1to9only wrote:These require 5 swordfishes, all at the beginning (so ED=3.8/3.8/3.8), a number of them are then stte.
- Code: Select all
..12....3..24....55....31..1....67.............83....4..58....16....79..8....95..
..12....3..24....54....62..7....86.............36....1..45....68....79..9....45..
..12....3..23....43....45..6....78.............31....5..54....78....96..9....34..
..12....3..23....43....45..6....78.............31....5..45....78....69..9....34..
..12....3..23....43....45..6....37.............38....1..84....57....69..9....84..
..12....3..23....43....45..6....37.............38....1..45....87....96..9....84..
..12....3..23....43....45..6....37.............38....1..45....87....69..9....84..
..12....3..21....45....61..4....78.............36....5..54....67....94..9....85..
..12....3..21....45....61..3....78.............46....5..53....67....93..9....85..
..12....3..21....43....52..5....67.............45....2..53....86....79..9....23..
..12....3..21....43....51..5....67.............45....1..53....86....79..9....13..
I didn't try all of them, but the first can be solved with only 4 swordfishes:
- Code: Select all
479 46789 1 2 56789 58 468 46789 3
379 36789 2 4 16789 18 68 6789 5
5 46789 4679 679 6789 3 1 246789 26789
1 23459 349 59 24589 6 7 23589 289
23479 2345679 34679 1579 1245789 12458 2368 1235689 2689
279 25679 8 3 12579 125 26 12569 4
23479 23479 5 8 2346 24 2346 23467 1
6 1234 34 15 12345 7 9 2348 28
8 12347 347 16 12346 9 5 23467 267
232 candidates, 1650 csp-links and 1650 links. Density = 6.16%
swordfish-in-columns: n7{c3 c4 c9}{r9 r5 r3} ==> r9c8 ≠ 7, r9c2 ≠ 7, r5c5 ≠ 7, r5c2 ≠ 7, r5c1 ≠ 7, r3c8 ≠ 7, r3c5 ≠ 7, r3c2 ≠ 7
swordfish-in-columns: n2{c1 c6 c7}{r7 r6 r5} ==> r7c8 ≠ 2, r7c5 ≠ 2, r7c2 ≠ 2, r6c8 ≠ 2, r6c5 ≠ 2, r6c2 ≠ 2, r5c9 ≠ 2, r5c8 ≠ 2, r5c5 ≠ 2, r5c2 ≠ 2
swordfish-in-columns: n9{c3 c4 c9}{r5 r4 r3} ==> r5c8 ≠ 9, r5c5 ≠ 9, r5c2 ≠ 9, r5c1 ≠ 9, r4c8 ≠ 9, r4c5 ≠ 9, r4c2 ≠ 9, r3c8 ≠ 9, r3c5 ≠ 9, r3c2 ≠ 9
swordfish-in-columns: n6{c3 c4 c9}{r5 r3 r9} ==> r9c8 ≠ 6, r9c5 ≠ 6, r5c8 ≠ 6, r5c7 ≠ 6, r5c2 ≠ 6, r3c8 ≠ 6, r3c5 ≠ 6, r3c2 ≠ 6
stte
The -1 difference may come from the order in which they are applied.
It would be interesting to see your solution with 5 swordfishes.
[Edit:]
I tried to apply diagonal symmetry before solving:
(diagonal-symmetry-9x9 "..12....3..24....55....31..1....67.............83....4..58....16....79..8....95..")
"..51...68.........12...85..24...38.............36...79..17...95.........35...41.."
I now find a solution with only 3 swordfishes, starting from the symmetric resolution state:
- Code: Select all
479 379 5 1 23479 279 23479 6 8
46789 36789 46789 23459 2345679 25679 23479 1234 12347
1 2 4679 349 34679 8 5 34 347
2 4 679 59 1579 3 8 15 16
56789 16789 6789 24589 1245789 12579 2346 12345 12346
58 18 3 6 12458 125 24 7 9
468 68 1 7 2368 26 2346 9 5
46789 6789 246789 23589 1235689 12569 23467 2348 23467
3 5 26789 289 2689 4 1 28 267
swordfish-in-rows: n7{r3 r4 r9}{c9 c5 c3} ==> r8c9 ≠ 7, r8c3 ≠ 7, r5c5 ≠ 7, r5c3 ≠ 7, r2c9 ≠ 7, r2c5 ≠ 7, r2c3 ≠ 7, r1c5 ≠ 7
swordfish-in-rows: n9{r3 r4 r9}{c5 c4 c3} ==> r8c5 ≠ 9, r8c4 ≠ 9, r8c3 ≠ 9, r5c5 ≠ 9, r5c4 ≠ 9, r5c3 ≠ 9, r2c5 ≠ 9, r2c4 ≠ 9, r2c3 ≠ 9, r1c5 ≠ 9
swordfish-in-rows: n6{r3 r4 r9}{c5 c3 c9} ==> r8c9 ≠ 6, r8c5 ≠ 6, r8c3 ≠ 6, r7c5 ≠ 6, r5c9 ≠ 6, r5c3 ≠ 6, r2c5 ≠ 6, r2c3 ≠ 6
stte
This should not be a total surprise: the maximum complexity of required Subsets is an invariant under isomorphism, the number of resolution rules applied is not.
[/Edit]
[Edit2:]
I wondered if I could do the same with the other puzzles.
#2 also gives a solution with 3 swordfishes:
(solve (diagonal-symmetry-9x9 "..12....3..24....54....62..7....86.............36....1..45....68....79..9....45.."))
swordfish-in-rows: n7{r3 r4 r9}{c9 c3 c5} ==> r8c9 ≠ 7, r8c5 ≠ 7, r8c3 ≠ 7, r7c5 ≠ 7, r5c5 ≠ 7, r5c3 ≠ 7, r2c9 ≠ 7, r2c3 ≠ 7
swordfish-in-rows: n9{r3 r4 r9}{c5 c4 c3} ==> r8c5 ≠ 9, r8c4 ≠ 9, r8c3 ≠ 9, r6c5 ≠ 9, r5c5 ≠ 9, r5c4 ≠ 9, r5c3 ≠ 9, r2c5 ≠ 9, r2c4 ≠ 9, r2c3 ≠ 9
swordfish-in-rows: n8{r3 r4 r9}{c5 c3 c9} ==> r8c9 ≠ 8, r8c5 ≠ 8, r8c3 ≠ 8, r7c5 ≠ 8, r5c9 ≠ 8, r5c3 ≠ 8, r2c5 ≠ 8, r2c3 ≠ 8
stte
After diagonal symmetry:
#3 has also 3 swordfishes in rows but it requires two whips[1] for the end
#4 has also 3 swordfishes in rows and then stte
#5 has also 3 swordfishes in rows and then stte
...
[/Edit2]
I don't think the word "require" is correct. They may have 5 swordfishes, depending on which order we apply them; but solutions with fewer ones can be found.