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sudokuEd
Grandmaster

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

Posted: Thu Oct 19, 2006 11:03 am    Post subject: X-treme October 19 - Help

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    | '------------------'------------------'------------------'
sudokuEd
Grandmaster

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

Posted: Tue Oct 24, 2006 11:20 am    Post subject:

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]
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