Steve R and
tarek, that's a very elegant solution IMO, so elegant that I've decided to not publish mine, even though it has a kraken house step (just for Mike).
Mike Barker wrote:Its funny that the reason I chose this puzzle was because I didn't like the large ALS that my solver used to find a solution and wanted to what alternatives existed. So what happens? Both tarek and Steve R find solutions with large ALS. I guess I need to rethink my ALS complexity factor. Any hints on how to find and use such big guys?
Finding the smallest ALS is easy; just pick a bivalued cell. Finding the largest ALS (in a house) is also easy; first take all the cells (in the house) and then remove one cell. Finding the almost largest ALS is still pretty easy; first take all the cells and then remove the only two cells that contain one of the candidates, if any. This can, of course, be extended to three cells and two candidates, etc., but that soon becomes more difficult than starting a search with small ALSs and working up.
Except for set D, the bivalue, all of the ALSs in Steve R's solution -- to which tarek contributed -- are of this latter "all minus two" type: for set A, the 2 cells in c1 containing digit 3 are removed; for set B, digit 2 in b3; for set C, digit 4 in r1; and for set E, digit 9 in c9 (see Steve's second pencilmarks for this one).
- Code: Select all
A1238 5 1378 | 13789 3789 1379 |C478 6 BC12479
168 1678 4 | 156789 2 15679 | 3 189 1789
9 2678 1378 | 13678 4 1367 | 78 B128 5
----------------------+----------------------+---------------------
248 9 6 | 1237 37 1237 | 5 248 2478
A123 127 137 | 4 5 8 | 67 F29 F2679
248 2478 5 | 2679 79 2679 | 1 3 2478
----------------------+----------------------+---------------------
7 148 189 | 589 6 459 | 2 1458 3
4568 468 2 | 37 1 37 | 9 458 468
456 3 189 | 2589 89 2459 | 468 7 1468
Without looking too hard I see another in r5 by removing cells r5c89, the only cells containing digit 9 in r5. Thus, F = {r5c1237} = {12367} and voilà, we have the derived strong inference r5c7 =6|9= r5c8.
Finding
an ALS that is useful is the difficult part, and it's more difficult to find a useful large ALS than a useful small ALS -- for two reasons. Firstly, an ALS larger than a bivalue frequently has fewer "directionality" options. One bivalue candidate sees twenty other cells, i.e., its "domain" includes a row, a column and a box. The domain of one large ALS candidate is often only one house.
Secondly, a useful ALS is also chainable even though we might think of them in sets. The larger the ALS, the more difficult it is to find a chain, simply because there are fewer cells (or cell sets) to which the large ALS may be chained.
But I preach to the choir, I'm sure.
[
edit: The "all" above also applies to smaller naked n-tuples (subsets) within a house, e.g., the naked quintuple in b8.]