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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Sat May 14, 2011 5:43 pm Post subject: A tough Menneske |
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I found this in my folder of unsolved puzzles. | Code: | Puzzle: M4738962sh(28)
+-------+-------+-------+
| . . . | 1 . . | . . . |
| 3 . 7 | . 2 . | 6 . 8 |
| 5 . . | . . 8 | . . 7 |
+-------+-------+-------+
| . . 2 | 4 9 . | . . 5 |
| . 7 . | . . . | . 4 . |
| 6 . . | . 7 3 | 1 . . |
+-------+-------+-------+
| 7 . . | 3 . . | . . 1 |
| 4 . 9 | . 6 . | 5 . 2 |
| . . . | . . 2 | . . . |
+-------+-------+-------+ |
It's a tough one - you may enjoy it. I have been chipping away at it with some interesting moves.
The grid has lots of bivalue cells. Can we solve it without resorting to long chains?
Keith |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sat May 14, 2011 6:42 pm Post subject: |
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Don't know if this fits the "no long chain" criteria! Flightless xy-wing with transport...
| Quote: | | xy-wing(14-5)r2c2 - (5)r2c4=r9c4 - r7c5=r7c3 ; r6c3<>5 |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat May 14, 2011 7:12 pm Post subject: |
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I started with a W-Wing on 38 and after that I saw nothing. An AIC finished it. The 5 in r2c8 proved a 1 in r3c3.
I looked at Peter's XY-Wing twice and couldn't find the transport. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat May 14, 2011 7:33 pm Post subject: |
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After basics:
| Code: | +-------------------+-------------------+-------------------+
| 2 68 68 | 1 345 7 | 349 359 34 |
| 3 14 7 | 59 2 49 | 6 15 8 |
| 5 9 14 | 6 34 8 | 234 123 7 |
+-------------------+-------------------+-------------------+
| 18 138 2 | 4 9 6 | 378 378 5 |
| 9 7 38 | 28 1 5 | 238 4 6 |
| 6 45 45 | 28 7 3 | 1 28 9 |
+-------------------+-------------------+-------------------+
| 7 2 568 | 3 458 49 | 489 689 1 |
| 4 38 9 | 7 6 1 | 5 38 2 |
| 18 13568 13568 | 59 458 2 | 34789 36789 34 |
+-------------------+-------------------+-------------------+ |
I have now solved it, and will post my solution later. It does NOT involve the -38 W-wing in C23 and the two URs it reveals.
A fun time.
Keith
Last edited by keith on Sat May 14, 2011 7:43 pm; 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: Sat May 14, 2011 7:36 pm Post subject: |
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I did it again on Draw/Play. I still don't see Peter's transport but there's a transport which zaps the 5 from r9c2 which finishes it. (Shame on me for not seeing it on paper the first time).
P.S. Never mind, I see it now, but the transport could've stopped at r9c4.
| Code: |
+----------------+-----------+----------------+
| 2 68 68 | 1 345 7 | 349 359 34 |
| 3 14 7 | 59 2 49 | 6 15 8 |
| 5 9 14 | 6 34 8 | 234 123 7 |
+----------------+-----------+----------------+
| 18 138 2 | 4 9 6 | 378 378 5 |
| 9 7 38 | 28 1 5 | 238 4 6 |
| 6 45 45 | 28 7 3 | 1 28 9 |
+----------------+-----------+----------------+
| 7 2 568 | 3 458 49 | 489 689 1 |
| 4 38 9 | 7 6 1 | 5 38 2 |
| 18 13568 13568 | 59 458 2 | 34789 36789 34 |
+----------------+-----------+----------------+
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Play this puzzle online at the Daily Sudoku site |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun May 15, 2011 12:11 am Post subject: |
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Peter's wing, with Marty's truncated transport, makes the same elimination I eventually got to.
R2C4 and R1C7 are M-wing pincers on 9. With transport, eliminating 9 in R7C7.
I'd now like to introduce an amazing breakthrough, the "Almost Single". Some might prefer: "Finned Single".
| Code: | +-------------------+-------------------+-------------------+
| 2 68 68 | 1 345 7 | 349 359 34 |
| 3 14 7 | 59 2 49 | 6 15 8 |
| 5 9 14 | 6 34 8 | 234 123 7 |
+-------------------+-------------------+-------------------+
| 18 138 2 | 4 9 6 | 378 378 5 |
| 9 7 38 | 28 1 5 | 238 4 6 |
| 6 45 45 | 28 7 3 | 1 28 9 |
+-------------------+-------------------+-------------------+
| 7 2 568 | 3 458 49 | 48 69 1 |
| 4 38 9 | 7 6 1 | 5 38 2 |
| 18 13568 13568 | 59 458 2 | 79 679 34 |
+-------------------+-------------------+-------------------+ |
Note the single 3 in R1C9, with fin 4 in R1C9. If the single is true, R3C5 is 3. If the fin is true, R3C5 is 3. R3C5 <>4.
(OK, I know it is a W-wing, but this is so much more fun!  )
Along the same lines, there is the "Inverse Finned Pair"  starting in R7C7. If R7C7 is 4, R9C5 is 4. If R7C7 is 8, R9C5 is 8. Ergo, the inverse fin is not true: R9C5 is not 5.
I am now finding these kinds of eliminations more often, as a by-product of searching for M-wings and W-wings.
After the 34 W-wing: | Code: | +-------------------+-------------------+-------------------+
| 2 68 68 | 1 45 7 | 349 359 34 |
| 3 14 7 | 59 2 49 | 6 15 8 |
| 5 9 14 | 6 3 8 | 24 12 7 |
+-------------------+-------------------+-------------------+
| 18 138 2 | 4 9 6 | 378 378 5 |
| 9 7 38 | 28 1 5 | 238 4 6 |
| 6 45 45 | 28 7 3 | 1 28 9 |
+-------------------+-------------------+-------------------+
| 7 2 568 | 3 458 49 | 48 69 1 |
| 4 38 9 | 7 6 1 | 5 38 2 |
| 18 13568 13568 | 59 458 2 | 79 679 34 |
+-------------------+-------------------+-------------------+ |
A 45 W-wing, with Marty's transport, eliminates 5 in R9C2, solving the puzzle.
Keith  |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun May 15, 2011 5:48 am Post subject: |
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A 5-cell XY-Chain is a reasonable single-stepper:
| Code: | (5=4)r6c2 - (4=1)r2c2 - (1=5)r2c8 - (5=9)r2c4 - (9=5)r9c4 => r9c2<>5
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But not nearly as much fun as:
| Code: | External SIS on <18> UR r49c12: ( r1c2=8 or r8c2=8 ) and/or r2c2=1
(8)r8c2 - (8)r9c1
(8)r1c2 - (38=1)r84c2 - (1=8)r4c1 - (8)r9c1
(1)r2c2 - ( 1=4)r3 c3 - (4=5)r6c3 - r7c3 = (5-8)r7c5 = r9c5 - (8)r9c1
+-----------------------------------------------------------------------+
| 2 68 68 | 1 345 7 | 349 359 34 |
| 3 14 7 | 59 2 49 | 6 15 8 |
| 5 9 14 | 6 34 8 | 234 123 7 |
|-----------------------+-----------------------+-----------------------|
| 18 138 2 | 4 9 6 | 378 378 5 |
| 9 7 38 | 28 1 5 | 238 4 6 |
| 6 45 45 | 28 7 3 | 1 28 9 |
|-----------------------+-----------------------+-----------------------|
| 7 2 568 | 3 458 49 | 489 689 1 |
| 4 38 9 | 7 6 1 | 5 38 2 |
| 18 13568 13568 | 59 458 2 | 34789 36789 34 |
+-----------------------------------------------------------------------+
# 66 eliminations remain
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Unfortunately, the third stream is lengthy. _  _ |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun May 15, 2011 11:00 am Post subject: |
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Keith,
Your "Inverse Finned Pair", besides being a continuous AIC loop, is once again a Sue de Coq (4589 in r7c567|r9c4). Besides eliminating <5> in r9c5, it eliminates <8> in r7c3. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun May 15, 2011 12:00 pm Post subject: |
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| daj95376 wrote: | A 5-cell XY-Chain is a reasonable single-stepper:
| Code: | (5=4)r6c2 - (4=1)r2c2 - (1=5)r2c8 - (5=9)r2c4 - (9=5)r9c4 => r9c2<>5
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But not nearly as much fun as:
| Code: | External SIS on <18> UR r49c12: ( r1c2=8 or r8c2=8 ) and/or r2c2=1
(8)r8c2 - (8)r9c1
(8)r1c2 - (38=1)r84c2 - (1=8)r4c1 - (8)r9c1
(1)r2c2 - ( 1=4)r3 c3 - (4=5)r6c3 - r7c3 = (5-8)r7c5 = r9c5 - (8)r9c1
+-----------------------------------------------------------------------+
| 2 68 68 | 1 345 7 | 349 359 34 |
| 3 14 7 | 59 2 49 | 6 15 8 |
| 5 9 14 | 6 34 8 | 234 123 7 |
|-----------------------+-----------------------+-----------------------|
| 18 138 2 | 4 9 6 | 378 378 5 |
| 9 7 38 | 28 1 5 | 238 4 6 |
| 6 45 45 | 28 7 3 | 1 28 9 |
|-----------------------+-----------------------+-----------------------|
| 7 2 568 | 3 458 49 | 489 689 1 |
| 4 38 9 | 7 6 1 | 5 38 2 |
| 18 13568 13568 | 59 458 2 | 34789 36789 34 |
+-----------------------------------------------------------------------+
# 66 eliminations remain
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Unfortunately, the third stream is lengthy. _ _ |
Danny,
I don't buy it. R2C2=1 makes a pair 38 in B4, R3C1=1 and R9C1=8.
Keith |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun May 15, 2011 12:17 pm Post subject: |
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| Asellus wrote: | Keith,
Your "Inverse Finned Pair", besides being a continuous AIC loop, is once again a Sue de Coq (4589 in r7c567|r9c4). Besides eliminating <5> in r9c5, it eliminates <8> in r7c3. |
Asellus,
Thank you. When you pointed this out before I looked at the Sue de Coq, and decided it was not something I could reasonably look for on pencil and paper. I'll have another look, if only to understand the extra elimination.
Keith |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Sun May 15, 2011 2:01 pm Post subject: |
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| keith wrote: | | I don't buy it. R2C2=1 makes a pair 38 in B4, R3C1=1 and R9C1=8. |
Different paths may have different inferences, so I don't see an issue. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun May 15, 2011 2:48 pm Post subject: |
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Edit: Please skip down. Danny's solution is correct
| Code: | +-------------------+-------------------+-------------------+
| 2 68 68 | 1 345 7 | 349 359 34 |
| 3 1* 7 | 59 2 49 | 6 5a 8 |
| 5 9 4b | 6 34 8 | 234 123 7 |
+-------------------+-------------------+-------------------+
| 18 38 2 | 4 9 6 | 378 378 5 |
| 9 7 38 | 28 1 5 | 238 4 6 |
| 6 45 45 | 28 7 3 | 1 28 9 |
+-------------------+-------------------+-------------------+
| 7 2 568 | 3 458 49 | 489 689 1 |
| 4 38 9 | 7 6 1 | 5 38 2 |
| 18 3568d 13568 | 59c 458 2 | 34789 36789 34 |
+-------------------+-------------------+-------------------+ |
R2C2=1 forces 5 in a and 4 in b. a quickly forces 5 in c (look at C4), while b quickly forces 5 in d (look at C3 and B7).
This is a contradiction, two 5s in R9. Thus, R2C2<>1 and you have a one-stepper.
What I have said up to here has nothing to do with the UR.
Danny got unlucky, doing something we all do (and get away with) every day. In my view, the UR says that R128C2 are a pseudocell <18>, but I do not see that that leads anywhere.
Maybe someone else can explain this better than I can:
If A and / or B is true, then <AB> is a pseudocell. Nothing wrong with that. It allows that A or B may be false.
If A is true, then C is true.
If B is true, then C is true.
A and / or B is true, therefore C is true.
This is not correct, though we do it all the time. If, for example, B is in fact false, then any inference we draw by assuming it is true is suspect.
Keith
(Edited to fix a typo.)
Last edited by keith on Sun May 15, 2011 5:18 pm; edited 2 times in total |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun May 15, 2011 3:21 pm Post subject: |
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Edit: Please skip down. Danny's solution is correct
| ronk wrote: | | keith wrote: | | I don't buy it. R2C2=1 makes a pair 38 in B4, R3C1=1 and R9C1=8. |
Different paths may have different inferences, so I don't see an issue. |
I may have misunderstood Danny's post. If so, I apologize. The way I read his post is that to avoid the DP (R8C2 <8> or R1C2 <8>) and / or R2C2 <1>.
His three chains say:
1. If R8C2 <8> then R9C1 <>8.
2. If R1C2 <8> then R9C1 <>8.
3. If R2C2 <1> then R9C1 <>8.
Therefore, as I read it, R9C1 <1>, and the puzzle is solved. The logic is not correct.
1. If R8C2 <8> then R9C1 <>8. Correct.
2. If R1C2 <8> then R9C1 <>8. Maybe, but this is suspect.
3. If R2C2 <1> then R9C1 <>8. Incorrect. You can also infer that R9C1 <8>.
The correct conclusion is that R2C2 <>1
| Quote: | | Different paths may have different inferences |
I don't disagree, but what if the inferences contradict each other?
(Edit: I think ronk is implying, if you have different and even contradictory inferences, pick the one that suits your needs. I am coming around to that.)
Keith
Last edited by keith on Sun May 15, 2011 5:19 pm; edited 1 time in total |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun May 15, 2011 4:19 pm Post subject: |
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| keith wrote: | [(Edit: I think ronk is implying, if you have different and even contradictory inferences, pick the one that suits your needs. I am coming around to that.)
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Keith, that's exactly my understanding! Choose the one that works for you!
The fact that there is a contradiction forcing chain from (1)r2c2 just proves that in fact r2c2<>1 - its does not detract from the logic on showing one consequence of assuming r2c2=1.
Fwiw by putting your chain fragment and Danny's together you get an AIC that shows that r2c2<>1
| Code: | (1=38)r4c2|r5c3 - (8=1)r4c1 - (1=8)r9c1 - r9c5=(8-5)r7c5=r7c3 - (5=4)r6c3 - (4=1)r3c3 ; r2c2<>1
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun May 15, 2011 4:49 pm Post subject: |
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Hmmm! Of all the things I didn't expect, my solution leading to a discussion on logical deduction was at the top of the list. I'm hoping that Keith is just having an "off day".
When one of three conditions must be true, then I say:
Now, let's assume that A is in fact true and we have:
I hope you don't have a problem with a true assumption leading to a true conclusion.
What about the case where B is false and C is false. The first thing to remember is that nothing conclusive can be solely based on a false assumption. A false assumption may lead to only false conclusions. However, it's also possible that a false assumption can (eventually) lead to a true conclusion ... and subsequent true conclusions. In particular, it's possible for multiple paths from a single false assumption to lead to multiple conclusions -- some of which are contradictory.
Now, let's assume, even if neither of B nor C is true, that at least one path from each exists for:
| Code: | *) if B -> D
*) if C -> D
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Since one of {A,B,C} must be true, and they have a common conclusion, then the conclusion must be true.
In my solution, I knew that at least one of three assumptions was true. I then showed that paths existed from each that led to a common conclusion. I never claimed that all paths from all three assumptions would lead to a common conclusion.
Regards, Danny
===== ===== ===== ===== ===== Addendum
Keith sent a pm saying that he'd experienced an off moment. If I had a US $1 for every off moment that I'd experienced, I would be in France chasing pretty women instead of posting messages here. _  _ |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun May 15, 2011 6:03 pm Post subject: |
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| Quote: | | Keith sent a pm saying that he'd experienced an off moment. If I had a US $1 for every off moment that I'd experienced, I would be in France chasing pretty women instead of posting messages here. |
Actually, I said I was wrong.
Are there any volunteers to establish the "Send Danny and Keith to France" fund?
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun May 15, 2011 6:27 pm Post subject: |
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| Quote: | | If I had a US $1 for every off moment that I'd experienced, I would be in France chasing pretty women instead of posting messages here. |
| Quote: | | Are there any volunteers to establish the "Send Danny and Keith to France" fund? |
With you two guys chasing, I suspect most grandmothers could outrun you. |
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