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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sat Apr 10, 2010 12:15 am Post subject: |
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tlanglet wrote: | daj95376 wrote: | Ted: I just awoke and I'm still fuzzy-headed -- no wisecracks about that always being the case -- but it seems to me that your conclusion doesn't work. The easiest explanation is that r2c1=4 is true in the solution. If you shorten your chain, then there is a valid conclusion.
HP46[(46)r2c13 = (1)r2c3]r2c13 - r79c3 = r7c2 - (1=4)r7c8 - r4c8 = r4c2; => r1c2<>4
This forces one of r2c13 to be <4> and advances the solution a bit.
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Danny, I agree that the solution is that r2c1=4, but that does not help me prove or dis-prove my original step. Again, assuming r2c3=1 then r2c13=4 so the only I see way to meet this condition is for r2c1=4. In my simple view of an AIC, the initial premise is retain in order to deduce any eliminations.
Do you have any suggestions where I could go to get further help on this questions?
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The part in red is not your original premise.
Regards, Danny
[Edit: truncated reply.]
Last edited by daj95376 on Sat Apr 10, 2010 7:46 pm; edited 6 times in total |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sat Apr 10, 2010 12:59 am Post subject: |
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keith wrote: | I see that I am quoted, but I do not understand the discussion.
Can you please post the diagram, using:
X for cells that contain a candidate
/ for cells that do not contain X
* for cells where X can be eliminated
. for cells where the candidates do not matter
Since a skyscraper is a single-digit pattern, do we need more?
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It appears that my trying to help has backfired. Please pretend that I never mentioned Skyscraper. |
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arkietech
Joined: 31 Jul 2008 Posts: 1834 Location: Northwest Arkansas USA
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Posted: Sat Apr 10, 2010 1:08 am Post subject: |
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There are several combinations. Here is two of many: Code: | +-----------------------+
| . . . | . . . | . . . |
| X . . | X . . | . . . |
| . . . | . . . | . . . |
|-------+-------+-------|
| . . . | . . . | . . . |
| . . . | X . . | . . . |
| . . y | . . y | . . . |
|-------+-------+-------|
| . . * | . . . | . . . |
| X . * | . . . | . . . |
| . . * | . . . | . . . |
+-----------------------+
+-----------------------+
| . . . | . . . | . . . |
| X . . | X . . | . . . |
| . . . | . . . | . . . |
|-------+-------+-------|
| . Y . | . Y . | . . . |
| . . . | X . . | . . . |
| . . . | . . . | . . . |
|-------+-------+-------|
| . * . | . . . | . . . |
| X * . | . . . | . . . |
| . * . | . . . | . . . |
+-----------------------+ |
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storm_norm
Joined: 18 Oct 2007 Posts: 1741
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Posted: Sat Apr 10, 2010 5:54 am Post subject: |
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tlanglet wrote: | A one step solution IF IT IS VALID.
Using the code posted by Marty, notice the hidden pair 46 in r2c13. Either r2c13=46 or r2c3=1.
HP46[(46)r2c13 = (1)r2c3]r2c13 - r79c3 = r7c2 - (1=4)r7c8 - r4c8 = r4c2 - r1c2 = r2c13.
Thus if r2c3=1, then either r2c1=4 or r2c3=4 but r2c3<>4 since the condition was based on r2c3=1 so r2c1=4.
Ted |
your chain went too far.
all you had to do was end your chain at (4)r4c2 and it would eliminate the 4 in r1c2.
why?
because your chain says neither the naked pair {4,6} at r2c13 nor the 4 at r4c2 can both be false.
there isn't a need to go any further.
notice that its a naked pair that you are using in your first link, not a hidden pair. so this structure is a Almost locked set, not a almost hidden set. |
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Sat Apr 10, 2010 3:20 pm Post subject: |
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Danny & Norm,
Thanks for the feedback.
Danny: I understood my complete original premise was r2c13=46 or r2c3=1; both can not be false.
Norm: I understood that by stopping the chain at r4c2=4 I could delete 4 in r1c2. However by continuing the chain to r2c13=4, then I realize an additional deletion in r2c7<>4. In addition, IF THE COMPLETE CHAIN IS VALID, then r2c1=4 is forced which completed the puzzle.
Also, thanks for the comments about Hidden vs Naked; terminology is always an issue for me.
Ted |
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