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George Woods · Joined: 28 Mar 2006 Posts: 304 Location: Dorset UK

At the "crunch point" a 6 in row 5 cols 1 2 or 3 leads to a conflict that row 4 col 7 must hold both 6 or 9. Henace row 5 col1 must be 5 BUT does anyone out there have a logical line of analysis?

Steve R · Posted: Thu Apr 13, 2006 11:15 pm Post subject:

If we have the same crunch point, all candidates but 5 are excluded from r5c1 ... as long as you spot that column 3 splits (35)(469) and that the 3 or 5 in r2c3 forces the 6 of box1 into the first column.

Steve

George Woods · Joined: 28 Mar 2006 Posts: 304 Location: Dorset UK

Thanks - My fault - most puzzles can be done on doublets so I ignored the triplet 469 - that lead to the 35 etc etc... At least I realised that there must be a more direct route!!!!

Guest · Posted: Fri Apr 14, 2006 4:27 am Post subject:

George Woods · Joined: 28 Mar 2006 Posts: 304 Location: Dorset UK

The "crunch point is
008 200 397
070 000 182
201 008 465
700 005 008
000 987 014
180 420 750
017 000 006
002 000 040
360 802 570

The first observation is that the 3 in row 5 must lie in box 4
so in col 3 the cells in r4,r6,and r9 are all "attacked" by 35 so the only places left for 35 are r2 and r5! etc.......

dotdot · Joined: 07 Dec 2005 Posts: 29 Location: oberseen

Sarah · Joined: 04 Apr 2006 Posts: 6 Location: Brooklyn

OK, I'm feeling like a dummy, but I have to continue this until I get it.

I understand that you can isolate the 5 in box 4 to lie in r5. But I still find the options 3,5,6 for r5c3, and 3,5,6, 9 for r2c3. I am just not getting how the 3,5 pair in c3...

Help!
Thanks

dotdot · Joined: 07 Dec 2005 Posts: 29 Location: oberseen

Sarah,
Seeing what can be in r2c3 and r5c3 isn't conclusive here.
The argument goes in two steps:
- see what can't go in certain cells
- what can't go there must be accomodated in the other cells.

In col3 we already know 4 cells so the other five are free.
- 3 and 5 are seen to be excluded from r4c3, r6c3 and r9c3 (cf George above)
- so the 3 and 5 must occur in the remaining free cells (of col3).

Sarah · Joined: 04 Apr 2006 Posts: 6 Location: Brooklyn

Thanks dotdot. Sometimes it seems the logic comes almost naturally, and other times I have an unexplained block that all of sudden gets unclogged ... as happened here.
Thanks for the patient explanation.

keith · Posted: Fri Apr 14, 2006 8:10 pm Post subject: A simpler way?

Here is the situation posted by George, with the possibilities:

George Woods · Joined: 28 Mar 2006 Posts: 304 Location: Dorset UK

"Quelle Domage" Keith--- I had filled in my crunch point in ink, finishing the puzzle in pencil--- erased the pencil to create the "crunch point" that I published above. The 1 in the bottom right hand corner was erroneously entered (incompletely erased) so your beautifully simple solution would not have worked on a puzzle started from the beginning!

keith · Posted: Sat Apr 15, 2006 12:15 am Post subject:

Dang!

Keith

TKiel · Joined: 22 Feb 2006 Posts: 292 Location: Kalamazoo, MI

(George Woods--You can edit your post above to make the posted grid correct.)

With the 1 changed from a placement to a candidate in r9c9, the grid looks like this:

PGordon · Joined: 13 Apr 2006 Posts: 2 Location: New York City

Having reached the same crunch point as George, I managed to miss the obvious point that Box 1 "owns" the 6 in column 1 , so r5c1 can only be 5. So in a total fury I chose to apply the old rule "don't force it, get a bigger hammer" and decided to "see what happens if" r8c9 is a 3 rather than the 1 which was its only other possibility. This with over 40 cells still empty. Oddly enough, this actually led to the correct solution, using only logic and no further guesswork. How embarrassing!

keith · Posted: Thu Apr 20, 2006 9:18 pm Post subject:

Would you rather be good, or lucky?

Keith