dailysudoku.com

[Alan]

I normally manage to solve these puzzles comfortably in my lunchbreak, and even when time pressure means I resort to 'cheating' I understand the logic behinf a hint once I've seen it.

Today I'm obviously having an off day. I have got the puzzle this far

010 346 089
803 275 000
054 981 200

940 607 800
300 109 007
075 438 029

037 064 090
000 012 304
400 093 060

The hint says that the bottom row, cell3 is a 1.
Anyone enlighten me why the 1 cannot be between the 4 and 6 on row 4, or between the 3 and 6 on the bottom row?

As I said, I'm obviously having a thick day.

[Kim]

<The hint says that the bottom row, cell3 is a 1.
Anyone enlighten me why the 1 cannot be between the 4 and 6 on row 4, or between the 3 and 6 on the bottom row? >

Try this...
in r7c7 it can be a (1 5) r7c9 (1 2
in r8c8 is can be a (5 7)
in r9c7 it can be a (1 5 7) and r9c9 (1 2 8)

see the pattern the (1 5 7) in r7c7, r8c8 and r9c7???

That will elimatine the 1 in r7c9 and r9c9 giving you fixed pairs of (2 8)

In doing so, it gives you a fixed pair of (2 in r9c2 and r9c9 making the only number left in r9c3 a 1....

Understand??? I have a tough time explaining this stuff like the pros do.

Kim

[Guest]

<<Try this...
in r7c7 it can be a (1 5) r7c9 (1 2
in r8c8 is can be a (5 7)
in r9c7 it can be a (1 5 7) and r9c9 (1 2

see the pattern the (1 5 7) in r7c7, r8c8 and r9c7??? >>

Darn it...It left out the 8's.

It should look like ....
In r7c7, it can be a (1 5) and in r7c9 (1 2
In r8c8, it can be a (5 7)
In r9c7 it can be a (1 5 7) and in r9c9 (1 2 8)

See the pattern?

Hope that clarifies!!!
Kim

[Kim]

Darn smileys are everywhere....

Wherever you see the stupid smiley with dark glasses, replace that stupid piece of crap with an 8 or eight!!!

Note to self...Never put an 8 right next to a ) or you will get a smiley with glasses!!!

Nothing like 15 posts to answer your question....Good luck
Kim

[Alan]

Doh!!!

I'd done most of the hard work, then missed the easy bit.
I had the 157 triplet and the matched 28's in c9.
I'd then been too blind to see the 5 in r3c2, leaving 258 as possibilities for r9c2 so didn't get the matched pair of 28's in the bottom row.

Again, I say, Doh!!!

and thanks.

[Susie22]

Can someone please explain to me about the triplet (1 5 7) in r7c7, r8c8 and r9c7??? Why will this elimatine the 1 in r7c9 and r9c9 giving you fixed pairs of 28's.

[David Bryant]

[Guest]

[someone_somewhere]

[someone_somewhere]

Or we could have the situation for a 3x3>


157 9 12

3 57 4

157 6 12

Now, if the numer 1 would be in r1c3 than we would habe the pair of numbers 1 and 5 left to be in 3 cells: r1c1 r2c2 and r3c1 and this would be impossible !
So, we can eliminate / exclude number 1 from cells r1c3 and r3c3.

have fun,

[David Bryant]

[Guest]

column 3 should be 128, so after eliminating 1 should make it 28

[someone_somewhere]

Ok, Ok, if you want to correct things ...
Let's do it proper:

The initial position has a 9 in r1c9 and r6c9. At least if you take a look at the first message posted here.

. . .

see u,