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 12 Jan X-treme - MOAX! Goto page 1, 2  Next
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sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Fri Jan 12, 2007 11:02 am    Post subject: 12 Jan X-treme - MOAX!

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: .---------------------.---------------------.---------------------. | 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: .---------------------.---------------------.---------------------. | 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   | '---------------------'---------------------'---------------------'

Last edited by sudokuEd on Sat Jan 13, 2007 4:50 am; edited 1 time in total
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Fri Jan 12, 2007 6:38 pm    Post subject: 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.
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Fri Jan 12, 2007 9:08 pm    Post subject:

 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]

Last edited by sudokuEd on Sat Jan 13, 2007 4:53 am; edited 1 time in total
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

Posted: Fri Jan 12, 2007 10:35 pm    Post subject:

Ok this is how far i got without you. Let's compare

 Code: .---------------------.---------------------.---------------------. | 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: .---------------------.---------------------.---------------------. | 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

Last edited by Para on Fri Jan 12, 2007 10:55 pm; edited 1 time in total
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

Posted: Fri Jan 12, 2007 10:48 pm    Post subject:

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: .---------------------.---------------------.---------------------. | 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
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Sat Jan 13, 2007 4:46 am    Post subject:

 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
rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

 Posted: Sat Jan 13, 2007 11:57 am    Post subject: 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
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Sat Jan 13, 2007 3:15 pm    Post subject: 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
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Sat Jan 13, 2007 8:50 pm    Post subject: 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
rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

 Posted: Sat Jan 13, 2007 10:12 pm    Post subject: 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
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

Posted: Sun Jan 14, 2007 12:03 pm    Post subject:

 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
rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

Posted: Sun Jan 14, 2007 3:38 pm    Post subject:

 Para wrote: 7 in R8C2? Para

Sorry - you're right. Just get's rid of the 6
rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

 Posted: Sun Jan 14, 2007 5:49 pm    Post subject: 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  . . . and that leaves 22. the 1 in D\ must be at r8c8
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Sun Jan 14, 2007 6:22 pm    Post subject: 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
rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

 Posted: Sun Jan 14, 2007 7:12 pm    Post subject: 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
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