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Set H Puzzle 38

 
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Mon Jan 05, 2009 9:32 pm    Post subject: Set H Puzzle 38 Reply with quote

Code:
 +-----------------------+
 | 9 . . | 8 5 7 | . . . |
 | . . . | 9 . . | . . . |
 | . . 8 | . 2 . | . . 1 |
 |-------+-------+-------|
 | 6 4 . | . . 8 | 9 . . |
 | 5 . 9 | . 6 2 | . 1 8 |
 | 1 . . | 5 9 . | 7 . . |
 |-------+-------+-------|
 | . . . | 1 . 5 | . . 6 |
 | . . . | . 4 . | . 8 . |
 | . . 5 | . 8 . | 1 . 9 |
 +-----------------------+

Play this puzzle online at the Daily Sudoku site
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Tue Jan 06, 2009 4:29 am    Post subject: Reply with quote

going go skip the x-wing and something interesting comes up.
Code:
.---------------------.---------------------.---------------------.
| 9     1-23    134   | 8      5      7     | 6     *23     234   |
|*24     256    46    | 9      3      1     | 8      257    2457  |
| 37     357    8     | 4      2      6     | 35     9      1     |
:---------------------+---------------------+---------------------:
| 6      4      237   | 37     1      8     | 9      235    235   |
| 5      37     9     | 37     6      2     | 4      1      8     |
| 1      8      23    | 5      9      4     | 7      6      23    |
:---------------------+---------------------+---------------------:
| 8      9      34    | 1      7      5     | 2     *34     6     |
| 237    12367  1367  | 26     4      9     | 35     8      357   |
|*247    267    5     | 26     8      3     | 1     *47     9     |
'---------------------'---------------------'---------------------'

the strong links on 4's in row 9 say that the 2 in r2c1 is true or the 3 in r7c8 is true, and extending the 3 up to the {2,3} cell in r1c8, makes this chain
(2=4)r2c1 - (4)r9c1 = (4)r9c8 - (4=3)r7c8 - (3=2)r1c8; r1c2 <> 2

now something interesting happens, and I think this goes along with what Danny has said in one other post about using the UR to determine inferences outside of UR's. but in this example it helps us make an elimination (I think) directly...
Code:
.---------------------.---------------------.---------------------.
| 9      13     134   | 8      5      7     | 6      23     234   |
| 24 [2]5[6]    46    | 9      3      1     | 8      57     457   |
| 37     357    8     | 4      2      6     | 35     9      1     |
:---------------------+---------------------+---------------------:
| 6      4      237   | 37     1      8     | 9      235    235   |
| 5      37     9     | 37     6      2     | 4      1      8     |
| 1      8      23    | 5      9      4     | 7      6      23    |
:---------------------+---------------------+---------------------:
| 8      9      34    | 1      7      5     | 2      34     6     |
| 237   U12367  1367  |U26     4      9     | 35     8      357   |
| 247   U267    5     |U26     8      3     | 1      47     9     |
'---------------------'---------------------'---------------------'

in order to avoid the UR marked on {2,6} r89c24, we can't have a 2 or a 6 left in column 2 in box 7, right?

that means either of the 2 or the 6 in r2c2 has to be true...
in other words if the 5 is true in r2c2, then the DP would be forced to exist. NO NO.
I believe this eliminates the 5 in r2c2 and solves the puzzle

any thoughts?


Last edited by storm_norm on Tue Jan 06, 2009 6:05 am; edited 2 times in total
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keith



Joined: 19 Sep 2005
Posts: 3178
Location: near Detroit, Michigan, USA

PostPosted: Tue Jan 06, 2009 6:02 am    Post subject: Reply with quote

storm_norm wrote:
going go skip the x-wing and something interesting comes up.
Code:
.---------------------.---------------------.---------------------.
| 9     1-23    134   | 8      5      7     | 6     *23     234   |
|-24     256    46    | 9      3      1     | 8      257    2457  |
| 37     357    8     | 4      2      6     | 35     9      1     |
:---------------------+---------------------+---------------------:
| 6      4      237   | 37     1      8     | 9      235    235   |
| 5      37     9     | 37     6      2     | 4      1      8     |
| 1      8      23    | 5      9      4     | 7      6      23    |
:---------------------+---------------------+---------------------:
| 8      9      34    | 1      7      5     | 2     *34     6     |
| 237    12367  1367  | 26     4      9     | 35     8      357   |
|*247    267    5     | 26     8      3     | 1     *47     9     |
'---------------------'---------------------'---------------------'

the strong links on 4's in row 9 say that the 2 in r2c1 is true or the 3 in r7c8 is true, and extending the 3 up to the {2,3} cell in r1c8, makes the 2 there true. eliminates the 2 in r1c2...

now something interesting happens, and I think this goes along with what Danny has said in one other post about using the UR to determine inferences outside of UR's. but in this example it helps us make an elimination (I think) directly...
Code:
.---------------------.---------------------.---------------------.
| 9      13     134   | 8      5      7     | 6      23     234   |
| 24 [2]5[6]    46    | 9      3      1     | 8      57     457   |
| 37     357    8     | 4      2      6     | 35     9      1     |
:---------------------+---------------------+---------------------:
| 6      4      237   | 37     1      8     | 9      235    235   |
| 5      37     9     | 37     6      2     | 4      1      8     |
| 1      8      23    | 5      9      4     | 7      6      23    |
:---------------------+---------------------+---------------------:
| 8      9      34    | 1      7      5     | 2      34     6     |
| 237   U12367  1367  |U26     4      9     | 35     8      357   |
| 247   U267    5     |U26     8      3     | 1      47     9     |
'---------------------'---------------------'---------------------'

in order to avoid the UR marked on {2,6} r89c24, we can't have a 2 or a 6 left in column 2 in box 7, right?

that means either of the 2 or the 6 in r2c2 has to be true...
in other words if the 5 is true in r2c2, then the DP would be forced to exist. NO NO.
I believe this eliminates the 5 in r2c2 and solves the puzzle

any thoughts?

Norm,

In your first diagram, what does the -24 in R2C1 mean? I agree with the elimination of <2> in R1C2

In your second diagram, this is a Type 3 UR: The UR cells in C2 are 26+137 and 26 +7. They make a pseudocell <137> which makes a triple with <13> and <37> in the same column, solving R3C2 <357> as <5>, eliminating <5> in R2C2.

Your logic is sort of like a Type 4: One of the UR cells in C2 cannot be <26>, so R2C2 must be <26>, so it cannot be <5>.

Different views of the same elephant.

Keith
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Tue Jan 06, 2009 6:15 am    Post subject: Reply with quote

Quote:
In your first diagram, what does the -24 in R2C1 mean

corrected.

Quote:
In your second diagram, this is a Type 3 UR

oh yes, very true.
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