in r8c6. I think this xyzwing should only be able to kill candidates in box 8. 

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Chuck B New kid on the Grid
Joined: 09 Jul 2006 Posts: 4

Posted: Sun Aug 06, 2006 1:27 pm Post subject: 


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 xyzwing cells, right? r4c5 cannot see the 4 in r8c6. I think this xyzwing should only be able to kill candidates in box 8. 
Duh!!!! Binary thinking about a trivalent cell  brain cramp on my part?
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 New kid on the Grid
Joined: 09 Jul 2006 Posts: 4

Posted: Mon Aug 07, 2006 8:37 pm Post subject: Plan 'B' 


Let me try to recover from my XYZ wing SNAFU. 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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