SudoCue Users

sudokuEd · Posted: Sat Feb 03, 2007 6:29 am Post subject: 23 January X-treme - JOAX!

Been lots of very high rating X-files for January - this X-treme from 23 Jan is the highest at 37000+. So have dubbed this JOAX.

History demands we give it a whirl shakng2

.

From here: [edited to fix mistake in code referred to in following posts]

Para · Posted: Sat Feb 03, 2007 11:07 am Post subject:

37.000+??? how the hell do you get a puzzle to a 37000+ rating? That can seriously not be good for our health. Let's see what we can make of this.

Para

rcbroughton · Expert Joined: 15 Nov 2006 Posts: 143 Location: London

Hi guys

taking a break from Ed's killer - not really sure what 37000 means, except we're probably expecting some nasty moves ahead?.

Anyway here's a couple of quick moves for you:

[Edited to reflect Para picking up I'd missed a "5" at r5c5]

no longer valid
6. No 5 in r3c4
6a. r3c4=5 -> r3c3<>5 => r4c4=5 (D\) -> r3c4<>5

7. 5 lnow locked in N2 for c5

8. no 5 at r4c1
8a. r4c1=5 -> r7c1<>6 -> r7c3=5 -> r3c3<>5 => r4c4=5 -> r4c1<>5

no longer valid
9. no 5 at r4c3
9a. r4c3=5 -> r3c3<>5 => r4c4=5 (D\) -> r4c3<>5

10. no 8 at r3c3
10a Placing 8 at r3c3 eliminates all candidates for 8 in column 8
10b. r3c3=8-> r3c8<>8
10c r3c3=8-> r3c3=8 -> r5c5=4 -> r6c6=9 -> r8c8=7 -> r7c7=3 -> r4c4=5 -> r7c6=4 -> r7c8=2 -> r4c6=2 -> r8c2=8 -> r2c8<>8
10d. no other cells with 8 in column 8

11. 8 now locked for D\ in n5

12. Placing 8 at r3c4 forces 2 cells to 9 in row 8
12a. r3c4=8 -> r8c4={79}, r2c8=8 -> r5c5=4 -> r8c8 = {79}
12b. r3c4=8 -> r1c6=6, r2c8=8 -> r5c5=4 -> r4c6=2 ->r8c2=7 -> r8c48 both forced to 9

13. placing 2 at r3c7 leaves no 2 in n5 or c6
13a. r3c7=2 -> r4c6<>2, r8c27<>2 -> r8c5=2 -> r6c5<>2, r7c6<>2

14. no 2 at r4c2
14a. r4c2=2 -> r4c6<>2 => r8c2=2 -> r4c2<>2

Richard

Para · Posted: Sat Feb 03, 2007 4:19 pm Post subject:

There seem to be some typo's in your code Ed. There are still 9's in R1, C9, N3 and D/.

And i wondered about the 5 in R4C1 but i see how it works over D\ now.

Now to find somethings useful.

Para

Para · Posted: Sat Feb 03, 2007 4:26 pm Post subject:

Para · Posted: Sat Feb 03, 2007 4:58 pm Post subject:

15. No 3 in R3C8 because of 3's in C4
15a. R3C4 = 3 --> no 3 in R4C8
15b. R4C4 = 3 -->> R7C8 = 4 --> no 3 in R4C8
15c. R5C4 = 3 -->> R23C9 = 3 --> no 3 in R4C8

rcbroughton · Expert Joined: 15 Nov 2006 Posts: 143 Location: London

Para

thanks - I clearly transposed something wrong from Ed's code - I ended up without a 5 at r5c5.

I'll edit my post - only steps 8, 13 and 14 valid at this point.

I also noted a few odditied in Ed's codes - so, to summarise up to end of move 15 - but not including 6,7,9,10,11,12 - I think that leaves us here.

Para · Posted: Sat Feb 03, 2007 7:37 pm Post subject:

12 still works if you use the edit i posted. so that one should remain valid. Because the edit eliminates the 5 from R5C5 so the rest of the steps are still valid so the 8 can be removed from R3C4.

12. Placing 8 at r3c4 forces 2 cells to 9 in row 8
12a. r3c4=8 -> r8c4={79} -->> R7C4 = 6 -->> R45C4= {35} -->> r2c8=8 -> r5c5=4 -> r8c8 = {79}
12b. r3c4=8 -> r1c6=6, r2c8=8 -> r5c5=4 -> r4c6=2 ->r8c2=7 -> r8c48 both forced to 9

Para

rcbroughton · Expert Joined: 15 Nov 2006 Posts: 143 Location: London

Para

agreed! 8 comes out of r3c4 - my move 12 doesn't work but your move 12 does. Credit to you for seeing a more complex move.