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DB Puzzle Dec 24

 
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keith



Joined: 19 Sep 2005
Posts: 3150
Location: near Detroit, Michigan, USA

PostPosted: Sat Dec 24, 2005 1:06 pm    Post subject: DB Puzzle Dec 24 Reply with quote

Here is today's Bodycombe puzzle.

Code:


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



Can anyone find this online? In my local newspaper, the online puzzle is not the same as the one printed in the paper.

Merry Christmas!

Keith
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keith



Joined: 19 Sep 2005
Posts: 3150
Location: near Detroit, Michigan, USA

PostPosted: Wed Dec 28, 2005 12:22 am    Post subject: This is a tough one! Reply with quote

56 views, and no solution yet? Not even any replies?

This is a tough, but very interesting puzzle! Some hints:

Normal methods lead to a point where the puzzle seems almost to be solved. The bottom three blocks are complete, as are most of the other squares. Only 21 squares are incomplete.

Then, there is one critical square. It has three possible values. (There are only three such squares. The other 18 unsolved squares have only two possible values.) If you can exclude one value in this square, the puzzle is solved!

Of course, I am hoping others will find other paths to the solution!

Best wishes,

Keith
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Steve R



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

PostPosted: Wed Dec 28, 2005 6:25 am    Post subject: This is a tough one! Reply with quote

Keith:

Yes (twice). It is an interesting puzzle. And it can be solved in a different way.

At least, it may be different. I did not manage to reach your position with the bottom three boxes complete. At this stage I took a different route:

900 000 830
704 108 005
008 002 740

046 800 007
890 000 062
070 001 408

587 300 004
600 714 589
419 080 073

There are only two places for a 9 in box (1, 2), r3c5 and its neighbour to the left. The same applies to box (3, 2), in r7c5 and its neighbour to the right. So 9 can be eliminated from the rest of column 5; in particular it can be eliminated from r4c5.
This discloses an XY-Wing for 5, 1 and 2 in r4c1, r4c5 and r5c3. Thus 5 can be excluded from r5c4 and the only entry left for that cell is 4. The rest is elementary.

It would be useful to know of the abnormal method you had in mind.

Thanks for introducing me to the puzzle.

Steve.
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Steve R



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

PostPosted: Wed Dec 28, 2005 5:14 pm    Post subject: This is a tough one! And embarrassing for some. Reply with quote

Be kind. Ignore my previous note.

The shameful truth is I attempted the puzzle after a great deal of burgundy. So I copied it incorrectly and composed a note of complete gibberish.

No suggestions for a New Year resolution are required, thank you.

Apologies.

Steve
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David Bryant



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

PostPosted: Sat Dec 31, 2005 12:23 am    Post subject: I used double implication chains Reply with quote

I might have replied to your post earlier, Keith, but was gone to Tucson to visit relatives for Christmas and just got back last night.

After 21 moves the grid looks like this.
Code:
  9    56   13   45   47   37   8    2    16
  7    2    4    1    6    8    9    3    5
  13   56   8    9    35   2    7    4    16
 123   4    6    8    23   9    13   5    7
  8    9   135   45   47   37   13   6    2
  23   7    35   6   235   1    4    9    8
  5    8    7    3    9    6    2    1    4
  6    3    2    7    1    4    5    8    9
  4    1    9    2    8    5    6    7    3

I rather stupidly traced out a couple of forcing chains rooted in r5c6 and arrived at the following result:

r5c6 = 3 ==> r1c6 = 7 ==> r1c5 = 4 ==> r1c4 = 5 ==> r3c5 = 3
r5c6 = 3 ==> r4c5 = 2
r5c6 = 3 ==> r5c7 = 1 ==> r4c7 = 3
r4c7 = 3 and r4c5 = 2 ==> r4c1 = 1 ==> r3c1 = 3

This contradiction proves that r5c6 <> 3. Therefore r5c6 = 7, and the rest of the puzzle is a piece of cake. dcb

PS There might be a shorter set of "double-implication chains" in this puzzle somewhere, but I didn't spend a lot of time looking -- I just grabbed the first one I found that worked and ran with it. Any old port in a storm!

PPS Here's a bone for the "logical purism" crowd. We have already traced out a chain showing that if r5c6 = 3 then r3c1 =- 3. What happens if r5c6 = 7? Then we have

r5c6 = 7 ==> r5c5 = 4 ==> r5c4 = 5 ==> r3c5 = 5 ==> r3c2 = 6 ==> r3c9 = 1 ==> r3c1 = 3

So r3c1 = 3 whichever value we place at r5c6, and that piece of information is again enough to resolve the rest of the puzzle.
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keith



Joined: 19 Sep 2005
Posts: 3150
Location: near Detroit, Michigan, USA

PostPosted: Sat Dec 31, 2005 2:15 pm    Post subject: My solution Reply with quote

David,

From the position you have, there is a forcing chain from R5C3 that proves R5C3 is not 1. Therefore it is <35>, there is a pair in C3, and the rest is trivial.

(Assume R5C3 = 1. Then, eventually, R5C7 = 1, which is a contradiction in R5.)

Happy New Year!

Keith
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