dailysudoku.com

storm_norm · Joined: 18 Oct 2007 Posts: 1741

http://www.sudoku.org.uk/PrintWeeklySudoku.asp?number=59

this is the puzzle

9 x x x x 1 x x 2

x x 5 x x x 3 x x

3 x x 8 x x x x 6

x 4 x x 5 x x 9 x

1 x x x x x x x 7

x 6 x x 8 x x 4 x

5 x x 3 x 6 x x x

x x 2 x x x 6 x x

x x x x x 9 x x 8

this is as far as I can get, then it gets over my head:

9 x 6 x 3 1 x x 2

x x 5 x 6 x 3 x x

3 x x 8 x 5 x x 6

x 4 x 6 5 x x 9 x

1 5 x x x x x 6 7

x 6 9 1 8 x x 4 x

5 x x 3 x 6 x x x

x x 2 x x 8 6 x x

6 x x x x 9 x x 8

probably a variety of ALS moves and finned moves from here, and most likely a couple long chains.

and longer chains!

I can see why no one has posted any of these puzzles in the past. just way too many steps to complete them in a logical way. however, I would love to find out what others think of them

Norm

Marty R. · Posted: Wed Nov 14, 2007 10:21 pm Post subject:

I put it into grid form for those who might want to play with it.

re'born · Joined: 28 Oct 2007 Posts: 80

Here is an opening salvo:
Potential deadly pattern in r46c16<27> implies r4c1 = 8, r4c6 = 3 or r6c6 = 3. So we get the short AIC
(3)r46c6 = (8)r4c1 - (8=3)r5c3, => r5c6 <> 3.

keith · Posted: Thu Nov 15, 2007 12:00 am Post subject:

Sudoku Susser says it needs:

9 x Comprehensive Forcing Chains
4 x Simple Forcing Chains
1 x XYZ-Wing
6 x Intersection Removal
1 x Simple Hidden Sets
3 x Simple Naked Sets
20 x Pinned Squares

Keith

Asellus · Posted: Thu Nov 15, 2007 8:16 am Post subject:

I like re'born's UR-based AIC. Nice!

I found one that does the same without the UR:
(8=2)R5C7-(2=5)R6C7-(5={47})R1C47-(7=8)R1C2-(8=8)R7C23; R5C3<>8

Asellus · Posted: Thu Nov 15, 2007 8:28 am Post subject:

After either of the previously posted AICs, there is a {78} W-Wing in R1C2 and R4C3 that removes <7> from R3C3.

It's not that much help. The grid now looks like:

keith · Posted: Thu Nov 15, 2007 9:21 am Post subject:

An XYZ-wing <147> takes out <4> in R7C3.

Keith

re'born · Joined: 28 Oct 2007 Posts: 80

re'born · Joined: 28 Oct 2007 Posts: 80

(5)r8c9 = (5)r6c9 - (5=2)r6c7 - (2=7)r6c1 - (7=4)r8c1, => r8c9<>4

(naked triple in column 9)

I then noticed a useless wxyz-wing r2c9, r3c378 with pivot r3c7, which implies at least one of those cells is a 4. Transporting r2c9 to r3c3 via r2c1, we then get that r3c5<>4.
This takes here:

Asellus · Posted: Thu Nov 15, 2007 10:09 am Post subject:

Yeah... well while you were doing all that, I found a way to get rid of <7> in R4C1 by exploiting the strong inferential link between the W-Wing <7>s:

(7)R4C3=(7-8)R1C2=(8-5)R1C8=(5)R1C7-(5)R6C7=(5-3)R6C9=(3-7)R6C6=(7)R4C6; R4C1<>7

PS: I noticed the XYZ Wing but didn't want to post again until I found something more powerful. Don't know if I succeeded.

Asellus · Posted: Thu Nov 15, 2007 10:44 am Post subject:

Okay... another multi-coloring inspired AIC chips away at this beast:

(1)R8C5=(1-5)R8C9=(5)R6C9-(5=2)R6C7-(2=7)R6C1-(7=4)R8C1; R8C5<>4

re'born · Joined: 28 Oct 2007 Posts: 80

Asellus · Posted: Thu Nov 15, 2007 11:23 am Post subject:

Asellus · Posted: Thu Nov 15, 2007 11:35 am Post subject:

re'born,

That AIC loop makes sense to me. So, with the additional eliminations, this is the grid I now have:

re'born · Joined: 28 Oct 2007 Posts: 80

An AIC finishes it from here:

(7=5)r8c4 - (5)r8c9 = (5)r6c9 - (5=2)r6c7 - (2=7)r6c1 - (7)r2c1 = (7)r1c2, => r1c4<>7, solving the puzzle.

Asellus · Posted: Thu Nov 15, 2007 11:58 am Post subject:

Okay... I didn't rest. I got the exact same AIC: it's a fairly straightforward Medusa trap.

[Edit: Not quite the same... my chain went via R6C6 rather than R6C7.]

re'born · Joined: 28 Oct 2007 Posts: 80

Asellus · Posted: Thu Nov 15, 2007 8:58 pm Post subject: