dailysudoku.com Forum Index dailysudoku.com
Discussion of Daily Sudoku puzzles
 
 FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

Coloring or Trial and Error?

 
Post new topic   Reply to topic    dailysudoku.com Forum Index -> General discussion
View previous topic :: View next topic  
Author Message
AZ Matt



Joined: 03 Nov 2005
Posts: 63
Location: Hiding under my desk in Phoenix AZ USA

PostPosted: Thu Nov 03, 2005 10:41 pm    Post subject: Coloring or Trial and Error? Reply with quote

David Bryant posted this puzzle in a different topic in this forum, but that string is getting so long I thought I'd post a new topic for comments (and I finally got a chance to look at it). I am new to this, so forgive my inaccurate lexicon:

This was the puzzle as posted:

050400000
000030800
000000001
300080700
060000050
000200000
000506040
108000300
000000000

I solved to here fairly easily:

051408036
600135804
483060501
305680710
860301450
010250683
030506048
108000365
506803007

Now I throw up my hands. It's like a perfect storm of perfectly matched candidates. By that I mean that the puzzle has not been solved for the 1247 and 9, but no cell has a "stray candidate" (one that dosen't match itself in its row, column, or box) -- or, put another way, there is no logical way to "guess" where you might start an x-z wing or a set for forcing chains.

So this is how I solved the puzzle (but what I really want to know is what is the "right" way to solve the puzzle, because whatever the definition of "trial and error" is, I think this really pushes the envelope):

There are a slew of cells with 7 as a candidate -- 20 to be exact. There are six "pairs" where only two 7s can fit in a column, row, or box. So the odds that there is some coloring going on are pretty high, and this "method" (I use the term losely) only takes two minutes.

I throw the sevens on a grid and get this:

700070000
077000070
000707070
000000000
007070000
707007000
707070000
070777000
000000000

The connected sevens are in r1; r5; c2; c4; c8/box 3; and box 5.

If you look at that (I drew arrows on the grid) you see that there is only one 7 that is linked twice and only one "box only" connection -- an "imbalance"? -- and both are in the center of the puzzle. That leads me to believe (based on my experience in with x-y wings and forcing chains) that there is hope for this attempt. But you can immediately see that r5c5=7 gets you nowhere.

So I decide, based on a gut assessment of the pattern, that r1c1 is the place to start. From there it is easy to determine the if r1c1=7, r5c5= not 7 (or r5c3=7); and if r1c1= not 7, r5c5 = not 7 (or r5c3=7).

So now I have solved two cells (r5c3 and r6c6), and the puzzle crumbles.

BTW -- to test this puzzle (for kicks) I plugged in the numbers from where I stalled (second from the top) into the draw program, and it said "Unsolv able..." I added the 7 at r5c3, and the program said "Easy."

Anyway, what did I miss, and any thoughts on the method?
Back to top
View user's profile Send private message
someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Fri Nov 04, 2005 7:46 am    Post subject: Reply with quote

Hi,

From the grid of 7's that you have shown us,
we can "see" the:
r2c2 r8c2 r3c4 r8c4 r2c8 r3c8 (Swordfish on Column)

and this implies:
7 not in r8c5,
7 not in r8c6,
7 not in r3c6

and after this, the puzzle can be easy finished.

see u,
Back to top
View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    dailysudoku.com Forum Index -> General discussion All times are GMT
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum


Powered by phpBB © 2001, 2005 phpBB Group