ronk wrote:Allan Barker wrote:Final Set-Linkset Group, SLG

24 Sets = {3R24568 5R2 7R24568 1C2 4C6 5C124568 6N56 357B4}

25 Links = {1r6 4r7 5r169 3c1567 7c3567 567n2 28n5 28n6 46n8 357b5}

Elimination--> (4r7*7n2) => r7c2<>4 Permutations 3614

Note: this is a rank 1 elimination by the overlap of 4r7 and 7n2.

Leaving 5r245c1 without a cover set seems bizaare, but it's certainly less bizaare than the following, with no cover sets at all.

Also, my solver does require linksets, the follow will work as well. I checked, the number of permutations is really 9999!

As a Set Group

49 Sets = {1R6 3R24568 4R7 5R1269 7R24568 1C2 3C1567 4C6 5C124568 7C3567 2N56 4N8 5N2 6N2568 7N2 8N56 357B4 357B5}

0 Links = {}

Elimination--> [4R7*7N2] => r7c2<>4

RE: "Leaving 5r245c1 without a cover set". This is un-common but can happen for two reasons I have seen. 1) The embedding of broken wing type logic, i.e., loops of odd numbers of strong sets. In this case row set 5r1 is a "strong" cover set, as well as several others. 2) Two "out of phase" AICs that travel through (partially or all) the same candidates.

RE: "the following, with no cover sets at all". (Edit: my solver does

not require cover sets, I have changed my original post). The solver does not use a base/cover set model, rather some of the logic it finds can also be expressed as a base/cover set model. This accounts for about 95% of all the logic it does find.

Champagne wrote:I am slighly diappointed that you could not do with a smaller number of additions.

I hoped that one would be small enough to be attractive.

There may be, or probably are, several other logical arrangements. This is only the one I got by filling in extra linksets.