Tough Sudoku for 2/September/2009

Above tree line? Pretty high!

Beautiful, Wilodene!

Enjoy your day/night, everyone!

Oh!!! Just beautiful - I would like to have a small cabin beside the lake! Thank you, Wilodene!

34 minutes

Pretty! My first thought was Lake Louise, but this
is just as beautiful.

1. Note triples 789 at ade6;136 at efi4.Unique possibilities to 35.
2. Note triple 267 at g9/i89;pair 89 at h78.If c8=4,i8=7,b8=2,a9=3=h5=d1,e1=5,i4=1,i1=4 and column a is devoid of 4.So e8=4.UPs to 40.
Here I thought I found a proof that g3=3, but it was a mistake.Maybe someone can take it from here.

I shall eventually blog this one.

Hi Alfred, in your step 2, has the possibility of (4)a2 been resolved?

Rather than the usual very detailed blog post, for today's puzzle I choose a sparsely illustrated solution path. This of course saved me much time..... - is lazy a virtue?

35:48, a bit slow today. Lovely photo!

Hi Steve.a2 is not determined yet in step 2.Step 2. resolves e789 and h78. Regards,Alfred.

Of course lazy is a virtue. If we weren't lazy we'd still be living in caves.

Thanks Alfred, I think. What I meant, perhaps unclearly, is that from your forcing chain, I still have (4)a2 as a possibility. Thus, column a is not quite yet devoid of (4). Perhaps I missed something easy?

Thanks for the blog, Steve. Wow.

Brenda

48:02

Steve's blog was great but I couldn't help wondering if the puzzle couldn't possibly be solved by more elementary means. Here’s the “lesser mortal” solution I’ve worked out with some help and polishing from Alfred (Thanks, Alfred!):

Starting at 22, given UPs to 23 --

1) ghi6=456, so d6=9, etc. to UP35.

2) c8=4 or e8=4. If c8=4, i8=7, b8=2, a9=3, h5=3, i4=1, d5=1, d1=3, e1=5, i1=4, a1=8, a7=5, b4=8, a4=2, c4=5, c3=2, & all candidates are eliminated from e3. So e8=4 & UPs to 40.

3)Note gi7,h9=134. a9=3 or h9=3. If h9=3, g7<>3.
If a9=3, h9=4, h6=5, h2=1, h5=3, i4=1, i7=3 & g7<>3. So g7<>3.

4) g7=1 or 4. If g7=1, g13<>1. If g7=4, g6=6, g9=7, g2=1, and g13<>1. So g13<>1.

5a) i4=1 or h5=1. If i4=1, g7=1, f4=6, e4=3, d1=3, e1=5, g2 & i1=47 & gi3<>4.
If h5=1, i4=3, e4=6, f4=1, f3=4, & gi3<>4. So gi3<>4. (Note then that FC3 in g13 so i3<>3 & g13=38.)

5b) i4=1 or h5=1. If i4=1, i3<>1 so ghi123 has FC gh2=1, so f2<>1.
If h5=1, i4=3, e4=6, f4=1 & f2<>1. So f2<>1.

6) (Alfred’s) f3=1 or 4. If f3=1, f2=4, i1=4, d1=3, b1=7, and row 1 is devoid of 1.
So f3=4, UP 41. Then SSTs to UP=57.

8) b4= 2 or 5. If b4=2, c4=8, a4=5, a2=4, a1=8, g1=3, i1=4, h2=5, i3=1 and all candidates are eliminated from i7. So b4=5, UP 58. SSTs to end.

SE=9.0

possible solution similar to Steve's:
1.SST to UP35
2.K column(1)f234 > -1h2
(1)f4-d5=g5
||
(1)f3-b3=(17')b12
(1-4)f3=f2-(4=17)g2
||
(1)f2
3.(4=5)h2-(5=134)i147 > -4i1,FXW(3),-3i1
4.(3)h9=a9-(3=258)ac7b8-(2=7)i8-i1=YW(153)i14e1-(3)e4=d5 > -3h5, UP81