ronk wrote:We see things a little differently here. It seems OK to me to consider the endo-fin-cell r1c8 and exo-fin-cell r2c8 as two elements of one fin.
So do I. But the other version with the endo-fin r1c8 and the exo-fin r6c2 I would consider as two fins.
ronk wrote:Is the same size fish with two exo-fins "simpler" than the fish with one exo-cell and one endo-cell in the same fin?
Both versions have one endo- and one exo-fin cell.
[edit] Here's the alternative pattern:
- Code: Select all
. . *9 | . . . | . @9 .
. *9 . | . . 9 | . *9 .
. . . | . *9 . | . . *9
---------+---------+---------
9 . 9 | . . 9 | 9 . 9
9 *9 . | . *9 9 | . . .
9 #9 9 | . . . | 9 -9 .
---------+---------+---------
. . . | 9 . . | . 9 9
9 . 9 | 9 . . | . . .
9 . 9 | 9 . . | 9 . .
Sashimi Mutant Jellyfish r1c25b3/r35c8b1 with fins r1c8 and r6c2
I was wrong. Since r1c8 must be treated as fin, there's only one vertex for r1 left. Hence this is also degenerating.
And now for something completely different ...
I'd like to share some thoughts about equivalent fishes. Two different fishes are considered equivalent if they can both be transformed into the same hidden pattern (which describes the empty cells and the exclusions that follow thereby).
Now the terms we use to describe a fish (e.g. r1c25b3/r35c8b1) are a bit like algebraic equations. As I mentioned some posts ago, if the same sector appears on both sides (in the base and in the cover set) this sector can be subtracted from both sets without changing the hidden pattern. In the same way it is a valid operation to add the same sector to both sets. Now this wouldn't be very useful if there wasn't some equal subsets of sectors: for example {r123} = {b123}, because both sets of sectors contain exactly the same candidates. Or even {r123456789} = {c123456789}.
You can use this to explain the equivalence of the classical fishes you're probably familiar with. Take for example a classic jellyfish r1357/c2468. You can add c123456789 to the base set and r123456789 to the cover set without changing the overall pattern. The result would be r1357c123456789/r123456789c2468. Now you can subtract all sectors that appear on both sides and get: c13579/r24689. This is the equivalent classic squirmbag for our example.
You can use this technique, too, if you want to transform a mutant fish into a franken one. If you take our Fig. 4C for example: Mutant Jellyfish r58c58/r2c2b68. To remove the columns from the base set you can add c4679 to both sides first: r58c456789/r2c24679b68. Now replace c456 by b258 and c789 by b369: r58b235689/r2c24679b68. And subtraction of b68 gives r58b2359/r2c24679. Now to remove the row from the cover set add r13 first: r1358b2359/r123c24679. Replace r123 by b123: r1358b2359/c24679b123. And subtraction of b23 from both sides gives: r1358b59/c24679b1. Et voilá, an equivalent franken whale.
It may happen that you're adding the same sector multiple times to the base or the cover set. While this doesn't hurt the conclusions the resulting fish pattern is very abstract and probably of no practical use.
Forgive me if all this is already an old hat. I find these transformations to be very useful to search for equivalent fishes like you do for your catalogue.