The New Sudoku Players' Forum

by **jovi_al01** » Sun Jan 30, 2022 12:13 am

Code: Select all: +-----------------------------------+ | 4 5 . 6 . . . 8 9 | | | | 6 . 7 . . . 4 . 5 | | | | . 8 9 . . . 6 7 . | | | | 8 . . . . 9 . . 6 | | | | . . . . 1 . . . . | | | | . . . 2 . 3 7 . . | | | | . 4 5 . . 8 . 6 . | | | | 7 . 8 . . . 9 . 4 | | | | 9 6 . 7 . . . 5 . | +-----------------------------------+ 45.6...896.7...4.5.89...67.8....9..6....1.......2.37...45..8.6.7.8...9.496.7...5.

by **yzfwsf** » Sun Jan 30, 2022 12:40 am

Triplet Oddagon :b1p357,b3p159,b7p159,b9p159 => r9c9 <> 123
many singles
Skyscraper : 1 in r1c3,r7c1 connected by r17c7 => r3c1,r9c3 <> 1
stte

Website · by **denis_berthier** » Sun Jan 30, 2022 6:38 am

.

Code: Select all: Resolution state after Singles and whips[1]: +----------------+----------------+----------------+ ! 4 5 123 ! 6 237 127 ! 123 8 9 ! ! 6 123 7 ! 1389 2389 12 ! 4 123 5 ! ! 123 8 9 ! 1345 2345 1245 ! 6 7 123 ! +----------------+----------------+----------------+ ! 8 1237 1234 ! 45 457 9 ! 1235 1234 6 ! ! 235 2379 2346 ! 458 1 4567 ! 2358 2349 238 ! ! 15 19 146 ! 2 4568 3 ! 7 149 18 ! +----------------+----------------+----------------+ ! 123 4 5 ! 139 239 8 ! 123 6 7 ! ! 7 123 8 ! 135 2356 1256 ! 9 123 4 ! ! 9 6 123 ! 7 234 124 ! 1238 5 1238 ! +----------------+----------------+----------------+ 158 candidates.

We find a solution in W6, consistent with SER = 9.1.

1) Simplest first solution in W6: Show

2) Let's see if we can simplify this (other than by looking for fewer steps).
In the resolution state after Singles (and whips[1]), one can see the diagonals of 123 in the 4 corner blocks (with an additional 8 in the 4th). Moreover, digits 1,2,3 appear as givens only in the central block. This suggests using eleven's digit replacement method.
But I'll do it in a way that doesn't require introducing any symbolic variables - so that I can still use SudoRules all the way to the solution.

First step: replace any occurrence of 1, 2 or 3 (including the givens) by 123:

Code: Select all: +-------------------+-------------------+-------------------+ ! 4 5 123 ! 6 1237 1237 ! 123 8 9 ! ! 6 123 7 ! 12389 12389 123 ! 4 123 5 ! ! 123 8 9 ! 12345 12345 12345 ! 6 7 123 ! +-------------------+-------------------+-------------------+ ! 8 1237 1234 ! 45 457 9 ! 1235 1234 6 ! ! 1235 12379 12346 ! 458 123 4567 ! 12358 12349 1238 ! ! 1235 1239 12346 ! 123 4568 123 ! 7 12349 1238 ! +-------------------+-------------------+-------------------+ ! 123 4 5 ! 1239 1239 8 ! 123 6 7 ! ! 7 123 8 ! 1235 12356 12356 ! 9 123 4 ! ! 9 6 123 ! 7 1234 1234 ! 1238 5 1238 ! +-------------------+-------------------+-------------------+

Second step: in any of the 4 corner blocks, replace each of the 3 diagonal 123s by one of 1, 2, 3 (all different).
Here is what it gives for the south-west corner:

Code: Select all: +-------------------+-------------------+-------------------+ ! 4 5 123 ! 6 1237 1237 ! 123 8 9 ! ! 6 123 7 ! 12389 12389 123 ! 4 123 5 ! ! 123 8 9 ! 12345 12345 12345 ! 6 7 123 ! +-------------------+-------------------+-------------------+ ! 8 1237 1234 ! 45 457 9 ! 1235 1234 6 ! ! 1235 12379 12346 ! 458 123 4567 ! 12358 12349 1238 ! ! 1235 1239 12346 ! 123 4568 123 ! 7 12349 1238 ! +-------------------+-------------------+-------------------+ ! 1 4 5 ! 1239 1239 8 ! 123 6 7 ! ! 7 2 8 ! 1235 12356 12356 ! 9 123 4 ! ! 9 6 3 ! 7 1234 1234 ! 1238 5 1238 ! +-------------------+-------------------+-------------------+

Third step: solve as any Sukaku puzzle:

Code: Select all: (solve-sukaku-grid +-------------------+-------------------+-------------------+ ! 4 5 123 ! 6 1237 1237 ! 123 8 9 ! ! 6 123 7 ! 12389 12389 123 ! 4 123 5 ! ! 123 8 9 ! 12345 12345 12345 ! 6 7 123 ! +-------------------+-------------------+-------------------+ ! 8 1237 1234 ! 45 457 9 ! 1235 1234 6 ! ! 1235 12379 12346 ! 458 123 4567 ! 12358 12349 1238 ! ! 1235 1239 12346 ! 123 4568 123 ! 7 12349 1238 ! +-------------------+-------------------+-------------------+ ! 1 4 5 ! 1239 1239 8 ! 123 6 7 ! ! 7 2 8 ! 1235 12356 12356 ! 9 123 4 ! ! 9 6 3 ! 7 1234 1234 ! 1238 5 1238 ! +-------------------+-------------------+-------------------+ ) hidden-pairs-in-a-row: r2{n8 n9}{c4 c5} ==> r2c5≠3, r2c5≠2, r2c5≠1, r2c4≠3, r2c4≠2, r2c4≠1 z-chain[3]: r5n9{c8 c2} - c2n7{r5 r4} - r4n3{c2 .} ==> r5c8≠3 z-chain[3]: b5n3{r6c6 r5c5} - c9n3{r5 r3} - c1n3{r3 .} ==> r6c2≠3 z-chain[3]: b5n3{r6c6 r5c5} - c1n3{r5 r3} - c9n3{r3 .} ==> r6c8≠3 z-chain[3]: r4n3{c8 c2} - c1n3{r6 r3} - c9n3{r3 .} ==> r5c7≠3 t-whip[3]: c8n9{r5 r6} - r6c2{n9 n1} - b5n1{r6c6 .} ==> r5c8≠1 whip[3]: c1n3{r6 r3} - c9n3{r3 r6} - b5n3{r6c4 .} ==> r5c2≠3 biv-chain[4]: b1n1{r1c3 r2c2} - c2n3{r2 r4} - b4n7{r4c2 r5c2} - c6n7{r5 r1} ==> r1c6≠1 z-chain[4]: c2n3{r4 r2} - b1n1{r2c2 r1c3} - r1c7{n1 n2} - r7c7{n2 .} ==> r4c7≠3 biv-chain[3]: r2c2{n1 n3} - r4n3{c2 c8} - r8c8{n3 n1} ==> r2c8≠1 biv-chain[3]: r2c8{n2 n3} - r2c2{n3 n1} - r1c3{n1 n2} ==> r1c7≠2 biv-chain[3]: r1c7{n3 n1} - r1c3{n1 n2} - r3c1{n2 n3} ==> r3c9≠3 whip[1]: c9n3{r6 .} ==> r4c8≠3 singles ==> r4c2=3, r2c2=1, r1c3=2, r3c1=3, r6c2=9, r5c2=7, r4c5=7, r1c6=7, r5c8=9 whip[1]: r4n2{c8 .} ==> r5c7≠2, r5c9≠2, r6c8≠2, r6c9≠2 finned-x-wing-in-rows: n3{r1 r7}{c7 c5} ==> r8c5≠3 biv-chain[2]: c9n2{r9 r3} - r2n2{c8 c6} ==> r9c6≠2 biv-chain[3]: b3n1{r1c7 r3c9} - c9n2{r3 r9} - b9n8{r9c9 r9c7} ==> r9c7≠1 biv-chain[3]: c8n3{r2 r8} - b9n1{r8c8 r9c9} - c9n2{r9 r3} ==> r2c8≠2 stte

4th step: if the digits 123 are not correctly placed in the given places (central block), make the relevant permutation of 1, 2, 3 everywhere (not necessary here). I chose this corner block for two reasons:
- I get the correct result without needing any permutation;
- it has the lowest rating (Z4, same as northeast block; northwest block is in Z5)
Notice that one may not do this in the SE block, because r9c9 could be 8 (which it is indeed).

Finally, the method allowed to significantly decrease the difficulty (from W6 to Z4).

by **champagne** » Sun Jan 30, 2022 7:45 am

yzfwsf wrote:Triplet Oddagon :b1p357,b3p159,b7p159,b9p159 => r9c9 <> 123

Hi yzfwsf
could you tell more about this, Including the way your solver came to it

by **champagne** » Sun Jan 30, 2022 10:30 am

champagne wrote:
yzfwsf wrote:Triplet Oddagon :b1p357,b3p159,b7p159,b9p159 => r9c9 <> 123

Hi yzfwsf
could you tell more about this, Including the way your solver came to it

As I like very much such findings compared to long boring sequences of chains, I investigated a little more.

No doubt and confirmed using XSUDO, the four box are leading to r9c9=8.

As the logic in Xsudo is far from being rank 0, I tried to figure out what could be the proof.

First of all, due to the high symmetry in the four box, we can apply any permutaion of the digit 123 in the main diagonal box 9 and show that this is not valid;
We then reach something as

Code: Select all: ! 4 5 123 !! 12 8 9 ! ! 6 123 7 !! 4 13 5 ! ! 123 8 9 !! 6 7 23 ! +-------------------++-------------------+ ! 12 4 5 !! 3 6 7 ! ! 7 13 8 ! 9 2 4 ! ! 9 6 23 !! 8 5 1 !

can not be valid

to stay in symmetric mode, then we can show that each of the 3 possibilities in r2c2 fails.

Nice shot yzfwsf, and again, How did you get it??

by **yzfwsf** » Sun Jan 30, 2022 11:38 am

champagne wrote:Nice shot yzfwsf, and again, How did you get it??

Actually, my solver doesn't implement this technique yet.
Since this technique is easily discovered by a manual solver, I haven't implemented it yet.

The technique origin of this is here：
http://forum.enjoysudoku.com/post312166.html?hilit=triplet%20oddagon#p312166

by **StrmCkr** » Sun Jan 30, 2022 1:05 pm

hey champ

that's the 3 digit 3 overlapping "sword fish" pattern that results in 6 solutions when its present in full
the cell that completes the pattern for the section if true leaves a grid with 6 solutions~.

R9C9 = 8

Code: Select all: +---------------------+------------------+---------------------+ | 4 5 (123) | 6 237 127 | (123) 8 9 | | 6 (123) 7 | 1389 2389 12 | 4 (123) 5 | | (123) 8 9 | 1345 2345 1245 | 6 7 (123) | +---------------------+------------------+---------------------+ | 8 1237 1234 | 45 457 9 | 1235 1234 6 | | 235 2379 2346 | 458 1 4567 | 2358 2349 238 | | 15 19 146 | 2 4568 3 | 7 149 18 | +---------------------+------------------+---------------------+ | (123) 4 5 | 139 239 8 | (123) 6 1237 | | 7 (123) 8 | 135 2356 1256 | 9 (123) 4 | | 9 6 (123) | 7 234 124 | 1238 5 8-123 | +---------------------+------------------+---------------------+

one of the key properties i used to build my "stormdoku" puzzles years ago but i could never really explain it in a coherent context... still cant really.

mostly i used this form but it also applies to the above

Code: Select all: +-----------+-----------+-----------+ | . . . | . . . | . . . | | . 123 . | . 123 . | . 123 . | | . . . | . . . | . . . | +-----------+-----------+-----------+ | . . . | . . . | . . . | | . 123 . | . 123 . | . 123 . | | . . . | . . . | . . . | +-----------+-----------+-----------+ | . . . | . . . | . . . | | . 123 . | . 123 . | . 123 . | | . . . | . . . | . . . | +-----------+-----------+-----------+

for this problem i draw three squares:
i call it the
9 - 4- 1 reduction ( 3^3) > (3+1) >( 1)

Code: Select all: x | | | | x | | | | x |

the first digit placed regardless of where leaves 4 valid choices and then 1
the 2nd digit placed lets call it J has 4 spots left of which 2 leaves no spots for the last digit
like this:

Code: Select all: x | | J | | x | | j | | x |

so J actually has only 1 arrangement based on x placed (3 choice with 1 out come)
1 - 1 - 1

Code: Select all: x | | J | J | x | | | J | x |

which leaves exactly 1 spot for each of z candidates
1

ie muti solution state...

by **champagne** » Sun Jan 30, 2022 2:17 pm

This "deadly pattern", as the SK loop, has heavy constraints.
A kind of challenge to beat, is there any very high rating (say 11.x) having this property??

by **999_Springs** » Sun Jan 30, 2022 2:54 pm

yes, it features a lot in mith's 11+ puzzles with high clue counts posted in the hardest sudokus thread and elsewhere. i remember that almost every one of these (non minimal) puzzles in the "maximum clues per se rating" thread has a tri value oddagon, but it could be the case that those puzzles are very "close" to each other. it can help for manual solving, such as in this SE 11.7 puzzle

edit: strmckr, the point is that the pattern if it is completed would have 0 solutions, not 6 solutions - it behaves like an oddagon and not a UR, so does not rely on uniqueness

by **StrmCkr** » Sun Jan 30, 2022 3:21 pm

edit: strmckr, the point is that the pattern if it is completed would have 0 solutions, not 6 solutions - it behaves like an oddagon and not a UR, so does not rely on uniqueness

thanks i'll look at it again.. mine would needs R9C7 and R7C9 as the parts that interlock for the full 3 square loop.

with R7C9 solving as a single the multi state is avoided

However the pattern it's self is still valid and is reduced to 1 solution instead of 6 when removing 1 cell thus the others are locked. R9c7 <>8 thus r9c9 =8 ... Probably too presumptive for mosts tastes.

the way the last box loops its self back on to different row/cols @ R7C7 results in a more complicated internal knot {a zero state}
proving R9C9 eliminatee 1,2,3

by **champagne** » Sun Jan 30, 2022 4:10 pm

999_Springs wrote:yes, it features a lot in mith's 11+ puzzles with high clue counts posted in the hardest sudokus thread and elsewhere. i remember that almost every one of these (non minimal) puzzles in the "maximum clues per se rating" thread has a tri value oddagon, but it could be the case that those puzzles are very "close" to each other. it can help for manual solving, such as in this SE 11.7 puzzle

so it should be added to the list of exotic patterns, solving in a nice way puzzles classified as extremely hard;
BTW, I am to day less involved in this field and I completely missed this :oops:

by **yzfwsf** » Sun Jan 30, 2022 6:00 pm

champagne wrote:Nice shot yzfwsf, and again, How did you get it??

Choose three cells from each of the four boxes that make up the rectangle, and the three cells are located
in different mini rows and different mini columns.
At least 2 of the three cells in each box will be adjacent, and the line segment connecting them is either parallel to the diagonal or parallel to the anti-diagonal, so they are divided into 2 categories, if these 2 categories form 1+3 or 3+ 1, then there is no solution Triplet Oddagon structure.

by **champagne** » Sun Jan 30, 2022 7:52 pm

yzfwsf wrote:
champagne wrote:Nice shot yzfwsf, and again, How did you get it??

Choose three cells from each of the four boxes that make up the rectangle, and the three cells are located
in different mini rows and different mini columns.
At least 2 of the three cells in each box will be adjacent, and the line segment connecting them is either parallel to the diagonal or parallel to the anti-diagonal, so they are divided into 2 categories, if these 2 categories form 1+3 or 3+ 1, then there is no solution Triplet Oddagon structure.

Hi yzfwsf,
Some difficulties to follow the wording, but no problem with the proof after what I wrote earlier.
Anyway, once the proof done, this is a pattern to use as you did it, my problem was that I did not see it earlier.
As you say, it is so easy to spot (surely easier than the sk loop) that any player aware of it will see it.
Applying this "exotic pattern" we end with an easy puzzle. This will be likely the case for many of the very hard puzzles having this pattern.

by **mith** » Wed Feb 02, 2022 12:14 am

rangsk and I touched on this in our youtube series, but later rangsk did a follow-up on a cleaner way to see this involving parity:

https://www.youtube.com/watch?v=V7RC1hJ8vZ8

For the 11+ puzzles I've looked at, the simplest application (such as here ruling out 123 from r9c9) is usually not enough to break open the puzzle, but there are more complex parity arguments that can be used. It can be generalized to determining whether a graph of the relevant cells is 3-colorable or not (and in some cases, the triples don't span all three rows and columns of a box, and the parity argument discussed in the video may not apply).

shye's Patto Patto is another example of a technique can be viewed in the same way - the relevant cells are not 3-colorable.

by **StrmCkr** » Wed Feb 02, 2022 2:43 am

Code: Select all: +----------------------+-----------------------+-----------------------+ | (147) 2 3 | (17) 6 5 | (47-1) 8 9 | | 9 167 18 | 278-1 137 4 | 1267 367 5 | | 5 1467 148 | 9 37-1 278-1 | 12467 3467 (17-4) | +----------------------+-----------------------+-----------------------+ | 6 457 2459 | 3 47 27 | 579-4 1 8 | | 3 8 14 | 5 9 17 | 67-4 467 2 | | (147) 157-4 1259-4 | (+2-147) 8 6 | 3 579-4 (47) | +----------------------+-----------------------+-----------------------+ | 2 3 145 | 478-1 147 1789 | 14579 4579 6 | | 8 145 7 | 46-1 2 19 | 1459 459 3 | | (14) 9 6 | (147) 5 3 | 8 2 (147) | +----------------------+-----------------------+-----------------------+

same thing... with box 3 being a weak link for +1 cell instead of r1c9 {which matches my original pattern in full} => R6C5 = 2 or we have 6 solutions.

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