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12 Jan X-treme - MOAX!
Posted: Fri Jan 12, 2007 11:02 am
by sudokuEd
This must be the Mother Of All X-tremes. SudoCue scores it at 16909! Even the horrible Oct 23 puzzle was 'only' 11845. That took a week to solve.
OK, that's the end of the MOAM
Got to here OK, then found some good moves to get started.
Code: Select all
.---------------------.---------------------.---------------------.
| 3 1 2689 | 6789 5 6789 | 46789 246789 2679 |
| 269 2689 25689 | 4 13678 136789| 16789 126789 125679|
| 469 45689 7 | 689 2 1689 | 3 1689 1569 |
:---------------------+---------------------+---------------------:
| 1269 2569 123569| 2689 13678 126789| 1679 15679 4 |
| 169 7 13569 | 69 4 1369 | 2 1569 8 |
| 8 269 4 | 5 167 1269 | 1679 3 1679 |
:---------------------+---------------------+---------------------:
| 124679 24689 12689 | 3 678 2678 | 5 126789 12679 |
| 5 68 1268 | 2678 9 4 | 1678 1268 3 |
| 2679 3 2689 | 1 678 5 | 46789 246789 269 |
'---------------------'---------------------'---------------------'
1.2 cannot be in r2c2. Here's how:
r2c2 = 2 =>r4c6 -> no 2 left for n4
-> r2c2<>2
2.2 not in r8c8. Here's how:
r8c8 = 2 =>r7c6 -> no 2 possible for n5
-> r8c8<>2
3.
7 in r9c1 =>r46c5 -> no 7 r7c5
7 r7c1 -> no 7 r7c5
->no 7 r7c5
4.
1 in r6c6 =>r2c5=>D/r7c3->no 1 in r4c3 or r2c8
1 in r8c8 -> no 1 in r2c8 -> 1 in D/ in r4c6 or r7c3 -> cross-over -> no 1r4c3
-> no 1 r4c3 or r2c8
5.
1 in r6c6-> 1 in D/ in r7c3 -> no 1 r7c9
1 in r8c8 -> no 1 r7c9
-> no 1 r7c9 -> no 1 r8c3
6.1 in r8c7 ->1 in D\in r6c6 -> 1 in n6 in c8 -> no 1 r3c8
1 in r8c8-> no 1 in r3c8
-> no 1 r3c8
7.2 cannot go in r2c8. Here's how:
r2c8=2=>r7c9=>r8c4=>r9c3 but this leaves no 2 for n1
->r2c8<>2
That should get to here. I'll let someone else take over from here - havn't even looked at this weeks Assassin yet
Code: Select all
.---------------------.---------------------.---------------------.
| 3 1 2689 | 6789 5 6789 | 46789 246789 2679 |
| 269 689 25689 | 4 13678 136789| 16789 6789 125679|
| 469 45689 7 | 689 2 1689 | 3 689 1569 |
:---------------------+---------------------+---------------------:
| 1269 2569 23569 | 2689 13678 126789| 1679 15679 4 |
| 169 7 13569 | 69 4 1369 | 2 1569 8 |
| 8 269 4 | 5 167 1269 | 1679 3 1679 |
:---------------------+---------------------+---------------------:
| 124679 24689 12689 | 3 68 2678 | 5 26789 2679 |
| 5 68 268 | 2678 9 4 | 1678 168 3 |
| 2679 3 2689 | 1 678 5 | 46789 246789 269 |
'---------------------'---------------------'---------------------'
Posted: Fri Jan 12, 2007 6:38 pm
by Para
I'll start at it tonight and see where it gets me. Then i'll let you see how far i got compared to you. Was kinda tempting to read yours first but i want to have a go at it first.
Para
p.s. yes also have to start the assasin, do todays nightmare and one trick pony and going to play tennis. Now to find time for sleep.
Posted: Fri Jan 12, 2007 9:08 pm
by sudokuEd
sudokuEd wrote:I'll let someone else take over from here
Well, Para's post sort of counts as someone else's turn
Here are a few more. I
know Para won't be impressed with my little ALS, but I know emm will be. Oh emm, I miss you so
WARNING - these steps not valid since 2 should still be in r4c6
8.
2 in r8c4 -> 2 in r6c6 -> no 2 in r9c9
2 in r8c3 -> 2 in D/ must be in r1c9 -> no 2 in r9c9
-> no 9 r9c9
9.
2 in r7c89 -> 2 in D/ must be in r9c1 -> no 2 in r9c3
2 in r9c8 -> no 2 r9c3
-> no 2 r9c3
10.
xyfin thing. I'll just write it as little ALS. Sounds more pro.
little ALS A[r2c2r9c9] B[r8c2] X = 8 Z = 6 r8c8 <>6
11.
2 in r9c8 -> 2 in r1 in c3 or c9 -> crossover -> no 2 r7c3
2 in r9c1 -> no 2 in r7c3
-> no 2 r7c3
On a roll here - if any more come soon, will edit this
[edit - deleted faulty marks pic]
Posted: Fri Jan 12, 2007 10:35 pm
by Para
Ok this is how far i got without you. Let's compare
Code: Select all
.---------------------.---------------------.---------------------.
| 3 1 2689 | 6789 5 6789 | 46789 246789 2679 |
| 269 689 25689 | 4 13678 136789| 16789 26789 125679|
| 469 45689 7 | 689 2 1689 | 3 689 1569 |
:---------------------+---------------------+---------------------:
| 1269 2569 23569 | 2689 13678 126789| 1679 15679 4 |
| 169 7 13569 | 69 4 1369 | 2 1569 8 |
| 8 269 4 | 5 167 1269 | 1679 3 1679 |
:---------------------+---------------------+---------------------:
| 124679 24689 12689 | 3 678 2678 | 5 26789 2679 |
| 5 68 268 | 2678 9 4 | 1678 168 3 |
| 2679 3 2689 | 1 678 5 | 46789 246789 269 |
'---------------------'---------------------'---------------------'
Code: Select all
.---------------------.---------------------.---------------------.
| 3 1 2689 | 6789 5 6789 | 46789 246789 2679 |
| 269 689 25689 | 4 13678 136789| 16789 6789 125679|
| 469 45689 7 | 689 2 1689 | 3 689 1569 |
:---------------------+---------------------+---------------------:
| 1269 2569 23569 | 2689 13678 16789 | 1679 15679 4 |
| 169 7 13569 | 69 4 1369 | 2 1569 8 |
| 8 269 4 | 5 167 1269 | 1679 3 1679 |
:---------------------+---------------------+---------------------:
| 124679 24689 1689 | 3 68 2678 | 5 26789 2679 |
| 5 68 268 | 2678 9 4 | 1678 18 3 |
| 2679 3 689 | 1 678 5 | 46789 246789 69 |
'---------------------'---------------------'---------------------'
Nope nothing removed that you haven't. K gonna use your moves and hope i can get any further.
sudokuEd wrote:sudokuEd wrote:I'll let someone else take over from here
Well, Para's post sort of counts as someone else's turn
Here are a few more. I
know Para won't be impressed with my little ALS, but I know emm will be.
Nice ALS
. I love that move btw. I always see that as a diagonal xyz wing. i love all moves over the diagonal.
Always impressed by ALS's as i only was able to understand them for about a month and can't seem to master them since.
I only often seem to find some way not to use and i like anything over ALS or Medusa colouring.
Para
p.s. this doesn't count as input by me, just me rambling cause i can't get any further
Posted: Fri Jan 12, 2007 10:48 pm
by Para
Ummm Ed
One quick question. How did you eliminate the 2 in R4C6. It is gone but don't see how you did it? Just curious.
This is the code after your first post. And i am missing the reason why the 2 there is gone. I am sure you got it fogured out but could you share
?
Code: Select all
.---------------------.---------------------.---------------------.
| 3 1 2689 | 6789 5 6789 | 46789 246789 2679 |
| 269 689 25689 | 4 13678 136789| 16789 6789 125679|
| 469 45689 7 | 689 2 1689 | 3 689 1569 |
:---------------------+---------------------+---------------------:
| 1269 2569 23569 | 2689 13678 16789 | 1679 15679 4 |
| 169 7 13569 | 69 4 1369 | 2 1569 8 |
| 8 269 4 | 5 167 1269 | 1679 3 1679 |
:---------------------+---------------------+---------------------:
| 124679 24689 12689 | 3 68 2678 | 5 26789 2679 |
| 5 68 268 | 2678 9 4 | 1678 168 3 |
| 2679 3 2689 | 1 678 5 | 46789 246789 269 |
'---------------------'---------------------'---------------------'
Para
Posted: Sat Jan 13, 2007 4:46 am
by sudokuEd
Para wrote:One quick question. How did you eliminate the 2 in R4C6. It is gone but don't see how you did it? Just curious....I am sure you got it
fogured out but could you share
?
Great typo Para - that's how I eliminated that 2 - been very hot and humid here in Sydney.
Whew, I'm glad you picked that up. Thanks. So, that 2 needs to go back into r4c6 - so practice ALS will have to stay in the box.
So, we are your final marks pic with the 2 in r4c6. I'll edit my marks pics too.
Thought I'd found a contradiction move - now its gone missing too. Goin Crazy! Hope you can find something Para. Have to go and relax with an Assassin for a while till then
Posted: Sat Jan 13, 2007 11:57 am
by rcbroughton
Okay, I'll add a couple of moves I can see
12. 1 cannot be in r4c6 - ALS using value 3 between [r4c3 r1c3 r2c3 r7c3 r8c3 r9c3] and [r4c5 r6c5 r7c5 r9c5]
13. 1 has to be in r7c3 - hidden single in the diagonal
Posted: Sat Jan 13, 2007 3:15 pm
by Para
I am resulting to raw moves nothing beautiful.
14. No 7 in R1C9. This is how.
R1C9 and R8C4 contain at least one 2. (colour the 2's from N8)
Placing a 7 in R1C9 forces a 7 in R4C8 and this interferes with previous statement. So no 7 in R1C9.
Para
Posted: Sat Jan 13, 2007 8:50 pm
by Para
Ok, this step is dedicated to Ed who used it before in the December 29th X-file.
15. no 8 in R4C5.
15a. 8 on D\ in R2C2 --->>> 8 in D/ in R4C6 --->> no 8 in R4C5
15b. 8 on D\ in R4C4 --->>> no 8 in R4C5
15c. 8 on D\ in R8C8 --->>> 8 in D/ in R4C6 --->> no 8 in R4C5
Para
Posted: Sat Jan 13, 2007 10:12 pm
by rcbroughton
Another ALS following the last elimination:
16. 8 cannot be in r8c4 - ALS using 7 on [r79c5] and [r4c45 r5c46 r6c56]
17. xy chain (2)r8c4->r4c4->(8)r4c4->r4c6->r8c2->(6)r8c2. Cannot have 6 in r8c4.
18. xy chain (7)r8c7->r8c4->(2)r8c4->r4c4->(8)r4c4->r4c6->r8c2->(6)r8c2. Cannot have 6 in r8c7 or 7 in r8c2
19. xy chain (1)r8c8->r8c7->(7)r8c7->r8c4->(2)r8c4->r4c4->(8)r4c4->r4c6->r8c2->(6)r8c2. Cannot have 6 in r8c8
20. 6 is now locked in n7 for row 8 - cannot appear anywhere else in n7
Posted: Sun Jan 14, 2007 12:03 pm
by Para
rcbroughton wrote:
18. xy chain (7)r8c7->r8c4->(2)r8c4->r4c4->(8)r4c4->r4c6->r8c2->(6)r8c2. Cannot have 6 in r8c7 or 7 in r8c2
7 in R8C2?
Para
Posted: Sun Jan 14, 2007 3:38 pm
by rcbroughton
Para wrote:
7 in R8C2?
Para
Sorry - you're right. Just get's rid of the 6
Posted: Sun Jan 14, 2007 5:49 pm
by rcbroughton
Getting desperate here - not particlarly elegant next step but I can't see anything else.
Anybody got anything better?
21. 1 cannot be in r6c6
my reasoning. if we place a 1 there we force the following.
21a r6c9(1) -> r8c8(8) -> r8c2(6)
21b r8c2(6) -> r2c2(9) -> r2c8(7)
21c r8c2(6) -> r8c3(2) -> r9c1(9) & r8c4(7) & r9c4(8)
21d r9c1(9) & r2c8(7) -> r1c9(2) -> r9c9(6)
This leads to a contradiction at r9c5 as r8c4(7) r9c3(8) and r9c9(6) remove all possibilities at that cell
[edit]
. . . and that leaves
22. the 1 in D\ must be at r8c8
Posted: Sun Jan 14, 2007 6:22 pm
by Para
Funny.
I just made exactly the same elimination in exactly the same desperate way.
Was still thinking on how to write it down.
If we are resulting to these measures
23. R2C8 can't be 7. This is the path.
23a. R2C8= 7 -->> R9C7=7 -->> R8C7=8
23b. R8C2= 6 -->> R8C3=2 -->> R9C1=9
23c. R9C3= 8 -->> R9C5=6 -->> R1C9 and R9C9 both 2
Para
Posted: Sun Jan 14, 2007 7:12 pm
by rcbroughton
Ok - getting somewhere.
24. r6c5 cannot be 6.
24a. r6c5(6) -> r5c4(9), r4c4(8), r6c6(2), r4c6(7), r6c2(9) and r4c6(7)
24b. r6c2(9) -> r2c2(6) -> r9c9(9), r8c2(8) -> r9c1=2 and r9c3=2
Richard