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re'born
Joined: 28 Oct 2007
Posts: 80
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 Posted: Wed Nov 07, 2007 2:09 am Post subject: Nov 7 vh- one step alternatives |
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| Code: | .------------------.------------------.------------------.
| 169 126 8 | 29 7 5 | 69 3 4 |
| 5 3 7 | 489 6 48 | 89 1 2 |
| 69 26 4 | 1 3 28 | 689 5 7 |
:------------------+------------------+------------------:
| 167 4 256 | 2568 259 2678 | 3 26 1569 |
| 67 8 256 | 3 259 1 | 27 4 569 |
| 3 156 9 | 256 4 267 | 127 8 156 |
:------------------+------------------+------------------:
| 8 56 3 | 25 125 9 | 4 7 16 |
| 2 7 1 | 46 8 46 | 5 9 3 |
| 4 9 56 | 7 15 3 | 12 26 8 |
'------------------'------------------'------------------' |
Here are two one-step solutions that bypass the 2 xy-wing solution.
1. There is a potential deadly pattern in r13c17<69>. The strong links on 9 and 6 in columns 1 and 7, respectively, imply that r1c1<>6 and r3c7<>9, solving the puzzle.
2. There is a useless xyz-wing in r4c38|r9c3<256> which implies that one of those three cells is a 6. However r9c3=6 <=> r4c8=6 and so we can eliminate 6 from r4c1469 and r5c3, which reduces the puzzle to basics. |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Wed Nov 07, 2007 9:24 am Post subject: |
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re'born,
My approach on the potential <69> DP in R13C17 was diffirent, but the result is the same.
To avoid the <69> DP there are 3 possible solutions.
1. R1C1=1
2. R3C7=8
3. OR both (R1C1=1 AND R3C7=8 )
Suppose R3C7=8 => R3C6=2 => R1C4=9 => R1C7=6 => R1C1=1, which means that R1C1 must be <1>, and the puzzle is solved. |
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sdq_pete
Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL
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| Posted: Wed Nov 07, 2007 11:03 am Post subject: |
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I had:
| Code: | XY 267 R5C7
XY 156 R7C2
XY 256 R9C3 |
Peter |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Wed Nov 07, 2007 2:27 pm Post subject: |
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| Johan wrote: | re'born,
My approach on the potential <69> DP in R13C17 was diffirent, but the result is the same.
To avoid the <69> DP there are 3 possible solutions.
1. R1C1=1
2. R3C7=8
3. OR both (R1C1=1 AND R3C7=8 )
Suppose R3C7=8 => R3C6=2 => R1C4=9 => R1C7=6 => R1C1=1, which means that R1C1 must be <1>, and the puzzle is solved. |
Very clever, Johan. An alternative way using both of our ideas is the following: If r3c7=8, then r1c7=6 (as there are only two 6's in column 7) and r3c1=9 (as there are only two 9's in row 3) and hence r1c1=1.
| sdq_pete wrote: | I had:
| Code: | XY 267 R5C7
XY 156 R7C2
XY 256 R9C3 |
Peter |
Peter, if you note the locked candidates after your first xy-wing, then you can eliminate your second xy-wing. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Wed Nov 07, 2007 5:12 pm Post subject: |
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| Quote: |
XY 267 R5C7
XY 156 R7C2
XY 256 R9C3
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I stared at this puzzle for so long that I decided to try another techique and look up the solution on this website. However, I'm still staring. Sorry Peter but I can't see significance of your cells - as far as I can see they do not create xy or xyz wings. |
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duffy
Joined: 13 Sep 2007
Posts: 26
Location: Toronto Canada
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| Posted: Wed Nov 07, 2007 5:22 pm Post subject: |
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For me, the two alternatives given by re'born are not one-step solutions. However, that was an interesting analysis of the DP by re'born and Johan. (Should we give this a UR Type number?) Anyway, my simpler mind stuck to the xy-wing route. 
After Peter's first xy-wing (which removes three 6's, not counting the one in r6c2 which also quickly goes), there is an interesting combination of 2,5 and 6 in pairs at the corners where r49 and c38 intersect. Peter's last xy-wing is one of two possible on those corners at this point; I used the one pivoted on r4c9. 
Don D. |
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duffy
Joined: 13 Sep 2007
Posts: 26
Location: Toronto Canada
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| Posted: Wed Nov 07, 2007 5:31 pm Post subject: |
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Sorry, I keep making notation errors: the pivot I used was r4c3. 
At the risk of making another such error, I will add a note for cgordon: the first xy-wing is 26-27-67 pivoted on r5c7. The three 6's I noted are two in box 4 and one in box six.
Don |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Wed Nov 07, 2007 7:27 pm Post subject: |
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| Quote: | | ...as far as I can see they do not create xy or xyz wings |
Yeah well I see it now ... I'm just a bit slow ..takes me forever to spot wings - which is why I prefer recognisable patterns like ERs, URs and skyscrapers. Didn't see any of them here though. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Thu Nov 08, 2007 12:17 am Post subject: |
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Here's another one-step solution:
| Code: | .------------------.------------------.------------------.
| 169 12r6 8 | 2g9 7 5 | 69 3 4 |
| 5 3 7 | 489 6 48 | 89 1 2 |
| 69 2g6 4 | 1 3 28 | 689 5 7 |
:------------------+------------------+------------------:
| 167 4 256 |-2568 259 2678 | 3 26 1569 |
| 67 8 256 | 3 259 1 | 27 4 569 |
| 3 156 9 |-256 4 267 | 127 8 156 |
:------------------+------------------+------------------:
| 8 56 3 | 2R5 125 9 | 4 7 16 |
| 2 7 1 | 46 8 46 | 5 9 3 |
| 4 9 56 | 7 15 3 | 12 26 8 |
'------------------'------------------'------------------' |
Call it pincer transport, or coloring from a pincer, the otherwise useless XY Wing (pivot R7C2) does the trick. |
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