SudoCue Users Forum Index SudoCue Users
A forum for users of the SudoCue programs and the services of SudoCue.Net
 
 FAQFAQ   SearchSearch   UsergroupsUsergroups   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

X-treme October 19 - Help

 
Post new topic   Reply to topic    SudoCue Users Forum Index -> X-Files
View previous topic :: View next topic  
Author Message
sudokuEd
Grandmaster
Grandmaster


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

PostPosted: Thu Oct 19, 2006 11:03 am    Post subject: X-treme October 19 - Help Reply with quote

No Killer pairs, no cage sums, no innies-outies - help!

Can someone show the way? Thanks.
Code:
.------------------.------------------.------------------.
| 378   4     138  | 6     5     2    | 1378  378   9    |
| 5     378   9    | 4     378   1    | 6     2     378  |
| 238   236   13678| 9     378   38   | 5     4     1378 |
:------------------+------------------+------------------:
| 9     1     378  | 2     4     38   | 378   6     5    |
| 23678 2378  378  | 1     368   5    | 4     9     378  |
| 4     5     368  | 7     36    9    | 38    1     2    |
:------------------+------------------+------------------:
| 368   9     4    | 5     2     7    | 138   38    1368 |
| 367   36    2    | 8     1     4    | 9     5     367  |
| 1     78    5    | 3     9     6    | 2     78    4    |
'------------------'------------------'------------------'
Back to top
View user's profile Send private message
sudokuEd
Grandmaster
Grandmaster


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

PostPosted: Tue Oct 24, 2006 11:20 am    Post subject: Reply with quote

Finally worked out how to unlock this one.
1. From puzzle pic above - two 7's in N7.
-7 in r9c2=>r8c9=>r1c78-> no 7 in r1c1 [edit:typo]
-7 in r8c1 -> no 7 in r1c1
->no 7 in r1c1

2. each 8 on D\ eliminates 8 from r1c3
-8 in r7c7=>r5c9=>r46c3 -> no 8 in r1c3
-8 in r5c5 =>r46c3-> no 8 in r1c3
-8 in r1c1, r2c2 or r3c3 -> no 8 in r1c3
-> no 8 in r1c3 [edit:typo]
r1c1 = {38}, r1c3 ={13}

3. 7 in r1 locked in n3 -> no 7 elsewhere n3-> 7 in c9 in r58c9.
3a. 7 in r8c9 =>r9c2 => r3c3 -> no 7 in r5c23
3b. 7 in r5c9 -> no 7 in r5c23
-> no 7 in r5c23 [edit2:this step added]

This leaves the puzzle in the following position

Code:
.------------------.------------------.------------------.
| 38    4     13   | 6     5     2    | 1378  378   9    |
| 5     378   9    | 4     378   1    | 6     2     38   |
| 238   236   13678| 9     378   38   | 5     4     138  |
:------------------+------------------+------------------:
| 9     1     378  | 2     4     38   | 378   6     5    |
| 23678 238   38   | 1     368   5    | 4     9     378  |
| 4     5     368  | 7     36    9    | 38    1     2    |
:------------------+------------------+------------------:
| 368   9     4    | 5     2     7    | 138   38    1368 |
| 367   36    2    | 8     1     4    | 9     5     367  |
| 1     78    5    | 3     9     6    | 2     78    4    |
'------------------'------------------'------------------'

4. A contradiction chain shows that 3 cannot go in r1c3.
3 in r1c3=>8 r1c1=>1 r1c7 => 7 r1c8 also 3 in r7c7 (from D\ and r1c7) but this leaves 8 in both r79c8
->r1c3 = 1 [edit2: typo]

From here
Code:
.------------------.------------------.------------------.
| 38    4     1    | 6     5     2    | 378   378   9    |
| 5     378   9    | 4     378   1    | 6     2     38   |
| 238   236   3678 | 9     378   38   | 5     4     1    |
:------------------+------------------+------------------:
| 9     1     378  | 2     4     38   | 378   6     5    |
| 23678 238   38   | 1     368   5    | 4     9     378  |
| 4     5     368  | 7     36    9    | 38    1     2    |
:------------------+------------------+------------------:
| 368   9     4    | 5     2     7    | 1     38    368  |
| 367   36    2    | 8     1     4    | 9     5     367  |
| 1     78    5    | 3     9     6    | 2     78    4    |
'------------------'------------------'------------------'


5.r5c5<>8 since leaves 8 for n12 in r3 -> r4c6 = 8

6.8 in D\ only in n1 -> no 8 in r3c1

7. two 3's in D/: cross-over -> no 3 in r5c2 or r2c2

the rest unravels from there

[edit2:steps 5-7 added]
Back to top
View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    SudoCue Users Forum Index -> X-Files All times are GMT
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum


Powered by phpBB © 2001, 2005 phpBB Group