hard one

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hard one

Postby oberon » Sun Mar 19, 2006 4:00 pm

Having solved some simple-sudoku-puzzles in expert-level I now dared to try some "extreme" ones ... First 3 went fine - now I am confronted with this:

Code: Select all
*-----------------------------------------------------------*
 | 4     36    1369  | 6789  5     689   | 367   1267  1237  |
 | 8     7     16    | 46    3     2     | 5     146   9     |
 | 2     5     369   | 1     47    69    | 3467  8     37    |
 |-------------------+-------------------+-------------------|
 | 7     36    5     | 468   124   368   | 9     126   123   |
 | 136   4     2     | 679   17    369   | 8     5     137   |
 | 136   9     8     | 67    127   5     | 367   1267  4     |
 |-------------------+-------------------+-------------------|
 | 9     2     7     | 5     8     4     | 1     3     6     |
 | 5     1     4     | 3     6     7     | 2     9     8     |
 | 36    8     36    | 2     9     1     | 47    47    5     |
 *-----------------------------------------------------------*

Any suggestions?
oberon
 
Posts: 7
Joined: 18 March 2006

Re: hard one

Postby foxglove » Sun Mar 19, 2006 4:19 pm

oberon wrote:Any suggestions?



!+4@3/5=>-4@4/5<=>+4@4/4=>-8@4/4<=>+8@1/4=>-7@1/4<=>+7@3/5<=>-4@3/5!

or

!+4@3/5=>-4@2/4<=>+6@2/4=>-6@6/4<=>+7@6/4=>-7@1/4<=>+7@3/5<=>-4@3/5!

eliminates 4@3/5 and you're (almost) done.
foxglove
 
Posts: 12
Joined: 04 February 2006

Re: hard one

Postby oberon » Sun Mar 19, 2006 4:26 pm

foxglove wrote:!+4@3/5=>-4@4/5<=>+4@4/4=>-8@4/4<=>+8@1/4=>-7@1/4<=>+7@3/5<=>-4@3/5!

or

!+4@3/5=>-4@2/4<=>+6@2/4=>-6@6/4<=>+7@6/4=>-7@1/4<=>+7@3/5<=>-4@3/5!

eliminates 4@3/5 and you're (almost) done.

Very impressive! And it took you only 19 min since my posting!
oberon
 
Posts: 7
Joined: 18 March 2006

Postby Carcul » Sun Mar 19, 2006 4:40 pm

Hi Oberon.

I have one suggestion. Let's start by note the possible Unique Rectangle in cells r1c36/r3c36, and we could write:

[r4c6](-8-[r4c4])-8-[r1c6|r3c6]-6-[r2c4](-4-[r4c4])-4-[r3c5]{-7-[r3c9]-3-[r3c3]-(Unique Rectangle:r1c3/r1c6/r3c3/r3c6)-6,9-[r1c3]}=4=[r3c7]-4-[r9c7]-7-[r1c2|r1c7]-3-[r1c3]-1-[r2c3]-6-[r1c2]=6=[r4c2]-6-[r4c4],

which means that r4c6=8 would make r4c4 an empty cell. So, r4c6 cannot be "8" and the puzzle is solved.

Regards, Carcul
Carcul
 
Posts: 724
Joined: 04 November 2005

Re: hard one

Postby Havard » Sun Mar 19, 2006 5:44 pm

Being a bit of a fish-monger lately, I could not help pointing out this little finned swordfish:
Code: Select all
*-----------------------------------------------------------*
 | 4     36*   1369  | 6789  5     689*  | 367-  1267  1237* |
 | 8     7     16    | 46    3     2     | 5     146   9     |
 | 2     5     369   | 1     47    69    | 3467  8     37#   |
 |-------------------+-------------------+-------------------|
 | 7     36*   5     | 468   124   368*  | 9     126   123*  |
 | 136   4*    2     | 679   17    369*  | 8     5     137*  |
 | 136   9     8     | 67    127   5     | 367   1267  4     |
 |-------------------+-------------------+-------------------|
 | 9     2     7     | 5     8     4     | 1     3     6     |
 | 5     1     4     | 3     6     7     | 2     9     8     |
 | 36    8     36    | 2     9     1     | 47    47    5     |
 *-----------------------------------------------------------*
* marks the swordfish in columns 2,6 and 9
# marks the fin
- shows the elimination


The same elimination can also be done with an easier two-string kite! See if you can find it!:)

Havard
Havard
 
Posts: 377
Joined: 25 December 2005


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