## Need Help!!!!!!

Post the puzzle or solving technique that's causing you trouble and someone will help
Thanks a lot rep'nA
Kent

Posts: 98
Joined: 28 February 2006

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.
Neunmalneun

Posts: 52
Joined: 22 December 2005

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.
Ruud

Posts: 664
Joined: 28 October 2005

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.
Neunmalneun

Posts: 52
Joined: 22 December 2005

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
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Previous