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storm_norm
Joined: 18 Oct 2007 Posts: 1741
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Posted: Sat Mar 06, 2010 3:21 pm Post subject: MM 1699 3-6-10 |
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009401500000000010024000390608042005000000000400310608047000250010000000006207400
Code: |
+-------+-------+-------+
| . . 9 | 4 . 1 | 5 . . |
| . . . | . . . | . 1 . |
| . 2 4 | . . . | 3 9 . |
+-------+-------+-------+
| 6 . 8 | . 4 2 | . . 5 |
| . . . | . . . | . . . |
| 4 . . | 3 1 . | 6 . 8 |
+-------+-------+-------+
| . 4 7 | . . . | 2 5 . |
| . 1 . | . . . | . . . |
| . . 6 | 2 . 7 | 4 . . |
+-------+-------+-------+
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Mogulmeister
Joined: 03 May 2007 Posts: 1151
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Posted: Tue Mar 30, 2010 5:42 am Post subject: |
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Another example of what we were discussing Norm.
(ie that some of the Diabolical MMs were -er- lacking bite) |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Tue Mar 30, 2010 6:25 am Post subject: |
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Hm, I must be getting old ...
I felt the diabolical xx alright
if only in the number of steps it took me to complete this puzzle.
coloring, m-wing, xy-wing, coloring again ... I've lost track what exactly it was that I did, but it seems I used the whole arsenal.
Of course, no computer assist (so I might have missed a simple step), no multi-candidate AIC, no ALS ...
Curious how you guys did it ... |
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Mogulmeister
Joined: 03 May 2007 Posts: 1151
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Posted: Tue Mar 30, 2010 7:42 am Post subject: |
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I'm no expert on categories Nat - just that recently the diabolicals from MM seemed to me sometimes more VHard or perhaps "fiendish". Do they still use fiendish as a category ? The recent diabolical for 28/3 was considered a "medium" by Norm and I do agree. I also posted 1720 as an example.
With this one I concentrated on the bivalues, and it was 2 xy chains and a swordfish then a final xy chain for me.
No fins, no multi-xx AIC (which I do like but is a "diabolical" technique perhaps) not even a UR.
Last edited by Mogulmeister on Tue Mar 30, 2010 2:16 pm; edited 1 time in total |
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Mogulmeister
Joined: 03 May 2007 Posts: 1151
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Posted: Tue Mar 30, 2010 8:34 am Post subject: |
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Might your path have been different if you'd started (after initial eliminations) with the removal of 7 in r4c8 pincers at r4c4 and r6c8 ? |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Tue Mar 30, 2010 4:43 pm Post subject: |
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Mogulmeister wrote: | Might your path have been different if you'd started (after initial eliminations) with the removal of 7 in r4c8 pincers at r4c4 and r6c8 ? |
Hi there Mogulmeister, traveller in the Alps
I re-did the puzzle (this time with some computer help), and found this:
The m-wing you mention (7)r4c4=r6c8 using the strong link on 9 in column 2 was my second step. It can be done first, then my steps 1 and 3, both coloring on 3 can be combined for a rich harvest.
After that, I had an xy-wing 35-39-59 (in r2c3,r2c6,r6c6) that set r6c3=2
I used a few more steps after that, but instead the xy-wing can be extended to an xy-chain that I think solves the puzzle right away.
I'm just not that into xy-chains, I find it easier to look for m- and w-wings. |
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Mogulmeister
Joined: 03 May 2007 Posts: 1151
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Posted: Tue Mar 30, 2010 5:21 pm Post subject: |
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Nice one.
Being visual I like to start with bivalues and see how they might connect and am getting less concerned at seeing them as wings, m,xy or w. They are all variations on the AIC theme which I now see as the meta-language for all these moves.
Then, hard to find are the meaningful ALS's from where eliminations can be made but rewarding when you spot one. |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Tue Mar 30, 2010 6:07 pm Post subject: |
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(playing with my solver) I found three steps that should appeal to Marty.
Code: | gM-Wing: (7=9)r4c4 - r6c6 = (9-7)r6c2 = (7)r6c8 => r4c8<>7
<39+5> XY-Wing r2c6/r2c3+r6c6 <> 5 r6c3
extended/transported/??? ...
gM-Wing: (7=3)r1c1 - r5c1 = (3-7)r5c2 = r4c2 - r4c4 = (7)r2c4 => r1c5,r2c1<>7
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Mogulmeister
Joined: 03 May 2007 Posts: 1151
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Posted: Wed Mar 31, 2010 6:56 am Post subject: |
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Alternatively, Danny your gM-Wing can be changed to a standard xy chain if you go to r6c2 instead of r5c2.
(7=3)r1c1 - (3=5)r5c1 - (5=9)r6c2 - (9=7) r4c2 - r4c4 = (7)r2c4 => r1c5,r2c1<>7
(Pincers at r1c1 r2c4 in green, eliminated 7s in blue for those who don't get on with notation.)
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