The New Sudoku Players' Forum

by **ArkieTech** » Sat Dec 07, 2019 11:19 am

Code: Select all: *-----------* |7..|85.|96.| |...|.2.|...| |...|61.|...| |---+---+---| |.58|9..|6..| |.4.|38.|..9| |9..|..1|.8.| |---+---+---| |.29|...|...| |.73|..6|.14| |..4|...|...| *-----------* 7..85.96.....2.......61.....589..6...4.38...99....1.8..29.......73..6.14..4......

by **Cenoman** » Sat Dec 07, 2019 3:45 pm

Code: Select all: +-----------------+------------------+-----------------------+ | 7 1 2 | 8 5 4 | 9 6 3 | | 34 89 6 | 7 2 39 | 1 zB45 z58 | | 34 89 5 | 6 1 39 | 278 d27-4 278 | +-----------------+------------------+-----------------------+ | 2 5 8 | 9 4 7 | 6 3 1 | | 6 4 1 | 3 8 b25* | 257 c257 9 | | 9 3 7 | 25* 6 1 | 4 8 25* | +-----------------+------------------+-----------------------+ | 1 2 9 | 4 37 58 | 3578 B57 6 | | 58 7 3 | 25 9 6 | 258 1 4 | | 58 6 4 | 1 37 a258* | 23578 9 yA2578* | +-----------------+------------------+-----------------------+

Bivalue oddagon (25)r69, c69, b5
(8-2)r9c6 = r5c6 - r5c8 = (2)r3c8
(7)r9c9 - (7=54)r27c8
(8)r9c9 - (8=54)r2c89
=> -4 r3c8; ste

by **eleven** » Sat Dec 07, 2019 5:17 pm

Code: Select all: *----------------------------------------------------------* | 7 1 2 | 8 5 4 | 9 6 3 | | 34 89 6 | 7 2 39 | 1 45 b58 | | 34 89 5 | 6 1 39 | a278 247 278 | |---------------+------------------+-----------------------| | 2 5 8 | 9 4 7 | 6 3 1 | | 6 4 1 | 3 8 25 |ba257 257 9 | | 9 3 7 | d25 6 1 | 4 8 c25 | |---------------+------------------+-----------------------| | 1 2 9 | 4 37 58 | 3578 57 6 | | 58 7 3 | e25 9 6 | 58-2 1 4 | | 58 6 4 | 1 37 258 | 23578 9 2578 | *----------------------------------------------------------*

2r35c7 = 5r5c7|85b3p76 - (5=2)r6c9 - r6c4 = r8c4 => -2r8c7, stte

by **Mauriès Robert** » Sat Dec 07, 2019 8:09 pm

Hi,
Resolution with TDP.
With anti-track : (Th2 TDP part 1)
E = {2r35c7, 2r8c4}
P'(E) = {578r358c7, 3r7c7, 2r9c7, ...} => -2r8c7 because 2r8c7 sees 2r35c7, 2r8c4 and 2r9c7, stte.
With conjugated tracks: (Th2 TDP part 2)
P(2r5c6) = {2r5c6, 2r6c9, 2r9c7, 2r3c8, ...}
P(2r6c4) = {2r6c4, 2r8c7, 2r5c8, 2r3c9, 7r5c7, 7r3c8, ...}
=> -4r3c8, stte.
Robert

by **SpAce** » Sat Dec 07, 2019 10:46 pm

Code: Select all: .-----------.-------------------.--------------------------. | 7 1 2 | 8 5 4 | 9 6 3 | | 34 89 6 | 7 2 39 | 1 45 58 | | 34 89 5 | 6 1 39 | 278 247 b78#2 | :-----------+-------------------+--------------------------: | 2 5 8 | 9 4 7 | 6 3 1 | | 6 4 1 | 3 8 25^ | d2(5)+7^ 257 9 | | 9 3 7 | 25*^ 6 1 | 4 8 25* | :-----------+-------------------+--------------------------: | 1 2 9 | 4 37 58 | c3578 c57 6 | | 58 7 3 | d5(2)*^ 9 6 | c58-2 1 4 | | 58 6 4 | 1 37 258* | ac3578[#2] 9 b2578* | '-----------'-------------------'--------------------------'

5-link Oddagon+2 (2)r69,c49,b8 (*), Almost-RP (25)r5c76,r68c4 (^)

(2)r9c7 == (2,7)r39c9 - (7,5)b9p2147 == (5,2)r5c7,r8c4 => -2 r8c7; stte

--
Btw, I finally found proof that David used the comma notation:

David P Bird wrote:Consider the right hand pairs (15)r79c2 & (15)13c8 in the SK loop which must hold the same number of truths.
a) If they both hold no truths then (1,5)r4,5c2 & (5,1)r5,6c8 will be forced producing a contradiction for (5).
b) If they both hold two truths then (1,5)r2c4,5 & (5,1)r8c5,6 will be forced giving a second (5) contradiction.

The obvious difference is that he put a comma in the cell list as well. I do that too if it's absolutely necessary to avoid ambiguity, but otherwise I think it makes it (even) uglier.

--
Added. A bit simpler solution using the same oddagon:

Code: Select all: .-----------.------------------.--------------------------. | 7 1 2 | 8 5 4 | 9 6 3 | | 34 89 6 | 7 2 39 | 1 45 58 | | 34 89 5 | 6 1 39 | 278 247 a8[#2]-7 | :-----------+------------------+--------------------------: | 2 5 8 | 9 4 7 | 6 3 1 | | 6 4 1 | 3 8 25 | 257 257 9 | | 9 3 7 | 25* 6 1 | 4 8 25* | :-----------+------------------+--------------------------: | 1 2 9 | 4 37 58 | 3578 57 6 | | 58 7 3 | 25* 9 6 | 258 1 4 | | 58 6 4 | 1 b(37) 258* | b58(37#2) 9 b25(7)8* | '-----------'------------------'--------------------------'

(2)r3c9 == (23,7)r9c759 => -7 r3c9; stte

by **StrmCkr** » Sun Dec 08, 2019 8:22 am

Code: Select all: +-----------+---------------+----------------------+ | 7 1 2 | 8 5 4 | 9 6 3 | | 34 89 6 | 7 2 39 | 1 4-5 58 | | 34 89 5 | 6 1 39 | 278 247 8(27) | +-----------+---------------+----------------------+ | 2 5 8 | 9 4 7 | 6 3 1 | | 6 4 1 | 3 8 5(2) | 257 (257) 9 | | 9 3 7 | 25 6 1 | 4 8 5(2) | +-----------+---------------+----------------------+ | 1 2 9 | 4 37 58 | 3578 (57) 6 | | 58 7 3 | 25 9 6 | 258 1 4 | | 58 6 4 | 1 37 58(2) | 23578 9 58(27) | +-----------+---------------+----------------------+

singles to the end.
ahs ) 2,7 @ R369C9
Als ) R57C8 @ 257
Strong Link digit 2 @ R59C6
and
2 is @ R9C6 is a {restricted Common}? digit & cell between als and ahs => R2C8 <> 5
2 is in als reduces the strong link and the ahs has 1 cell left for digit 7 which restricts the als by 1 digit => R2C8 <> 5
2 is @ r5C5 then als is a naked pair => R2C8 <> 5

2 is R9C6 then the (restricted cell/digit) between the two sets has 3 affects its in either als|ahs sets or neither which is the same as the ones listed above

either way R2C8 <> 5 and singles to the end.

by **SpAce** » Sun Dec 08, 2019 10:08 am

Hi StrmCkr,

I don't understand your solution. What does the AHS do? What singles chain?

If I understood correctly, your AHS is this almost hidden pair:

(27)r39c9 = (2)r6c9

...but I don't see how it helps at all. I think you need this:

(2,7)r39c9 = (2)r69c9

...but that's not an AHS. It's an almost-almost ordered 2-tuple. With that I can see a chain like this:

(57=2)r75c8 - r5c6 = r6c4&r9c6 - r69c9 = (2,7)r39c9 - (7=5)r7c8 => -5 r2c8; stte

by **StrmCkr** » Sun Dec 08, 2019 10:29 am

Code: Select all: I don't understand your solution. What does the AHS do? What singles chain?

consider the bi local is also reduced when the als contains the 2: which means the ahs is reduced by 2 placements.

the ahs ) 2,7 @ R369C9 goes from {2 digits in 3 cells} reduced by 2 cells => 1 digit in 1 cells: 2 in R3C9 and R9C9 = 7

Code: Select all: Two [53,70] 12 Candidates, 5 Truths = {2C69 7C9 57N8} 7 Links = {2r59 57c8 3n9 2b6 7b9} 1 Elimination --> r2c8<>5

Code: Select all: (5c8) (2b6) (2r5) (7c8) (7b9) (3n9) (2r9) 5N8: 5r5c8==2r5c8A=2r5c8A=7r5c8 | | | | 7N8: 5r7c8====|======|====7r7c8B=7r7c8B | | | | 7C9: | | | 7r9c9==7r3c9 | | | | 2C9: | 2r6c9====|==================2r3c9==2r9c9 | | | 2C6: | 2r5c6=======================2r9c6 | Elim 5r2c8

alternative with pure als.

Code: Select all: +-----------+---------------+----------------------+ | 7 1 2 | 8 5 4 | 9 6 3 | | 34 89 6 | 7 2 39 | 1 4-5 58 | | 34 89 5 | 6 1 39 | 278 247 (278) | +-----------+---------------+----------------------+ | 2 5 8 | 9 4 7 | 6 3 1 | | 6 4 1 | 3 8 5(2) | 257 (257) 9 | | 9 3 7 | 25 6 1 | 4 8 (25) | +-----------+---------------+----------------------+ | 1 2 9 | 4 37 58 | 3578 (57) 6 | | 58 7 3 | 25 9 6 | 258 1 4 | | 58 6 4 | 1 37 58(2) | 23578 9 (2578) | +-----------+---------------+----------------------+

Code: Select all: Two [53,70] 16 Candidates 8 Truths = {2C69 7C9 57N8 369N9} 6 Links = {2r59 57c8 2b6 7b9} 1 Elimination --> r2c8<>5

Code: Select all: (5c8) (2b6) (2r5) (7c8) (7b9) (***) (2r9) (***) (***) (***) (***) 5N8: 5r5c8==2r5c8A=2r5c8A=7r5c8 | | | | 7N8: 5r7c8====|======|====7r7c8C=7r7c8C | | | | 7C9: | | | 7r9c9E=7r3c9 | | | | | 9N9: | | | 7r9c9E========2r9c9D=5r9c9==8r9c9 | | | | | | 2C6: | | 2r5c6=======================2r9c6 | | | 2C9: | 2r6c9B=============================2r9c9D===============2r3c9 | | | 6N9: | 2r6c9B=========================================================5r6c9 | | 1R1: ==|========================================================================== ==|========================================================================== Elim 5r2c8 =======================================================================

by **SpAce** » Sun Dec 08, 2019 11:40 am

StrmCkr wrote:the ahs ) 2,7 @ R369C9 goes from {2 digits in 3 cells} reduced by 2 cells => 1 digit in 1 cells: 2 in R3C9 and R9C9 = 7

Exactly. Which seems to me that you're not using it as a normal AHS. That would be when 2 digits in 3 cells becomes 2 digits in 2 cells. Your operation requires removing two distinct spoilers (even though they're the same digit), which I'd rather call almost-almost because it has a different result than removing just one. On the other hand, the corresponding ANS (2=587)r269c9 works as a normal almost-pattern, even though it has the same two spoiler candidates -- because together they're a single spoiler digit (which is all that counts for an ANS).

In any case, your verbose description of the logic is hard to follow. It can be expressed much less ambiguously with a short chain or a 3-way kraken. Why make it so complicated?

by **eleven** » Sun Dec 08, 2019 2:27 pm

SpAce wrote:With that I can see a chain like this:

(57=2)r75c8 - r5c6 = r6c4&r9c6 - r69c9 = (2,7)r39c9 - (7=5)r7c8 => -5 r2c8; stte

Here you use r6c4, which is not part of StrmCkr's solution.

What StrmCkr did not mention (at least in the first post) is, that the strong link for 7c9 (or the als 2578 r269c9) is also needed to get this elimination from the marked cells only.
We could write somewhat like this:
(57=2)r57c8 - 2r5c6|r6c9 = 52r59c6 & 5r6c9 - 2r69c9 = (2-7)r3c9 = 7r9c7 - (7=5)r7c9

by **SpAce** » Sun Dec 08, 2019 5:41 pm

eleven wrote:
SpAce wrote:With that I can see a chain like this:

(57=2)r75c8 - r5c6 = r6c4&r9c6 - r69c9 = (2,7)r39c9 - (7=5)r7c8 => -5 r2c8; stte

Here you use r6c4, which is not part of StrmCkr's solution.

True. Only because I chose not to write:

(57=2*)r75c8 - r5c6 = r9c6 - r9*6c9 = (2-7)r3c9 = r9c9 - (7=5)r7c8 => -5 r2c8; stte

or:

Code: Select all: (2-7)r3c9 = r9c9 - (7=5)r7c8 || (2)r6c9 - (2=75)r57c8 || (2)r9c9 - r9c6 = r5c6 - (2=75)r57c8 => -5 r2c8; stte

Both of which are strictly equivalent to this:

StrmCkr wrote:5 Truths = {2C69 7C9 57N8}
7 Links = {2r59 57c8 3n9 2b6 7b9}

Btw, that's an example of set logic that is not simple to understand. Its global rank is 2 (7-5), yet the victim is only covered by one link 5c8 (instead of three required by rank 2). Try explaining that! It requires seeing that the local rank in link 5c8 drops to zero due to two link triplets (2r5c8 and 7r7c8). I think that's a weakness of Allan Barker's system. It leaves it up to the reader to figure out why it works when triplets are used. Here's my alternative to make it explicit:

Alien ObiFish 8\10 (Rank 2): {2C69 7C9 55777N8 \ 2r59 77c8 2b6 7b9 3n9 [555c8]} => -5 r2c8

Now 5r2c8 is covered thrice, as it should in Rank 2 logic. No mixed rank logic needed.

matrix: Show

We could write somewhat like this:
(57=2)r57c8 - 2r5c6|r6c9 = 52r59c6 & 5r6c9 - 2r69c9 = (2-7)r3c9 = 7r9c7 - (7=5)r7c9

That works too, though in my view its even farther from StrmCkr's stated solution than mine, because neither 5r5c6 nor 5r6c9 is part of it (only the cells are the same). You could get rid of 5r5c6 by simplifying the blue part: 2r9c6 & 5r6c9 or (2,5)r9c6,r6c9, but I see no easy way to lose 5r6c9.

Added. The second piece of set logic makes no sense:

StrmCkr wrote:
Code: Select all
Two [53,70] 16 Candidates 8 Truths = {2C69 7C9 57N8 369N9} 6 Links = {2r59 57c8 2b6 7b9} 1 Elimination --> r2c8<>5

Note that you have more truths than links. What were you trying to do?

by **StrmCkr** » Tue Dec 10, 2019 1:25 pm

Added. The second piece of set logic makes no sense:

StrmCkr wrote:

Code: Select all
Two [53,70] 16 Candidates
8 Truths = {2C69 7C9 57N8 369N9}
6 Links = {2r59 57c8 2b6 7b9}
1 Elimination --> r2c8<>5

Note that you have more truths than links. What were you trying to do?

same cells using all digits in the cells instead : same elimination
it has missing "links" from the triples another problem with xsudoku set solver.

but we'll toss that one away and focus back on the original fun one

Code: Select all: +-----------+---------------+----------------------+ | 7 1 2 | 8 5 4 | 9 6 3 | | 34 89 6 | 7 2 39 | 1 4-5 58 | | 34 89 5 | 6 1 39 | 278 247 8(27) | +-----------+---------------+----------------------+ | 2 5 8 | 9 4 7 | 6 3 1 | | 6 4 1 | 3 8 5(2) | 257 (257) 9 | | 9 3 7 | 25 6 1 | 4 8 5(2) | +-----------+---------------+----------------------+ | 1 2 9 | 4 37 58 | 3578 (57) 6 | | 58 7 3 | 25 9 6 | 258 1 4 | | 58 6 4 | 1 37 58(2) | 23578 9 58(27) | +-----------+---------------+----------------------+

Code: Select all: Ahs 2 @ R59C5 | {cells} \ 2 AAHS 2 @ R369C9 {Cells} ALS 257 @ R87C8 | {cells} / 7 Ahs 7 @ R3C9 =>> R2C8 <> 5 singles to the end.

better?

by **SpAce** » Tue Dec 10, 2019 3:01 pm

StrmCkr wrote:
Code: Select all
Ahs 2 @ R59C5 | {cells} \ 2 AAHS 2 @ R369C9 {Cells} ALS 257 @ R87C8 | {cells} / 7 Ahs 7 @ R3C9 =>> R2C8 <> 5 singles to the end.

better?

Yes, it's better (smart to separate the 2 and 7 in c9). You should have two (overlapping) ALSs, though. In terms of linked ALS/A*HS I would see it like this:

Code: Select all: ALS (57'2)r75c8 - AAHS (2)r69'3c9 - AHS (7)r3'9c9 - ALS (7'5)r7c8 => -5 r2c8; stte \ / AHS (2)r5'9c6

Would you agree with something like that? Either way, it kind of highlights my original point that this is too complex a pattern to express reasonably in those terms. It's much simpler as a chain or a kraken.

Btw, the above has a natural mapping to my 3D notation:

Code: Select all: ALS AHS AAHS AHS ALS (57=2*)r75c8 - r5=9c6 - r9*6=3c9 - 7r3=9c9 - (7=5)r7c8 => -5 r2c8; stte

--

Btw, about the other one you tossed away (good choice)...

alternative with pure als.
...
same cells using all digits in the cells instead : same elimination
it has missing "links" from the triples another problem with xsudoku set solver.

I wouldn't blame XSudo if your chosen sets don't make sense in the first place

If you want a pure A*LS alternative then you should have only cell truths, like this:

8\10 (Mixed Rank 2/0): {579N6 57N8 269N9 \ 2r59 58c6 7c8 58c9 2b6 7b9 [5c8]} => -5 r2c8

or as an ObiFish with a fixed rank:

11\13 (Rank 2): {579N6 55777N8 269N9 \ 2r59 58c6 77c8 58c9 2b6 7b9 [555c8]} => -5 r2c8

as a memory chain:

Code: Select all: ALS ALS AALS ALS ALS (57=2*)r75c8 - (2=58)r57c6 - (5|8=2)r9c6 - (*2=587)r629c9 - (7=5)r7c8 => -5 r2c8; stte

Obviously your original mixed ALS/A*HS was more efficient.

by **StrmCkr** » Wed Dec 11, 2019 10:14 am

Code: Select all: ALS AHS AAHS AHS ALS (57=2*)r75c8 - r5=9c6 - r9*6=3c9 - 7r3=9c9 - (7=5)r7c8 => -5 r2c8; stte

works for me

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