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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Sun Mar 28, 2010 7:04 pm Post subject: Puzzle 10/03/28 (B) |
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| Code: | +-----------------------+
| . 2 . | . 1 3 | . . . |
| 5 . 7 | 4 . . | . . . |
| . 4 8 | . 5 . | . . 6 |
|-------+-------+-------|
| . 7 . | 5 . . | . 6 . |
| 4 . 6 | . . . | . . 2 |
| 1 . . | . . 6 | . . 5 |
|-------+-------+-------|
| . . . | . . . | 2 . . |
| . . . | 2 . . | . 1 4 |
| . . 4 | . 8 1 | . 9 . |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Mar 28, 2010 8:04 pm Post subject: |
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Two steps... seemed to go down lots of deadends with other UR and HR
| Quote: |
x-wing(7) - I did this but I dont think it is needed for the next steps to solve the puzzle...
colouring 8
type 1 UR (23)
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sun Mar 28, 2010 11:52 pm Post subject: |
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| I used two steps also: |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Mar 29, 2010 1:01 am Post subject: Nice One danny |
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A 1 stepper ?
| Quote: | A one stepper that is slightly different:
Checking the bivalues after basics I took a look at r7c8 which is <78> and noticed:
Set it to 7 thus:
(8=7)r7c8-r7c1=(7-8)r8c1=r8c2 so r8c2 becomes 8
Then set it it to 8 thus:
(7=8)r7c8-r7c9=r1c9-r1c4=r6c4-r6c2=r8c2 so r8c2 is still 8 ! |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Mon Mar 29, 2010 1:47 am Post subject: Re: Nice One danny |
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| Mogulmeister wrote: | A 1 stepper ?
| Quote: | Checking the bivalues after basics I took a look at r7c8 which is <78> and noticed:
Set it to 7 thus:
(8=7)r7c8-r7c1=(7-8)r8c1=r8c2 so r8c2 becomes 8
Then set it it to 8 thus:
(7=8)r7c8-r7c9=r1c9-r1c4=r6c4-r6c2=r8c2 so r8c2 is still 8 ! |
[ |
Nice find! Is this right?
(8)r8c2=r6c2-r6c4=r1c4-r1c9=r7c9-(8=7)r7c8-r7c1=(7-8)r8c1=(8)r8c2; r8c2=8
edited to correct |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Mar 29, 2010 2:04 am Post subject: |
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| Spot on Dan! |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Mar 29, 2010 10:29 pm Post subject: |
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| Nothing new, I used five steps. Two Ws, Type 6 UR, X (7) and XY (893), flightless with pincer transport. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Mar 30, 2010 6:58 pm Post subject: |
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I worked with the UR23 in r46c35. Looking at the implications outside the UR we find that:
Digit 2: we have a x-wing so no outside implications exist,
Digit 3: r4c7=3 or r6c4=3 or r6c7=3 to prevent the UR.
(3)r6c4 = (3)r46c7 - r5c78 = r5c5; r46c5<>3.
Next I found a finned Remote Pair 78 from r1c4 to r7c8 with fin 3 in r7c9.
If Remote Pair true, then r7c4<>7
If fin true, (3)r7c9 - (3=7)r9c9 - r1c9 = r1c4; r7c4<>7,
Thus both conditions delete 7 in r7c4. (and yes, I realize the x-wing on 7 deletes the same cell plus several others as well, but it was not as much fun.)
And finally, w-wing 78 in r1c9 & r8c1 with grouped strong link on 8 in r16c4 & b4 with pincer 7 in r8c1 transported to delete 7 in r9c9.
A fun puzzle looking for some unusual moves.
Ted |
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