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keith · Posted: Sat Jan 21, 2006 1:20 pm Post subject: Saturday Puzzle Jan 21

Here is today's puzzle:

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

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

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

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,

keith · Posted: Sat Jan 28, 2006 7:07 pm Post subject:

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.

Steve R · Posted: Mon Jan 30, 2006 2:57 am Post subject: Saturday Puzzle Jan 21

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

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

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.