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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Fri Apr 22, 2011 2:25 pm Post subject: Free Press April 22, 2011 |
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Not yet started. Seems to be on the hard side.
Code: | Puzzle: FP042211
+-------+-------+-------+
| 8 7 . | 1 . . | . . . |
| . . 2 | . . . | 1 . 4 |
| . . . | . 5 9 | 7 8 . |
+-------+-------+-------+
| 3 . . | 4 . 6 | . . . |
| . . 7 | . . . | 9 . . |
| . . . | 8 . 3 | . . 6 |
+-------+-------+-------+
| . 4 5 | 9 . . | . . . |
| 2 . 3 | . . . | 4 . . |
| . . . | . . 7 | . 5 9 |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Keith |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Fri Apr 22, 2011 11:12 pm Post subject: |
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Quote: | Seems to be on the hard side. |
You'll get no argument from me. All I could find was a worthless W-Wing (or Skyscraper) that eliminated a couple of 2s. After that, Medusa traps and a wrap set up a puzzle-ending Multi-coloring; r7c5<>6.
Medusa's fun to watch develop while doing it, but I consider it largely trial-and-error, so I'm not very satisfied when I have to resort to it. |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sat Apr 23, 2011 12:48 am Post subject: |
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Marty,
I have not solved it, but noticed a couple of bizarre moves. After basics (Grid A): Code: | +-------------------+-------------------+-------------------+
| 8 7 69d | 1 236 4 |-2356 2369 235 |
| 59 35-6 2 | 7 36 8 | 1 -369e 4 |
| 146 13-6 146c | 236@ 5 9 | 7 8 23g |
+-------------------+-------------------+-------------------+
| 3 1258 189 | 4 79 6 | 258 127 12578 |
| 46 68a 7 | 25 12 125 | 9 34 38 |
| 59 125 149 | 8 79 3 | 25 1247 6 |
+-------------------+-------------------+-------------------+
| 7 4 5 | 9 -1-2368 12 | 2368 1236 1-238 |
| 2 9 3 | 56 168 15 | 4 167 178 |
| 16 168 168b | 23@ 4 7 | 23f 5 9 |
+-------------------+-------------------+-------------------+ |
Some kind of M-wing: If a is 8, b is 8. If b is not 6, one of cd is 6. Eliminating 6 in R23C2.
If d is 6, e is 6. If d is 9, e is 9. Eliminating 3 in e.
f and g are a W-wing, taking out 2 as shown in C79.
There is a type 6 UR 12 taking out 12 in R7C5.
Leaving us here (Grid B): Code: | +-------------------+-------------------+-------------------+
| 8 7 69 | 1 236 4 | 356 2369 235 |
| 59 35 2 | 7 36 8 | 1 69 4 |
| 146 13 146 | 236 5 9 | 7 8 23 |
+-------------------+-------------------+-------------------+
| 3 1258 189 | 4 79 6 | 258 127 12578 |
| 46 68 7 | 25 12 125 | 9 34 38 |
| 59 125 149 | 8 79 3 | 25 1247 6 |
+-------------------+-------------------+-------------------+
| 7 4 5 | 9 368 12 | 2368 1236 138 |
| 2 9 3 | 56 168 15 | 4 167 178 |
| 16 168 168 | 23 4 7 | 23 5 9 |
+-------------------+-------------------+-------------------+ |
More to come, if I make progress.
Keith |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Sat Apr 23, 2011 3:42 am Post subject: |
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Keith, those first two moves are very impressive. In the remaining grid I see an M-Wing and Kite offhand, but don't know how far they take you. |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Sat Apr 23, 2011 7:33 am Post subject: |
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A UR move takes it down...
Code: | ur(25)r46c27[(8)r4c7=(5)r6c1] - (5=9)r2c1 - (9=6)r1c3 - r1c7=r7c7 ; r7c7<>8 |
(Using an internal and external strong inference) |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sat Apr 23, 2011 1:10 pm Post subject: |
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peterj wrote: | A UR move takes it down...
Code: | ur(25)r46c27[(8)r4c7=(5)r6c1] - (5=9)r2c1 - (9=6)r1c3 - r1c7=r7c7 ; r7c7<>8 |
(Using an internal and external strong inference) |
I don't see this. Can someone spell it out?
Keith |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Sat Apr 23, 2011 1:52 pm Post subject: |
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keith wrote: | I don't see this. Can someone spell it out? |
To stop the potential DP(25) in r46c28 at least one of...
(1) r4c7=8 (external strong inference)
(2) r6c1=5 (internal strong inference)
For (2) consider that (25) is almost a hidden pair in block 4 i.e. the only thing stopping it becoming a naked (25) and so part of the DP is the (5)r6c1. (You can get the same result by considering the external DP-stoppers (18)r4c2 and (1)r6c2 - it's just more painful!)
Now, if (5)r6c1 then (9)r2c1 and (6)r1c3 => r1c7<>6 and r7c7 must be 6.
So in both case (1) and (2), r7c7<>8 |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sat Apr 23, 2011 2:03 pm Post subject: |
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Thanks.
Keith |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sat Apr 23, 2011 4:33 pm Post subject: |
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An alternative suggestion for the first strong inference:
Code: | (8)r4c7 = UR_4[(25)r46c7 - (5)r46c2] = (5)r6c1
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This way, the UR includes a necessary weak inference!
[Edit: decided to move the initial assumption outside the UR_4.]
Last edited by daj95376 on Sat Apr 23, 2011 8:33 pm; edited 2 times in total |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sat Apr 23, 2011 4:54 pm Post subject: |
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My solver missed the UR_4 option and found an awkward network, instead. I turned it into :
Code: | Kraken Cell [r7c9]
(1)r7c9 - (1=2)r7c6 - r9c4 = r9c7 - (25=8)r64c7 - (8)r7c7
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(2)r7c9 - r3c9 = r3c4 - r9c4 = r9c7 - (25=8)r64c7 - (8)r7c7
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(3)r7c9 - (3=2)r9c7 - (25=8)r64c7 - (8)r7c7
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(8)r7c9 - (8)r7c7
+-----------------------------------------------------------------------+
| 8 7 69 | 1 236 4 | 2356 2369 235 |
| 59 356 2 | 7 36 8 | 1 369 4 |
| 146 136 146 | 236 5 9 | 7 8 23 |
|-----------------------+-----------------------+-----------------------|
| 3 1258 189 | 4 79 6 | 258 127 12578 |
| 46 68 7 | 25 12 125 | 9 34 38 |
| 59 125 149 | 8 79 3 | 25 1247 6 |
|-----------------------+-----------------------+-----------------------|
| 7 4 5 | 9 12368 12 | 2368 1236 1238 |
| 2 9 3 | 56 168 15 | 4 167 178 |
| 16 168 168 | 23 4 7 | 23 5 9 |
+-----------------------------------------------------------------------+
# 87 eliminations remain
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sat Apr 23, 2011 5:41 pm Post subject: |
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The first three blocks from my grid B: Code: | +-------------------+-------------------+-------------------+
| 8 7 69 | 1 23-6k 4 | 356 2369 235 |
| 59 35 2 | 7 36 8 | 1 69 4 |
| 146 13 146 | 236 5 9 | 7 8 23j |
+-------------------+-------------------+-------------------+ | We can eliminate 6 from k, since j and k must be the same value.
Keith |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Sun Apr 24, 2011 12:12 am Post subject: |
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keith wrote: | The first three blocks from my grid B: Code: | +-------------------+-------------------+-------------------+
| 8 7 69 | 1 23-6k 4 | 356 2369 235 |
| 59 35 2 | 7 36 8 | 1 69 4 |
| 146 13 146 | 236 5 9 | 7 8 23j |
+-------------------+-------------------+-------------------+ | We can eliminate 6 from k, since j and k must be the same value.
Keith |
Dare I ask what says j and k must be the same value? |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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ronk
Joined: 07 May 2006 Posts: 398
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Posted: Sun Apr 24, 2011 1:28 am Post subject: |
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Marty R. wrote: | keith wrote: | The first three blocks from my grid B: Code: | +-------------------+-------------------+-------------------+
| 8 7 69 | 1 23-6k 4 | 356 2369 235 |
| 59 35 2 | 7 36 8 | 1 69 4 |
| 146 13 146 | 236 5 9 | 7 8 23j |
+-------------------+-------------------+-------------------+ | We can eliminate 6 from k, since j and k must be the same value. |
Dare I ask what says j and k must be the same value? |
It is due to one of several possible m-rings.
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sun Apr 24, 2011 1:50 am Post subject: |
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Marty R. wrote: | keith wrote: | The first three blocks from my grid B: Code: | +-------------------+-------------------+-------------------+
| 8 7 69 | 1 23-6k 4 | 356 2369 235 |
| 59 35 2 | 7 36 8 | 1 69 4 |
| 146 13 146 | 236 5 9 | 7 8 23j |
+-------------------+-------------------+-------------------+ | We can eliminate 6 from k, since j and k must be the same value.
Keith |
Dare I ask what says j and k must be the same value? |
S-Loop/Ring:
Code: | (2)r1c5 = r1c89 - (2=3)r3c9 - r1c789 = (3)r1c5 - loop => r1c5<>6
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