The New Sudoku Players' Forum

[shye]

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+-------+-------+-------+
| . . 1 | 3 5 7 | 9 . . |
| . . 9 | 2 . 1 | 7 . . |
| . . . | . 6 . | . . . |
+-------+-------+-------+
| 3 . . | . . . | . . 2 |
| . 1 . | . 2 . | . 4 . |
| 5 . . | . . . | . . 6 |
+-------+-------+-------+
| . 2 . | 1 . 4 | . 8 . |
| 1 . 6 | . 9 . | 4 . 7 |
| . 8 . | . 7 . | . 1 . |
+-------+-------+-------+
..13579....92.17......6....3.......2.1..2..4.5.......6.2.1.4.8.1.6.9.4.7.8..7..1.

estimated rating: 6.6
spent maybe too much time fiddling with the little details of this ヽ(´▽`)/

[totuan]

estimated rating: 6.6
spent maybe too much time fiddling with the little details of this ヽ(´▽`)/

I do not know what you meant, my path's one step - eliminated 2r1c8, and stte.

Thanks for the puzzle.
totuan

[denis_berthier]

.

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Resolution state after Singles and whips[1]:
   +-------------------+-------------------+-------------------+
   ! 2468  46    1     ! 3     5     7     ! 9     26    48    !
   ! 468   3456  9     ! 2     48    1     ! 7     356   3458  !
   ! 2478  3457  34578 ! 489   6     89    ! 2358  235   1     !
   +-------------------+-------------------+-------------------+
   ! 3     4679  478   ! 45678 148   568   ! 158   579   2     !
   ! 678   1     78    ! 56789 2     35689 ! 358   4     358   !
   ! 5     479   2     ! 478   148   38    ! 138   379   6     !
   +-------------------+-------------------+-------------------+
   ! 79    2     57    ! 1     3     4     ! 6     8     59    !
   ! 1     35    6     ! 58    9     258   ! 4     235   7     !
   ! 49    8     345   ! 56    7     256   ! 235   1     359   !
   +-------------------+-------------------+-------------------+

There is a 1-step solution in W8:

Code: Select all

whip[8]: r1n2{c1 c8} - r1n6{c8 c2} - r2c1{n6 n4} - r9n4{c1 c3} - c3n3{r9 r3} - c2n3{r3 r8} - r8c8{n3 n5} - r3c8{n5 .} ==> r1c1≠8
stte

But that seems unreasonably complicated, considering there's an elementary simplest-first solution in S+BC3:

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hidden-pairs-in-a-column: c8{n7 n9}{r4 r6} ==> r6c8≠3, r4c8≠5
finned-x-wing-in-rows: n3{r8 r2}{c2 c8} ==> r3c8≠3
biv-chain[3]: r1c2{n4 n6} - r1c8{n6 n2} - b1n2{r1c1 r3c1} ==> r3c1≠4
biv-chain[3]: r2c5{n8 n4} - b3n4{r2c9 r1c9} - r1n8{c9 c1} ==> r2c1≠8
naked-pairs-in-a-block: b1{r1c2 r2c1}{n4 n6} ==> r3c3≠4, r3c2≠4, r2c2≠6, r2c2≠4, r1c1≠6, r1c1≠4
singles ==> r3c4=4, r2c5=8, r3c6=9, r5c4=9
whip[1]: r5n7{c3 .} ==> r4c2≠7, r4c3≠7, r6c2≠7
hidden-single-in-a-column ==> r3c2=7
naked-pairs-in-a-block: b1{r1c1 r3c1}{n2 n8} ==> r3c3≠8
whip[1]: c3n8{r5 .} ==> r5c1≠8
x-wing-in-columns: n3{c2 c8}{r2 r8} ==> r2c9≠3
finned-x-wing-in-columns: n5{c2 c8}{r8 r2} ==> r2c9≠5
stte

Shye, why do you always write "estimated rating" instead of "rating" or "SER"... Are you using another software than SE for your ratings?

[marek stefanik]

Most likely similar to totuan's:

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.-------------------.--------------------.-----------------.
| 2468  46    1     | 3       5    7     | 9     6–2  48   |
| 468   3456  9     | 2       48   1     | 7     356  3458 |
| 2478  3457 *35–478| 489     6    89    | 2358 *235  1    |
:-------------------+--------------------+-----------------:
| 3     4679  478   | 456789  148  5689  | 158   579  2    |
| 6789  1     78    | 56789   2    35689 | 358   4    3589 |
| 5     479   2     | 4789    148  389   | 138   379  6    |
:-------------------+--------------------+-----------------:
| 79    2    *57    | 1       3    4     | 6     8    59   |
| 1    *35    6     | 58      9    258   | 4    *235  7    |
| 49    8    *345   | 56      7    256   | 235   1    359  |
'-------------------'--------------------'-----------------'

Whichever digit appears in r8c2 is forced into r3c3 (–478r3c3) and absent from r38c8 => remote triple, –2r1c8, stte

5 Truths = {35C3 3N8 8N28}
9 Links = {2c8 35r3 35r8 35c8 35b7}
(add link 3n3 for –478r3c3)
(if we use 2-links 3* and 5* instead of 4 links for each of them, we get rank0)

Marek

[shye]

I do not know what you meant, my path's one step - eliminated 2r1c8, and stte.

just meant that i was being picky while making the puzzle lol, the main idea of it took a few minutes to set up but i spent the best of an hour or two being fussy over the resolution stages lol. looks like you got the elimination i was going for though! care to share how?

Shye, why do you always write "estimated rating" instead of "rating" or "SER"... Are you using another software than SE for your ratings?

i use yzf_sudoku to create my puzzles, so its convenient for me to use the approximate rating value on that software since i dont use SE. im aware its not precise so i note it as an estimation, just so people have a vague idea of what difficulty to expect before trying the puzzle :>
im a fan of this step in your simplest-first path: biv-chain[3]: r2c5{n8 n4} - b3n4{r2c9 r1c9} - r1n8{c9 c1} ==> r2c1≠8
an m-wing i believe? the pair it gives in b1 is lovely

Whichever digit appears in r8c2 is forced into r3c3 (–478r3c3) and absent from r38c8 => remote triple, –2r1c8, stte

always just what i was going for ✦ ヮ ✦ the equivalence between r8c2 & r3c3 was how i thought of it too, completing the [235] als in c8
another very similar way to view it, (which is probably the same as the rank0 pattern with 2-links) is a bivalue oddagon on [35] using the empty rectangle in b1 to ensure r8c2 and r3c8 are different. guardians 2r38c8

[totuan]

just meant that i was being picky while making the puzzle lol, the main idea of it took a few minutes to set up but i spent the best of an hour or two being fussy over the resolution stages lol. looks like you got the elimination i was going for though! care to share how?

Again, thanks for your puzzle. My path’s not nice as yours and marek’s

Code: Select all

 *--------------------------------------------------------------------*
 | 2468   46     1      | 3      5      7      | 9      6-2    48     |
 | 468   *3456   9      | 2      48     1      | 7     *356   *3458   |
 | 2478   3457   34578  | 489    6      89     | 2358  %235    1      |
 |----------------------+----------------------+----------------------|
 | 3      4679   478    | 45678  148    568    | 158    79     2      |
 | 678    1      78     | 56789  2      35689  | 358    4      358    |
 | 5      479    2      | 478    148    38     | 138    79     6      |
 |----------------------+----------------------+----------------------|
 | 79     2      57     | 1      3      4      | 6      8      59     |
 | 1    *35     6       |#58     9     #258    | 4     *235    7      |
 | 49     8      345    | 56     7      256    | 235    1      359    |
 *--------------------------------------------------------------------*

Look at 3’s and 5’s on R28, if r8c46<>5 => pair(35)R28 lead to eliminate (35)r3c8.

(2)r8c8=(28-5)r8c46=(35)[r8c8=r8c2-r2c2=r2c89]-(35=2)r3c8 => r1c8<>2, stte

totuan

[jco]

I found a solution using an almost discontinuous loop as follows.

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.---------------------------------------------------------.
|B2468 C46    1     | 3      5    7     | 9    D26   48   |
|B468  c3456  9     | 2      48   1     | 7   eE356 d3458 |
|B2478  3457 b34578 | 489    6    89    | 2358 d235  1    |
|-------------------+-------------------+-----------------|
| 3     4679  478   | 45678  148  568   | 158   79   2    |
| 678   1     78    | 56789  2    35689 | 358   4    358  |
| 5     479   2     | 478    148  38    | 138   79   6    |
|-------------------+-------------------+-----------------|
| 79    2     57    | 1      3    4     | 6     8    59   |
| 1    g35    6     | 58     9    258   | 4    f235  7    |
|A49    8   ha35-4  | 56     7    256   | 235   1    359  |
'---------------------------------------------------------'

If r2c8 <> 3, we have a discontinuous loop (DL) that places 3 at r9c3:

(3)r9c3 = r3c3 - r2c2 *=* (3)r2c9 - (3)r23c8 = r8c8 - r8c2 = (3)r9c3

But the chain (4)r9c1 = r123c1 - (4=6)r1c2 - r1c8 = (6-3)r2c8 = DL

shows that (4)r9c1 == (3)r9c3, so -4 r9c3; ste

[P.O.]

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after singles:

f246+8   4×6     1       3       5       7       9      l-2+6    e+48               
c(4-68) j-3456   9       2      c(48)    1       7      k*356   kd*3-458             
g+2478   3457    34578   489     6       89      2358   l2-35     1               
 3      a4-679   478     456789  148     5689    158     579      2               
b+6789   1       78      56789   2       35689   358     4        3589             
 5       479     2       4789    148     389     138     379      6               
h+79     2      i+5-7    1       3       4       6       8        59               
 1      i+3-5    6       58      9       258     4      l2-35     7               
 49      8       345     56      7       256     235     1        359             

depth: 8  candidate: 6  from cell
(((1 2 1) (4 6)))

((6 0) (4 2 4) (4 6 7 9))
((6 0) (5 1 4) (6 7 8 9))
((4 1 1 201) ((2 1 1) (4 6 8)) ((2 5 2) (4 8)))
((4 2 12) (1 9 3) (4 8))
((8 3 10) (1 1 1) (2 4 6 8))
((2 4 10) (3 1 1) (2 4 7 8))
((7 5 10) (7 1 7) (7 9))
((3 6 103) (8 2 7) (3 5))
((3 7 1 2) ((2 8 3) (3 5 6)) ((2 9 3) (3 4 5 8)))
((6 8 152) (1 8 3) (2 6))

ste.


[shye]

Look at 3’s and 5’s on R28, if r8c46<>5 => pair(35)R28 lead to eliminate (35)r3c8.

(2)r8c8=(28-5)r8c46=(35)[r8c8=r8c2-r2c2=r2c89]-(35=2)r3c8 => r1c8<>2, stte

ooh this is lovely, the resulting 35 pair giving finned x-wings ☆ ヮ ☆
reminds me of this

[eleven]

Whichever digit appears in r8c2 is forced into r3c3 (–478r3c3) and absent from r38c8 => remote triple, –2r1c8

Thanks for this example, shye (they are rare).
I see r8c2 as the base cell with 35, and r38c8 as the target cells.
Both base digits (via r23c2 or column 3) also must be in r3c3 and therefore missing in both r38c8, forcing a pair there with 2 (and the other base digit).

[Added:] For the 35 oddagons with guardians 2r38c8 there are 2 ways, either using the strong link between r2c2 and r2c89 for both 3 and 5, or the one between r23c2 and r3c3.

[denis_berthier]

im a fan of this step in your simplest-first path: biv-chain[3]: r2c5{n8 n4} - b3n4{r2c9 r1c9} - r1n8{c9 c1} ==> r2c1≠8
an m-wing i believe?

Hi Shye,
Like you, I use a solver and can find only what it finds.
SudoRules doesn't have XYWZTUVLGBTQF wings. Bivalue-chains are so simple chains that I didn't feel useful to code zillions of special cases.
Maybe someone else more knowledgeable about these particular cases can answer whether this is (equivalent to) and m-wing.

[marek stefanik]

My path’s not nice as yours [shye's] and marek’s

I think they're more similar than you think.
In your path r3c8 and r8c2 are connected via 35r2, also creating a remote triple with r8c8.
In our paths you can find similar almost-fishes, in mine almost-kites 35r8c3\r3c8b7 and in shye's almost-ERs 35r8b1\r3c28 with the same guardians.
The only difference is that the fishes use 5r8c46 as fins, instead of 2r8c8 (externals of the 35r8c28, instead of the equivalent internal).

Thanks for this example, shye (they are rare).

Yes, very rare in computer-generated puzzles, but we've seen some great hand-crafted puzzles with them already.
Just a few examples for anyone interested:
Shye's Sparkler somehow contains two of these gems.
Jovi_al01's Cobra Roll has actually a remote quadruple.
I wonder if one day we're gonna see a remote octuple creating a single in an untouched cell.

biv-chain[3]: r2c5{n8 n4} - b3n4{r2c9 r1c9} - r1n8{c9 c1} ==> r2c1≠8
an m-wing i believe? the pair it gives in b1 is lovely

Yes, it's an m-wing.

Marek

[shye]

Like you, I use a solver and can find only what it finds.
SudoRules doesn't have XYWZTUVLGBTQF wings. Bivalue-chains are so simple chains that I didn't feel useful to code zillions of special cases.
Maybe someone else more knowledgeable about these particular cases can answer whether this is (equivalent to) and m-wing.

was asking moreso to verify that im understanding the notation right ｡•́ < •̀｡ i do agree somewhat that theres perhaps too much names for things that could be simplified, especially the wxyz-type wings that SE has, but i think the basic wings (y, w, m, l, auau) having names makes sense, especially when ALS and other extensions of them are brought in. also, i may use a solver to construct puzzles but i like to solve by hand

Thanks for this example, shye (they are rare).

my pleasure! theyre definitely very particular to set up which is probably why theyre not so common in generated puzzles, like marek mentioned

I wonder if one day we're gonna see a remote octuple creating a single in an untouched cell.

im not gonna be the one to try and set that nightmare up but id definitely be amazed to see even higher levels of these remote sets

[denis_berthier]

also, i may use a solver to construct puzzles but i like to solve by hand

I still solve some puzzles by hand, but it is mainly to preserve my ability to find my chains in a real puzzle. I think I'm really good at it, because I don't have to memorise zillions of specific rules. And that was my precise purpose in defining my chain rules:
- there should be few enough of them to make them easy to memorise; in practice, only 5 useful types of chains (with clear relations between them): bivalue-chains, z-chains, t-whips, whips, g-whips;
- they should be simple enough to be easily found on a grid; in practice, one only needs very short chains for any human solvable puzzle;
- and they should be powerful enough to solve any puzzle a human solver would consider solving; this is largely proven by my stats.

As you can see, this approach is totally opposed to exhibiting as many abstruse rules as possible.