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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Tue Dec 23, 2008 4:14 am Post subject: Set H Puzzle 17 |
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This puzzle just keeps on giving!
Besides presenting a difficult traditional solution, it can be cracked with any of the following:
1) Either of 2x W-Wings
2) A monster M-Wing
3) An XY-Wing extended to a 5-cell XY-Chain
Heck, there may even be URs that I didn't catch.
| Code: | +-----------------------+
| 5 . . | . . . | . . . |
| . 8 3 | . 6 5 | . . 2 |
| . 6 7 | . . . | . . . |
|-------+-------+-------|
| . . . | . . . | 9 . 4 |
| . 5 . | . 4 . | 7 2 . |
| . 7 . | . . 2 | . 5 . |
|-------+-------+-------|
| . . . | 9 7 . | 8 4 . |
| . . . | . 5 4 | 6 3 . |
| . 3 . | 8 . . | . . 9 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site
BTW: There are M-Wings and W-Wings that don't crack the puzzle!  |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Dec 23, 2008 5:20 am Post subject: |
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| Maybe I made a lucky mistake, but it solved with one UR. I'll wait and see what others say before trying it again. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Dec 23, 2008 7:04 am Post subject: |
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After basics:
| Code: | +----------------+----------------+----------------+
| 5 4 29 | 13 289 89 | 13 6 7 |
| 19 8 3 | 7 6 5 | 4 19 2 |
| 129 6 7 | 4 239 19 | 5 189 38 |
+----------------+----------------+----------------+
| 2368 12 126 | 5 38 7 | 9 18 4 |
| 389 5 19 | 136 4 189 | 7 2 368 |
| 389 7 4 | 136 389 2 | 13 5 368 |
+----------------+----------------+----------------+
| 26 12 126 | 9 7 3 | 8 4 5 |
| 7 9 8 | 2 5 4 | 6 3 1 |
| 4 3 5 | 8 1 6 | 2 7 9 |
+----------------+----------------+----------------+ |
The <13> rectangle is across four blocks and is not a UR. Not a way to get lucky, because <6> in R6C4 is incorrect.
| Quote: | | According to Sudoku Susser, it takes 1 X-wing and four XY-wings. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Dec 23, 2008 7:27 am Post subject: |
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| keith wrote: | After basics:
| Code: | +----------------+----------------+----------------+
| 5 4 29 | 13c 289 89 | 13d 6 7 |
| 19 8 3 | 7 6 5 | 4 19 2 |
| 129 6 7 | 4 239b 19 | 5 189f 38e |
+----------------+----------------+----------------+
| 2368 12 126 | 5 38a 7 | 9 1-8 4 |
| 389 5 19 | 136 4 189 | 7 2 368 |
| 389 7 4 | 136 389 2 | 13 5 368 |
+----------------+----------------+----------------+
| 26 12 126 | 9 7 3 | 8 4 5 |
| 7 9 8 | 2 5 4 | 6 3 1 |
| 4 3 5 | 8 1 6 | 2 7 9 |
+----------------+----------------+----------------+ |
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Following <3> through abcde shows the implication that if a is <3>, e is also <3>. Strong link ef on <8>. af is an M-wing with pincers <8>. Solving R4C8 and the puzzle. 
Danny, is this your "monster M-wing"? (Actually, this ia a half M-wing.)
Edit: a and e are also a W-wing, strong link <8> in C8. Taking out <3> in b, also solving the puzzzle.
Keith
Last edited by keith on Wed Dec 24, 2008 12:34 am; edited 1 time in total |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue Dec 23, 2008 8:08 am Post subject: |
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| If r4c1 is 26, a 126 Deadly Pattern results in r47c123. (Note that this would make r1c3=2.) Thus, r4c1 is 38, which solves the puzzle. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Dec 23, 2008 4:50 pm Post subject: |
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| keith wrote: | After basics:
| Code: | +----------------+----------------+----------------+
| 5 4 29 | 13 289 89 | 13 6 7 |
| 19 8 3 | 7 6 5 | 4 19 2 |
| 129 6 7 | 4 239 19 | 5 189 38 |
+----------------+----------------+----------------+
| 2368 12 126 | 5 38 7 | 9 18 4 |
| 389 5 19 | 136 4 189 | 7 2 368 |
| 389 7 4 | 136 389 2 | 13 5 368 |
+----------------+----------------+----------------+
| 26 12 126 | 9 7 3 | 8 4 5 |
| 7 9 8 | 2 5 4 | 6 3 1 |
| 4 3 5 | 8 1 6 | 2 7 9 |
+----------------+----------------+----------------+ |
The <13> rectangle is across four blocks and is not a UR. Not a way to get lucky, because <6> in R6C4 is incorrect.
| Quote: | | According to Sudoku Susser, it takes 1 X-wing and four XY-wings. |
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In boxes 47, the 12 UR is both a Type 2 and Type 4. As a Type 4, the 6 is a strong link in c3 and after removing the 1s, just one 1 remains in c3. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Dec 23, 2008 4:57 pm Post subject: |
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| keith wrote: | Coloring on <3> abcde shows that if a is <3>, e is <3>. Strong link ef on <8>. af is an M-wing with pincers <8>. Solving R4C8 and the puzzle.
Danny, is this your "monster M-wing"?
Edit: a and e are also a W-wing, strong link <8> in C8. Taking out <3> in b, also solving the puzzzle.
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No, that's not my monster M-Wing. While reviewing the puzzle before posting, my solver stopped at your W-Wing elimination.
| Code: | (9=1)a - (1)b = (1)c - (1)d = (1)e - (1)ff = (1-9)g = (9)h
+--------------------------------------------------------------+
| 5 4 h29 | b13 28-9 8-9 | c13 6 7 |
| 19 8 3 | 7 6 5 | 4 19 2 |
| 12-9 6 7 | 4 239 a19 | 5 189 38 |
|--------------------+--------------------+--------------------|
| 2368 f12 f126 | 5 38 7 | 9 e18 4 |
| 389 5 g19 | 136 4 189 | 7 2 368 |
| 389 7 4 | 136 389 2 | d13 5 368 |
|--------------------+--------------------+--------------------|
| 26 12 126 | 9 7 3 | 8 4 5 |
| 7 9 8 | 2 5 4 | 6 3 1 |
| 4 3 5 | 8 1 6 | 2 7 9 |
+--------------------------------------------------------------+
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue Dec 23, 2008 6:53 pm Post subject: |
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| Marty R. wrote: | | In boxes 47, the 12 UR is both a Type 2 and Type 4. |
I'm afraid the 12UR is of no help. It would be a Type 2 if the <6>s were not conjugate (in which case the Type 2 would eliminate other <6>s in c3). It would be a Type 4 if either the <1>s or the <2>s in r47c3 were conjugate (in which case whichever of the two was NOT conjugate would be eliminated from those cells). The conjugate <6>s actually destroy the 12UR.
It happens that r5c3 must be <1> due to the 6-cell 126 DP I mentioned previously. I can't see any way to get to that with just a UR. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Dec 23, 2008 10:26 pm Post subject: |
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| Asellus wrote: | | Marty R. wrote: | | In boxes 47, the 12 UR is both a Type 2 and Type 4. |
I'm afraid the 12UR is of no help. It would be a Type 2 if the <6>s were not conjugate (in which case the Type 2 would eliminate other <6>s in c3). It would be a Type 4 if either the <1>s or the <2>s in r47c3 were conjugate (in which case whichever of the two was NOT conjugate would be eliminated from those cells). The conjugate <6>s actually destroy the 12UR.
It happens that r5c3 must be <1> due to the 6-cell 126 DP I mentioned previously. I can't see any way to get to that with just a UR. |
I'm sure you're right even though I don't follow the explanation. It just looks to an amateur like me that if one of the cells has to be = 6, then a 1 can't be present as a candidate in either. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue Dec 23, 2008 11:42 pm Post subject: |
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| Marty R. wrote: | | It just looks to an amateur like me that if one of the cells has to be = 6, then a 1 can't be present as a candidate in either. |
Marty: My turn to try and clear it up for you.
You are correct that one of [r4c3]=6 or [r7c3]=6 must be true. However, this still leaves [r7c3]=12 or [r4c3]=12, respectively. With only this information, we can not say for certain that <1> or <2> has been eliminated from both cells. The best we can say is that we could eliminate other <6s> in [c3] if they existed. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Wed Dec 24, 2008 12:09 am Post subject: |
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Marty,
What can you say about this UR?
1. One of R47C3 is <6>. True, but not useful. There are no other candidates <6> that see both of R47C3.
2. The solutions for R47C3 are <12> and <6>. Let's try them:
| Code: | +----------------+----------------+----------------+
| 5 4 29 | 13 289 89 | 13 6 7 |
| 19 8 3 | 7 6 5 | 4 19 2 |
| 129 6 7 | 4 239 19 | 5 189 38 |
+----------------+----------------+----------------+
| 2368 12 6 | 5 38 7 | 9 18 4 |
| 389 5 19 | 136 4 189 | 7 2 368 |
| 389 7 4 | 136 389 2 | 13 5 368 |
+----------------+----------------+----------------+
| 26 12 12 | 9 7 3 | 8 4 5 |
| 7 9 8 | 2 5 4 | 6 3 1 |
| 4 3 5 | 8 1 6 | 2 7 9 |
+----------------+----------------+----------------+
OR
+----------------+----------------+----------------+
| 5 4 29 | 13 289 89 | 13 6 7 |
| 19 8 3 | 7 6 5 | 4 19 2 |
| 129 6 7 | 4 239 19 | 5 189 38 |
+----------------+----------------+----------------+
| 2368 12 12 | 5 38 7 | 9 18 4 |
| 389 5 19 | 136 4 189 | 7 2 368 |
| 389 7 4 | 136 389 2 | 13 5 368 |
+----------------+----------------+----------------+
| 26 12 6 | 9 7 3 | 8 4 5 |
| 7 9 8 | 2 5 4 | 6 3 1 |
| 4 3 5 | 8 1 6 | 2 7 9 |
+----------------+----------------+----------------+ |
The candidate lists are valid, and both avoid the DP.
If you play them both out, you will find that one is incorrect, but you cannot decide that here and now.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Dec 24, 2008 1:21 am Post subject: |
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| I know that the more I ask, the deeper the hole I'll dig for myself. Just one question: is it a generality that a UR with a Type 2 pattern can't be used as a Type 4? |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Dec 24, 2008 1:48 am Post subject: |
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| Marty R. wrote: | | I know that the more I ask, the deeper the hole I'll dig for myself. Just one question: is it a generality that a UR with a Type 2 pattern can't be used as a Type 4? |
I should wait for more knowledgeable members to answer your question.
What I do know is that I've seen it mentioned more than once that searching for a UR Type 4 should occur after searching for other Types.
I know for a fact that many UR solvers search for all URs and eliminations before performing any eliminations. I recall a puzzle that had two overlapping URs. Performing only one of the URs left a difficult solution. Performing both URs as one step resulted in a reasonable solution.
What you need to remember is that a UR Type 4 also requires that one of the DP candidates be locked in an X-Wing pattern using these four cells. That's not the case above for either <1> or <2>. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Wed Dec 24, 2008 4:12 am Post subject: |
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Marty,
I believe you can make both Type 2 and/or Type 4 eliminations on the same UR. Take a look at this:
http://www.brainbashers.com/sudokuuniquerectangles.asp
Danny is correct, that since the Type 4 takes out one of the DP candidates, it might obscure a Type 2 (or 3?).
Danny is not quite correct, in that a Type 4 only requires that one of the DP candidates be locked in the two cells with extra candidates.
A UR on an X-wing is a Type 6.
(I hope this doesn't further confuse the issue.)
Keith |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Dec 24, 2008 4:54 am Post subject: |
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| daj95376 wrote: | | What you need to remember is that a UR Type 4 also requires that one of the DP candidates be locked in an X-Wing pattern using these four cells. |
Example from Sudopedia:
| Code: | <27> UR Type 4 in [r79c58] => [r9c58]<>2
+-----------------------------------------------------------+
| 2 6 8 | 4 3 7 | 1 9 5 |
| 4 15 19 | 29 6 58 | 7 3 28 |
| 59 3 7 | 29 1 58 | 6 28 4 |
|-------------------+-------------------+-------------------|
| 1 458 34 | 37 9 2 | 358 6 78 |
| 59 25 6 | 37 8 4 | 235 1 279 |
| 7 28 239 | 1 5 6 | 238 4 289 |
|-------------------+-------------------+-------------------|
| 8 9 5 | 6 *27 1 | 4 *27 3 |
| 6 7 24 | 8 24 3 | 9 5 1 |
| 3 124 124 | 5 *27+4 9 | 28 *27+8 6 |
+-----------------------------------------------------------+
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| Code: | X-Wing r79\c58 pattern on <7> in UR cells
+-----------------------------------+
| . . . | . . 7 | . . . |
| . . . | . . . | 7 . . |
| . . 7 | . . . | . . . |
|-----------+-----------+-----------|
| . . . | 7 . . | . . 7 |
| . . . | 7 . . | . . 7 |
| 7 . . | . . . | . . . |
|-----------+-----------+-----------|
| / / / | / 7 / | / 7 / |
| . 7 . | . . . | . . . |
| / / / | / 7 / | / 7 / |
+-----------------------------------+
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| keith wrote: | Danny is not quite correct, ...
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keith: Do you have an example of a UR Type 4 w/o this X-Wing pattern? |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Dec 24, 2008 4:58 am Post subject: |
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| I did look at that BB UR of 35-35-356-356 and it doesn't look that different from the one being discussed here, except for the other candidates present in the box. Regardless, if I came across the UR here of 12-12-126-126, I'd never recognize it as an invalid Type 4 and would make the same mistake 100 out of 100 times. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Dec 24, 2008 5:32 am Post subject: |
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| Marty R. wrote: | | Regardless, if I came across the UR here of 12-12-126-126, I'd never recognize it as an invalid Type 4 and would make the same mistake 100 out of 100 times. |
| Code: | after basics: check [r47c23] for possible UR Type 4
+--------------------------------------------------------------------+
| 5 4 #29 | 13 289 89 | 13 6 7 |
| 19 8 3 | 7 6 5 | 4 19 2 |
| 129 6 7 | 4 239 19 | 5 189 38 |
|----------------------+----------------------+----------------------|
| 2368 *12 *126 | 5 38 7 | 9 18 4 |
| 389 5 #19 | 136 4 189 | 7 2 368 |
| 389 7 4 | 136 389 2 | 13 5 368 |
|----------------------+----------------------+----------------------|
| 26 *12 *126 | 9 7 3 | 8 4 5 |
| 7 9 8 | 2 5 4 | 6 3 1 |
| 4 3 5 | 8 1 6 | 2 7 9 |
+--------------------------------------------------------------------+
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| Code: | can not form X-Wing c23\r47 because [r5c3] contains <1>
+-----------------------------------+
| . / / | 1 . . | 1 . . |
| 1 / / | . . . | . 1 . |
| 1 / / | . . 1 | . 1 . |
|-----------+-----------+-----------|
| . 1 1 | . . . | . 1 . |
| . / #1 | 1 . 1 | . . . |
| . / / | 1 . . | 1 . . |
|-----------+-----------+-----------|
| . 1 1 | . . . | . . . |
| . / / | . . . | . . 1 |
| . / / | . 1 . | . . . |
+-----------------------------------+
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| Code: | can not form X-Wing c23\r47 because [r1c3] contains <2>
+-----------------------------------+
| . / #2 | . 2 . | . . . |
| . / / | . . . | . . 2 |
| 2 / / | . 2 . | . . . |
|-----------+-----------+-----------|
| 2 2 2 | . . . | . . . |
| . / / | . . . | . 2 . |
| . / / | . . 2 | . . . |
|-----------+-----------+-----------|
| 2 2 2 | . . . | . . . |
| . / / | 2 . . | . . . |
| . / / | . . . | 2 . . |
+-----------------------------------+
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UR Type 4 does not exist in [r47c23] ! |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Wed Dec 24, 2008 7:04 am Post subject: |
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| Marty R. wrote: | | Regardless, if I came across the UR here of 12-12-126-126, I'd never recognize it as an invalid Type 4 and would make the same mistake 100 out of 100 times. |
Let me try...
The thing to remember about a Type 4 is that one of the UR digits in the "roof" cells must be conjugate; this is what allows us to remove the other UR digit from those roof cells:
| Code: | The <y>s in the top row are conjugate:
xya xyb ya yb
--becomes-->
xy xy xy xy |
"a" and "b" are any number of other digits.
What is throwing you off is that in the puzzle above a=b=6 and those <6>s are conjugate. That is fine, but it's not Type 4. It is only Type 4 if either the <1>s or the <2>s in the "roof" are conjugate. And, if one of them were to be conjugate, then the <6>s couldn't be (or you have a hidden pair and haven't completed basic reductions).
As for a Type 4 also being a Type 2... When the non-UR digits in the roof are the same extra single digit and one of the UR digits in the roof is conjugate, then a Type 4 is also a Type 2 and vice versa. Say the roof cells are both {xya}, where "a" is a single digit and the <y>s are conjugate. If you apply Type 4 reasoning first, you get a naked {ya} pair that eliminates all other <a>s in that row or column. If you apply Type 2 reasoning first, you eliminate all the other <a>s in that row or column and now have a hidden {ya} pair. That's the difference: hidden or naked. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Wed Dec 24, 2008 7:34 am Post subject: |
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how about this one guys?
| Code: | .------------------.------------------.------------------.
| 5 4 29 |-13 289 89 |*13 6 7 |
| 19 8 3 | 7 6 5 | 4 19 2 |
| 129 6 7 | 4 239 19 | 5 189 *38 |
:------------------+------------------+------------------:
| 2368 12 126 | 5 38 7 | 9 18 4 |
| 389 5 19 |U136 4 189 | 7 2 U368 |
| 389 7 4 |U136 389 2 | 13 5 U368 |
:------------------+------------------+------------------:
| 26 12 126 | 9 7 3 | 8 4 5 |
| 7 9 8 | 2 5 4 | 6 3 1 |
| 4 3 5 | 8 1 6 | 2 7 9 |
'------------------'------------------'------------------' |
the UR on (3,6) says...
(1)r56c4 = (8)r56c9...
in other words... if r56c4 is NOT 1's then r56c9 has to be 8's
now you can form this chain
36URr56c49[(1)r56c4 = (8)r56c9] - (8=3)r3c9 - (3=1)r1c7; r1c4 <> 1
and if you look really closely at that chain why by golly gee, its a xy-wing {1,3,8}
{1=8} - (8=3) - (3=1)
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Wed Dec 24, 2008 9:20 am Post subject: |
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| Quote: | | how about this one guys? |
Well, if you note that the <6>s are an X-Wing, then external <3>s destroy the DP. This means that the UR creates the following strong inference directly:
36UR[(3)r1c4=(3)r3c9]
Thus, r1c7|r3c5<>3.
You can see that as an XY-Wing also (using your UR[1=8]). However, the logic holds even if those <3>s aren't in bivalue cells that happen to match the strong inference digits of the UR. It is also possible to see this as akin to a Skyscraper if you need to take the inferences through the UR explicitly:
(3)r1c4=36UR[(3)r56c4-(3)r56c9]=(3)r3c9
The grouped <3>s within the UR have a weak inference due to the X-Wing <6>s (if a <3> is true in both columns of the UR, then the DP results).
Once you start understanding inferences, the whole sudoku ballgame changes! |
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