denis_berthier wrote:arcilla wrote:A while ago, when learning about FISHes, it occurred to me that one could look at them in a different way.
[…]
The idea is, that for most 'normal' fishes (for instance where boxes do not come into play, as in 'franken' types, if I understand them correctly), if you administer them in a certain way, they are equivalent to naked or hidden pairs, triples, quads, etc. The administration is simple. Make a list of nine cells, like a Sudoku row. In cell n (1..9) write the (row-)numbers (1..9) of the cells in column n that contain the candidate.
[…]
An X-wing shows up as a naked pair […] and a sword fish as a hidden triple.
Arcilla, I was not aware of your remark when I wrote my book, "The Hidden Logic of Sudoku". Otherwise, I would have cited you. You had a great idea. Unfortunately, as I can see from the posts in this thread, nobody seems to have really understood it or pushed it further.
Based on more general ideas of symmetry, I came upon quite the same idea, formalised it. More generally, I introduced rn- and cn- spaces and an associated extended sudoku board to deal with them. I also extended the idea to chains, where it allows introducing completely new types of chains (hidden chains).
Actually I understood quite well and even was writing down a version of your rn and cn spaces. Here are a couple of my missing posts from this thread:
rep'nA (now re'born) wrote:arcilla,
This is a very creative approach. Kudos! It gives me hope that I might find a jellyfish or finned swordfish on my own someday.
I have a rather remedial question for you (or anyone else kind enough to point out the obvious to me).arcilla wrote:As far as I can see finned fish are equivalent to almost locked sets (and can therefore be attacked analogously).
Here is an example from the newly formed Ultimate Fish Guide:
- Code: Select all
5.....1..2...16.8...35....7...4.8.6...6....9..85..32...5.64.9....91.....3......14
*--------------------------------------------------------------------*
| 5 479-6 478 | 23789 2789 2479 | 1 234 2369 |
| 2 479 47 | 379 1 6 | 345 8 359 |
|#14689 *1469 3 | 5 289 249 |*46 24 7 |
|----------------------+----------------------+----------------------|
| 179 12379 127 | 4 2579 8 | 357 6 35 |
| 47 2347 6 | 27 257 1 | 34578 9 358 |
| 479 8 5 | 79 6 3 | 2 47 1 |
|----------------------+----------------------+----------------------|
| 178 5 1278 | 6 4 27 | 9 237 238 |
| 4678 2467 9 | 1 3 27 | 678 5 268 |
| 3 *267 278 | 2789 2789 5 |*678 1 4 |
*--------------------------------------------------------------------*
The row placements of 6 (which should be viewed as a column vector) are
R_6: [(29)(6)(127)], [(8)(3)(5)], [(4)(1279)(27)]
I added the []'s to help denote the blocks which seems important to me.
R_6[3,9] (the third and ninth entries) form an almost locked set, with the 1 in R_6[3] being the obstruction to a naked pair (and hence the corresponding x-wing). The conclusion, it seems to me, is that in general you could remove a 2 from any of R_6[1,2] since any elimination must occur in the same block as the obstruction and must also be in the same triple as the obstruction (the triples are {1,2,3}, {4,5,6}, {7,8,9}, though this is just a numerical consequence of needing to be in the same block as the obstruction).
First, is this the correct way to view the elimination of the 2 in R_6[1] (or equivalently of the 6 from r1c2)?
Second, if instead you take the column placements of 6:
C_6: [(38)(1389)(5)], [(7)(6)(2)], [(389)(4)(18)],
then we see an almost locked set in C_6[1,7] with 9 being the obstruction to the naked pair. Again, I would think that this implies that one could remove an 8 from any other entry in the third block of C_6 (since that is where the obstruction is), namely remove the 8 from C_6[9].
- Code: Select all
*--------------------------------------------------------------------*
| 5 4679 478 | 23789 2789 2479 | 1 234 2369 |
| 2 479 47 | 379 1 6 | 345 8 359 |
|*14689 1469 3 | 5 289 249 |*46 24 7 |
|----------------------+----------------------+----------------------|
| 179 12379 127 | 4 2579 8 | 357 6 35 |
| 47 2347 6 | 27 257 1 | 34578 9 358 |
| 479 8 5 | 79 6 3 | 2 47 1 |
|----------------------+----------------------+----------------------|
| 178 5 1278 | 6 4 27 | 9 237 238 |
|*4678 2467 9 | 1 3 27 | *678 5 -268 |
| 3 267 278 | 2789 2789 5 | #678 1 4 |
*--------------------------------------------------------------------*
Of course, that is exactly what the theory of finned x-wings tells you can be done and while it is not mentioned in that thread, it very well could have been (of course, it isn't needed after the first elimination). So this is very nice and was terribly easy for me to spot using your numerology. My question is how should I see the first elimination using C_6 (or the second elimination using R_6)? It seems I could use almost hidden sets and get the deduction, but is there a faster way?
To be more precise about how I would see it, consider again
R_6: [(29)(6)(127)], [(8)(3)(5)], [(4)(1279)(27)].
R_6[3,8,9] is an almost hidden set (with hidden candidates 1 and 7) with the obstruction to a hidden pair being the 7 in R_6[9]. Then you can remove any 9 from an entry in the third block, i.e., a 9 from R_6[8]. Is that how you would see it?
and
rep'nA (now re'born) wrote:Here is another example taken from the Ultimate Fish Guide.# Sashimi Swordfish digit 3
- Code: Select all
003400000000025009040700060801000090070050010060000703080006020600170000000003500
.------------------.------------------.------------------.
| 259 259 3 | 4 6 189 | 128 578 12578|
| 7 1 6 | 38 2 5 | 348 #348 9 |
| 259 4 8 | 7 *139 19 |-123 6 125 |
:------------------+------------------+------------------:
| 8 *235 1 | 236 *34 7 | 246 9 245 |
| 2349 7 249 | 23689 5 2489 | 2468 1 248 |
| 2459 6 2459 | 289 1489 12489| 7 458 3 |
:------------------+------------------+------------------:
| 349 8 479 | 5 49 6 | 1349 2 147 |
| 6 *2359 2459 | 1 7 2489 | 3489 *348 48 |
| 1 29 2479 | 289 489 3 | 5 478 6 |
'------------------'------------------'------------------'
It took me quite a while to make this deduction initially, not because of the logic, that is pretty straight forward, but because even if somebody tells you that the puzzle solves with a Sashimi Swordfish and even if somebody tells you "hint, hint, look at 3", it may still be difficult to reel in the fish. On the other hand, converting the '3' placements into the Arcilla matrix, we get:
C_3: [(57),(48),(1)], [(245),(34),(9)], [(2378),(28),(6)]
It is reasonably easy to spot the ALS in C_3[2,5,8] where the 2 in C_3[8] is an obstruction to the ALS being a (degenerate) naked triple. Therefore, we can remove any 1 or 3 from different entries in the same block, i.e., we can remove the 3 from C_3[7]. This translates to deducing r3c7 <> 3. On the other hand, the not-so-bright examiner (e.g., me) will ask why we didn't take the 3 in C_3[5] to be the obstruction. This is perfectly legitimate and implies that we can remove any 1 or 2 from different entries in the same block, i.e., we can remove the 2 from C_3[4]. This translates to deducing r2c4 <> 3 and corresponds to the sashimi swordfish:
# Sashimi Swordfish digit 3
- Code: Select all
.------------------.------------------.------------------.
| 259 259 3 | 4 6 189 | 128 578 12578|
| 7 1 6 |-38 2 5 | 348 *348 9 |
| 259 4 8 | 7 #139 19 | 123 6 125 |
:------------------+------------------+------------------:
| 8 *235 1 | 236 *34 7 | 246 9 245 |
| 2349 7 249 | 23689 5 2489 | 2468 1 248 |
| 2459 6 2459 | 289 1489 12489| 7 458 3 |
:------------------+------------------+------------------:
| 349 8 479 | 5 49 6 | 1349 2 147 |
| 6 *2359 2459 | 1 7 2489 | 3489 *348 48 |
| 1 29 2479 | 289 489 3 | 5 478 6 |
'------------------'------------------'------------------'
Naturally, either deduction implies the other so you don't need both, but it is very pleasing to see both fish constructed so easily, especially for someone like me who historically bought his fish at the fish market, pre-gutted, sliced and diced.
