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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Sat Oct 09, 2010 9:58 am Post subject: Free Press Oct 8, 2010 |
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| Code: |
Puzzle: FP100810
+-------+-------+-------+
| . . . | 8 . 4 | 2 . . |
| . . . | . . 1 | . 7 4 |
| . . 7 | . 3 . | 5 . 9 |
+-------+-------+-------+
| . 6 1 | . 8 . | 7 . . |
| . . . | . . . | . . . |
| . . 9 | . 6 . | 3 5 . |
+-------+-------+-------+
| 6 . 3 | . 4 . | 9 . . |
| 1 5 . | . . . | . . . |
| . . 4 | 7 . 5 | . . . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Not yet started.
Keith |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sat Oct 09, 2010 4:24 pm Post subject: |
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Two steps - needing an ANT/AIC combo ...
| Quote: | ER(2) ; (2)r5c5=r9c5 - r9c12=r8c3 ; r5c3<>2
AIC with ANT ; (5=8)r5c3 - ANT(247)[(8)r6c12=(1)r6c4]r6c126 - r7c4=r7c8 - r3c8=r3c2 - (1=9)r1c2 - (9=5)r1c1 ; r1c3<>5, r45c1<>5
(I had an x-wing(1) and xy-wing also - but extraneous)
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[Edit] Clarified (!) AIC |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Oct 09, 2010 7:27 pm Post subject: |
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| elsewhere, Keith wrote: | | Code: | after basics and X-Wing on <1>
+--------------------------------------------------------------------------------+
| 59 19 56 | 8 7 4 | 2 136 136 |
| 238 238 268 | 9 5 1 | 68 7 4 |
| 48 148 7 | 26 3 26 | 5 18 9 |
|--------------------------+--------------------------+--------------------------|
| 345 6 1 | 345 8 39 | 7 49 2 |
| 234578 2348 258 | 2345 12 2379 | 1468 4689 68 |
| 2478 248 9 | 124 6 27 | 3 5 18 |
|--------------------------+--------------------------+--------------------------|
| 6 7 3 | 12 4 8 | 9 12 5 |
| 1 5 28 | 36 9 36 | 48 248 7 |
| 289 289 4 | 7 12 5 | 168 2368 368 |
+--------------------------------------------------------------------------------+
# 79 eliminations remain
(1)r1c89 = UR Type 3 => r9c5=1
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Continuing that analysis:
| Code: | r9c5=1 r7c4<>1 r7c8=1 => r3c8<>1
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sat Oct 09, 2010 8:01 pm Post subject: |
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| Danny, that's an elegant move! |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Oct 09, 2010 8:07 pm Post subject: |
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| Thanks. Unfortunately, all I did was add a little icing to a cake that Keith had already prepared. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Oct 09, 2010 9:56 pm Post subject: |
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| I eliminated a couple of 1s via coloring. Then looked at the 36 DP. A couple of cells were forced and that broke it open. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Oct 10, 2010 3:42 pm Post subject: |
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I looked at the AUR(36)r19c89 before taking the x-wing 1, and found a useless deletion:
External SIS:r2c7=6, r9c7=6;
(6)r2c7
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(6)r9c7; r5c7<>6
At this point I moved on to look for other moves including the x-wing 1. What I failed to pursue again was the AUR(36)r19c89. Using the identical SIS, we get
(6)r2c7-(8)r2c7=(8)r3c8
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(6)r9c7-(1)r9c7=r7c8-(1=8)r3c8; r3c8=8 as posted by Keith/Danny
Ted |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Oct 10, 2010 4:07 pm Post subject: |
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| peterj wrote: | Two steps - needing an ANT/AIC combo ...
| Quote: | ER(2) ; (2)r5c5=r9c5 - r9c12=r8c3 ; r5c3<>2
AIC with ANT ; (5=8)r5c3 - ANT(247)[(8)r6c12=(1)r6c4]r6c126 - r7c4=r7c8 - r3c8=r3c2 - (1=9)r1c2 - (9=5)r1c1 ; r1c3<>5, r45c1<>5
(I had an x-wing(1) and xy-wing also - but extraneous)
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[Edit] Clarified (!) AIC |
Peter,
I am confused about your use of the ANT(247). I see a valid ANT(247)r6c126 but you included (1)r6c4 in your post which makes it a ANQ.
As an ANT, I think it would need to be
(5=8)r5c3 - ANT(247)[(8)r6c12=(247)r6c126]r6c126 - (24=1)r6c4-r7c4=r7c8 - r3c8=r3c2 - (1=9)r1c2 - (9=5)r1c1 ; r1c3<>5, r45c1<>5
It is a great move in any case!
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Oct 10, 2010 4:32 pm Post subject: |
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Ted's comment on ANQ() caused me to review the grid.
| Code: | after basics and Peter's ER(2)
*-----------------------------------------------------------------------------*
| 59 19 56 | 8 7 4 | 2 136 136 |
| 238 238 268 | 9 5 1 | 68 7 4 |
| 48 148 7 | 26 3 26 | 5 18 9 |
|-------------------------+-------------------------+-------------------------|
| 345 6 1 | 345 8 39 | 7 49 2 |
| 234578 2348 58 | 12345 12 2379 | 1468 14689 168 |
| 2478 248 9 | 124 6 27 | 3 5 18 |
|-------------------------+-------------------------+-------------------------|
| 6 7 3 | 12 4 8 | 9 12 5 |
| 1 5 28 | 36 9 36 | 48 248 7 |
| 289 289 4 | 7 12 5 | 168 12368 1368 |
*-----------------------------------------------------------------------------*
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What I saw was an ANQ()/ALS consuming four of five unsolved cells in [r6]. Whenever something like this happens, there always seems to be a simpler interpretation using the unused cell:
(5=8)r5c3 - r6c12 = (8-1)r6c9 = r6c4 - r7c4=r7c8 - r3c8=r3c2 - (1=9)r1c2 - (9=5)r1c1 ; r1c3<>5, r45c1<>5
As Ted mentioned, nice find ... no matter what the interpretation. |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Oct 10, 2010 4:48 pm Post subject: |
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| tlanglet wrote: | | I am confused about your use of the ANT(247) |
Ted, you are quite right. I originally had exactly your notation, and went back to edit it as I thought it was a bit clunky and the link 8=1 was "clear"! Cut one too many corners Thanks. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Oct 10, 2010 8:32 pm Post subject: |
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| peterj wrote: | | tlanglet wrote: | | I am confused about your use of the ANT(247) |
Ted, you are quite right. I originally had exactly your notation, and went back to edit it as I thought it was a bit clunky and the link 8=1 was "clear"! Cut one too many corners Thanks. |
Peter, it is very interesting that I messed up yesterday the same way by changing my steps when I prepared a post. Maybe we both have learned a small lesson..........
Ted |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Oct 11, 2010 3:06 am Post subject: |
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| daj95376 wrote: | | elsewhere, Keith wrote: | | Code: | after basics and X-Wing on <1>
+--------------------------------------------------------------------------------+
| 59 19 56 | 8 7 4 | 2 136 136 |
| 238 238 268 | 9 5 1 | 68 7 4 |
| 48 148 7 | 26 3 26 | 5 18 9 |
|--------------------------+--------------------------+--------------------------|
| 345 6 1 | 345 8 39 | 7 49 2 |
| 234578 2348 258 | 2345 12 2379 | 1468 4689 68 |
| 2478 248 9 | 124 6 27 | 3 5 18 |
|--------------------------+--------------------------+--------------------------|
| 6 7 3 | 12 4 8 | 9 12 5 |
| 1 5 28 | 36 9 36 | 48 248 7 |
| 289 289 4 | 7 12 5 | 168 2368 368 |
+--------------------------------------------------------------------------------+
# 79 eliminations remain
(1)r1c89 = UR Type 3 => r9c5=1
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Continuing that analysis:
| Code: | r9c5=1 r7c4<1> r3c8<>1
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Danny,
I disagree. This is a very nice observation that I did not see.
What I did see is that 6 in R9C7 forces 6 in R2C7 via R7C8 and R3C8, so R9C7 <>6.
Which, after a couple of simplifications, reveals Marty's UR in R19C89 here: | Code: | +----------------------+----------------------+----------------------+
| 59 19 56 | 8 7 4 | 2 136 136 |
| 238 238 268 | 9 5 1 | 68 7 4 |
| 48 148 7 | 26 3 26 | 5 18 9 |
+----------------------+----------------------+----------------------+
| 345 6 1 | 345 8 39 | 7 49 2 |
| 234578 2348 258 | 2345 12 2379 | 146 469 68 |
| 2478 248 9 | 124 6 27 | 3 5 18 |
+----------------------+----------------------+----------------------+
| 6 7 3 | 12 4 8 | 9 12 5 |
| 1 5 28 | 36 9 36 | 48 248 7 |
| 289 289 4 | 7 12 5 | 18 36 36 |
+----------------------+----------------------+----------------------+ |
Keith |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Mon Oct 11, 2010 3:41 am Post subject: |
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| keith wrote: | I disagree. This is a very nice observation that I did not see.
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I credited you with the first part of the solution. It came from your statement here.
| keith wrote: | I see it as either one of R1C89 is 1, and / or there is a pseudo-triple 289 in R9C12(8+9), which forces R9C5=1 and R9C7=6. (R9C89 are a pseudocell 28.)
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All I did was transport r9c5=1 using the strong link (1)r7c4=(1)r7c8.
Regards, Danny |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Oct 11, 2010 3:53 am Post subject: |
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Danny,
Yes. But I did not connect to a common implication of a 1 in R1B3 and a pseudo-triple in R9.
Keith |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Mon Oct 11, 2010 3:56 am Post subject: |
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| keith wrote: | Yes. But I did not connect to a common implication of a 1 in R1B3 and a pseudo-triple in R9.
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That's why I only gave you credit for "part of the solution". _  _ |
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