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competition # 922, not really diabolical???

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storm_norm

Joined: 18 Oct 2007
Posts: 1741

Posted: Sun Jan 20, 2008 8:03 pm Post subject: competition # 922, not really diabolical???

Code:

. . 8 | . . 1 | 9 . .
. . . | 4 . . | . 5 .
. . . | . 3 . | 8 . 6
---------------------
. 5 2 | . . . | 7 9 .
. . . | 1 . 3 | . . .
. 6 3 | . . . | 2 4 .
---------------------
3 . 4 | . 7 . | . . .
. 8 . | . . 5 | . . .
. . 6 | 9 . . | 3 . .

after basics, this one boils down to one "advanced" move, and visually, I think this particular advanced move is the easiest to see, and IMHO isn't much of an advanced move after one learns it.
nonetheless, a very powerful technique when you see it.

http://www.sudoku.org.uk/DailySudoku.asp?day=18/01/2008

keith

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

Posted: Sun Jan 20, 2008 8:36 pm Post subject:

Well, since this came up yesterday: If you want more of a challenge, solve it as a BUG+3. Keith

keith

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

Posted: Sun Jan 20, 2008 11:07 pm Post subject: A BUG+3

Here is how the BUG+n works.

Basics get you to here:

Code:

+----------------+----------------+----------------+
| 267 27 8 | 267 5 1 | 9 3 4 |
| 2679 3 79 | 4 8 26 | 1 5 27 |
| 4 1 5 | 27 3 9 | 8 27 6 |
+----------------+----------------+----------------+
| 1 5 2 | 68 4 68 | 7 9 3 |
| 79 4 79 | 1 2 3 | 5 6 8 |
| 8 6 3 | 5 9 7 | 2 4 1 |
+----------------+----------------+----------------+
| 3 9 4 | 28 7 28 | 6 1 5 |
| 27 8 1 | 3 6 5 | 4 27 9 |
| 5 27 6 | 9 1 4 | 3 8 27 |
+----------------+----------------+----------------+

The BUG deadly pattern (DP) is as follows:
1. Every unsolved cell has only two candidates.
2. In each row, column, and block, each candidate appears exactly twice.

Look at B2: The candidates are <267>, <26>, and <27>. In the DP, R1C4 must be <67>.

Check in C4: <67>, <27>, <68>, <28> meets the conditions of the DP.

Look at R1: The candidates are <267>, <27>, <67>. In the DP, R1C1 must be <26>.

Look at B1: The candidates are <26>, <27>, <2679>, <79>. In the DP, R2C1 must be <69>.

Check in C1: The candidates are <26>, <69>, <79>, <27>. We have a DP! Each candidate occurs twice in each row, column, and box.

So, we have:

Code:

+----------------+----------------+----------------+
|26(7) 27 8 |(2)67 5 1 | 9 3 4 |
|(27)69 3 79 | 4 8 26 | 1 5 27 |
| 4 1 5 | 27 3 9 | 8 27 6 |
+----------------+----------------+----------------+
| 1 5 2 | 68 4 68 | 7 9 3 |
| 79 4 79 | 1 2 3 | 5 6 8 |
| 8 6 3 | 5 9 7 | 2 4 1 |
+----------------+----------------+----------------+
| 3 9 4 | 28 7 28 | 6 1 5 |
| 27 8 1 | 3 6 5 | 4 27 9 |
| 5 27 6 | 9 1 4 | 3 8 27 |
+----------------+----------------+----------------+

To avoid the DP,
1. R1C4 is <2>, and/or
2. R1C1 is <7>, and or
3. R2C1 is <2> or <7>.

Now, what? There is no recipe, you have to figure out each case on its own.

Well, if R1C4 or R2C1 is <2>, R1C2 is <7>. The other possibilities are R1C1 or R2C1 is <7>. In B1, R2C3 cannot be <7>.

Brilliant!

Except, it does not solve the puzzle. We get to here:

Code:

+--------------+--------------+-------------+
|26(7) 27 8 |(2)67 5 1 | 9 3 4 |
|(2)67 3 9 | 4 8 26 | 1 5 27 |
| 4 1 5 | 27 3 9 | 8 27 6 |
+--------------+--------------+-------------+
| 1 5 2 | 68 4 68 | 7 9 3 |
| 9 4 7 | 1 2 3 | 5 6 8 |
| 8 6 3 | 5 9 7 | 2 4 1 |
+--------------+--------------+-------------+
| 3 9 4 | 28 7 28 | 6 1 5 |
| 27 8 1 | 3 6 5 | 4 27 9 |
| 5 27 6 | 9 1 4 | 3 8 27 |
+--------------+--------------+-------------+

Another BUG+3! So:
1. R1C4 is <2> (and R1C2 is <7>), and/or
2. R1C1 is <7>, and/or
3. R2C1 is <2>.

In any event, R2C1 is not <7>, and the puzzle is solved.

Keith

(Yes, I know. There is a UR or a zillion remote pair eliminations that also solve the puzzle. This is more fun. It's the journey, not the destination.)

PS: I think Green Bay will win tonight, and lose the Super Bowl in a blowout.

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