The New Sudoku Players' Forum

by **mith** » Fri Jun 11, 2021 8:37 pm

Code: Select all: +-------+-------+-------+ | . 1 2 | . 7 . | 3 5 . | | 3 . . | 2 . 1 | 4 . 7 | | 4 . . | 5 . . | . 1 6 | +-------+-------+-------+ | 2 . . | . . . | . 7 . | | 1 6 . | . 8 . | . . 2 | | . 4 8 | . . . | 6 3 . | +-------+-------+-------+ | . . . | 8 . . | . . 4 | | . . . | 1 5 . | . . 3 | | . . . | . 4 3 | 1 2 . | +-------+-------+-------+ .12.7.35.3..2.14.74..5...162......7.16..8...2.48...63....8....4...15...3....4312.

Posted with permission. This has something I hadn't seen before, and we were trying to figure out what to call it.

by **marek stefanik** » Fri Jun 11, 2021 11:55 pm

Code: Select all: +---------------------+---------------------+---------------------+ | 689 1 2 | 46–9 7 4689 | 3 5 89 | | 3 589 569 | 2 69 1 | 4 89 7 | | 4 789 79 | 5 3 89 | 2 1 6 | +---------------------+---------------------+---------------------+ | 2 59 359 | 3469 169 4569 | 589 7 1589 | | 1 6 3579 | 379 8 579 | 59 4 2 | | 579 4 8 | 79 129 2579 | 6 3 15–9 | +---------------------+---------------------+---------------------+ | 5679 3 1 | 8 269 2679 | 579 69 4 | | 6789 2 4 | 1 5 679 | 789 689 3 | | 789–56 5789 5679 | 679 4 3 | 1 2 589 | +---------------------+---------------------+---------------------+

Column 1 and row 9 seem interesting.

7r6c4 => 7r9c1
8r1c9 => 8r9c1
9r1c9 and 9r6c4 => 9r9c1
–56r9c1, –9r1c4, –9r6c9

Then solvable with fish patterns.

Hidden Text: Show

There could be a pattern that covers both the main pattern and the fishes (at least the kites). I'm curious whether the intended solution does it.

Marek

Website · by **denis_berthier** » Sat Jun 12, 2021 7:15 am

Hi Mith,

As any puzzle has lots of possible resolution paths, it's difficult to guess what you consider as new.

Code: Select all: Resolution state after Singles and whips[1]: +-------------------+-------------------+-------------------+ ! 689 1 2 ! 469 7 4689 ! 3 5 89 ! ! 3 589 569 ! 2 69 1 ! 4 89 7 ! ! 4 789 79 ! 5 3 89 ! 2 1 6 ! +-------------------+-------------------+-------------------+ ! 2 59 359 ! 3469 169 4569 ! 589 7 1589 ! ! 1 6 3579 ! 379 8 579 ! 59 4 2 ! ! 579 4 8 ! 79 129 2579 ! 6 3 159 ! +-------------------+-------------------+-------------------+ ! 5679 3 1 ! 8 269 2679 ! 579 69 4 ! ! 6789 2 4 ! 1 5 679 ! 789 689 3 ! ! 56789 5789 5679 ! 679 4 3 ! 1 2 589 ! +-------------------+-------------------+-------------------+

From here on, I first tried to activate only Subsets and Finned Fish - no one would consider them as new:

Code: Select all: finned-swordfish-in-columns: n7{c2 c3 c4}{r9 r3 r5} ==> r5c6 ≠ 7 naked-pairs-in-a-row: r5{c6 c7}{n5 n9} ==> r5c4 ≠ 9, r5c3 ≠ 9, r5c3 ≠ 5 finned-swordfish-in-columns: n5{c2 c3 c9}{r9 r2 r4} ==> r4c7 ≠ 5 Resolution state: +-------------------+-------------------+-------------------+ ! 689 1 2 ! 469 7 4689 ! 3 5 89 ! ! 3 589 569 ! 2 69 1 ! 4 89 7 ! ! 4 789 79 ! 5 3 89 ! 2 1 6 ! +-------------------+-------------------+-------------------+ ! 2 59 359 ! 3469 169 4569 ! 89 7 1589 ! ! 1 6 37 ! 37 8 59 ! 59 4 2 ! ! 579 4 8 ! 79 129 2579 ! 6 3 159 ! +-------------------+-------------------+-------------------+ ! 5679 3 1 ! 8 269 2679 ! 579 69 4 ! ! 6789 2 4 ! 1 5 679 ! 789 689 3 ! ! 56789 5789 5679 ! 679 4 3 ! 1 2 589 ! +-------------------+-------------------+-------------------+

Then, starting from this point, I added bivalue-chains (basic AICs) - not really new either. There is only one:

Code: Select all: biv-chain[3]: b1n6{r1c1 r2c3} - r2c5{n6 n9} - b3n9{r2c8 r1c9} ==> r1c1 ≠ 9 Resolution state: +-------------------+-------------------+-------------------+ ! 68 1 2 ! 469 7 4689 ! 3 5 89 ! ! 3 589 569 ! 2 69 1 ! 4 89 7 ! ! 4 789 79 ! 5 3 89 ! 2 1 6 ! +-------------------+-------------------+-------------------+ ! 2 59 359 ! 3469 169 4569 ! 89 7 1589 ! ! 1 6 37 ! 37 8 59 ! 59 4 2 ! ! 579 4 8 ! 79 129 2579 ! 6 3 159 ! +-------------------+-------------------+-------------------+ ! 5679 3 1 ! 8 269 2679 ! 579 69 4 ! ! 6789 2 4 ! 1 5 679 ! 789 689 3 ! ! 56789 5789 5679 ! 679 4 3 ! 1 2 589 ! +-------------------+-------------------+-------------------+

Next step was to add z-chains - not really new, as I defined them in [HLS 2007] - and they were enough to solve the puzzle:

Code: Select all: z-chain[3]: b4n5{r4c3 r6c1} - r7n5{c1 c7} - r5n5{c7 .} ==> r4c6 ≠ 5 z-chain[3]: r3c3{n9 n7} - b4n7{r5c3 r6c1} - c1n9{r6 .} ==> r9c3 ≠ 9 biv-chain[4]: r5c7{n9 n5} - c6n5{r5 r6} - b5n2{r6c6 r6c5} - r6n1{c5 c9} ==> r6c9 ≠ 9 z-chain[4]: c4n4{r1 r4} - c4n6{r4 r9} - c3n6{r9 r2} - r2c5{n6 .} ==> r1c4 ≠ 9 finned-x-wing-in-rows: n9{r1 r5}{c6 c9} ==> r4c9 ≠ 9 whip[1]: b6n9{r5c7 .} ==> r7c7 ≠ 9, r8c7 ≠ 9 z-chain[4]: b2n9{r3c6 r2c5} - r2n6{c5 c3} - r2n5{c3 c2} - r4c2{n5 .} ==> r4c6 ≠ 9 z-chain[4]: c9n9{r9 r1} - b3n8{r1c9 r2c8} - c2n8{r2 r3} - c2n7{r3 .} ==> r9c2 ≠ 9 whip[1]: b7n9{r9c1 .} ==> r6c1 ≠ 9 whip[1]: r6n9{c6 .} ==> r4c4 ≠ 9, r4c5 ≠ 9, r5c6 ≠ 9 singles ==> r5c6 = 5, r5c7 = 9, r4c7 = 8, r8c7 = 7, r7c7 = 5 hidden-pairs-in-a-column: c6{n2 n7}{r6 r7} ==> r7c6 ≠ 9, r7c6 ≠ 6, r6c6 ≠ 9 finned-x-wing-in-columns: n8{c9 c1}{r1 r9} ==> r9c2 ≠ 8 whip[1]: c2n8{r3 .} ==> r1c1 ≠ 8 stte

Curious to know what your new pattern is.

by **shye** » Sat Jun 12, 2021 8:31 am

hello all, shye here and im the author :>
decided to finally make an account here

marek was pretty spot on to the core step here. the way i thought about it while making the puzzle was by using row 9 and column 1 to "transform" the quintuple in box 7 and spread it across r1c1, r6c1, r9c1, r9c4, r9c9, from there the deductions can all be unified as somehow using this unorthodox quint. the first step like marek put it is to rule 5 and 6 out of r9c1, the way i considered this part was similar to exocet, if you consider r1c9 and r6c4 base cells of sorts then all combos for them place a base candidate in r9c1 (within the quintuple), and one option where the bases are 7 and 8 would cause a repeat in the quint so it can not only be ruled out as a possibility but it creates a strong link on 9 in the bases and reaps some elims. this is a bit to (grouped) kites how exocet is to fish. once 5 and 6 are out of r9c1 its pretty clear using the quintuple that 6 can be removed from r1c4 and 5 from r6c9 too, rather than using grouped kites. and instead of the ER to finish, you can eliminate 7 from r5c6 as it would place 7 in r6 and c4 both into the quintuple causing a repeat. these are all pretty different deductions but i think can be at least viewed as related which was the goal

unsure how denis' path works as ive not heard of z-chains or whips before, sorry. also not the best at reading the notation, i can see a number of cells important to the intended solve path occuring in there but also ones that dont seem relevant so it might be a different way through. this puzzle isnt completely bypass-proof, yzf can make do with als y-wings and a bit of other stuff, so this is probably another bypass

thanks for taking the time to look at it! im quite proud of the finished result. also as a side note, the title has since been changed to "Hanabi", just thought it fit better

by **yzfwsf** » Sat Jun 12, 2021 2:16 pm

Code: Select all: .------------------------.-----------------.----------------. | #689 1 2 | 4-69 7 4689 | 3 5 #89 | | 3 589 569 | 2 69 1 | 4 89 7 | | 4 789 79 | 5 3 89 | 2 1 6 | :------------------------+-----------------+----------------: | 2 59 359 | 3469 169 4569 | 589 7 1589 | | 1 6 3579 | 379 8 579 | 59 4 2 | | #579 4 8 | #79 129 2579 | 6 3 1-59 | :------------------------+-----------------+----------------: | #579-6 3 1 | 8 269 2679 | 579 69 4 | | #6789 2 4 | 1 5 679 | 789 689 3 | | #789-56 #5789 #679-5 | #679 4 3 | 1 2 #589 | '------------------------'-----------------'----------------'

by **marek stefanik** » Sat Jun 12, 2021 4:48 pm

After seeing yzfwsf's post, I realised the pattern can be written as a fish-reduced MSLS.

The eliminations are cool, some of them took me a while to prove.

Since you didn't provide any explanation, I'm assuming you brute-forced the #-marked cells.
If it's not the case, I'm curious about the way you spotted/detected them (I think the 9s require different links for different eliminations).

Marek

by **Cenoman** » Sun Jun 13, 2021 12:40 pm

Code: Select all: +------------------------+----------------------+---------------------+ | #689 1 2 | 4-69 7 4689 | 3 5 #89 | | 3 589 569 | 2 69 1 | 4 89 7 | | 4 789 79 | 5 3 89 | 2 1 6 | +------------------------+----------------------+---------------------+ | 2 59 359 | 3469 169 4569 | 589 7 1589 | | 1 6 3579 | 379 8 579 | 59 4 2 | | #579 4 8 | #79 129 2579 | 6 3 1-59 | +------------------------+----------------------+---------------------+ | #5679 3 1 | 8 269 2679 | 579 69 4 | | #6789 2 4 | 1 5 679 | 789 689 3 | | #789-56 #5789 #5679 | #679 4 3 | 1 2 #589 | +------------------------+----------------------+---------------------+

marek stefanik wrote:I'm assuming you brute-forced the #-marked cells.

No need of brute force.
Once spotted the fivefold Empty Rectangle in box 7 , it makes sense to consider the other four cells in c1 and r9. It also makes sense to consider cells r6c4 and r1c9, as Marek did, to close rectangles with digits 78.
The 11 marked cells are filled with 5 digits 56789. Digits 5,6,7,8 can have at most 2 occurrences each; digit 9 can have at most 3 occurrences. They all must be there at their maximum count.
Each digit has one occurrence in b7, the second (and third for digit 9) must be outside b7:
- for digit 5, at r6c1 OR r9c9 => -5 r9c1, r6c9
- for digit 6, at r1c1 OR r9c4 => -6 r9c1, r1c4
- for digit 9, two occurrences must be at the diagonal vertices of trapezes (r1c19, r4c14) OR (r49c4, r19c9)

Code: Select all: +---------------+---------------+---------------+ | X - - | O - - | - - X | | - - - | - - - | - - - | | - - - | - - - | - - - | +---------------+---------------+---------------+ | - - - | - - - | - - - | | - - - | - - - | - - - | | X - - | X - - | - - O | +---------------+---------------+---------------+ | X - - | - - - | - - - | | X - - | - - - | - - - | | X X X | X - - | - - X | +---------------+---------------+---------------+

In all configurations, there is a digit 9 at r1c9 OR r4c4, in sight of target cells r1c4, r6c9 => -9 r1c4, r6c9

Then, with these eliminations (as well as with yzfwsf's), comes the following resolution state:

Code: Select all: +--------------------+------------------+--------------------+ | 6 1 2 | 4 7 89 | 3 5 89 | | 3 589 59 | 2 6 1 | 4 89 7 | | 4 789 79 | 5 3 89 | 2 1 6 | +--------------------+------------------+--------------------+ | 2 59 3 | 6 1 4 | 589 7 589 | | 1 6 579 | 3 8 57 | 59 4 2 | | b57 4 8 | c79 29 257 | 6 3 1 | +--------------------+------------------+--------------------+ | a579 3 1 | 8 29 267 | 579 69 4 | | a789 2 4 | 1 5 67 | 789 689 3 | | a789 5-7 6 | d79 4 3 | 1 2 589 | +--------------------+------------------+--------------------+

Finned X-Wing: (7)r789c1 = r6c1 - r6c4 = r9c4 => -7 r9c2; ste

Acknowledgement: I had not seen that, before reading Marek's and yzfwsf's posts

Thanks a lot, mith, for sharing such an interesting puzzle

by **marek stefanik** » Mon Jun 14, 2021 5:58 am

Seeing Cenoman's map, I noticed a branchless way to cover the 9s:

9r9(c12349), 9c1(r1678), 9r1c9, 9r6c4

(initially I used 9r1, 9c4 for one elimination and 9c9, 9r6 for the other)

As for the other eliminations found by yzfwsf (5r9c3, 6r7c1), here is my proof:

7 truths: r1c19, r9c9, 5689b7
10 links: 5r9(c1239), (5)r7c1, 6c1(r1789), (6)r9c3, 8r1(c19), 8r9(c1239), 8c1(r78), 9r1(c19), 9r9(c1239), 9c1(r78)
contra.: (L)8r1c19, (L)8r9c1239, (L)8r78c1
contra.: (L)9r1c19, (L)9r9c1239, (L)9r78c1

Since the contadictions don't share any links, the rest of the pattern is of rank 1 (L – T – C).
Any two links not included in the contradictions then form a truth.
The links 5r7c1 and 6r9c3 have two common eliminations (5r9c3, 6r7c1)
–5r9c3, –6r7c1

About my notation:
Because of the contradictions I find it necessary to distinguish between the parts of links which cover the truths and those which don't.
In this spirit, I use the word 'link' for the part that covers the truths (such as 5r9c3), rather than for the truth that includes it (r9c3).
I imagine this modified notation being used without brackets.
The Ls in contradictions stand for 'link'. Sometimes truths are needed to express contadictions, I find it covenient to notate.

Marek

Edit1: Corrected a typo. Initially I wrote 5r9c7 instead of 5r9c3.
Edit2: Since we have now discovered three different patterns, none of which solves the puzzle by itself, I think there might be a tiny chance the puzzle was designed to showcase a completely different pattern that all of us missed. Just sayin'.

by **eleven** » Mon Jun 14, 2021 6:27 am

Nice way to show that, Cenoman !
Concerning the 9, it is obvious, that one must be in r1c9 or r6c4, because without them only 1 row and one column would be left (for 3 9's).

by **Cenoman** » Mon Jun 14, 2021 9:03 pm

With a heartful dedication to SteveC (sudtyro2), here is another way to show the eliminations.

Code: Select all: +------------------------+----------------------+---------------------+ | B689 1 2 | 4-69 7 4689 | 3 5 B89 | | 3 589 569 | 2 69 1 | 4 89 7 | | 4 789 79 | 5 3 89 | 2 1 6 | +------------------------+----------------------+---------------------+ | 2 59 359 | 3469 169 4569 | 589 7 1589 | | 1 6 3579 | 379 8 579 | 59 4 2 | | A579 4 8 | A79 129 2579 | 6 3 1-59 | +------------------------+----------------------+---------------------+ | #5679 3 1 | 8 269 2679 | 579 69 4 | | #6789 2 4 | 1 5 679 | 789 689 3 | | #789-56 #5789 #5679 | A679 4 3 | 1 2 B589 | +------------------------+----------------------+---------------------+

Another breakdown of the pattern:
- Marked #, the fivefold Empty Rectangle in b7
- Marked A, CoALS (5679)r6c14,r69c4
- Marked B, CoALS (5689)r1c19,r19c9

CoALS rule applied to CoALS A: (56)=(79)r6c14,r9c4
CoALS rule applied to CoALS B: (56)=(89)r1c19,r9c9
Myth Jellies' Coals reference here

Then using ERs in b7:
Inference a: (5)r6c1 - r79c1 = r9c23 - r9c9 => 5 can't be True in both A & B
(6)r1c1 - r789c1 = r9c3 - r9c4 => 6 can't be True in both A & B
Inference b: r6c1 ≠ 5 AND r9c4 ≠ 6 => +7 OR +9 at (r6c1 AND r9c4) => b7 void of 7, OR void of 9
r1c1 ≠ 6 AND r9c9 ≠ 5 => +8 OR +9 at (r1c1 AND r9c9) => b7 void of 8, OR void of 9
=> (5 & 6) can't be both False in A nor in B,

Subsequently, (5 & 6) can't be both True in A (from inference a, 5 & 6 would be both False in B, which leads to a contradiction from inference b).
Similarly they can't be both True in B.
So, 5 is True in one and only one, out of {A, B} AND 6 is True in the other; then (79) pair is True in remaining cells of A, AND (89) pair is True in remaining cells of B (from CoALS rule)

The pattern has two and only two verity configurations:
Config 1: (5)r6c1, (6)r1c1, (79)r69c4, (89)r19c9
Config 2: (6)r9c4, (5)r9c9, (79)r6c14, (89)r1c19
=> -56 r9c1, -69 r1c4, -59r6c9 in any configuration

Edit: redesigned wording and presentation

by **marek stefanik** » Mon Jun 14, 2021 9:18 pm

Hi shye,

Welcome to the forum.

The idea of using r1c1, r6c1, r9c1, r9c4, r9c9 as a quintuple is great. Thanks for sharing it.
It makes everything easier, including the extra eliminations found by yzfwsf (they can now be eliminated by a Groupped ALS S-Wing, as r1c19, r9c9 are somehow an ALS).
(I realised that this doesn't sound easy at all. It becomes a rank 1 pattern, rather than rank 3.

)

Thanks for creating the puzzle, it was an awesome experience to solve it (and to find this many exotic patterns in one puzzle).

Marek

by **shye** » Tue Jun 15, 2021 5:06 am

thank you for the warm welcome! glad to see this puzzle interested people here, and to see so much different takes on it!

lots of these approaches are super interesting, some of them i'm not sure i fully understand (bad at reading notation, but ill learn :lol:

), cenoman's first post especially i find really clear and identifiable.

if you strip down the structure to its bare bones, the computer solvers i know of cant get (all of) the deductions, not without brute force or anything that is.
something like this, to demonstrate:

Code: Select all: +---------------+---------------+---------------+ | - 1 2 | 7 - - | 5 - - | | 3 - - | - - - | - - - | | 4 - - | - - 8 | - - 7 | +---------------+---------------+---------------+ | - 3 4 | 6 - - | 8 - - | | 1 - - | - - - | - - - | | 2 - - | - - 5 | - - 6 | +---------------+---------------+---------------+ | - - - | - - 2 | - - 3 | | - - - | - - 1 | - - 4 | | - - - | 3 4 - | 1 2 - | +---------------+---------------+---------------+ .127..5..3........4....8..7.346..8..1........2....5..6.....2..3.....1..4...34.12.

have fun playing around with this yourself, im pretty certain brutally hard puzzles could be made with this setup, as i was trying to make this one as easy as possible (just the trick and as little follow-up as i can) and it ended up an 8.1 haha

i think other configs of the 4 "intersection cells" in b2356 could lead to interesting stuff, maybe a way to make all 4 required for something, thatd be interesting

by **jco** » Tue Aug 10, 2021 9:07 pm

Solution withdrawn: incorrect. My apologies.

Edit: I found a correction shown in my last post in this thread.

by **jco** » Wed Aug 11, 2021 3:21 am

jco wrote:I found this solution using eleven's replacement approach (curious about how many steps it would take).
Of course this solution comes very late, since this puzzle was thoroughly discussed a while ago.
(...)
Edit: re-posted solution with correction at the cost of more steps.

I could not find a shorter way to solve the puzzle using this approach.
I did not like specially step 4 (long chain). I may have missed a much simpler path.
Anyway, even being late, I really wanted to try this (very nice!) puzzle, so this is it.

Website · by **denis_berthier** » Wed Aug 11, 2021 8:45 am

jco wrote:I could not find a shorter way to solve the puzzle using this approach.
I did not like specially step 4 (long chain). I may have missed a much simpler path.

This is 6 steps (counting eleven's variable relabelling* as a step), including a long chain of length 8.
(*) replacement is a better term than relabelling => I've changed the name of my thread on this topic.

It gave me the idea to try my recent implementation of the fewer-steps algorithm. After only 2 tries, I get five non-W1 steps with no chain of length > 6. As I'm busy with other things, I din't try to find fewer steps.

Code: Select all: Resolution state after Singles and whips[1]: +-------------------+-------------------+-------------------+ ! 689 1 2 ! 469 7 4689 ! 3 5 89 ! ! 3 589 569 ! 2 69 1 ! 4 89 7 ! ! 4 789 79 ! 5 3 89 ! 2 1 6 ! +-------------------+-------------------+-------------------+ ! 2 59 359 ! 3469 169 4569 ! 589 7 1589 ! ! 1 6 3579 ! 379 8 579 ! 59 4 2 ! ! 579 4 8 ! 79 129 2579 ! 6 3 159 ! +-------------------+-------------------+-------------------+ ! 5679 3 1 ! 8 269 2679 ! 579 69 4 ! ! 6789 2 4 ! 1 5 679 ! 789 689 3 ! ! 56789 5789 5679 ! 679 4 3 ! 1 2 589 ! +-------------------+-------------------+-------------------+

=====> STEP #1
finned-swordfish-in-columns: n7{c2 c3 c4}{r9 r3 r5} ==> r5c6≠7

=====> STEP #2
whip[6]: r2c5{n9 n6} - c3n6{r2 r9} - c4n6{r9 r4} - c4n3{r4 r5} - r5n7{c4 c3} - r3c3{n7 .} ==> r3c6≠9
naked-single ==> r3c6=8
whip[1]: r3n9{c3 .} ==> r1c1≠9, r2c2≠9, r2c3≠9

=====> STEP #3
z-chain[6]: c2n8{r9 r2} - r2n5{c2 c3} - c3n6{r2 r9} - r9c4{n6 n9} - r6c4{n9 n7} - c1n7{r6 .} ==> r9c2≠7
hidden-single-in-a-column ==> r3c2=7
naked-single ==> r3c3=9

=====> STEP #4
whip[6]: r2c3{n5 n6} - r1c1{n6 n8} - r2n8{c2 c8} - r8n8{c8 c7} - r4c7{n8 n9} - r4c2{n9 .} ==> r4c3≠5
naked-single ==> r4c3=3
hidden-single-in-a-block ==> r5c4=3
hidden-single-in-a-row ==> r5c3=7

=====> STEP #5
z-chain[6]: b1n6{r1c1 r2c3} - r9c3{n6 n5} - r7n5{c1 c7} - c7n7{r7 r8} - r8n8{c7 c8} - r2n8{c8 .} ==> r1c1≠8
stte

The New Sudoku Players' Forum

Hanabi (formerly Cerithiidae) by shye (SER 8.1)

Hanabi (formerly Cerithiidae) by shye (SER 8.1)

Re: Cerithiidae by shye (SER 8.1)

Re: Cerithiidae by shye (SER 8.1)

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Re: Cerithiidae by shye (SER 8.1)

Re: Cerithiidae by shye (SER 8.1)

Re: Cerithiidae by shye (SER 8.1)

Re: Cerithiidae by shye (SER 8.1)

Re: Cerithiidae by shye (SER 8.1)

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Re: Hanabi (formerly Cerithiidae) by shye (SER 8.1)

Re: Hanabi (formerly Cerithiidae) by shye (SER 8.1)

Re: Hanabi (formerly Cerithiidae) by shye (SER 8.1)

Re: Hanabi (formerly Cerithiidae) by shye (SER 8.1)