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competition # 922, not really diabolical???

 
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Sun Jan 20, 2008 8:03 pm    Post subject: competition # 922, not really diabolical??? Reply with quote

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
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keith



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

PostPosted: Sun Jan 20, 2008 8:36 pm    Post subject: Reply with quote

Well, since this came up yesterday:

If you want more of a challenge, solve it as a BUG+3.

Keith
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keith



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

PostPosted: Sun Jan 20, 2008 11:07 pm    Post subject: A BUG+3 Reply with quote

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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