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Free Press June 17, 2011

 
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Fri Jun 17, 2011 1:30 pm    Post subject: Free Press June 17, 2011 Reply with quote

Not yet started. Requires advanced moves.
Code:
Puzzle: FP061711
+-------+-------+-------+
| . . . | 8 . 2 | 1 . . |
| . 2 . | . . . | . 5 . |
| . . . | 4 1 . | 2 . . |
+-------+-------+-------+
| 7 5 9 | . . . | . . 3 |
| . 3 . | . 4 . | . 8 . |
| 8 . . | . . . | 5 6 9 |
+-------+-------+-------+
| . . 5 | . 7 8 | . . . |
| . 6 . | . . . | . 3 . |
| . . 2 | . . 6 | . . . |
+-------+-------+-------+

Play this puzzle online at the Daily Sudoku site

Keith
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills

PostPosted: Fri Jun 17, 2011 2:35 pm    Post subject: Reply with quote

A quick two stepper..............

Quote:
Type 6 UR(79)r13c28; r1c8,r3c2<>9
Plus external sis r9c2=7,r9c8=7; r9c9<>7
kite (7) b9; r1c3<>7=3


A second pass uncovered a one step solution.

Quote:
Hidden aur(47)r19c89 with x-wing (4); r1c9<>7
Plus external sis: (7)r9c2=(7)r8c3-aur(47)[(7)r8c9=(7)r3c89]; r3c2<>7

Ted
[Edited to add second solution.]
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cgordon



Joined: 04 May 2007
Posts: 769
Location: ontario, canada

PostPosted: Fri Jun 17, 2011 6:46 pm    Post subject: Reply with quote

I used a Type 6 UR (79) and a Type 4 (47) and an ER on 7. Probably more than needed.

Don't know what a sis is short for. I assume it's not a sibling or a girlie guy.
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Sat Jun 18, 2011 1:20 am    Post subject: Reply with quote

What is there about certain puzzles that one makes consistent mechanical errors? And the frustrating thing is that I rarely get to find out what I did wrong. On this one I ended three times with an invalid solution. Thankfully, I got it on the 4th try or who knows how much more time I would've wasted.

Type 6 UR (79); r1c2, r3c8=9
Kite; r1c3<>7
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Sat Jun 18, 2011 3:45 am    Post subject: Reply with quote

Marty,

That was the solution I eventually found.

This one had a lot of naked / hidden pairs to get through basics.

Keith
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dejsmith



Joined: 23 Oct 2005
Posts: 42

PostPosted: Tue Jun 21, 2011 10:20 pm    Post subject: BUG +5? Reply with quote

I tried to do something a little different this time since I am pretty confident with URs. After basics & an XY Wing extension so that r8c5<>5; I noticed there were only 5 cells left that each had 3 candidates; i.e. the rest were all pairs. Based on my understanding of BUGs, either in Box 1 r3c23=7, or in Box 3 r1c89=7, or in Box 9 r9c9=7. Since they all see r3c9, then r3c9<>7; & the puzzle solves. Is that valid reasoning; & would that be called a BUG +5?

Dave
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Wed Jun 22, 2011 12:03 am    Post subject: Reply with quote

I can't comment on the reasoning without seeing a grid. However, I do believe it would properly be called a BUG+5. That move is rather unusual, as most of the BUG moves don't go higher than BUG+3.
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Wed Jun 22, 2011 12:21 am    Post subject: Re: BUG +5? Reply with quote

dejsmith wrote:
I tried to do something a little different this time since I am pretty confident with URs. After basics & an XY Wing extension so that r8c5<>5; I noticed there were only 5 cells left that each had 3 candidates; i.e. the rest were all pairs. Based on my understanding of BUGs, either in Box 1 r3c23=7, or in Box 3 r1c89=7, or in Box 9 r9c9=7. Since they all see r3c9, then r3c9<>7; & the puzzle solves. Is that valid reasoning; & would that be called a BUG +5?

Your grid as I've reconstructed it. A BUG+5 using a single extra value. Impressive!

Code:
 *--------------------------------------------------*
 | 56   79   37   | 8    35   2    | 1    49+7 46+7 |
 | 4    2    1    | 79   6    79   | 3    5    8    |
 | 56   89+7 38+7 | 4    1    35   | 2    79   6-7  |
 |----------------+----------------+----------------|
 | 7    5    9    | 6    8    1    | 4    2    3    |
 | 2    3    6    | 59   4    59   | 7    8    1    |
 | 8    1    4    | 27   23   37   | 5    6    9    |
 |----------------+----------------+----------------|
 | 9    4    5    | 3    7    8    | 6    1    2    |
 | 1    6    78   | 25   29   4    | 89   3    57   |
 | 3    78   2    | 1    59   6    | 89   47   45+7 |
 *--------------------------------------------------*

It would also have worked as a BUG+6 if you hadn't performed r8c5<>5:

Code:
(5)r8c5 - (5=7)r8c9 - (7)r3c9

Essentially, basics followed by a BUG+6.

Regards, Danny
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susume



Joined: 13 May 2011
Posts: 36
Location: Southeastern US

PostPosted: Wed Jun 22, 2011 3:25 am    Post subject: Reply with quote

Besides basics I needed an X-wing (9) r89c57 to get to daj's grid of mostly bivalues (with the 5 still in r8c5).

From there, those lovely bivalue cells can be used in a Nice Loop that solves it:
r1c9 -6- r1c1 -5- r1c5 -3- r1c3 -7- r8c3 =7= r8c9 -7- r3c9 -6- r1c9 => r1c9<>6, r3c9=6

It's essentially a W-wing of 7 in c39, using the bivalues in r1c15 to convert the r1c3 pincer to a 6.
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