The New Sudoku Players' Forum

by **JC Van Hay** » Sat Dec 05, 2015 5:17 pm

Here is the first(?) 9 cells rank 0 pattern at the intersection of 3 rows and 3 columns I saw here

The puzzle : ....73.6....6..4.96.....3..9.5.8.....1....5..3..4...9..8...2...4..7...3...2.....1
A double exocet solves the cells r1c4=9 and r2c8=7.

Code: Select all: +-----------------------+------------------------+-----------------------+ | 1258 245 148 | 9 7 3 | 128 6 258 | | 1258 235 138 | 6 125 158 | 4 7 9 | | 6 24579 14789 | 1258 1245 1458 | 3 1258 258 | +-----------------------+------------------------+-----------------------+ | 9 2467 5 | 12-3 8 167 | 1267 12-4 23467 | | (278) 1 467-8 | (23) 369-2 679 | 5 (248) 3467-28 | | 3 267 678 | 4 1256 1567 | 12678 9 2678 | +-----------------------+------------------------+-----------------------+ | (157) 8 3679-1 | (135) 3469-15 2 | 679 (45) 467-5 | | 4 569 169 | 7 1569 15689 | 2689 3 2568 | | (57) 3679-5 2 | (358) 3469-5 469-58 | 679-8 (458) 1 | +-----------------------+------------------------+-----------------------+

9 Truths = {579N1 579N4 579N8}
9 Links = {1r7 2r5 5r79 8r59 3c4 4c8 7c1}
15 Eliminations --> 1r7c35, 2r5c59, 5r7c59, 5r9c256, 8r5c38, 8r9c67, 3r4c4, 4r4c8

by **champagne** » Sat Dec 05, 2015 6:43 pm

JC Van Hay wrote:Here is the first(?) 9 cells rank 0 pattern at the intersection of 3 rows and 3 columns I saw here

The puzzle : ....73.6....6..4.96.....3..9.5.8.....1....5..3..4...9..8...2...4..7...3...2.....1
A double exocet solves the cells r1c4=9 and r2c8=7.

[/code]9 Truths = {579N1 579N4 579N8}
9 Links = {1r7 2r5 5r79 8r59 3c4 4c8 7c1}
15 Eliminations --> 1r7c35, 2r5c59, 5r7c59, 5r9c256, 8r5c38, 8r9c67, 3r4c4, 4r4c8

Hi,

nothing to object, remarkable as usual.

That puzzle is rated 10.3 10.3 2.6 by skfr, just below the cut off for the data base of potential hardest.

My solver finds another rank 0 logic

SLG rank 0
18 Truths = {1R68 1C14 2R68 2C14 5R68 5C14 8R68 8C14 1N9 3N9 }
18 Links = {1b578 2c9 2b45 5c9 5b578 8c9 8b48 1n1 2n1 3n4 6n7 8n7 }
19 elims 6r6c7 7r6c7 6r8c7 9r8c7 2r4c2 8r5c3 1r4c6 2r5c5 1r7c3 5r9c2 1r7c5 5r7c5 5r9c5 5r9c6 8r9c6 2r4c9 2r5c9 5r7c9 8r5c9
6r6c7 7r6c7 6r8c7 9r8c7 2r4c2 8r5c3 1r4c6 2r5c5 1r7c3 5r9c2 1r7c5 5r7c5 5r9c5 5r9c6 8r9c6 2r4c9 2r5c9 5r7c9 8r5c9

Once again, the cell rank 0 logic gives the smallest number of "Truths"

Congratulations

champagne

by **David P Bird** » Sat Dec 05, 2015 9:45 pm

JC, it's an interesting puzzle. What you've identified is a rank 0 pattern embedded in an double JExocet pattern.
However if all the JExocet eliminations are made immediately it isn't needed.

Code: Select all: *-----------------------------*-----------------------------*-----------------------------* | 1258 T 2459 1489 | 9-1258 <7> <3> | 128 b <6> 258 b | | 1258-7t 2357 1378 | <6> 125 B 158 B | <4> 7-1258 <9> | | <6> 79-25 79-18 | 1258-9t 12459 14589 | <3> 1258-7T 2578 | *-----------------------------*-----------------------------*-----------------------------* 12 | <9> 467-2 <5> | 123* <8> 67-1 | 67-12 1247* 3467-2 | 28 | 278* <1> 467-8 | 239* 369-2 679 | <5> 2478* 3467-28 | | <3> 267 678 | <4> 1256 1567 | 12678 <9> 2678 | *-----------------------------*-----------------------------*-----------------------------* 15 | 157* <8> 3679-1 | 1359* 34569-15 <2> | 679 457* 467-5 | | <4> 569 169 | <7> 1569 15689 | 2689 <3> 2568 | 58 | 57* 3679-5 <2> | 3589* 3469-5 469-58 | 679-8 4578* <1> | *-----------------------------*-----------------------------*-----------------------------* CL1 CL2 CL3 * = 'S' cells

In rows 579 the eliminations from the rank 0 pattern are all non-'S' cell instances of the base digits in their cover houses.
In row 4 the JE eliminations leave (12)r4c48 as a hidden pair.

So the hunt must be on to find your rank 0 pattern in isolation!

DPB
.

by **StrmCkr** » Fri Jul 22, 2016 10:01 am

http://forum.enjoysudoku.com/post220194.html#p220194

based on the above link for a "muti sector locked set" als-xy ring /generalized als-xy ring example
isn't it possible as the following idea: {if i'm reading/understanding it correctly}

a 2x2 muti fish has 4 digits potentially locked into 4 cells. {base x cover = intersection count ==>> # of digits locked into position }
there is exactly 9 digits known thus 5 free floating digits.

{USING THE ALS-XYRING {GENERALIZED OR NOT example}

formulate a base search sector for digts 56789 on Rows 25
{where
1 indicates a cell that contains these digits
0 indicates a cells that do not contain these digits}

Code: Select all: .1.|.1.|... .0.|.0.|... .1.|.1.|... ---------------- .1.|.1.|... .0.|.0.|... .1.|.1.|... ---------------- .1.|.1.|... .1.|.1.|... .1.|.1.|...

formulate a base search sector for digts 56789 on Cols 25

Code: Select all: ...|...|... 101|101|111 ...|...|... ---------------- ...|...|... 101|101|111 ...|...|... ---------------- ...|...|... ...|...|... ...|...|...

when
base x cover = [empty]
there is exactly zero intersections in the cover/base set containing the digit set 56789
indicating directly that there is exactly 4 cells holding digits [1234 ]
all peer cells containing digits 1234 that can see all copies of each digit [1,2,3,4] in the intersection cells <> [1,2,3,4]

by **David P Bird** » Fri Jul 22, 2016 1:01 pm

Chris, what you seem to describe (I haven't the time to study it deeply) seems to be a special case of the way I use complementary digit sets in what I call 'truth balancing'.

Here is the SK loop from the <Domino Loops thread> using (1259) for the rows and (34678) for the columns.

1.....9...3...7.4...5.....2.....6.7.....1.....4.3.8...9.......5.7.8...3...2...1..#8668;tax

Code: Select all: *-------------------------*-------------------------*-------------------------* 25 | <1> 268 47 | 2456 3468-25 2345 | <9> 568 37 | | 268 <3> 689 | 1259-6 25689 <7> | 568 <4> 168 | 19 | 47 689 <5> | 1469 3468-9 1349 | 37 168 <2> | *-------------------------*-------------------------*-------------------------* | 2358 1259-8 1389 | 259-4 2459 <6> | 23458 <7> 13489 | 259 | 3678-25 25689 3678-9 | 24579 <1> 2459 | 3468-25 25689 3468-9 | | 2567 <4> 1679 | <3> 2579 <8> | 256 1259-6 169 | *-------------------------*-------------------------*-------------------------* 12 | <9> 168 3468-1 | 12467 3467-2 1234 | 4678-2 268 <5> | | 456 <7> 146 | <8> 24569 1259-4 | 246 <3> 469 | 59 | 3468-5 568 <2> | 45679 3467-59 3459 | <1> 689 4678-9 | *-------------------------*-------------------------*-------------------------* 68 467 34 68

MS-NS:(25)r1,(19)r3,(259)r5,(12)r7,(59)r9,(68)c2,(467)c4,(34)c6,(68)c8 (20 digit instances, 20 fully covered cells)
=> 21 Elims: 25r1c5, 6r2c4, 9r3c5, 8r4c2, 4r4c4, 25r5c1, 9r5c3, 25r5c7, 9r5c9, 6r6c8, 1r7c3, 2r7c5, 2r7c7, 4r8c6, 5r9c1, 59r9c5, 9r9c9

I call the different digit sets the 'home' and 'away' sets as they are either both cover sets or both base sets <described here>

The cover sets shown above are a transformation (one of many possible) of those shown in the original post using using Obi-Wahn's fish transformations <repeated here>.

For SK loops identifying the set members to use is simple but for other puzzles I build a list of givens in each row and column. I then look for what appears to be locked sets or almost locked sets in the givens by row or by column.

Hope this helps - a real world example of your approach would be a great help in any further discussions.

David

by **champagne** » Mon Mar 09, 2020 12:38 pm

98.7.....7.6...8......5....4......3..9.6.........24..1.6.9..5......3..4......1..2;38692;GP;12_07

Comments on this puzzle extracted from the data base of potential hardest and heavily discussed in this thread pages 22-23.
the final solution for the puzzle is

985763214736412859214859367428195736391687425657324981162948573579236148843571692

Years ago, my code classified this puzzle as having a 2 rows 2 columns rank 0 logic. This should not have been. One condition to have a rank 0 logic of “free use” is to have no triple point in the truths, so, the crossing in the 2 rows 2 columns must be free, and this was not true in this sudoku.

However, as we will see, it works here despite one point 2 truths.
In this example, many solutions have been produced; I pick up part of them of general interest.

Several preliminary remarks.

remark 1: A valid SLG can not clear a candidate of the solution.

This is trivial. No logical rule can destroy the solution. If a Sets/Links group can be built, leading to eliminations, these eliminations are always candidates not belonging to the solution.

This evidence can be used in 2 ways:
to build an active SLG (by hand or with a computer), saving wrong racks,
to check without Xudo if a SLG is valid.

remark 2: Xsudo is a kind of “brute force” applied to a set/links group. As in a multi-fish analysis, the results must be used with care.
In a rank 0 logic, eliminations done within the matrix of truths should not be seen as a result of the rank 0 logic.

remark 3: in a rank 0 logic, triple points with 2 links don’t destroy the logic of the proof. This is not the case with triple point having 2 truths.
Using a rank 0 logic with such triple points need more validation.

The pm at the start is the following

Hidden Text: Show

the reduced pm for digits 1234 is the following

Hidden Text: Show

The multi floors analysis shows a high potential for eliminations in this multi floors.

I show in details several solutions. One post is open for each solution to have easier discussion on a specific one later.
we will see in next posts

mix rows+columns + added cells proposed by ronk
MSLS proposed by David
MSLS proposed by JC Van Hay
and to conclude a summary of the rank 0 logic as I see it to-day

by **champagne** » Mon Mar 09, 2020 12:38 pm

98.7.....7.6...8......5....4......3..9.6.........24..1.6.9..5......3..4......1..2;38692;GP;12_07

Hidden Text: Show

The first solution was posted by ronk

18 Truths = {123R5 1234R7 1234C4 1234C7 3N123}
18 Links = {1234r3 1n7 2n4 5n137 7n13 1b59 2b68 3b59 4b8}
19 Eliminations --> r5c3<>578, r3c68<>2, r3c69<>3, r7c13<>8, r5c1<>58, r1c7<>6, r3c8<>1, r3c9<>4, r4c5<>1, r5c7<>7, r7c3<>7, r8c6<>2, r9c5<>4,

This is typically a 2 rows 2 columns base with cells r3c123 added (containing only the digits of the multi floors);

In row 3, we have then 4 links + 3 cells truths replacing the 2 cells links r3c4 and r3c7.

The interesting point is in box 6. The cell r4c7 has the digit 2 belonging to 2 truths. In theory, it would be the same for the digit 4, but as in the row 4 the digit 4 is locked in the box 6, the digit 4 is not taken as truth in row 5.Instead, the digit 4 in the cell r5c7 is linked to 2r5c7.

At the end, 2r5c7 belongs to 2 truths and 2 links.
And the global situation is balanced giving a rank 0 logic.

if 2r5c7 is not assigned, we are left with a 18 truths 18 links situation with no triple point.
if 2r5c7 is assigned, we loose 2 truths and 2 links. The rest is a 16 truths 16 links logic with no triple point.

We see here that this “quad 2truths 2 links” works exactly as any other truth candidate for the proof allowing the cleaning of the links.

This would be valid for any number of “quad 2truths 2 links”. The door is open to other 2 rows 2 columns SLGs with more than one crossing occupied.
As far as I know, this has not been clearly identified before.

by **champagne** » Mon Mar 09, 2020 12:39 pm

98.7.....7.6...8......5....4......3..9.6.........24..1.6.9..5......3..4......1..2;38692;GP;12_07;

Hidden Text: Show

21 Truths = {1N35689 2N25689 3N123 5N5689 7N5689}
21 Links = {5r125 6r1 7r57 8r57 9r2 1c58 2c68 3c69 4c59 1234b1}
19 Eliminations --> r5c3<>578, r3c68<>2, r3c69<>3, r7c13<>8, r5c1<>58, r1c7<>6, r3c8<>1, r3c9<>4, r4c5<>1, r5c7<>7, r7c3<>7, r8c6<>2, r9c5<>4,

This SLG follows the general split rule in David MSLS
1234 links in rows
56789 links in columns

and the cells truths have
the most common form found when a 2 rows 2 column logic exists,
plus the box where are the added truths

Comparison of truths in this SLG and the previous one will help to see the point.
It seems to me that this is the right pattern to search in this situation

by **champagne** » Mon Mar 09, 2020 12:39 pm

My next step is the shorter MSLS proposed by JC Van Hay, but before, I find simpler to remind an example
recently posted by yzfwsf Sat Feb 29, 2020 12:28 am here

4...23...3..46..82...17..34....345...537...4..145..3..53..4.2.6.4.3.7...1.....4.3
after basic moves (including naked and hidden pair), we are here

Code: Select all: 4 6789 16789 |89 2 3 |1679 1569 159 3 79 1579 |4 6 59 |179 8 2 2689 2689 25689 |1 7 589 |69 3 4 ---------------------------------------------- 26789 2689 2689 |26 3 4 |5 1269 1789 2689 5 3 |7 189 12689 |689 4 89 26789 1 4 |5 89 2689 |3 269 789 ---------------------------------------------- 5 3 789 |89 4 189 |2 179 6 26 4 26 |3 159 7 |189 159 1589 1 789 789 |26 589 26 |4 579 3

here, yzfwsf shows a very simple rank0 logic on the digits 26:
truths r4c4 r8c1 26b4
links 26r3 26c4
this rank 0 logic has 2 triple point 2 links 2r4c1 and 6r4c1.

A triple point 2 links is not a problem in a rank 0 logic (see summary) but if we use Xsudo as check, Xsudo will also deliver some eliminations within the thruth matrix if they exist. Here, Xsudo kills 26 r3c1, this is the rank 0 logic, but also 26r4c1, not valid as “rank 0 effect”.

Here, the SLG is quite simple and it is easy to verify that 2r4c1 and 6r4c1 are not valid.
In a bigger SLG, the proof can be complex, as we will see in the next example.

we are now back to the puzzle 38692 and we consider the following MSLS proposed by JC Van Hay.

Code: Select all: 9 8 12345 |7 146 236 |1234 1256 345 7 12345 6 |1234 149 239 |8 1259 3459 123 1234 1234 |12348 5 23689 |1234679 12679 34679 1234 --------------------------------------------------------- 4 1257 12578 |158 1789 5789 |267 3 5678 12358 9 123578 |6 178 3578 |247 2578 4578 578 6 357 3578 |358 2 4 |79 5789 1 --------------------------------------------------------- 1238 6 123478 |9 478 278 |5 178 378 78 1258 1257 125789 |258 3 25678 |1679 4 6789 358 3457 345789 |458 4678 1 |3679 6789 2 58 679

22 Truths = {3N1 3N2 3N3 34689N4 57N5 57N6 345689N7 57N8 57N9}
22 Links = {1234r3 578r5 78r7 58c4 679c7 1b59 2b68 3b59 4b68}
18 Eliminations --> r5c3<>578, r3c68<>2, r3c69<>3, r7c13<>8, r5c1<>58, r3c8<>1, r3c9<>4, r4c5<>1, r5c7<>7, r7c3<>7, r8c6<>2, r9c5<>4

The MSLS of David had 21 cells, this one 22, but this is a side point.

In this MSLS, the links structure is relatively complex. for the digits 1234 some links are in rows, some in boxes, and for the digits 56789, they are partly in rows, partly in columns with a triple point 2 links in 7r5c7.

All eliminations given by xsudo are derived from the rank 0 logic, except for 7r5c7.

7r5c7 can not be considered here. In the small SLG seen before, it was easy to verify Xsudo clearing, here it is complex.

by **champagne** » Mon Mar 09, 2020 12:39 pm

Recent events and my work on this topic push me to clarify several points.

First of all, in the so called “rank 0 logic”,
we are looking for a {truths/links} group where,
if truths are assigned, all links are assigned.

truths must have one candidate and only one candidate assigned
links can only have one candidate assigned in the group when truths are assigned.
Each candidate of a truth must be “linked”.

When this is true, all candidates of the links not belonging to a truth can be cleared.

In the basic case, truths and links are taken in
the 81 cell sets
the 9*27 digit+row;column;box sets

This is the field studied by Allan Barker (reference in post 1) and processed in Xsudo, the program released by him.

In this field, if we can build a SLG without candidates linked to 2 truths nor 2 links and with a number of links equal to the number of truths, we have the condition to clear links. this is the typical rank 0 logic.

The proof is quite simple .
Each assignment in a truth kills one link and all candidates belonging to the truth or the link.
We are left with n-1 truths and n-1 links having the same property. By recurrence, when all truths are assigned, all links are assigned.

As this was the first clear case of the condition (if truths are assigned, all links are assigned) searched, the logic has been called “rank 0 logic”. No better name has been proposed.

In fact, as seen in various examples above, The rule “as many truths as links and no triple point” is too restrictive.

triple points in links usually don’t change the logic.We could face a problem if a link fully covers a truth?

For the same reason, less links than truths is acceptable.

a “truth/link quad” 2 truths and 2 links for the same candidate is valid if the number of truths is equal to the number of links. The proof is the same except that assigning a quad you kill 2 truths and 2 links.

These possibilities have been used, as we could see in the 38692 case, by several forum members relying on Xsudo to check whether a SLG was active.

However, many examples have a rank >0. They are out of the scope of the rank 0 logic.

So far, only rank 0 described here and rank <0 (surely exists, but not seen by me so far) can justify eliminations in the links of candidates not in the truths.

Note : in a rank<0 logic, we must have triple points 2 links, but as we have seen, this is not a problem. We surely have also some potential for eliminations in the truths to ignore.

by **champagne** » Tue Mar 10, 2020 3:24 am

This is an open point

Can we with a sudoku produce a truths/links logic of negative rank.
I have doubts for several reasons

AFAIK, nobody produced one so far
Trying to build simple examples, I get not valid patterns as this one

Code: Select all: .2.2 .22 22. 22.

this can produce a 3truths(columns) 2 links (boxes) logic but this is not possible in a sudoku.

If a negative rank is not possible, then, in a rank 0 with a triple point 2 links, the triple point can not be assigned, the remaining logic would have a negative rank.
And this is verifed in both examples above.

EDIT 1:
to reinforce my suspicion,
let us assume that we have a 15 truths 14 links logic.
we can assign the right candidate in truths having no triple point. We then have a similar situation with less truths. At the end, we should find a pattern with 2 truths and one link. This should not happen in a sudoku
EDIT 2:
I have added in the next page a restriction to cover one simple case of such a logic

by **Leren** » Tue Mar 10, 2020 6:05 am

Code: Select all: *--------------------------------------------------------------------------------* | 9 8 12345 | 7 146 236 | 1234-6 1256 3456 | | 7 12345 6 | 1234 149 239 | 8 1259 3459 | |*123 *1234 *1234 |*12348 5 689-23 |*1234679 679-12 679-34 | |--------------------------+--------------------------+--------------------------| | 4 *1257 *12578 |*158 789-1 5789 |*267 3 5678 | | 123-58 9 123-578 | 6 178 3578 | 24-7 2578 4578 | | 6 *357 *3578 |*358 2 4 |*79 5789 1 | |--------------------------+--------------------------+--------------------------| | 123-8 6 1234-78 | 9 478 278 | 5 178 378 | |*1258 *1257 *125789 |*258 3 5678-2 |*1679 4 6789 | |*358 *3457 *345789 |*458 678-4 1 |*3679 6789 2 | *--------------------------------------------------------------------------------*

23 Truths = {389N12347, 46N2347}
23 Links = {1234r3 12r4 12r8 34r9 & 3r6 ; 8c1 7c2 789c3 58c4 679c7 ; 5b4 5b7 }
19 Eliminations --> r1c7 <> 6, r5c3<>578, r3c68<>2, r3c69<>3, r7c13<>8, r5c1<>58, r3c8<>1, r3c9<>4, r4c5<>1, r5c7<>7, r7c3<>7, r8c6<>2, r9c5<>4

Hi Champagne, with some spare time on my hands I thought I'd try my old MSLS code on that puzzle 38692 and see what came up. What did was the above, which had 23 links and one extra elimination r1c7 <> 6, (which is also missing from your PM).
As I freely admit that I never understood triple points, all my MSLS should be pure Rank 0, although as you say, the link structure is quite complex.

Leren

by **champagne** » Tue Mar 10, 2020 7:51 am

Hi Leren,

So many variants have been given pages 23-24 that I don’t know if this one has been posted.
Seen by me, the link structure is not complex, but the cells thruths distribution is complex.

In fact, I wanted to revise my code in this field. As a first step, I fly over these 7 years old discussions to start with the best possible set of rules and patterns.

My main targets remain

identification of puzzles having a big rank 0 logic
and later use of it in the solver

so, basically, the first rank 0 logic will stop the process.

As you, I had a poor idea of the triple point properties. Although, it seems that in 2013 the properties of the “quad 2 truths 2 links” had been clearly seen. (see page 29)

I am now nearly sure that the conjecture (no logic of negative rank in a valid puzzle) is correct, so I intend to accept “triple point links” and “quad 2 links 2 truths” in the new code and to clear the triple point 2 links.
I have still 25 pages to digest. After only I’ll restart the coding

To-day, the last code can only cover the classical cases

by **Leren** » Tue Mar 10, 2020 9:06 am

Champagne wrote: ... the cells thruths distribution is complex.

Yes, it's been 7 years since I wrote my version of MSLS. It seems that I must have avoided the triple point issue by juggling the Truth cells to match the links.

So, for example, I managed to get yzfwsf's eliminations but my MSLS had 14 Truths/links, so I guess I've traded off one complication for another.

Anyway, best of luck in your future endeavours on this subject. Leren

by **creint** » Tue Mar 10, 2020 7:34 pm

champagne, about your post "p38692 MSLS david"

The truth cells in box 1 are not contributing to exclusions and can be ignored. So you are left with a normal MSLS. If adding 3 cell truths from r357c1 you can get the same eliminations with Xsudo as with post "p38692 2 rows 2 columns". However these extra eliminations are within those truths.

My solver can like xsudo give those eliminations when entering the truths. (ignoring rank and currently 5ms, but can still be optimized to 1ms)
Xsudo also fails/is very slow in finding any rank 0.
Enumerating all combinations is unfeasible, hope you find something that can speed this up.

The New Sudoku Players' Forum

Exotic patterns a resume

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rank 0 and puzzle 38692

p38692 2 rows 2 columns

p38692 MSLS david

p38692 MSLS JC Van Hy

Rank 0 logic update of the view

conjecture : no valid logic negative rank

Re: Exotic patterns a resume

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