ravel wrote:In this case a or b is part of the ntuple. Then you also can safely eliminate a or b from the 2 UR cells.
I dont know, if this variant already has a name.
"Cannibalistic UR Type 3?" Seriously, I view that as a continuous chain.
ravel wrote:In this case a or b is part of the ntuple. Then you also can safely eliminate a or b from the 2 UR cells.
I dont know, if this variant already has a name.
ravel wrote:I dont know, if this variant already has a name.
. 18 .
. 358 .
158 . 358Y

58 . 58
ronk wrote:ravel wrote:In this case a or b is part of the ntuple. Then you also can safely eliminate a or b from the 2 UR cells.
I dont know, if this variant already has a name.
"Cannibalistic UR Type 3?" Seriously, I view that as a continuous chain.
**
.4.5.1.8.
..2.6.9..
...9.7...
++
.17...69.
28.....51
.3.....4.
++
...789...
.........
..12.68..
**
**
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 
**
abW  abX

abY  abZ
b[w]
/
ab[z]  (ab&ab&ab)[w,x,y]=(W or X or Y)[w,x,y]  a[x] = a[w]  b[w]
\
a[y] = a[w]  b[w]
Myth Jellies wrote:Proof of the URC weak link
Say you have
 Code: Select all
abW  abX

abY  abZ
in cells w, x, y, and z; with strong links a[w]=a[x] and a[w]=a[y] forming a strong corner on w. Opposite cell is z.
deadly pattern avoidance tells you that ab[z]  (ab&ab&ab)[w,x,y]
continuing we get...
 Code: Select all
b[w]
/
ab[z]  (ab&ab&ab)[w,x,y]=(W or X or Y)[w,x,y]  a[x] = a[w]  b[w]
\
a[y] = a[w]  b[w]
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.
Myth Jellies wrote: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.
keith wrote:If you make this nonUR reduction and continue, you get to a point where basic techniques run out. However, UR's A and D survive, and your reductions solve the puzzle.
ronk wrote:Following the xyzwing without using any URs, a finned swordfish in r249c159 solves the puzzle too ... but then we wouldn't have had any fun with URs.
Here I am not exactly trying to show that b is excluded from the w cell. I am trying prove the URC link, i.e. to show that b in the w cell excludes a or b in the z cell. Once you establish that, then as you say, strong links involving b[z] easily exclude b from cell w. I showed that in the A, B, and C chains above.ronk wrote:Myth Jellies wrote:Proof of the URC weak link
Say you have
 Code: Select all
abW  abX

abY  abZ
in cells w, x, y, and z; with strong links a[w]=a[x] and a[w]=a[y] forming a strong corner on w. Opposite cell is z.
deadly pattern avoidance tells you that ab[z]  (ab&ab&ab)[w,x,y]
continuing we get...
 Code: Select all
b[w]
/
ab[z]  (ab&ab&ab)[w,x,y]=(W or X or Y)[w,x,y]  a[x] = a[w]  b[w]
\
a[y] = a[w]  b[w]
The exclusion of b from the w cell depends upon b[x]=b[z] and/or b[y]=b[z]. I don' t see that requrement showing up in the above notation ...
ronk wrote:probably because I don't understand the ab[z]  (ab&ab&ab)[w,x,y] thingie.
Would you please put that weak link into words?
TIA, Ron
I agree it sounds like I said all 7's move out, but that is not what I meant. Just to be clear, without uniqueness considerations the rectangle could contain two 7's, one in each pairing. With uniqueness and the locked 9's, only one 7 can be allowed in the rectangle. Therefore one of the pairings must lose its 7. We do not assume which pairing loses the 7, however, any cell which sees all possible nonUR landing zones for both pairings, cannot contain a 7.keith wrote:MJ,
Question:
You said: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.
I agree that the 7's have to move out, and I think I agree that two 7's have to move out. Also, the two that move out cannot be on a diagonal, since that leaves the UR as still possible with 7's on the other diagonal.
Since the pairings in columns are also constrained by a box, you are free to consider alternate nonUR locations for the 7 in either that box or that column.keith wrote:You have chosen the 7's to move out of one or other of the columns.
Why do you not consider the 7's to possibly move out of one of the rows?
Keith wrote:
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++
   # 
 12  123 
   # 
++
 12  124 
   # 
   34 
++
   # 
   # 
   # 
++
There are obvious variations when the extra candidates are in the same row and/or the same box. This is usually called a Type 3 Unique Rectangle. There is also a diagonal variant.
[cell #]<>1234
++
 12 . 123 
 # . . 
1234 . . 
+
 . . . 
 . . . 
 124 . 12 
++