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keith · Posted: Mon Jul 27, 2009 1:19 am Post subject:

Here's another one:

storm_norm · Joined: 18 Oct 2007 Posts: 1741

my two cents...

instead of looking for the 5 candidate locked set
you can make a really simple deduction when you think about the rules of a UR.

the candidates 2 and 3 cannot be locked into r89c1 and r89c4 no matter what.

now look at columns 1 and 4... ask yourself how many other occurences of candidates 2 and 3 there are in those columns?? because there has to be at least one in order to keep the 23 UR from being locked into its cells. which is a big no no.

the answer is one. one lonely 2 in r7c1.

it is important to realize that this lonely 2 is the only 2 in columns 1 and 4 that keeps the UR23 from being locked into r89c14. therefore that 2 can be safely placed in r7c7

then turn your attention to rows 8 and 9

ask yourself how many other occurences (besides the UR23) of candidates 2 and 3 there are in those rows.

we see two occurences of candidate 2. both in column 7, at least one of these 2's must exist in order to stop candidates 2 and 3 from being locked into columns 1 and 4 rows 8 and 9. both can't be false.

UR23[(2)r8c7 = (2)r9c7]; r7c7 <> 2

this allows us to remove the 2 from r7c7

keith · Posted: Mon Jul 27, 2009 12:41 pm Post subject:

Norm,

I think that's the Type 4 I mentioned ...

Keith