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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Mon Jan 05, 2009 9:32 pm Post subject: Set H Puzzle 38 |
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| 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 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Tue Jan 06, 2009 4:29 am Post subject: |
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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: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Jan 06, 2009 6:02 am Post subject: |
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| 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
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| Posted: Tue Jan 06, 2009 6:15 am Post subject: |
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| 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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