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DB Saturday Puzzle - August 12, 2006

 
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keith



Joined: 19 Sep 2005
Posts: 3165
Location: near Detroit, Michigan, USA

PostPosted: Sat Aug 12, 2006 3:12 pm    Post subject: DB Saturday Puzzle - August 12, 2006 Reply with quote

A tough one, compared to last week's ...

Keith
Code:
Puzzle: DB081206  ******
+-------+-------+-------+
| 7 . . | 8 . . | . 2 . |
| . 3 2 | . . 6 | . 7 . |
| . 4 8 | 3 . . | . . 5 |
+-------+-------+-------+
| . . . | . 7 9 | . 5 2 |
| . . . | 2 . 8 | 7 . . |
| 2 7 . | 5 1 . | . . . |
+-------+-------+-------+
| 9 . . | . . 4 | 5 8 . |
| . 5 . | 7 . . | 1 9 . |
| . 8 . | . . 5 | . . 7 |
+-------+-------+-------+


Last edited by keith on Sat Aug 19, 2006 9:08 am; edited 1 time in total
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Marty R.



Joined: 12 Feb 2006
Posts: 5157
Location: Rochester, NY, USA

PostPosted: Sat Aug 12, 2006 5:04 pm    Post subject: Reply with quote

I reached this position, where I got stuck. The only thing I could do other than the very basic stuff was to break up a deadly pattern by removing the "6" from r5c3.

Code:
----------------------------------------------------------
|7     69    569   |8     45    1     |346   2     346   |
|15    3     2     |49    459   6     |48    7     18    |
|16    4     8     |3     2     7     |9     16    5     |
----------------------------------------------------------
|8     1     346   |46    7     9     |346   5     2     |
|3456  69    3459  |2     46    8     |7     1346  19    |
|2     7     469   |5     1     3     |468   46    89    |
----------------------------------------------------------
|9     2     7     |1     36    4     |5     8     36    |
|346   5     346   |7     8     2     |1     9     346   |
|346   8     1     |69    369   5     |2     346   7     |
----------------------------------------------------------   


At this point it's easily solvable with--take your choice--trial and error or a chain. Starting a chain at r2c4 solves the puzzle with a "4" in that cell. However, I suspect there are other, less controversial, techniques available that I did not see.
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Sat Aug 12, 2006 5:23 pm    Post subject: I used a "DIC" on this one Reply with quote

In the position Marty reached I can only find one "easy" move -- there's a fork on the "6"s in row 3 and column 2, so r5c8 <> 6.

I ended up using a double-implication chain from r9c4 on this one:

A. r9c4 = 9 ==> r2c4 = 4 ==> r2c7 = 8
B. r9c4 = 6 ==> r5c5 = 6 ==> r1c2 = 6 ==> {3, 4} pair in r1c79 ==> r2c7 = 8

and the rest is straightforward. dcb
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Marty R.



Joined: 12 Feb 2006
Posts: 5157
Location: Rochester, NY, USA

PostPosted: Sat Aug 12, 2006 9:33 pm    Post subject: Reply with quote

Quote:
In the position Marty reached I can only find one "easy" move -- there's a fork on the "6"s in row 3 and column 2, so r5c8 <> 6.


David, I don't see it. Assuming a fork is what I'd call a strong link, I see strong links on "6" in r3 and c2, but I don't see where they are joined so as to use the fork technique.

Quote:
I ended up using a double-implication chain from r9c4 on this one:

A. r9c4 = 9 ==> r2c4 = 4 ==> r2c7 = 8
B. r9c4 = 6 ==> r5c5 = 6 ==> r1c2 = 6 ==> {3, 4} pair in r1c79 ==> r2c7 = 8


After all these months, I finally figured out how you get all those DICs and I never do: apparently you stop when you see a cell that is forced due to it having the same value regardless of the value in the base cell.

When I start a chain, I carry it out as far as I can, which is why my chains always result in contradictions. Had I started my chain where you did in r9c4, one of the possibilities would have probably led to a duplicate, so I would have solved it with the other value and proceeded from there.

Or had I started in the cell that was forced with the "8", the "4" would have led to a duplicate, the "8" would have been entered and I would have proceeded to solve it the way you did.

One less mystery to wonder about.
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keith



Joined: 19 Sep 2005
Posts: 3165
Location: near Detroit, Michigan, USA

PostPosted: Sat Aug 12, 2006 11:02 pm    Post subject: The fork (Turbot fish) Reply with quote

Marty, the weak link which connects the two strong links is in Box 1. R3C1 - R1C2.

The logic, which is easy to prove, is that at least one of R5C2 or R3C8 is <6>.

If you join the cells by drawing lines R5C8 - R3C8 - R3C1 - R1C2 - R5C2 - R5C8 you will perhaps "see" the fish.

Or, for an alternative, look at the chain(s) starting with <4> in R5C8.

Keith's contribution, which does not solve the puzzle, is the XYZ-wing on <468> in R6C7, which eliminates <4> in R4C7.

Keith


Last edited by keith on Sat Aug 12, 2006 11:45 pm; edited 2 times in total
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Sat Aug 12, 2006 11:16 pm    Post subject: Seeing the "fork" Reply with quote

Marty R wrote:
David, I don't see it. Assuming a fork is what I'd call a strong link, I see strong links on "6" in r3 and c2, but I don't see where they are joined so as to use the fork technique.

They're joined in box 1.
Code:
----------------------------------------------------------
|7     69+   569   |8     45    1     |346   2     346   |
|15    3     2     |49    459   6     |48    7     18    |
|16~   4     8     |3     2     7     |9     16=   5     |
----------------------------------------------------------
|8     1     346   |46    7     9     |346   5     2     |
|3456  69-   3459  |2     46    8     |7     1346  19    |
|2     7     469   |5     1     3     |468   46    89    |
----------------------------------------------------------
|9     2     7     |1     36    4     |5     8     36    |
|346   5     346   |7     8     2     |1     9     346   |
|346   8     1     |69    369   5     |2     346   7     |
----------------------------------------------------------

r1c2 = 6 ==> r3c1 <> 6 ==> r3c8 = 6
r1c2 <> 6 ==> r5c2 = 6

So no matter where we place the "6" in column 2, there can't possibly be a "6" at r5c8. This doesn't really help us much with the puzzle, but it is fairly simple logic.
Marty R wrote:
... I finally figured out how you get all those DICs and I never do: apparently you stop when you see a cell that is forced due to it having the same value regardless of the value in the base cell.

That's (almost) exactly right. The way I work with DIC's, there are three possibilities.

1. I may find a "target" cell that's forced to the same value regardless of the value chosen in the "root" cell.

2. I may find a value that can be eliminated from one or more cells in the puzzle. For example, if say r1c1 can be 1 or 4, I may find that placing a "1" at r1c1 forces a "6" at r5c1, while a "4" at r1c1 forces a "6" at r5c5. The implication is that the "6" in row 5 must either be at r5c1, or at r5c5, so "6" can be eliminated everywhere else in row 5.

3. I may find that choosing one value in the root cell leads to a contradiction.

Personally, I don't care which result I obtain. If I can't get one or more of these things to happen after concentrating on a particular bi-valued cell for a while, I just move to some other bi-valued cell and try working from there. Interestingly, the bi-valued cells often form networks within the puzzle, so that once I've decided that one cell won't work, I may be able to avoid 4 or 5 other bi-valued cells for my next choice, because of the way those 4 or 5 are linked to the one where I've already tried, and failed. dcb
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Marty R.



Joined: 12 Feb 2006
Posts: 5157
Location: Rochester, NY, USA

PostPosted: Sun Aug 13, 2006 12:55 am    Post subject: Reply with quote

Thanks guys. David's DIC-looking example is easy enough to follow. However, neither the weak link nor the fish concept is yet in my techniques arsenal.

Nice XYZ-Wing. I've probably missed a few of these because I forget that 468 doesn't have to be XYZ, but that it can also be XZY, ZYX, or whatever.
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