The New Sudoku Players' Forum

Website · by **denis_berthier** » Wed Jan 25, 2023 6:56 am

.
It is quite difficult to find a puzzle that:
- cannot be solved only with whips + Tridagon ORk-chains, all of length ≤ 8,
- can be solved if one adds eleven's pattern #97 (ex #37) in 15 cells and associated El97 ORk-chains,
- doesn't rely on the application of the elementary tridaagon elimination rule (only 1 guardian).

This is an example (unfortunately, the number of tridagon guardians quickly falls to 2). I wonder if RT can still simplify the solution.

Code: Select all: +-------+-------+-------+ ! . . . ! . 5 6 ! . . . ! ! . 5 7 ! 1 8 . ! . 3 . ! ! 6 8 . ! 7 . 3 ! 1 5 . ! +-------+-------+-------+ ! . 1 6 ! . 7 5 ! . . . ! ! . . 8 ! . . . ! . . 1 ! ! . 3 . ! 8 . 1 ! . . . ! +-------+-------+-------+ ! . . . ! . . . ! 9 . . ! ! . 7 . ! . . . ! 4 6 . ! ! . . . ! 5 . 7 ! . 1 8 ! +-------+-------+-------+ ....56....5718..3.68.7.315..16.75.....8.....1.3.8.1.........9...7....46....5.7.18;3275;274896 SER = 10.4

Code: Select all: Resolution state after Singles and whips[1]: +----------------------+----------------------+----------------------+ ! 12349 249 12349 ! 249 5 6 ! 278 24789 2479 ! ! 249 5 7 ! 1 8 249 ! 26 3 2469 ! ! 6 8 249 ! 7 249 3 ! 1 5 249 ! +----------------------+----------------------+----------------------+ ! 249 1 6 ! 2349 7 5 ! 238 2489 2349 ! ! 24579 249 8 ! 23469 23469 249 ! 23567 2479 1 ! ! 24579 3 2459 ! 8 2469 1 ! 2567 2479 24679 ! +----------------------+----------------------+----------------------+ ! 123458 246 12345 ! 2346 12346 248 ! 9 27 2357 ! ! 123589 7 12359 ! 239 1239 289 ! 4 6 235 ! ! 2349 2469 2349 ! 5 23469 7 ! 23 1 8 ! +----------------------+----------------------+----------------------+ 198 candidates.

by **totuan** » Wed Jan 25, 2023 8:58 am

Code: Select all: *-----------------------------------------------------------------------------* | 13 249 13 | 249 5 6 | 278 24789 2479 | | 249 5 7 | 1 8 249 | 26 3 2469 | | 6 8 249 | 7 249 3 | 1 5 249 | |-------------------------+-------------------------+-------------------------| | 249 1 6 | 2349 7 5 | 238 2489 2349 | |a24579 249 8 | 23469 23469 249 |a2567-3 b2479 1 | | 24579 3 2459 | 8 2469 1 | 2567 2479 24679 | |-------------------------+-------------------------+-------------------------| | 123458 246 12345 | 2346 12346 248 | 9 c27 2357 | | 123589 7 12359 | 239 1239 289 | 4 6 235 | | 2349 2469 2349 | 5 23469 7 |d23 1 8 | *-----------------------------------------------------------------------------*

My path for this one:
01: (57)r5c17=(7)r5c8-(7=2)r7c8-(2=3)r9c7 => r5c7<>3

Code: Select all: *-----------------------------------------------------------------------------* | 13 *249 13 |*249 5 6 | 278 *24789 2479 | |*249 5 7 | 1 8 *249 | 26 3 2469 | | 6 8 *249 | 7 *249 3 | 1 5 249 | |-------------------------+-------------------------+-------------------------| |*249 1 6 |*249 7 5 | 238 2489 *2349 | | 24579 *249 8 | 23469 23469 *249 | 2567 *2479 1 | | 24579 3 249-5 | 8 *2469 1 | 2567 *2479 24679 | |-------------------------+-------------------------+-------------------------| | 123458 246 12345 | 2346 12346 248 | 9 27 2357 | | 123589 7 12359 | 239 1239 289 | 4 6 235 | | 2349 2469 2349 | 5 23469 7 | 23 1 8 | *-----------------------------------------------------------------------------*

For this one, need simpler “impossible pattern” than eleven’s IP 15-cells.
Impossibe pattern (249) * marked cells => (6)r6c5=(7)r156c8=(8)r1c8=(3)r4c9

Hidden Text: Show

Code: Select all: *-----------------------------------------------------------* | . 249 . | 249 . . | . A249 . | | 249 . . | . . 249 | . . . | | . . 249 | . 249 . | . . . | |-------------------+-------------------+-------------------| | 249 . . | 249 . . | . . 249 | | . 249 . | . . 249 | . 249 . | | . . . | . 249 . | . 249 . | |-------------------+-------------------+-------------------| | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | *-----------------------------------------------------------* A=(2|4|9) => impossible Prove: A=2 => *-----------------------------------------------------------* | . *49 . |*49 . . | . 2 . | | 249 . . | . . 249 | . . . | | . . 249 | . 249 . | . . . | |-------------------+-------------------+-------------------| |*49 . . |*49 . . | . . 2 | | . *249 . | . . 249 | . 49 . | | . . . | . 249 . | . 49 . | |-------------------+-------------------+-------------------| | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | *-----------------------------------------------------------* r1c8=2 => r4c9=2 => oddagon(49) * marked cells => r5c2=2 => r6c5=2 => r2c6=2 => r3c3=2 => *-----------------------------------------------------------* | . *49 . |*49 . . | . 2 . | |*49 . . | . . 2 | . . . | | . . 2 | . 49 . | . . . | |-------------------+-------------------+-------------------| |*49 . . |*49 . . | . . 2 | | . 2 . | . . 49 | . 49 . | | . . . | . 2 . | . 49 . | |-------------------+-------------------+-------------------| | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | *-----------------------------------------------------------* Oddagon (49) * marked cells => impossible, the same for A=(4|9)

02: Present as diagram: => r6c3<>5

Code: Select all: (6)r6c5-r6c79=(6-5)r5c7=r6c7* || (8)r1c8-r1c7=(8-3)r4c7=r9c7--- || | (7)r156c8-(7=2)r7c8-(2=3)r9c7--(3=249)r249c1-(249=57)r56c1* || | (3)r4c9-r78c9=r9c7------------

Code: Select all: *--------------------------------------------------------------------* | 13 *249 13 |*249 5 6 | 278 24789 2479 | |*249 5 7 | 1 8 *249 | 26 3 2469 | | 6 8 *249 | 7 *249 3 | 1 5 249 | |----------------------+----------------------+----------------------| |*249 1 6 |*249 7 5 | 238 2489 2349 | | 57 *249 8 | 23469 23469 *249 | 2567 2479 1 | | 57 3 *249 | 8 *249+6 1 | 2567 2479 24679 | |----------------------+----------------------+----------------------| | 12348 246 12345 | 2346 12346 248 | 9 27 2357 | | 12389 7 12359 | 239 1239 289 | 4 6 235 | | 2349 2469 2349 | 5 23469 7 | 23 1 8 | *--------------------------------------------------------------------*

03: Tridagon (249) * marked cells: r6c5=6, some singles

Code: Select all: *-----------------------------------------------------------* | 13 a29 13 | 49-2 5 6 | 7 8 49 | | 49 5 7 | 1 8 49 | 2 3 6 | | 6 8 249 | 7 29 3 | 1 5 49 | |-------------------+-------------------+-------------------| |c249 1 6 |d29 7 5 | 8 49 3 | | 5 b29 8 | 2349 239 249 | 6 7 1 | | 7 3 49 | 8 6 1 | 5 49 2 | |-------------------+-------------------+-------------------| | 13 4 5 | 6 13 8 | 9 2 7 | | 8 7 13 | 239 1239 29 | 4 6 5 | | 29 6 29 | 5 4 7 | 3 1 8 | *-----------------------------------------------------------*

04: 2’s abcd => r1c4<>2, stte

Thanks for the puzzle!
totuan

Website · by **denis_berthier** » Wed Jan 25, 2023 9:51 am

totuan wrote:02: Present as diagram: => r6c3<>5

With this elimination (here simulated at the start of resolution), the solution becomes almost trivial:

Code: Select all: hidden-pairs-in-a-column: c1{n5 n7}{r5 r6} ==> r6c1≠9, r6c1≠4, r6c1≠2, r5c1≠9, r5c1≠4, r5c1≠2 hidden-pairs-in-a-row: r1{n1 n3}{c1 c3} ==> r1c3≠9, r1c3≠4, r1c3≠2, r1c1≠9, r1c1≠4, r1c1≠2 +-------------------+-------------------+-------------------+ ! 13 249 13 ! 249 5 6 ! 278 24789 2479 ! ! 249 5 7 ! 1 8 249 ! 26 3 2469 ! ! 6 8 249 ! 7 249 3 ! 1 5 249 ! +-------------------+-------------------+-------------------+ ! 249 1 6 ! 2349 7 5 ! 238 2489 2349 ! ! 57 249 8 ! 23469 23469 249 ! 23567 2479 1 ! ! 57 3 249 ! 8 2469 1 ! 2567 2479 24679 ! +-------------------+-------------------+-------------------+ ! 12348 246 12345 ! 2346 12346 248 ! 9 27 2357 ! ! 12389 7 12359 ! 239 1239 289 ! 4 6 235 ! ! 2349 2469 2349 ! 5 23469 7 ! 23 1 8 ! +-------------------+-------------------+-------------------+ OR2-anti-tridagon[12] for digits 4, 9 and 2 in blocks: b1, with cells: r1c2, r2c1, r3c3 b2, with cells: r1c4, r2c6, r3c5 b4, with cells: r5c2, r4c1, r6c3 b5, with cells: r5c6, r4c4, r6c5 with 2 guardians: n3r4c4 n6r6c5 Trid-OR2-whip[2]: OR2{{n3r4c4 | n6r6c5}} - b8n6{r7c5 .} ==> r7c4≠3 +-------------------+-------------------+-------------------+ ! 13 249 13 ! 249 5 6 ! 278 24789 2479 ! ! 249 5 7 ! 1 8 249 ! 26 3 2469 ! ! 6 8 249 ! 7 249 3 ! 1 5 249 ! +-------------------+-------------------+-------------------+ ! 249 1 6 ! 2349 7 5 ! 238 2489 2349 ! ! 57 249 8 ! 23469 23469 249 ! 23567 2479 1 ! ! 57 3 249 ! 8 2469 1 ! 2567 2479 24679 ! +-------------------+-------------------+-------------------+ ! 12348 246 12345 ! 246 12346 248 ! 9 27 2357 ! ! 12389 7 12359 ! 239 1239 289 ! 4 6 235 ! ! 2349 2469 2349 ! 5 23469 7 ! 23 1 8 ! +-------------------+-------------------+-------------------+ Trid-OR2-whip[3]: OR2{{n6r6c5 | n3r4c4}} - b6n3{r4c7 r5c7} - c7n5{r5 .} ==> r6c7≠6 biv-chain[3]: b6n6{r6c9 r5c7} - r5n5{c7 c1} - b4n7{r5c1 r6c1} ==> r6c9≠7 Trid-OR2-whip[3]: b6n6{r5c7 r6c9} - OR2{{n6r6c5 | n3r4c4}} - b6n3{r4c7 .} ==> r5c7≠7 Trid-OR2-whip[3]: b6n6{r5c7 r6c9} - OR2{{n6r6c5 | n3r4c4}} - b6n3{r4c7 .} ==> r5c7≠5 singles ==> r6c7=5, r6c1=7, r5c1=5, r5c8=7, r7c8=2, r9c7=3, r8c9=5, r7c9=7, r1c7=7, r1c8=8, r4c7=8, r7c3=5, r4c9=3 At least one candidate of a previous Trid-OR2-relation has just been eliminated. There remains a Trid-OR1-relation between candidates: n6r6c5 Trid-ORk-relation with only one candidate => r6c5=6 singles ==> r7c4=6, r7c2=4, r7c6=8, r8c1=8, r9c5=4, r9c2=6, r5c7=6, r2c7=2, r6c9=2, r2c9=6 whip[1]: r9n2{c3 .} ==> r8c3≠2 whip[1]: r9n9{c3 .} ==> r8c3≠9 whip[1]: r5n4{c6 .} ==> r4c4≠4 finned-x-wing-in-rows: n2{r1 r4}{c4 c2} ==> r5c2≠2 stte

However, this raises a general question: we now know 630 contradictory patterns not in T&E(1). Tridagon is known to be present in hundreds of thousands of puzzles and is undoubtedly worth looking for.
But how do we know which of the remaining 629 patterns are worth it?
How do you (totuan) find which of them to look for?

P.S.:I'm still interested in possible use of RT.
.

by **totuan** » Wed Jan 25, 2023 1:13 pm

denis_berthier wrote:How do you (totuan) find which of them to look for?

This is very hard question - by my too bad in English

I'll try to explain later.

denis_berthier wrote:P.S.:I'm still interested in possible use of RT.

Using RTs, explaination by words – not sure 100%

(needs someone check more, thanks):
The puzzle after eliminating 3r5c7

Code: Select all: *-----------------------------------------------------------------------------* | 13 B249 13 |B249 5 6 | 278 24789 2479 | |C249 5 7 | 1 8 C249 | 26 3 2469 | | 6 8 A249RT | 7 A249RT 3 | 1 5 #249 | |-------------------------+-------------------------+-------------------------| |C249 1 6 |B249 7 5 | 238 2489 2349 | | 24579 B249 8 | 23469 23469 C249 | 2567 2479 1 | | 24579 3 A2459RT | 8 A2469RT 1 | 2567 2479 24679 | |-------------------------+-------------------------+-------------------------| | 123458 246 12345 | 2346 12346 248 | 9 27 2357 | | 123589 7 12359 | 239 1239 289 | 4 6 235 | | 2349 2469 2349 | 5 23469 7 | 23 1 8 | *-----------------------------------------------------------------------------*

Tridagon (249) => (5)r6c3=(6)r6c5
Look at: fill r6c3=5 and r6c5=6
- (5)r6c3-r6c7=(5-6)r5c7=r6c79 => r6c5<>6
- (6)r6c5-r6c79=(6-5)r5c7=r6c7 => r6c3<>5
=> (5)r6c3 & (6)r6c5 can’t both true, only one of them true.
=> Form RT (249) at r36c35 => Whichever (2|4|9) in r6c3 or r6c5 lead to r3c9
Now, look at other RTs.
Based on the true value at 5r6c3 or 6r6c5 we have other RTs B&C that cells B&C in B4&5 have the same cells on RT at r1&2 in B1&2.

B3 => Whichever (2|4|9)r3c9 => lead to cells B&C on B1&2 => cells B&C on B4&5 is not contain that value on r3c9
=> RT (249)r3c9/r4c14 => r4c9<>249 then the puzzle is almost solved with r6c5=6

Edit:
I studied more and see that this path can work, but need more explain – relate to others cells in puzzle.

Code: Select all: *--------------------------------------------------------------------* | 13 2 13 |*49 5 6 | 78 4789 479 | |*49 5 7 | 1 8 2 | 6 3 *469 | | 6 8 *49 | 7 *49 3 | 1 5 2 | |----------------------+----------------------+----------------------| | 2 1 6 |*49 7 5 | 38 489 *349 | | 4579 49 8 | 3469 3469 49 | 2567 2479 1 | | 4579 3 459 | 8 2 1 | 567 479 4679 | |----------------------+----------------------+----------------------| | 13458 46 12345 | 2346 1346 48 | 9 27 357 | | 13589 7 12359 | 239 139 89 | 4 6 35 | | 349 469 2349 | 5 3469 7 | 23 1 8 | *--------------------------------------------------------------------* In case r3c9=2, r4c1=2, r6c5=2

r3c9=(2|4|9) =>
- If r6c3=(2|4|9) => r6c5=6 => r5c7=6 => r2c7=2 => r9c7=3 => r4c9=3
- If r6c5=(2|4|9) =>
+ If r4c1<>(2|4|9) => pair (49|29|24)r4c14 => r4c9=3
+ If r4c1=(2|4|9) => r2c6=(2|4|9) => r1c2=(2|4|9) => oddagon * marked cells above => (6)r2c9=(3)r4c9 => r4c9=3

Too complex for this path

totuan

by **marek stefanik** » Wed Jan 25, 2023 6:12 pm

Slightly easier way to get use the RT (same elim, after -3r4c4 and basics):

Code: Select all: .---------------------.-------------------.--------------------. | 13 c249 13 |c249 5 6 | 278 b24789 2479 | | 249 5 7 | 1 8 249 | 26 3 2469 | | 6 8 249 | 7 249 3 | 1 5 249 | :---------------------+-------------------+--------------------: |c249 1 6 |c249 7 5 | 238 2489 a2349 | | 24579 d249 8 | 23469 23469 249 | 2567 2479 1 | | 24579 3 2459 | 8 e2469 1 | 2567 2479 24679 | :---------------------+-------------------+--------------------: | 123458 246 12345 | 2346 12346 248 | 9 27 2357 | | 123589 7 12359 | 239 1239 289 | 4 6 235 | | 2349 2469 2349 | 5 23469 7 | 23 1 8 | '---------------------'-------------------'--------------------'

If r4c9 (a) is not 3, 3b6 takes r4c7, forcing 2r9c7.
Then the digit in r4c9 (a) takes r1c8 (b) in b3, resulting in a skyscraper in r14 (c) on the remaining digits, forcing it to take r5c2 (d) and so r6c5 (e), eliminating it from r3c35 by the RT and breaking r3, i.e. contra.
Therefore 3r4c9 and the puzzle solves with basics and direct TH+RT elims.

Marek

by **eleven** » Thu Jan 26, 2023 12:30 pm

Nice deduction.
I did not understand ".. and so r6c5 (e) despite not being the digit in r3c9 (f), i.e. contra", but i could see that the skyscraper also forces the digit to r2c1, which with r1c8 and r6c5 kills all in box 2 and row 3.

by **totuan** » Thu Jan 26, 2023 3:00 pm

eleven wrote:i could see that the skyscraper also forces the digit to r2c1, which with r1c8 and r6c5 kills all in box 2 and row 3.

Yes, that is other view

eleven wrote:Nice deduction.
I did not understand ".. and so r6c5 (e) despite not being the digit in r3c9 (f), i.e. contra"

I think that mean: whichever (2|4|9) in r6c35 must in r3c9 by RT, so when r4c9=(2|4|9) => r6c5=(2|4|9) but at that time r3c9 is not contain that digit => contra.

totuan

by **marek stefanik** » Thu Jan 26, 2023 5:09 pm

Yes, that's what I meant. I rephrased my other post.
It's interesting that you can do it without the RT, just with b2.
I wonder if B2B can often reduce the number of TH guardians just like here, it could be interesting to see which puzzles keep the most of them.

Marek

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