Need Help!!!!!!

Post the puzzle or solving technique that's causing you trouble and someone will help

Postby Kent » Wed Mar 15, 2006 6:41 am

Thanks a lot rep'nA
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Postby Neunmalneun » Wed Mar 15, 2006 2:23 pm

Sorry tso and Ronk, I should really have reread my post (more than three times). It had some mistakes, but I still think the general idea is correct (though I admit the technique is not very elegant and looks a bit like t&e). If a cell has more than two candidates (abcd...) and the assumtion of ab leads to a then b is not a valid candidate. Or in other words: If ab leads to a then abc can't lead to b.

First deduction (in the original example): If R4C5 (458) contains 48 it kills the 8 in that very cell. So it can't be 8 even if the first assumtion was not correct (if the cell is 5 it can't be 8 anyway). That leaves a single 8 in R5C5 and kills the other 8s in R5.


My second deduction meant: Assuming R1C3 (138) contains 18 this would lead to R1C3<>1, so R1C3 can't be 1 in any given case. The rest is singles and blocked candidates.

Sorry for the confusion I caused.
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Postby Ruud » Wed Mar 15, 2006 5:07 pm

Neunmalneun wrote:though I admit the technique is not very elegant and looks a bit like t&e

Are you sure?

My perception of T&E (or bifurcation) is to try a value and see if it causes an error. Then do the opposite.

It this case, you are looking for a cycle of implications that returns to it's origin. That is a very specific target. You can also make the search a little more specific, and look for a house containing 2 cells with candidates ab and abc. You can then follow the implications of forming a naked pair. (which is the equivalent of dropping c as a candidate)

This is a nice example. It is Mike Mepham's unsolvable #1, at a stage where you need a breakthrough:

Look at r3c2 & r9c2. When 2 is dropped as a candidate from r3c2, this forces r1c2=6 => r2c1=3 => r3c1=7 => r3c2<>7.

Code: Select all
.------------------.------------------.----------- -------.
| 1     46    2456 | 9     24568 7    | 258   258   3    |
| 36    8     2456 | 1346  23456 1235 | 9     7     124  |
| 37    247   9    | 134   23458 12358| 6     1258  1248 |
:------------------+------------------+----------- -------:
| 5     3     7    | 2     168   9    | 4     168   168  |
| 4     1     26   | 367   3678  38   | 28    9     5    |
| 69    269   8    | 5     16    4    | 3     126   7    |
:------------------+------------------+----------- -------:
| 689   469   3    | 14    2459  125  | 7     2568  268  |
| 6789  5     16   | 37    2379  23   | 18    4     268  |
| 2     47    14   | 8     457   6    | 15    3     9    |
'------------------'------------------'----------- -------'

This allows you to make 3 placements and continue with other techniques.

I really like this technique.:)

Ruud.
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Postby Neunmalneun » Wed Mar 15, 2006 5:18 pm

Thanks a lot, Ruud. That's exactly what I had in mind. A perfect example for this unnamed technique. I'm glad you like it.
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Postby ronk » Wed Mar 15, 2006 8:18 pm

Ruud wrote:This is a nice example. It is Mike Mepham's unsolvable #1, at a stage where you need a breakthrough:

Look at r3c2 & r9c2. When 2 is dropped as a candidate from r3c2, this forces r1c2=6 => r2c1=3 => r3c1=7 => r3c2<>7.

One reaches the same "elimination by contradiction" by merely asserting r3c2=7 instead of r3c2=47 ... so I'm unsure where the benefit is.

Does the pair POV make a potential elimination easier to spot? Does creating a naked pair via an assertion have a higher likelihood of finding a contradiction than some other approach?

Just wondering, Ron
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