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Louise56 · Posted: Sun Nov 13, 2005 1:29 am Post subject: Need help with newspaper puzzle

This puzzle was in the San Diego paper 12 November, but I know David Bryant also gets it in his Colorado paper. This is the puzzle:

603 001 080
084 300 200
520 000 000

070 000 004
000 657 000
200 000 010

000 000 045
008 003 170
050 400 803

I worked on it and added some more numbers, but now I'm stuck. Hopefully someone can point me in the right direction.

693 241 587
084 305 200
520 000 430

070 000 054
000 657 000
205 004 710

000 000 045
008 503 170
050 400 803

In row 2 there's a 1 in r2c1 and r2c9. In row 3 there's a 1 in r3c3 and r3c9. That is probably an X-Y wing, which I've had pointed out to me before, but I'm just not sure what to do with it. Please advise. Confused

David Bryant · Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Hi, Louise!

Yes, this puzzle was in the Rocky Mountain News this morning. I spent far too much time on it today.

I certainly couldn't find an XY-Wing in it anywhere. I was able to use "coloring" to eliminate a few possibilities, but I hit an impasse at the same point you reached. Here's what the grid looks like with as many possibilities eliminated as I was able to work out.

someone_somewhere · Joined: 07 Aug 2005 Posts: 275 Location: Munich

603001080
084300200
520000000
070000004
000657000
200000010
000000045
008003170
050400803

I am geting to your same candidate table.

(1) r7c2 <> 6 because of the Turbot Fish:
r7c2-r7c7-r4c7-r4c3-r6c2 (for digit "6")

(2) r7c2 = 3 leads to a contradiction because:
(2a) r6c2 <> 3; r6c2 = 6; r6c9 <> 6; r4c7 = 6; r4c3 <> 6; r4c3 = 9;
(2b) r7c2 <> 1; r5c2 = 1; r5c3 <> 1; r5c3 = 9;
and digit 9 can't be simultan in r4c3 and r5c3 !!!

(3) r7c2 = 1 becasue of (1) and (2)

(4) r5c2 = 3 leads to a contradiction because:
(4a) r5c2 <> 4; r5c1 = 4; r5c1 <> 8; r5c9 = 8;
(4b) r6c2 <> 3; r6c2 = 6; r6c9 <> 6;
(4c) r5c7 <> 3; r5c7 = 9; r6c9 <> 9;
(4d) looking at (4b) and (4c) we get r6c9 =8;
and digit 8 can't be simultan in r5c9 (4a) and r6c9 (4d)

and r5c2 <> 3 leads to r5c2 = 4 that breaks the nack of this puzzle:

9 in r1c2 - Sole Candidate
7 in r1c9 6 in r8c2 - Sole Candidate
3 in r6c2 - Sole Candidate
... and so on .... up to the solution:

6 9 3 2 4 1 5 8 7
1 8 4 3 7 5 2 6 9
5 2 7 8 6 9 4 3 1
8 7 6 1 3 2 9 5 4
9 4 1 6 5 7 3 2 8
2 3 5 9 8 4 7 1 6
3 1 9 7 2 8 6 4 5
4 6 8 5 9 3 1 7 2
7 5 2 4 1 6 8 9 3

same that David has found.

P.S. my coffee got also could till I solved it!
hope someone is finding a quicker death for it!

see u,

Louise56 · Posted: Sun Nov 13, 2005 4:52 pm Post subject:

Thanks David and Someone, this one was out of my league but I'm glad to know you found it difficult too.

Louise56 · Posted: Sun Nov 13, 2005 5:33 pm Post subject:

I decided to try the puzzle again from where I left off. In the upper left hand 3x3 box there was a 7/1 pair so I thought I would put a 1 in r2c1 and see what happened. If it didn't work, I would try again with a 7 in r2c1. But this route didn't help as I got stuck again. Next I looked at the board and realized there were a lot of nines. I looked at the bottom right 3x3 and decided to try a nine in one of the spaces there. I put a 9 in r9c8 (random choice) and the rest of the puzzle went rather quickly to its solution. I think I will stick to 5-star puzzles in the future!

someone_somewhere · Joined: 07 Aug 2005 Posts: 275 Location: Munich

Hi,

I have to thank you for posting it.

I tipp: in such positions you should count in the candidate table:
(a) for each digit, how many are left
(b) for each digit how often it appears in pairs

If a digit is appearing 17, 19 or 21 times, there is a higher probability of finding a Turbot Fish for this digit, if one exists, or equivalent to eliminate one of the occurence with coloring.

If you start from a pair, at least you should select one that has a digit that appears as often as possible and has as much as possible occurences.

good luck,

P.S. you should look for all the stars on the sky, not only for 5.

see u, nice lady