dailysudoku.com

[AZ Matt]

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?

[someone_somewhere]

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,