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Saturday Puzzle Jan 21

 
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keith



Joined: 19 Sep 2005
Posts: 3174
Location: near Detroit, Michigan, USA

PostPosted: Sat Jan 21, 2006 1:20 pm    Post subject: Saturday Puzzle Jan 21 Reply with quote

Here is today's puzzle:

Code:


Puzzle: DB012106
+-------+-------+-------+
| 4 . 8 | . . . | . . . |
| . 9 2 | 4 . . | . . . |
| 6 7 3 | . 9 8 | 2 . . |
+-------+-------+-------+
| . . . | . . 1 | . 6 7 |
| 1 6 . | . 4 . | . 2 5 |
| 8 2 . | 6 . . | . . . |
+-------+-------+-------+
| . . 4 | 3 6 . | 1 7 8 |
| . . . | . . 4 | 5 3 . |
| . . . | . . . | 6 . 4 |
+-------+-------+-------+



I'll post comments later, after you've had time to work on the solution.

Keith
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David Bryant



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

PostPosted: Sat Jan 21, 2006 8:41 pm    Post subject: Short "double-implication chain" solves this one. Reply with quote

This puzzle was in my local paper this morning.

After placing 25 values I arrived at a position where a very short "double-implication chain" allowed me to eliminate one possibility in r4c5, and that cracked the puzzle wide open. dcb
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someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Sun Jan 22, 2006 8:07 am    Post subject: Reply with quote

Hi,

After the first 25 digits, I detected a Naked Triple 3 5 7 in 3x3 Block 4.
Could so eliminate 5 from r4c5 and 7 from r5c4.
So I could pinpoint other 3 digits:
5 in r4c3 - Unique Horizontal
7 in r1c4 7 in r2c7 - Unique Vertical
before I had to apply the "double implication chains, starting from pairs".
Starting with digit 3 from r1c7, and because same digit is in positions:
r1c5, r6c5 as well as in r1c9, r6c9 it would lead to a contradiction for it in row 6.
After eliminating 3 from r1c7, the rest was a piece of cake.

Yes, the digit 3 is forming a "5 star" constallation [r1c7, r1c5, r6c5, r6c9, r1c9] with the aplha star [r1c7] that can be exploded/excluded !
If it's in the ocean, it would be a Turbot-fish.

see u,
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keith



Joined: 19 Sep 2005
Posts: 3174
Location: near Detroit, Michigan, USA

PostPosted: Sat Jan 28, 2006 7:07 pm    Post subject: Reply with quote

Do we have an opinion on whether some solutions are more "elegant" than others? If so, I submit the following:

It is fairly easy to get to the following position. Along the way you have to identify a triplet <357> in block 5.

Code:

4 1  8    7  23 236   39 5 369
5 9  2    4   1  36    7 8  36
6 7  3    5   9   8    2 4   1

3 4  5  289  28   1   89 6   7
1 6 79   89   4  37  389 2   5
8 2 79    6  35 357    4 1  39

2 5  4    3   6   9    1 7   8
9 8  6    1   7   4    5 3   2
7 3  1   28 258  25    6 9   4


Now, there is a rectangle in rows 1 and 2, but it is not very useful. Here is the explanation:

"Squares R2C6, R2C9, R1C6 and R1C9 form a Type-4B Unique Rectangle on <36>. Because they share two rows, two columns, and two blocks, if they all had possibilities <36> then the puzzle would have two solutions; you could simply exchange the <3>s with the <6>s in the squares to get the other solution, and their common rows, columns and blocks would still contain one of each value. If you look carefully, you'll see that the only squares in row 1 that can contain <6> are the "roof" squares -- R1C6 and R1C9. Since one of these squares must be <6>, the only way to avoid the "deadly pattern" is if neither of them can contain <3>."

Much more useful is the loop comprising R1C5 <23>, R1C7 <39>, R4C7 <89>, and R4C5 <28>. The solution in these four cells is either
Code:

2  3
8  9

or

3  9
2  8

So, we can clear the remaining possibilities of <3> in R1, <9> in C7, <8> in R4, and <2> in C5>.

Now, every unsolved square has two possibilities, except R6C6 is <357>. This is a BUG, and R6C6 must be <3>, which solves the puzzle. Here is a fuller explanation:

" In a BUG pattern, in each row, column and block, each unsolved possibility appears exactly twice. Such a pattern either has 0 or 2 solutions, so it cannot be part of a valid Sudoku." In other words, R6C6 cannot be <57> if the puzzle has a single, unique solution.

If you don't like to assume the puzzle is unique, it is easily solved with forcing chains. For example, assume R6C6 is <7> and look at the implications in C9.

Someone asked how to find chains and cycles. Here's what I do: Start in a square that has three possibilities. Assume a value, and trace the implications in squares that have only two possibilities. If you do not quickly find a cycle or a contradiction, choose another value or another starting square.

Keith
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Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Mon Jan 30, 2006 2:57 am    Post subject: Saturday Puzzle Jan 21 Reply with quote

This puzzle calls another to mind: the introductions to “advanced” methods I have seen do not mention the simple logical fork.

In the partial solution you set out there are only two places for 3 in row 2, namely in column 6 and column 9 (one tine of the fork). There are also only two places for 3 in row 6, column 6 and column 7 (the second tine). As the link, column 6, can hold at most one 3, at least one of the other tine ends must contain 3. 3 can therefore be deleted from any common associate of these two cells. In particular, cell (1, 7) cannot contain 3, so it must contain 9.

As someone_somewhere says you can call this a turbot fish if you wish to be fashionable. It plays no part in his constellations. In my opinion they represent a genuine advance, though I do not find them easy to spot. A 5-star constellation solves this puzzle equally well if you miss the elementary fork.

Steve
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David Bryant



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

PostPosted: Mon Jan 30, 2006 9:03 pm    Post subject: Still another way to solve it ... Reply with quote

Apparently there are quite a few ways through the logjam at the end of this particular puzzle. Here's an amplification on the message I left last week. Oh -- I guess I misspoke when I said 25 cells. I should have said 28 cells. Anyway, with 20 cells still unresolved my matrix looked like this.
Code:
  4    1    8    7   23   236  39    5   369
  5    9    2    4    1   36    7    8   36
  6    7    3    5    9    8    2    4    1
  3    4    5   289  28    1   89    6    7
  1    6   79   89    4   37   389   2    5
  8    2   79    6   35   357   4    1   39
  2    5    4    3    6    9    1    7    8
  9    8    6    1    7    4    5    3    2
  7    3    1   28   258  25    6    9    4

At this point I saw a nice "double-implication chain" rooted in r5c6:

If r5c6 = 3 then r2c6 = 6 ==> {2, 3} pair in r1c5&6 ==> r1c7=9 ==> r4c7 = 8 ==> r4c5 = 2
If r5c6 = 7 then r5c3 = 9 ==> r5c4 = 8 ==> r4c5 = 2

I concluded that r4c5 = 2 and the rest was easy. dcb
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