T&E(3) Puzzles (split from "hardest sudokus" thread)

Everything about Sudoku that doesn't fit in one of the other sections

Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

denis_berthier wrote:
mith wrote:
Code: Select all
`1.3....89.57.8.....68..2....8.....4.......3...34....123....14.....92...3....43..1;15082;482867;3`

The first has a relatively tame TOFC from the 6 guardian candidate choice which reduces it to a couple skyscrapers, but the second looks completely disgusting. Curious what your solver will make of this one, Denis.

SER = 11.7
Harder, but solved in W8+OR6W9
..............

hidden-pairs-in-a-column: c5{n1 n3}{r3 r4} ==> r4c5≠9, r4c5≠7, r4c5≠6, r4c5≠5

Code: Select all
` *-----------------------------------------------------------------------------* | 1       2-4     3       | 4567    567     567     | 257     8       9       | | 24      5       7       | 134     8       9       | 126     236     46      | | 9       6       8       | 13457   13      2       | 157     357     457     | |-------------------------+-------------------------+-------------------------| | 2567    8       12569   | 123     13      567     | 5679    4       567     | | 567     129     12569   | 12     #5679    4       | 3      #5679    8       | | 567     3       4       | 567     5679    8       |#5679    1       2       | |-------------------------+-------------------------+-------------------------| | 3       279     2569    | 8       567     1       | 4       25679   567     | | 4568    147     156     | 9       2       567     | 5678    567     3       | | 2568    279     2569    | 567     4       3       |#256789  25679   1       | *-----------------------------------------------------------------------------*`

I see that, from here can elimination 4r1c2 as below: Tridagon (567)B5689 => (9)r5c5=(9)r5c8/r69c7=(2)r9c7=(8)r9c7
Present by diagram: => r1c2<>4
Code: Select all
`(8)r9c7-r9c1=(8-4)r8c1=(4)r2c1* ||(9)r5c5-r6c5=r6c7--(9=567)r4c679-(567=2)r4c1--(2=4)r2c1* ||               |                          |(9)r5c8/r69c7-----                           | ||                                          | (2)r9c7-r12c7=r2c8---------------------------`

Thanks for the puzzle!
totuan
totuan

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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

.
It's a much longer chain.
denis_berthier
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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

I've added my results on trivalue oddagons in the second post of the thread.
mith

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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

.
I stated that all the puzzles in mith's database of 63,137 T&E(3) min-expands (http://forum.enjoysudoku.com/the-hardest-sudokus-new-thread-t6539-1304.html)
are indeed in T&E(W2, 2).

My recent calculations allow to extend the above results:
all the puzzles in mith's database of 158,276 T&E(3) min-expands (http://forum.enjoysudoku.com/t-e-3-puzzles-split-from-hardest-sudokus-thread-t40514.html)
are indeed in T&E(W2, 2).

And it is enough to prove:
All the known 9x9 sudoku puzzles in T&E(3) are indeed in T&E(W2, 2) or less.
denis_berthier
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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

Very nice result, Denis. I wonder if we will find any that go beyond T&E(W2,2).

I've been giving my computer a break from generating puzzles, but I will likely resume work in the new year. jovi_al provided a large list of puzzles generated on the trivalue oddagon pattern which I need to run a depth check on to see if any are depth 3 and need to be added.
mith

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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

mith wrote:Very nice result, Denis. I wonder if we will find any that go beyond T&E(W2,2).

Hi mith. Thanks.
The trivalue oddagon pattern requires more than T&E(W2, 2) to be proven contradictory. Even if it didn't, this wouldn't imply that puzzles with an anti-tridagon are in T&E(W2, 2). Finally, we don't know if all the puzzles in T&E(3) have an extended anti-tridagon.
That makes many reasons for being careful about our expectations.

For more than 10 years, all the known puzzles were at most in T&E(2) - indeed in T&E(B7, 1) = B7B. It was rational at that time to conjecture that all the puzzles were also at most in B7B. With the large number of puzzles in your database, it is now rational to conjecture that all the puzzles are (at most) in T&E(W2,2).

But a conjecture is a conjecture. The longer it lasts, the more interesting it is to find a counter-example.
It's great that my T&E(2) and B7B conjectures were disproved by your Loki puzzle and it's great to see that the new puzzles have led to develop a whole set of new powerful resolution rules (ORk-chains and ORk-g-chains).

I wonder about the other T&E(3) patterns that have been found. Would your techniques allow to develop similar databases concentrated on them?
denis_berthier
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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

Hi all! I've been taking a longer break from this than expected, for two reasons:

1. I started a puzzle patreon in January, and making puzzles and solve videos for that has taken up a good amount of my free time!
2. The cooler on my computer blew out, so I was without my computer for a couple weeks.

Not sure when I'll get back to running scripts, but I know that jovi_al has done a search on the trivalue oddagon pattern and generated a bunch of puzzles so I need to check those for any new depth 3 to add to the database.
mith

Posts: 950
Joined: 14 July 2020

Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

.
Hi mith,
glad to see you back here.
Take your time. With the current database, you've already given us much analysis work. We're still very far from having run out of examples with new interesting features.
What's now clear is, even though all the known puzzles in T&E(3) are indeed in T&E(W2, 2) (see previous post), that leaves a very broad spectrum of complexity, ranging from those that can be solved with simple chains after the simplest tridagon elimination (like Loki) to those that remain in T&E(2) even after applying all the known impossible patterns with long ORk-chains.
.
denis_berthier
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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

denis_berthier wrote:
mith wrote:b. The 3 digits are placeable in the three cells of each box (ruling out cases where one of the 12 cells already can't contain one of the digits, but also a case like 1 can go in either of r1c1 or r2c2, but 2 and 3 can only go in r3c3; each cell can contain a digit from 123, but not all three at the same time).

There must be some misunderstanding here. For me "The 3 digits are placeable in the three cells of each box" means that the 3 candidates are present in all the 12 cells - i.e. the pattern is non-degenerate.

What I mean is simply "there exists some way to place the 3 digits of the triple in the 3 cells of the box, for each box of the pattern".

For example:

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`,---------------------,-----------------,---------------------,| 1      248*    3    | 248* 5    6     | 279   24789   2489  || 248*   5       7    | 1    248* 9     | 26    2348    23468 || 6      9       248* | 3    7    248*  | 125   12458   12458 |:---------------------+-----------------+---------------------:| 248    2468    1    | 248  9    3     | 256   2458    7     || 5      248     9    | 6    248  7     | 3     1248    1248  || 23478  234678  2468 | 5    1    248   | 269   2489    24689 |:---------------------+-----------------+---------------------:| 23478  123478  248* | 9    6    12458*| 1257  12357   1235  || 23789* 123678  268  | 278* 238  1258  | 4     123579  12359 || 23479  12347*  5    | 247  234* 124   | 8     6       1239  |'---------------------'-----------------'---------------------'`

This is one of the min-expands from the Loki tree, with the non-degenerate TO in boxes 1245. What I've marked is a degenerate TO. It is possible to place 248 in the marked cells of box 7, when considering only those cells: for example 2r7c3, 8r8c1, 4r9c2. Likewise, it is possible to place 248 in the marked cells of box 8. However, in both boxes there are cells which are missing one of the candidates from the triple (and this will of course always be the case if the box selection is anything other than the four boxes which don't share a band/stack with the box containing those givens).

This particular degenerate TO has 5 guardian cells and 10 guardian candidates; in theory, you could have some OR-branching chain based on this (and there are some puzzles with non-degenerate TOs after basics which have more guardian cells/candidates).

The full version of the script finds all of these. (One other obvious filter that I neglected to mention - if there is a cell which can only contain digits from the triple in question, it must be part of the pattern. marek pointed out that I could also check whether placing the triple breaks other cells in the box, I will likely add this to the filter for this update.)

Code: Select all
`1.3.56....571.9...69.37......1.93..75.96.73.....51.......96..........4....5...86.;1;3;511;1;n248;b1p249+b2p159+b4p159+b5p1595;10;n248;b1p249+b2p159+b7p348+b8p3485;11;n248;b1p249+b2p159+b7p357+b8p35710;25;n239;b1p348+b3p168+b7p267+b9p1598;22;n359;b2p267+b3p159+b8p168+b9p1596;11;n248;b4p159+b5p159+b7p348+b8p3576;12;n248;b4p159+b5p159+b7p357+b8p34810;26;n289;b4p267+b6p159+b7p267+b9p35710;25;n289;b4p267+b6p267+b7p267+b9p35710;23;n249;b4p267+b6p267+b7p348+b9p24910;19;n246;b4p159+b6p159+b7p357+b9p34810;25;n289;b4p267+b6p267+b7p357+b9p26710;25;n289;b4p267+b6p168+b7p267+b9p26710;25;n279;b4p267+b6p357+b7p267+b9p15910;24;n279;b4p267+b6p357+b7p348+b9p15910;25;n289;b4p168+b6p267+b7p267+b9p2679;23;n258;b4p249+b6p159+b7p159+b9p2678;23;n456;b4p249+b6p159+b7p159+b9p3489;22;n258;b4p249+b6p267+b7p159+b9p2678;23;n456;b4p249+b6p267+b7p159+b9p34810;24;n128;b4p357+b6p159+b7p267+b9p26710;23;n128;b4p357+b6p159+b7p348+b9p26710;23;n128;b4p357+b6p267+b7p267+b9p26710;22;n128;b4p357+b6p267+b7p348+b9p26710;25;n289;b4p168+b6p168+b7p357+b9p2679;22;n258;b4p249+b6p168+b7p249+b9p26710;22;n128;b4p357+b6p168+b7p168+b9p26710;23;n128;b4p357+b6p168+b7p357+b9p26710;23;n249;b4p168+b6p168+b7p348+b9p2499;23;n245;b4p249+b6p168+b7p159+b9p2499;22;n258;b4p249+b6p168+b7p159+b9p35710;22;n124;b4p357+b6p168+b7p267+b9p24910;23;n128;b4p357+b6p168+b7p267+b9p35710;21;n124;b4p357+b6p168+b7p348+b9p24910;22;n128;b4p357+b6p168+b7p348+b9p35710;26;n289;b4p168+b6p159+b7p357+b9p35710;25;n289;b4p168+b6p267+b7p357+b9p3579;24;n245;b4p249+b6p159+b7p249+b9p2499;23;n258;b4p249+b6p159+b7p249+b9p3579;22;n258;b4p249+b6p267+b7p249+b9p35710;22;n124;b4p357+b6p159+b7p168+b9p24910;23;n128;b4p357+b6p159+b7p168+b9p35710;23;n124;b4p357+b6p159+b7p357+b9p24910;24;n128;b4p357+b6p159+b7p357+b9p35710;21;n124;b4p357+b6p267+b7p168+b9p24910;22;n128;b4p357+b6p267+b7p168+b9p35710;22;n124;b4p357+b6p267+b7p357+b9p24910;23;n128;b4p357+b6p267+b7p357+b9p3579;18;n127;b5p168+b6p357+b8p357+b9p1599;21;n235;b5p357+b6p249+b8p357+b9p1598;21;n358;b5p357+b6p249+b8p348+b9p357`
Last edited by mith on Sat Sep 02, 2023 4:21 pm, edited 1 time in total.
mith

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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

I don't understand: "each cell can contain a digit from 123, but not all three at the same time". Do you mean some cyclic condition like a 12 23 31 pattern in each block? I've tried this, but I found it difficult to manage the inter-block conditions.

What I mean here is that there could be cell A with candidates 1xxx, cell B with candidates 1xxx, and C with candidates 23xx. This would pass a filter that only checks for whether there is at least one candidate from 123 in each cell of the pattern, but it fails the filter that tries to place 123 in ABC.
mith

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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

.
Hi Mith
I think your filters amount to the following (in addition to the pattern of cells):
- either one of the 3 cells of the block is decided (say = 1) and the other two cells have the remains 23 candidates;
- or there is a cyclic 123-pattern in the 3 cells of the block: at least 12, 23, 31 candidates.
.
denis_berthier
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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

Paquita wrote:I get the impression that T&E(3) has a lot of 11.8 puzzles. In T&E(2), 11.8 is quite rare, with a lot more 11.7 and 11.6.
Now, when I find a 11.8, I am almost sure it is T&E(3) (they are all already in miths T&E(3) collection). And although I find only a few T&E(3) because I use T&E(2) seeds almost half of them is 11.8. Maybe another illustration of how the SE/PGX rating fails for T&E(3), also since I read the posts about how easy those puzzles sometimes are.

Such posts reflect a confusion between two different ways of rating puzzles. One way is the SER (that doesn't take tridagons into account). The other way uses tridagons. It's like comparing bananas and snails.

As for the SER 11.8:
- it is indeed rare in the current T&E(2) database (but remember it hasn't been searched with the same vicinity and expansion methods);
- the T&E(3) database of 847,778 minimals shows that 11.8 is very far from having any predominance in it (and it confirms an anomaly at SER 10.4 that I had already pointed out in [PBCS], plus one at SER 10.9 plus one in the 9.x range):

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`-       7 have SER 11.9-     316 have SER 11.8-  95,482 have SER 11.7-  63,706 have SER 11.6-     989 have SER 11.5-      13 have SER 11.4-     825 have SER 11.3-   5,525 have SER 11.2-  27,422 have SER 11.1-  63,306 have SER 11.0- 109,033 have SER 10.9-     369 have SER 10.8-   2,855 have SER 10.7-  51,816 have SER 10.6-  52,857 have SER 10.5- 214,743 have SER 10.4- 101,502 have SER 10.3-  35,108 have SER 10.2-   1,653 have SER 10.1-     900 have SER 10.0- 16,429 have SER 9.x-    842 have SER 8.x-    528 have SER 7.x-  1,005 have SER 6.x-     24 have SER 5.x-    523 have SER 4.x`

The small values of the rating are due to rules of uniqueness in SER.
denis_berthier
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Re: T&E(3) Puzzles (split from "hardest sudokus" thread)

Yes, in the T&E(2) database about 1 in 5000 is 11.8; in this T&E(3) collection it is about 1 in 2500. Not as much as I thought.
Paquita

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