The New Sudoku Players' Forum

by **Leren** » Wed Oct 28, 2020 8:47 am

For some reason in the second puzzle I found two Multifish in the same cells for a total of 27 eliminations. From the look of them I'll bet Phil's variant SK loop will apply. Leren

Website · by **denis_berthier** » Wed Oct 28, 2020 8:49 am

Leren wrote:For some reason in the second puzzle I found two Multifish in the same cells for a total of 27 eliminations. From the look of them I'll bet Phil's variant SK loop will apply. Leren

Great. Could you write the 27 eliminations? I'll see what comes after.

by **Ajò Dimonios** » Wed Oct 28, 2020 9:07 am

Using contradiction chains it is possible to solve the puzzle in 4 steps.

1)27 eliminations by crossing the conjugate tracks T (1r4c8) and T (1r7c8) see post by Mauriès Robert
2) T (1r7c8) .T (8r1c3) => contracdiction
3) T (1r7c8) .T (8r1c7) => contracdiction => - 1r7c8
4) T (7r7c8) => contracdiction => - 7r7c8; stte.
Is it possible to write these chains of contradiction in AIC?

Paolo

by **Leren** » Wed Oct 28, 2020 9:42 am

Code: Select all: *--------------------------------------------------------------------------------* | 1 359 245 | 349 236 2479 | 257 679 8 | | 239 7 289 | 1368 5 138 | 129 4 269 | | 245 589 6 | 1489 128 2479 | 3 179 257 | |--------------------------+--------------------------+--------------------------| | 2359 138 12589 | 7 1348 6 | 14589 138 3459 | | 3567 4 1578 | 138 9 138 | 1578 2 3567 | | 379 1368 1789 | 2 1348 5 | 14789 1368 3479 | |--------------------------+--------------------------+--------------------------| | 457 159 3 | 14589 128 249 | 6 789 247 | | 469 2 149 | 1368 7 138 | 489 5 349 | | 8 569 457 | 3459 236 249 | 247 379 1 | *--------------------------------------------------------------------------------*

This is the PM after a preliminary Swordfish and the 27 eliminations. Putting this into Hodoku it's first 5 moves are Brute Force and 4 forcing chains.

Leren

Website · by **denis_berthier** » Wed Oct 28, 2020 10:41 am

Leren wrote:
Code: Select all
*--------------------------------------------------------------------------------* | 1 359 245 | 349 236 2479 | 257 679 8 | | 239 7 289 | 1368 5 138 | 129 4 269 | | 245 589 6 | 1489 128 2479 | 3 179 257 | |--------------------------+--------------------------+--------------------------| | 2359 138 12589 | 7 1348 6 | 14589 138 3459 | | 3567 4 1578 | 138 9 138 | 1578 2 3567 | | 379 1368 1789 | 2 1348 5 | 14789 1368 3479 | |--------------------------+--------------------------+--------------------------| | 457 159 3 | 14589 128 249 | 6 789 247 | | 469 2 149 | 1368 7 138 | 489 5 349 | | 8 569 457 | 3459 236 249 | 247 379 1 | *--------------------------------------------------------------------------------*

This is the PM after a preliminary Swordfish and the 27 eliminations. Putting this into Hodoku it's first 5 moves are Brute Force and 4 forcing chains.

Leren

It's interesting to notice that, after the 27 eliminations, the puzzle is still in T&E(2). However, it has moved down the BpB sub-hierarchy of T&E(2): the original puzzle was in B2B, it is now in B1B=gB, i.e. it can be solved by g-braids. (I need to learn how to use Sukaku Explainer in order to compute the SER of your PM.)

I don't know what Brute Force means in Hodoku (probably some kind of T&E). It may be the case that Hodoku doesn't have any grouped variant of forcing chains and that's why it has to use "Brute Force" at this point. But anyway, the second step should be a mere forcing chain.

As is generally the case for puzzles solvable by g-braids, it is also solvable by (much simpler) g-whips.
The first step is a long g-whip[13], with a single right-linking g-candidate: n9r9c456, in the third csp-variable:
g-whip[13]: r1n6{c8 c5} - r9n6{c5 c2} - r9n9{c2 c456} - r7n9{c6 c2} - r8c1{n9 n4} - c3n4{r9 r1} - r1c4{n4 n3} - r1c2{n3 n5} - r3c1{n5 n2} - b2n2{r3c6 r1c6} - c6n7{r1 r3} - r3n9{c6 c4} - r3n4{c4 .} ==> r1c8 ≠ 9

For the full solution, see here:

Code: Select all: (solve-sukaku-grid *--------------------------------------------------------------------------------* ! 1 359 245 ! 349 236 2479 ! 257 679 8 ! ! 239 7 289 ! 1368 5 138 ! 129 4 269 ! ! 245 589 6 ! 1489 128 2479 ! 3 179 257 ! !--------------------------+--------------------------+--------------------------! ! 2359 138 12589 ! 7 1348 6 ! 14589 138 3459 ! ! 3567 4 1578 ! 138 9 138 ! 1578 2 3567 ! ! 379 1368 1789 ! 2 1348 5 ! 14789 1368 3479 ! !--------------------------+--------------------------+--------------------------! ! 457 159 3 ! 14589 128 249 ! 6 789 247 ! ! 469 2 149 ! 1368 7 138 ! 489 5 349 ! ! 8 569 457 ! 3459 236 249 ! 247 379 1 ! *--------------------------------------------------------------------------------* )

Hidden Text: Show

by **SpAce** » Wed Oct 28, 2020 12:44 pm

denis_berthier wrote:I don't know what Brute Force means in Hodoku (probably some kind of T&E).

I'm not sure, but I think it simply picks a useful placement from the DLX-precomputed solution if it can't progress logically.

It may be the case that Hodoku doesn't have any grouped variant of forcing chains and that's why it has to use "Brute Force" at this point. But anyway, the second step should be a mere forcing chain.

Hodoku would call it a forcing net contradiction (if it could find it), which is what your whip is. (Don't get started on that. You've said yourself that whips and braids are AND-nets, and of course they are. Any use of memories makes a net.)

That said, I don't know why Hodoku fails to find it as a contradiction net. (It's obvious that it can't find it as a verity, because it can't handle multi-krakens.) It can certainly use groups and subsets in its chains and nets (though it does have problems with groups in some contexts). Maybe the sheer number of z- and t-candidates (5+9) in this pattern overwhelms it, even though they don't add anything to the complexity according to you.

In any case, I count this as a failure for Hodoku, and a point for SudoRules.

The first step is a long g-whip[13], with a single right-linking g-candidate: n9r9c456, in the third csp-variable:
g-whip[13]: r1n6{c8 c5} - r9n6{c5 c2} - r9n9{c2 c456} - r7n9{c6 c2} - r8c1{n9 n4} - c3n4{r9 r1} - r1c4{n4 n3} - r1c2{n3 n5} - r3c1{n5 n2} - b2n2{r3c6 r1c6} - c6n7{r1 r3} - r3n9{c6 c4} - r3n4{c4 .} ==> r1c8 ≠ 9

As a matrix (without any group-breaking):

13x13 TM: Show

Out of its thirteen SISs only three are binary.

--
Edit. I forgot to count the t-candidates from the left-linking group nodes. That's three more (6 --> 9).

by **SpAce** » Wed Oct 28, 2020 2:36 pm

denis_berthier wrote:
Leren wrote:This has 7 fewer candidates than the previous one. Pasting this into Hodoku, it's first move is a (very tedious) forcing chain:

Forcing Chain Contradiction in c3 => r3c9=2r3c9<>2 r3c9=4 r2c79<>4 r2c13=4 r1c3<>4r3c9<>2 r8c9=2 r8c9<>1 r8c4=1 r2c4<>1 r2c3=1 r2c3<>4r3c9<>2 r3c9=4 r3c5<>4 r9c5=4 r9c3<>4

Is this supposed to be a single chain (of length 9)?

No. Its length is 7 (one more than necessary), and it's actually not very tedious. It just looks that way, because Hodoku's notation is so horrible. (I never even attempt to read it; I just look at the pattern.) Furthermore, it forces a placement (+2r3c9) for no reason, which lengthens it by one unnecessarily, and uses a dumb routing. I don't know why Hodoku sometimes offers such inferior choices by default. Here's a better one with the same effect (found by Hodoku as a Forcing Chain Verity):

6x6 TM: Show

That would be a very simple braid[6]. It's easy to write as an AIC:

(2=4)r3c9 - r3c56 = r1c6;r9c5 - r19c3 = (4-1)r2c3 = r2c4 - r8c4 = (1)r8c9 => -2 r8c9

Even easier as a kraken column:

Code: Select all: (4)r1c3 - r1c6 = r3c56 - (4=2)r3c9 || (4-1)r2c3 = r2c4 - r8c4 = (1)r8c9 || (4)r9c3 - r9c5 = r3c5 - (4=2)r3c9 => -2 r8c9

Not tedious at all.

Added:

If so, Hodoku is pretty bad at this point, as it misses a much shorter chain (length 5):

whip[5]: b9n1{r7c8 r8c9} - c4n1{r8 r2} - r2n3{c4 c7} - b3n7{r2c7 r2c9} - r5c9{n7 .} ==> r7c8 ≠ 7

Well, Hodoku's move gets an immediate placement. That's what I would have aimed for. It finds the equivalent move to yours if asked (as a Forcing Chain Verity), but its slightly different routing makes it a z-chain[6].

5x5 TM; SudoRules whip(5): Show

6x6 TM; Hodoku z-chain(6): Show

Notice that, after the first whip[5], none of the patterns depends on the target(s) and almost all of them are very simple bivalue-chains (basics AICs).

The same with Hodoku (after its own initial move), if only simple chains are allowed:

Code: Select all: biv (4) => -7 r79c8 (AIC) biv (4) => -4 r2c1 (DNL) biv (5) => -8 r13c8,r9c7 (AIC) biv (3) => -8 r9c6 (W-Wing) biv (6) => -9 r13c2,r8c3 (XY-Chain) biv (6) => -6 r359c6 (XY-Chain) biv (3) => -9 r1c6,r9c4 (XY-Wing) biv (3) => -6 r6c2 (X-Chain) biv (7) => -6 r3c2,r46c1 (XY-Chain); stte

Those are slightly longer than yours, but there's only 9 of them and 14 of yours. As a whole that's simpler, considering that a higher number of moves adds more overhead than longer chains do, all other things remaining equal.

by **SpAce** » Thu Oct 29, 2020 4:16 am

Leren wrote:For some reason in the second puzzle I found two Multifish in the same cells for a total of 27 eliminations. From the look of them I'll bet Phil's variant SK loop will apply. Leren

It's actually the plain old one. It also results in a perfect Jellyfish, which is kind of cool. (How rare are those, btw?)

1.......8.7..5..4...6...3.....7.6....4..9..2....2.5.....3...6...2..7..5.8.......1

Step 1. Swordfish/Loop (6) => 4 elims

Step 2. SK-Loop:

Code: Select all: \b1:38 \b7:16 \28n \28n \b3:16 \b9:38 *1368 *1368 .-------------------------.-------------------------.--------------------------. | 1^ 359 2459 | 3469 236 23479 | 2579 679 8^ | | 239 7 289 | \1368-9 5 \138-29 | 129 4 269 | *1368 | 2459 589 6^ | 1489 128 124789 | 3^ 179 2579 | :-------------------------+-------------------------+--------------------------: | 2359 \138-59 12589 | 7 1348 6^ | 14589 \138-9 3459 | \n28 | 3567 4 1578 | 138 9 138 | 1578 2 3567 | | 3679 \1368-9 1789 | 2 1348 5 | 14789 \1368-79 34679 | \n28 :-------------------------+-------------------------+--------------------------: | 4579 159 3^ | 14589 128 12489 | 6^ 789 2479 | | 469 2 149 | \1368-49 7 \138-49 | 489 5 349 | *1368 | 8^ 569 4579 | 34569 236 2349 | 2479 379 1^ | '-------------------------'-------------------------'--------------------------'

MF (1368 RC): 16x16 {1368R28 1368C28 \ 38b19 16b37 28n46 46n28} => 13 elims; NT (138) => 6 elimsStep 3. Perfect Jellyfish with perfectly symmetrical eliminations:

Code: Select all: .----------------------.-------------------.----------------------. | 1 359 245-9 | 349 236 2479 | 257-9 679 8 | | *239 7 *289 | 1368 5 138 | *129 4 *269 | | 245-9 589 6 | 1489 128 2479 | 3 179 257-9 | :----------------------+-------------------+----------------------: | *2359 138 *12589 | 7 1348 6 | *14589 138 *3459 | | 3567 4 1578 | 138 9 138 | 1578 2 3567 | | *379 1368 *1789 | 2 1348 5 | *14789 1368 *3479 | :----------------------+-------------------+----------------------: | 457-9 159 3 | 14589 128 249 | 6 789 247-9 | | *469 2 *149 | 1368 7 138 | *489 5 *349 | | 8 569 457-9 | 3459 236 249 | 247-9 379 1 | '----------------------'-------------------'----------------------' (9)r2468\c1379 => 8 elims

Together the total of steps 2-3 is 27 eliminations (13+6+8), as with your pair of multifishes. (Would you mind showing them, btw?)

Variant Step 2. SK-Loop as a non-rectangular MSLS:

Code: Select all: \b1:389 \b7:169 \b3:169 \b9:389 \5[9] \7[9] .-------------------------.-----------------------.--------------------------. | 1^ *359 245-9 | 349 236 23479 | 257-9 *679 8^ | | *239 7 *289 | 1368-9 5 138-29 | *129 4 *269 | \2[9] | 245-9 *589 6^ | 1489 128 124789 | 3^ *179 257-9 | :-------------------------+-----------------------+--------------------------: | 2359 138-59 12589 | 7 1348 6^ | 14589 138-9 3459 | | 3567 4 1578 | 138 9 138 | 1578 2 3567 | | 379 1368-9 1789 | 2 1348 5 | 14789 1368-79 3479 | :-------------------------+-----------------------+--------------------------: | 457-9 *159 3^ | 14589 128 12489 | 6^ *789 247-9 | | *469 2 *149 | 1368-49 7 138-49 | *489 5 *349 | \4[9] | 8^ *569 457-9 | 3459 236 2349 | 247-9 *379 1^ | '-------------------------'-----------------------'--------------------------'

MSLS: 16x16 {1379N28 28N1379 \ 1b37 2r2 3b19 4r8 5c2 6b37 7c8 8b19 [9r28 9c28|9b1379]} => 21 elims; NT (138) => 6 elimsThat gives the same 27 eliminations without needing the Jellyfish.

by **StrmCkr** » Thu Oct 29, 2020 6:03 am

Code: Select all: +---------------------------+------------------------------+--------------------------+ | 7 (159) 458-19 | 29-13 (13-24) 245-9 | 248-3 (39-8) 6 | | (469) 2 (149) | (13-679) 8 (679-4) | (37-4) 5 (479) | | 458-69 (569) 3 | 2679 (467-2) 24579-6 | 1 (789) 248-79 | +---------------------------+------------------------------+--------------------------+ | 2689 (1679) 2789-1 | 5 (267) 3 | 4678 (1679-8) 4789-1 | | 356-8 4 157-8 | (678) 9 (78-6) | 3567-8 2 157-8 | | 25689-3 (3679-5) 25789 | 4 (267) 1 | 5678-3 (3679-8) 5789 | +---------------------------+------------------------------+--------------------------+ | 245-3 (37-5) 6 | 278-13 (13-247) 248-7 | 9 (178) 258-17 | | (39-2) 8 (279) | (13-2679) 5 (679-2) | (267) 4 (127) | | 1 (579) 245-79 | 26789 (467-2) 24789-6 | 258-67 (678) 3 | +---------------------------+------------------------------+--------------------------+

is the opening move i did before basics... {which includes the basics and a few extra eliminations}

reduces the grid to this

Code: Select all: +----------------------+---------------------+----------------------+ | 7 (159) 458 | 29 (13) 245 | 248 (39) 6 | | (469) 2 (149) | (13) 8 (679) | (37) 5 (479) | | 458 (569) 3 | 2679 (467) 24579 | 1 (789) 248 | +----------------------+---------------------+----------------------+ | 2689 (1679) 2789 | 5 (267) 3 | 4678 (1679) 4789 | | 356 4 157 | (678) 9 (78) | 3567 2 157 | | 25689 (3679) 25789 | 4 (267) 1 | 5678 (3679) 5789 | +----------------------+---------------------+----------------------+ | 245 (37) 6 | 278 (13) 248 | 9 (178) 258 | | (39) 8 (279) | (13) 5 (679) | (267) 4 (127) | | 1 (579) 245 | 26789 (467) 24789 | 258 (678) 3 | +----------------------+---------------------+----------------------+

aals 2rc rule

Code: Select all: +--------------------+------------------------+--------------------+ | 7 159 458 | (29) 13 245 | 248 (39) 6 | | 469 2 149 | 13 8 (679) | (37) 5 (479) | | (458) 569 3 | (2679) (467) (24579) | 1 (789) 28-4 | +--------------------+------------------------+--------------------+ | 2689 1679 2789 | 5 267 3 | 4678 1679 4789 | | 356 4 157 | 678 9 78 | 3567 2 157 | | 25689 3679 25789 | 4 267 1 | 5678 3679 5789 | +--------------------+------------------------+--------------------+ | 245 37 6 | 278 13 248 | 9 178 258 | | 39 8 279 | 13 5 679 | 267 4 127 | | 1 579 245 | 26789 467 24789 | 258 678 3 | +--------------------+------------------------+--------------------+

still needs ~4 more chains {maybe less but its getting easier at this point} as the chains are mostly just 2 digit short loops.

Website · by **denis_berthier** » Thu Oct 29, 2020 7:49 am

Ajò Dimonios wrote:Using contradiction chains it is possible to solve the puzzle in 4 steps.
1)27 eliminations by crossing the conjugate tracks T (1r4c8) and T (1r7c8) see post by Mauriès Robert
2) T (1r7c8) .T (8r1c3) => contracdiction
3) T (1r7c8) .T (8r1c7) => contracdiction => - 1r7c8
4) T (7r7c8) => contracdiction => - 7r7c8; stte.

I think you can make it still shorter:

Code: Select all: ttte

ttte = tracks to the end

by **Mauriès Robert** » Thu Oct 29, 2020 9:28 am

Hi Denis,

denis_berthier wrote:
Ajò Dimonios wrote:Using contradiction chains it is possible to solve the puzzle in 4 steps.
1)27 eliminations by crossing the conjugate tracks T (1r4c8) and T (1r7c8) see post by Mauriès Robert
2) T (1r7c8) .T (8r1c3) => contracdiction
3) T (1r7c8) .T (8r1c7) => contracdiction => - 1r7c8
4) T (7r7c8) => contracdiction => - 7r7c8; stte.

I think you can make it still shorter:
Code: Select all
ttte

ttte = tracks to the end

Questionable humour!
I know that on this site the technique (TDP) that I defend, which Paolo is a user, is not well seen by many participants. I hope that you are not one of them.
Certainly, Paolo could have detailed it a little, but I understand his reluctance to waste time, because I see that on the remark we made on this thread of 27 eliminations by crossing two tracks, there was no intervention, neither from you nor from anyone else. However, I have given the details.
Robert

Website · by **denis_berthier** » Thu Oct 29, 2020 10:02 am

Mauriès Robert wrote:Hi Denis,
denis_berthier wrote:
Ajò Dimonios wrote:Using contradiction chains it is possible to solve the puzzle in 4 steps.
1)27 eliminations by crossing the conjugate tracks T (1r4c8) and T (1r7c8) see post by Mauriès Robert
2) T (1r7c8) .T (8r1c3) => contracdiction
3) T (1r7c8) .T (8r1c7) => contracdiction => - 1r7c8
4) T (7r7c8) => contracdiction => - 7r7c8; stte.

I think you can make it still shorter:
Code: Select all
ttte

ttte = tracks to the end

Questionable humour!
I know that on this site the technique (TDP) that I defend, which Paolo is a user, is not well seen by many participants. I hope that you are not one of them.
Certainly, Paolo could have detailed it a little, but I understand his reluctance to waste time, because I see that on the remark we made on this thread of 27 eliminations by crossing two tracks, there was no intervention, neither from you nor from anyone else. However, I have given the details.
Robert

Glad you saw the humour in it. But I was serious also. There's an obvious abuse of notation and this solution from Paolo illustrates it very clearly.
I'm still in an undecided state about tracks.
About details, let's rather speak of them in the other thread, about a much simpler track.
BTW, do you have news of Defise?

by **Ajò Dimonios** » Thu Oct 29, 2020 12:08 pm

Hi Denis

After basic

Code: Select all: +--------------------+-------------------+-------------------+ | 7 159 14589 | 1239 13 2459 | 2348 389 6 | | 469 2 149 | 13679 8 4679 | 347 5 479 | | 45689 569 3 | 2679 467 245679 | 1 789 24789 | +--------------------+-------------------+-------------------+ | 2689 1679 12789 | 5 267 3 | 4678 16789 14789 | | 356 4 157 | 678 9 678 | 3567 2 157 | | 235689 35679 25789 | 4 267 1 | 35678 36789 5789 | +--------------------+-------------------+-------------------+ | 2345 357 6 | 12378 13 2478 | 9 178 12578 | | 239 8 279 | 123679 5 2679 | 267 4 127 | | 1 579 24579 | 26789 467 246789 | 25678 678 3 | +--------------------+-------------------+-------------------+

In fact I think you are right.
A solution can be achieved in just three steps. The first 27 eliminations are superfluous as they are contained in track T (1r7c8).
Track T (1r7c8) leads to this result using only TB:

Code: Select all: +---------------+-----------------+----------------+ | 7 59 4589 | 29 1* 2459 | 248 3* 6 | | 49 2 1* | 3* 8 6* | 47 5 479 | | 4589 6* 3 | 279 47 24579 | 1 89 2489 | +---------------+-----------------+----------------+ | 2689 1* 2789 | 5 267 3 | 478 6789 4789 | | 56 4 57 | 678 9 78 | 3* 2 1* | | 2689 3* 2789 | 4 267 1 | 578 6789 5789 | +---------------+-----------------+----------------+ | 245 57 6 | 278 3* 2478 | 9 1* 2578 | | 3* 8 29 | 1* 5 279 | 6* 4 27 | | 1 579 2459 | 26789 467 24789 | 2578 78 3 | +---------------+-----------------+----------------+

It can be seen that in r1 there are only 2 candidates 8 in r1c3 and 8 in r1c7.
The extension of T (1r7c8) with T (8r1c3) and with T (8r1c7) produces the following two results:

Code: Select all: +--------------+----------------+---------------+ | 7 59 8 | 29 1* 259 | 4* 3 6 | | 4* 2 1* | 3* 8 6 * | 7 5 79 | | 59 6* 3 | 79 4* 579 | 1 8* 2* | +--------------+----------------+---------------+ | 2689 1* 279 | 5 267 3 | 78 679 4789 | | 56 4 57 | 678 9 78 | 3* 2 1* | | 2689 3* 279 | 4 267 1 | 578 679 5789 | +--------------+----------------+---------------+ | 25 57 6 | 278 3* 4* | 9 1 578 | | 3* 8 29 | 1* 5 279 | 6* 4 7 | | 1 579 4* | 26789 67 2789 | 2578 7 3 | +--------------+----------------+---------------+

Code: Select all: +----------+--------+--------+ | 7 5 4 | 9 1 2 | 8 3 6 | | 9 2 1 | 3 8 6 | 7 5 4 | | 8 6 3 | 7 4 5 | 1 9 2 | +----------+--------+--------+ | 2 1 789 | 5 2 3 | 4 6 89 | | 5 4 . | 8 9 7 | 3 2 1 | | 26 3 789 | 4 26 1 | 5 7 89 | +----------+--------+--------+ | 4 7 6 | 2 3 8 | 9 1 5 | | 3 8 2 | 1 5 9 | 6 4 7 | | 1 9 5 | 6 7 4 | 2 8 3 | +----------+--------+--------+

Clearly the two sudokus are invalid 2 candidates 7 in b9 in the first extension and empty r5c3 in the second. At this point it is obvious that r7c8 ≠ 1.
The T (7r7c8) produces the following result after applying the TB:

Code: Select all: +------------+----------------+----------+ | 7 1* 589 | 29 3* 259 | 4* 89 6 | | 6* 2 49 | 1* 8 49 | 3* 5 7*| | 48 59 3 | 679 467 4579 | 1 89 2*| +------------+----------------+----------+ | 28 679 289 | 5 267 3 | 678 1* 4*| | 3* 4 1* | 678 9 78 | 67 2 5*| | 28 567 258 | 4 267 1 | 678 3* 9*| +------------+----------------+----------+ | 5* 3* 6 | 2 1* 24 | 9 7* 8*| | 9* 8 7* | 3* 5 6* | 2* 4 1*| | 1 . 24 | 2789 47 24789 | 5* 6* 3 | +------------+----------------+----------+

Also in this case the result is a contradiction r9c2 is empty, so r7c8≠7. At this point r7c8 = 8 which leads directly to the solution.
Surely to be clearer it is useful to use "*" instead of "." as a symbol for the boolean operator AND.

Paolo

by **mith** » Thu Oct 29, 2020 11:38 pm

SpAce wrote:It also results in a perfect Jellyfish, which is kind of cool. (How rare are those, btw?)

I'm assuming by perfect jellyfish you mean the digit can appear in all 16 cells? If so... quite rare, in my experience. The first one in Tatooine Reflections is one (though the eliminations aren't symmetric), and The Final Countdown has two (with symmetric eliminations) on the same set of cells. (The harder consecutive puzzle I posted in variants has a pair on the same digit, with different eliminations... and four more almost perfect.

)

by **Leren** » Fri Oct 30, 2020 12:54 am

For the avoidance of any doubt here is the Jellyfish in 9's r2468 / c1379 => - 9 r19c37, r37c19

Code: Select all: *---------------------------------------------------------* | 1 359 245-9 | 349 236 2479 | 257-9 679 8 | |*239 7 *289 | 1368 5 138 |*129 4 *269 | | 245-9 589 6 | 1489 128 2479 | 3 179 257-9 | |-------------------+-----------------+-------------------| |*2359 138 *12589 | 7 1348 6 |*14589 138 *3459 | | 3567 4 1578 | 138 9 138 | 1578 2 3567 | |*379 1368 *1789 | 2 1348 5 |*14789 1368 *3479 | |-------------------+-----------------+-------------------| | 457-9 159 3 | 14589 128 249 | 6 789 247-9 | |*469 2 *149 | 1368 7 138 |*489 5 *349 | | 8 569 457-9 | 3459 236 249 | 247-9 379 1 | *---------------------------------------------------------*

Leren

PS While on the subject of Jellyfish here is a perfect one from Hodoku that has 17 eliminations. Is that a record ?

.............1.....123.456...........27...38..36.5.42...........682.715..74...83.

Code: Select all: *---------------------------------------------------------------------* | 345678-9 *4589 *359 | 568-9 2678-9 2568-9 |*279 *1479 123478-9 | | 345678-9 *4589 *359 | 568-9 1 2568-9 |*279 *479 23478-9 | | 789 1 2 | 3 789 4 | 5 6 789 | |----------------------+------------------------+---------------------| | 1458-9 *4589 *159 | 14678-9 23 23 |*679 *179 1567-9 | | 1459 2 7 | 1469 469 169 | 3 8 1569 | | 189 3 6 | 1789 5 189 | 4 2 179 | |----------------------+------------------------+---------------------| | 1235-9 *59 *1359 | 14568-9 3468-9 13568-9 |*2679 *479 2467-9 | | 39 6 8 | 2 349 7 | 1 5 49 | | 1259 7 4 | 1569 69 1569 | 8 3 269 | *---------------------------------------------------------------------*

Jellyfish in 9's c2378 / r1247=> - 9 r17c14569, r2c1469, r4c149.

Leren

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