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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Wed Mar 10, 2010 12:47 am Post subject: Puzzle 10/03/09 (B) BBDB as VH+ puzzle |
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Last puzzle from a file of puzzles containing XYZ-Wing.
| Code: | +-----------------------+
| 5 . . | 3 . 8 | . 9 . |
| . 8 . | . . 9 | . . . |
| . . . | 4 . . | . 8 . |
|-------+-------+-------|
| 9 . 7 | 8 2 . | . . 6 |
| . . . | 9 4 . | 2 . . |
| 2 1 . | . . . | . 7 9 |
|-------+-------+-------|
| . . . | . 9 . | 7 3 . |
| 7 . 3 | . . 4 | 9 . 1 |
| . . . | 7 . 3 | . 6 4 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Wed Mar 10, 2010 4:13 pm Post subject: |
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This one took three steps: | Code: | xy-wing 485
skyscraper 8
*--------------------------------------------------------------------*
| 5 27 146 | 3 16 8 | 146 9 27 |
| 134 8 1246 | 125 1567 9 | 13456 245 2357 |
| 13 279 1269 | 4 1567 25 | 1356 8 2357 |
|----------------------+----------------------+----------------------|
| 9 345 7 | 8 2 1 | 345 45 6 |
| 6 35 58 | 9 4 7 | 2 1 358 |
| 2 1 48 | 56 3 56 | 48 7 9 |
|----------------------+----------------------+----------------------|
| 148 456 145 | 1256 9 256 | 7 3 258 |
| 7 56 3 | 256 8 4 | 9 25 1 |
| 18 29 29 | 7 15 3 | 58 6 4 |
*--------------------------------------------------------------------*
(8=4)6c7 - r1c7 = r1c3 - r2c1 = (4-8)r7c1 = (8)r7c9; r5c9,r9c7<>8
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Wed Mar 10, 2010 7:53 pm Post subject: |
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I know that all URs are not in the VH+ group, but this puzzle seemed like a good opportunity to exploit them.
However, I started with a xy-wing 4-58 with vertex in r6c7 since no potential URs were immediately usable(?), resulting in the following code.
| Code: |
*--------------------------------------------------------------------*
| 5 27 146 | 3 16 8 | 146 9 27 |
| 1346 8 1246 | 12 1567 9 | 13456 45 2357 |
| 136 279 1269 | 4 1567 25 | 1356 8 2357 |
|----------------------+----------------------+----------------------|
| 9 345 7 | 8 2 1 | 34 45 6 |
| 68 356 568 | 9 4 7 | 2 1 358 |
| 2 1 48 |#56 3 #56 | 48 7 9 |
|----------------------+----------------------+----------------------|
| 1468 #456 14568 | 12 9 #256 | 7 3 58 |
| 7 #56 3 |#56 8 4 | 9 2 1 |
| 18 29 29 | 7 15 3 | 58 6 4 |
*--------------------------------------------------------------------* |
Then, a 6-cell DP on 56 in r678c246 marked #provides a strong inference
(5=2)r3c6 - DP56[(2)r7c6 = (4)r7c2]r678c246 - r7c1 = r2c1 - (4=5)r2c8; r2c5|r3c79<>5 plus three additional transports results in the following code.
| Code: | *-----------------------------------------------------------*
| 5 27 146 | 3 #16 8 |#146 9 27 |
| 1346 8 1246 | 12 167 9 | 1456 45 237 |
| 136 279 1269 | 4 #1567 25 |#16 8 237 |
|-------------------+-------------------+-------------------|
| 9 45 7 | 8 2 1 | 3 45 6 |
| 68 3 568 | 9 4 7 | 2 1 58 |
| 2 1 48 | 56 3 56 | 48 7 9 |
|-------------------+-------------------+-------------------|
| 148 46 1458 | 12 9 26 | 7 3 58 |
| 7 56 3 | 56 8 4 | 9 2 1 |
| 18 29 29 | 7 15 3 | 58 6 4 |
*-----------------------------------------------------------* |
Type 6 UR 16 in r13c57 marked #; one of three conditions must be true to prevent the deadly pattern.
(4)r1c7 - (4=8)r6c7 - (8=5)r9c7 - (5=1)r9c5 - (1=2)r7c4 - (2=6)r7c6 - (6=5)r6c6 - (5=2)r3c6,
(5)r3c5 - (5=2)r3c6,
(7)r3c5 - r3c2 = r1c2 - (7=2)r1c9.
Thus r3c6 and r1c9 are pincers on 2; r3c9<>2
(A lot of work but it helps complete the puzzle given the following move)
UR29 in r39c23 with x-wing 9 overlay deletes 2 in r3c23.
Ted  |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Wed Mar 10, 2010 8:54 pm Post subject: |
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almost skyscraper on 8, given a way out via the 4
| Code: | +--------------------+-----------------+----------------------+
| 5 27 146 | 3 16 8 | 16(4) 9 27 |
| 136(4) 8 1246 | 125 1567 9 | 1356(4) 25(4) 2357 |
| 136 279 1269 | 4 1567 25 | 1356 8 2357 |
+--------------------+-----------------+----------------------+
| 9 345 7 | 8 2 1 | 345 45 6 |
| 6(8) 356 568 | 9 4 7 | 2 1 35-8 |
| 2 1 4-8 | 56 3 56 | (48) 7 9 |
+--------------------+-----------------+----------------------+
| 16(48) 456 14568 | 1256 9 256 | 7 3 258 |
| 7 56 3 | 256 8 4 | 9 25 1 |
| 1(8) 29 29 | 7 15 3 | 5(8) 6 4 |
+--------------------+-----------------+----------------------+ |
[skyscraper(8)r569c17] = (8-4)r7c1 = (4)r2c1 - (4)r2c78 = (4)r12c7 - (4=8)r6c7;
so either the skyscraper is true or if not then it loops back around to the 8 in r6c7.
either way the 8 in r5c9 is toast. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Wed Mar 10, 2010 9:38 pm Post subject: |
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| storm_norm wrote: | almost skyscraper on 8, given a way out via the 4
| Code: | +--------------------+-----------------+----------------------+
| 5 27 146 | 3 16 8 | 16(4) 9 27 |
| 136(4) 8 1246 | 125 1567 9 | 1356(4) 25(4) 2357 |
| 136 279 1269 | 4 1567 25 | 1356 8 2357 |
+--------------------+-----------------+----------------------+
| 9 345 7 | 8 2 1 | 345 45 6 |
| 6(8) 356 568 | 9 4 7 | 2 1 35-8 |
| 2 1 4-8 | 56 3 56 | (48) 7 9 |
+--------------------+-----------------+----------------------+
| 16(48) 456 14568 | 1256 9 256 | 7 3 258 |
| 7 56 3 | 256 8 4 | 9 25 1 |
| 1(8) 29 29 | 7 15 3 | 5(8) 6 4 |
+--------------------+-----------------+----------------------+ |
[skyscraper(8)r569c17] = (8-4)r7c1 = (4)r2c1 - (4)r2c78 = (4)r12c7 - (4=8)r6c7;
so either the skyscraper is true or if not then it loops back around to the 8 in r6c7.
either way the 8 in r5c9 is toast. |
Great find Norm 
I have occasionally looked for "almost" or "finned" patterns. I have found a couple of up-down skyscraper and some finned wings, but never a skyscraper
like yours. 
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Mar 10, 2010 10:45 pm Post subject: |
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| Norm: I'm glad that you and Ted understand what you're doing, but it doesn't even come close to making sense to me. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Mar 10, 2010 10:55 pm Post subject: |
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| storm_norm wrote: | almost skyscraper on 8, given a way out via the 4
| Code: | +--------------------+-----------------+----------------------+
| 5 27 146 | 3 16 8 | 16(4) 9 27 |
| 136(4) 8 1246 | 125 1567 9 | 1356(4) 25(4) 2357 |
| 136 279 1269 | 4 1567 25 | 1356 8 2357 |
+--------------------+-----------------+----------------------+
| 9 345 7 | 8 2 1 | 345 45 6 |
| 6(8) 356 568 | 9 4 7 | 2 1 35-8 |
| 2 1 4-8 | 56 3 56 | (48) 7 9 |
+--------------------+-----------------+----------------------+
| 16(48) 456 14568 | 1256 9 256 | 7 3 258 |
| 7 56 3 | 256 8 4 | 9 25 1 |
| 1(8) 29 29 | 7 15 3 | 5(8) 6 4 |
+--------------------+-----------------+----------------------+ |
[skyscraper(8)r569c17] = (8-4)r7c1 = (4)r2c1 - (4)r2c78 = (4)r12c7 - (4=8)r6c7;
so either the skyscraper is true or if not then it loops back around to the 8 in r6c7.
either way the 8 in r5c9 is toast. |
Could someone explain this in English? It looks to me if the Skyscraper is false, that means r7c1=8 which looks to force 8 in r5c9. If r7c1=8, r7c9<>8 and r5c9=8. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Wed Mar 10, 2010 11:54 pm Post subject: |
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alright, english.
and algebra
the next logical step in creating chains is the use of substitution.
kind of like in math when we learn
D = A + (C*B)
then we can substitute that expression for D everywhere we see D in this equation
(D/3) - (100 *D) + (5*D) = F
to find F.
right?
no different with nodes in a chain.
all i am doing is using a AIC as a node in a chain.
the node
[(8)r5c1 = (8)r9c1 - (8)r9c7 = (8)r6c7]
is being used like any other candidate.
it can be seen as being either true or false.
the key is to realize that this works because if the skyscraper and the 8 at r7c1 are both false then no more 8's exist in column 1.
that is why the above chain is strongly linked to the 8 at r7c1.
that is why I can write this
[(8)r5c1 = (8)r9c1 - (8)r9c7 = (8)r6c7] = (8)r7c1
| Code: | skyscraper is a node
|(8)r5c1 = (8)r9c1 - (8)r8c7 = (8)r6c7|
------------------|--------------------
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(node) = (8-4)r7c1 = (4)r2c1 - (4)r1c2 = (4)r1c7 - (4=8)r6c7 |
as clearly as I can
hopefully the chain makes sense in either direction.
either the 8 is true at r6c7 (makes 8 false at r5c9) or
the node is true.
the node is a skyscraper.
that skyscraper true makes the 8 at r5c9 false.
either way the 8 at r5c9 is false. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Thu Mar 11, 2010 12:58 am Post subject: |
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[Withdrawn: I should have stayed out of it.]
Last edited by daj95376 on Thu Mar 11, 2010 5:42 am; edited 1 time in total |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Mar 11, 2010 1:11 am Post subject: |
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| Thank you both. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Mar 11, 2010 1:51 am Post subject: |
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there is another way to show this elimination and its in the form of a diagram.
(8)r5c1
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(8-4)r7c1 = (4)r2c1 - (4)r1c3 = (4)r1c7 - (4=8)r6c7
||
(8)r9c1 - (8)r9c7 = (8)r6c7
because this shows more clearly that at least one of the 8's in column 1 is true.
the part in red would be the main structure, with the (8)r5c1 being the fin cell.
just like a finned x-wing,
either the x-wing is true or the fin is true.
[x-wing(A)] = finA |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Thu Mar 11, 2010 4:26 am Post subject: |
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[Withdrawn: I should have stayed out of it.]
Last edited by daj95376 on Thu Mar 11, 2010 5:28 am; edited 1 time in total |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Thu Mar 11, 2010 4:55 am Post subject: |
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Norm has provided a logical explanation of this condition, but my view is more simplistic. I view it as a "finned" condition where the pattern is true or the fin(s) is true; anything deleted under both conditions may be deleted. This approach is used and accepted for many patterns such as x-wings, xy-wings, swordfish, etc.
Ted |
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oaxen
Joined: 10 Jul 2006
Posts: 96
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| Posted: Fri Mar 12, 2010 8:11 am Post subject: |
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| "1" in r2c4. One stepper. |
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