The New Sudoku Players' Forum

[champagne]

Hi leren,

Thank you very much for the clue.
I'll work in that direction, but I have not the adequate infrastructure, so it can take time.

Regarding the "V loop", here is the list of puzzles resisting to the current code.

They have similarities, so it's likely the same reason (and it can be a bug in my code).
I have not yet studied the path proposed by the solver

..8...5...6...3.7.4.......1.7..32......9.5......67..2.5.......4.2.3...6...1...8..;7654;TkP;3138
9.......1.3...4.7...6...2...5.3.2.......6........78.5...2...6...4.7...3.1.......9;8400;tax;tarek-ultra-0204
98.7.....7.6.......54...7..4..6..9......5..3......2..1..94..6......1..5......3..2;8402;GP;H2040
1....6.8....7....2....3.5.....5....7.6...4.1.....2.3...14..8...6.2....4.89.......;8405;elev;H235
1.....9...3...7.4...5.....2.....6.7.....1.....4.3.8...9.......5.7.8...3...2...1..;8668;tax;tarek071223170000- 95895
..7...9...3...5.4.6.......8.4..51......7.2......34..1.9.......6.1.5...3...8...7..;9712;TkP;3164
..2...3...7...1.5.6.......8.4...5......43.......1.7.4.8.......2.5.7...1...3...6..;9717;TkP;3101
..9...7...4...1.2.8.......5.2...3......9.6.......12.3.7.......8.3.1...4...5...9..;9762;TkP;3715
..1...2...3...4.5.6.......7.....3.8.....1.....4.9.8...7.......1.5.8...4...2...6..;38022;ig;champagne

[ronk]

Because the number of links exceeds the number of truths by 1 then exactly 1 of these 17 eliminations is false ( ie the relevent candidate is true)
Scenario testing is then carried out using this information (ie setting in turn all but one to false and the remaining one to true) plus the possible Exocet Base and Target cell values.
...
I've also implemented the JC Van Hay method for SK loop puzzles. It applies to all the ones I've tested in your file so far.

Very interesting, converting a 1-rank** to 0-rank for numerous exclusions, but something doesn't seem right here. Isn't this the same as saying fins of single-digit fish must always be false? {**edit: It didn't seem right because it's not truly 1-rank logic. The term "almost 0-rank" is much more accurate and appropriate.]

Moreover, both methods make me think of Ariadne on steroids, hardly methods for players which champagne claims as his goal.

[champagne]

Multifish 5.1: 16 Truths = { 1R1347 2R1347 4R1347 7R1347 } 17 Links = { 1c24 2c14 4c26 7c16 1n357 3n89 4n78 7n79 }

This leads to the following potential eliminations: r1c3 <> 56, r1c5 <> 6, r2c1 <> 27, r2c2 <> 4, r2c4 <> 2, r3c8 <> 6, r4c7<> 35,
r5c4 <>1, r6c7<> 7, r7c7 <> 3, r8c1 <> 7, r8c4<> 1, r8c6<> 4, r9c2 <> 1.

I studied your rank 1 logic.

Contrary to the remark I have seen, I think it is a very good track. a rank 1 logic with no overlapping is very easy to translate in weak links and to use in the standard process.

Nevertheless, in the case of Golden Nugget, I would use instead that rank 1 logic

Code: Select all

     19 Truths = {1R1347 2R12347 4R12347 7R12347}
     20 Links = {1c24 2c14 4c26 7c16 12n3 12n5 47n7 4n8 7n9 1247b3}


This is done of the band containing the base plus the 2 rows containing the target. It is likely valid in most of the Jexocet patterns.


BTW I am much in favour of scenarios attached to a specific pattern. It's much easier to work out than a blind search of chains or nets.

[ronk]

Contrary to the remark I have seen, I think it is a very good track. a rank 1 logic with no overlapping is very easy to translate in weak links and to use in the standard process.

Using some derived weak link(s) in chains/networks is not what Lerner did IMO. Let's just wait for Lerner's response to my question ... to him.

[daj95376]

Regarding the "V loop", here is the list of puzzles resisting to the current code.

They have similarities, so it's likely the same reason (and it can be a bug in my code).
I have not yet studied the path proposed by the solver

..8...5...6...3.7.4.......1.7..32......9.5......67..2.5.......4.2.3...6...1...8..;7654;TkP;3138
9.......1.3...4.7...6...2...5.3.2.......6........78.5...2...6...4.7...3.1.......9;8400;tax;tarek-ultra-0204
98.7.....7.6.......54...7..4..6..9......5..3......2..1..94..6......1..5......3..2;8402;GP;H2040
1....6.8....7....2....3.5.....5....7.6...4.1.....2.3...14..8...6.2....4.89.......;8405;elev;H235
1.....9...3...7.4...5.....2.....6.7.....1.....4.3.8...9.......5.7.8...3...2...1..;8668;tax;tarek071223170000- 95895
..7...9...3...5.4.6.......8.4..51......7.2......34..1.9.......6.1.5...3...8...7..;9712;TkP;3164
..2...3...7...1.5.6.......8.4...5......43.......1.7.4.8.......2.5.7...1...3...6..;9717;TkP;3101
..9...7...4...1.2.8.......5.2...3......9.6.......12.3.7.......8.3.1...4...5...9..;9762;TkP;3715
..1...2...3...4.5.6.......7.....3.8.....1.....4.9.8...7.......1.5.8...4...2...6..;38022;ig;champagne

I guess you weren't interested in the solution from my solver for the first puzzle.

post220771.html#p220771

[champagne]

Regarding the "V loop", here is the list of puzzles resisting to the current code.

They have similarities, so it's likely the same reason (and it can be a bug in my code).
I have not yet studied the path proposed by the solver

..8...5...6...3.7.4.......1.7..32......9.5......67..2.5.......4.2.3...6...1...8..;7654;TkP;3138

I guess you weren't interested in the solution from my solver for the first puzzle.

post220771.html#p220771

My solver finds the "V loop" as well, my point is that using the JC VAN HAY strategy and the set of rules I described earlier, it does not find the final solution.

In your post, we have only the start of the path.
I'll study that puzzle to-day

Thanks anyway.

[Leren]

*-c1--------c2------------------*-c4-----------------c6--------*---------------------------------*
r1| 27-568 14-568 1247-56A | 2-68 247-6B 47-68 | 124C 3 9 |r1
| 27-689 4-689 247-6 | 2-3689 247-36 1 | 247 247-6 5 |
r3| 27-69 14-69 3 | 2-69 5 47-69 | 8 1247-6D 1247E |r3
*--------------------------------*-------------------------------*---------------------------------*
r4| 2-35 4-35 8 | 1-35 9 7-35 | 1247-35F 1247G 6 |r4
| -3569 7 4-56 | 1-3568 1-36 2 | 14-359 14-89 14-38 |
| 1 -3569 2-56 | 4 7-36 7-3568 | 27-359 27-89 27-38 |
*-------------------------------*--------------------------------*---------------------------------*
r7| 7-36 1-36 9 | 12-36 8 4-36 | 1247-3H 5 1247J |r7
| 7-358 2 17-5 | 1-359 14-3 4-359 | 6 147-89 147-38 |
| 4 1-3568 1-56 | 7 12-36 -3569 | 12-39 12-89 12-38 |
*-c1-----c2-------------------*-c4----------------c6---------*---------------------------------*

I'll try and answer the queries to my previous post by providing a more detailed description of the Almost Multifish method.

My apologies in advance if I get a bit too detailed or basic at times but I am mindful that this post may be read by others.

Recapping my previous post, the above PM is the Golden Nugget puzzle state following exocet eliminations.

The following Almost Multifish is found:

16 Truths = { 1R1347 2R1347 4R1347 7R1347 }
17 Links = { 1c24 2c14 4c26 7c16 1n357 3n89 4n78 7n79 }

In the PM I've tried to make this clearer by indicating the Set rows and Link columns and I've labelled the 9 link cells A - H & J.

If this were a Rank 0 logic set then the following 17 eliminations would apply: r1c3 <> 56, r1c5 <> 6, r2c1 <> 27, r2c2 <> 4, r2c4 <> 2, r3c8 <> 6, r4c7<> 35,
r5c4 <>1, r6c7<> 7, r7c7 <> 3, r8c1 <> 7, r8c4<> 1, r8c6<> 4, r9c2 <> 1. But it isn't.

Nevertheless, consider what would apply if one of the potential elimination (PE) candidates were in fact, true, say r5c4 = 1. Then the truths required in column 4 would reduce from 2 to 1
and the remnant logic set would be:

16 Truths = { 1R1347 2R1347 4R1347 7R1347 }
16 Links = { 1c2 2c14 4c26 7c16 1n357 3n89 4n78 7n79 }

ie a Rank 0 logic set. In this case the other 16 eliminations would apply. A similar argument applies for all the other PE's.

Thus if any one of the PE's is True then all the others must be False.

On the other hand the PE's can't all be False. Why not? Well, consider what would happen if they were. In that case all 9 link cells would resolve to base digits
(1,2,4 or 7) and that leaves only 7 base digits to fill the remaining positions in Rows 1,3,4 & 7 Cols 1,2,4 & 6. However in those columns the only positions that
the base digits can go is in Rows1,3,4 & 7 and there are a total of 8 required. We are 1 short - can't have that!

The net result of all this is that exactly 1 of the PE's is True and all but 1 are False.

Where to from here? At the moment I use this information in 2 ways.

1. Scenario testing - this is analagous to the Scenario testing of SK Loops described in recent pages of this thread. Each Scenario consists of
assuming one of the PE's true and all the others false - in concert with other information: 1) Where an Exocet has been found the scenarios also
include the assignment of Exocet base and target cell outcomes . This is usually is quite effective because the base and target cells must resolve
to the same 2 digit set - thus greatly reducing the number of Scenarios. 2) In the absence of an Exocet I use a set of bi-value cells and test against their
possible outcomes. This is less than ideal and somewhat less effective but does extend the method somewhat.

2. If Scenario testing is inconclusive then at the very least Weak links exist between all pairs of PE's and this can be used in any downstream move.

Is this an ideal solution method ?

Well it's clearly T&E so some would say not, although the choice of scenarios and the eliminations and assignments therein is far from arbitrary.

How does this method compare with Fins of single digit Fish ?

In answer to Ronk's question, the role of the PE's is more or less the converse of the role of Fins. If a Fin is False you could make the
Fish eliminations whereas you could make the Multifish eliminations if the PE is True. The converse is to assume that the Fin is True to see if you can make one or more Fish eliminations. Unfortunately the converse for the PE's
- assuming that they are False - is a weaker assumption than that for FIns, because most of the PE's are False. This has so far proven to be unproductive but I live in hope.

Can this method lead to constructive (nonT&E) eliminations?

Maybe. One thing I have explored is to collect all the ALmost Multifish to see if something can be deduced from all the resulting PE relationships.

For example in the above puzzle I've found the following Almost Multifish :

16 Truths = { 1R1347 2R1347 4R1347 7R1347 } 17 Links = { 1c24 2c14 4c26 7c16 1n357 3n89 4n78 7n79 }
19 Truths = { 1R1347 2R12347 4R12347 7R12347 } 20 Links = { 1c24 2c14 4c26 7c16 1n35 2n35 4n78 7n79 1247b3 }
19 Truths = { 12467R1 2467R2 12467R3 12467R7 } 20 Links = { 1n7 2n78 3n89 7n124679 12467b1 2467b2 }
19 Truths = { 35689R5 35689R6 3589R8 35689R9 } 20 Links = { 3c579 5c37 6c35 8c89 9c78 5n14 6n26 8n146 9n26 }
23 Truths = { 3R25689 5R5689 6R2569 8R25689 9R25689 } 24 Links = { 3c579 5c37 6c358 8c89 9c78 2n124 5n14 6n26 8n146 9n26 }
20 Truths = { 1C35789 2C35789 4C35789 7C35789 } 21 Links = { 1r589 2r269 4r258 7r268 1n357 3n89 4n78 7n79 }
19 Truths = { 12347C5 12347C7 1247C8 12347C9 } 20 Links = { 1n5 2n5 5n5 6n5 8n5 9n5 1247b3 12347b6 12347b9 }

No doubt may others can be found. I haven't checked these particuar sets but in general its true to say that the logic sets don't always produce the same set of PE's so
it may be possible to deduce eliminations from all this information. So far on other puzzles I've had no luck.

leren

[Leren]

Champagne,

After that long post just a quick note to let you know that I ran those SK/V Loop puzzles and they all passed my version of the JCVH method.

Leren

[David P Bird]

Leren, from your post I assume you also have played with this:

When we have two Almost Multi-fish each one with one invalid potential elimination, the combined set of PEs can be split into two subsets, those that are common to both and those that unique to one or other of them. Either the common set will contain one truth or the unique set will contain two truths. Hence either all the common or all the unique PEs are false.

Restricting myself to only looking for immediate contradictions, my trials have produced nothing. Are you saying that even when you follow the consequences to exhaustion you still get nowhere?

[Leren]

David Bird Said: Are you saying that even when you follow the consequences to exhaustion you still get nowhere?

David I'd say my experience is similar to your's on this topic. I took a good look at one of the potential hardest puzzles and found
14 PE sets. 4 of these were trivial and the other 10 reduced to 3 different sets with every possible overlap for 3 sets, that's 7 different areas on a
Venn diagram. From memory their were 2 truths, one in the common area of sets A and B that was not common to set C and the other was in the
part of set C that was not common. I coudn't make any other deductions so I left it at that.

You could make eliminations from just Set theory if you found 3 PE sets A, B and C with A& B having common elements
B&C having common elements but no elements common to A,B &C. Then you could make eliminations of the non-common elements of A, B and C.

Leren

[champagne]

Champagne,

After that long post just a quick note to let you know that I ran those SK/V Loop puzzles and they all passed my version of the JCVH method.

Leren

I had a look at the first "not cleared" scenario of the puzzle 7654. Seen by me, clearing would require a "Nishio" elimination, what is done by JC Van Hay, but not in my limited set of rules.
Do you apply such techniques on your side?

(you could also post your clearings, but I am not sure you can do it easily)

It this is the reason, it is quite normal that my program failed and I don't want so far pass the kite, sky scrapper level.

[ronk]

When we have two Almost Multi-fish each one with one invalid potential elimination, the combined set of PEs can be split into two subsets, those that are common to both and those that unique to one or other of them.

If the only premise is the existence of two almost 0-rank multi-fish, we don't even know if ... if ... if one of the PEs is an invalid exclusion in each case. What are your other premises?

[edit: "almost 0-rank" was "1-rank"]

[champagne]

Thinking on how to implement the rank 1 logic shown by Leren, I looked in my data base for examples of puzzles having both an exocet and a rank 0 logic.

Usually, these are combining an exocet and a “V loop” or an “almost “V loop”.

That one is much more in line with Leren work

98.7.....7.6...8......5.....7.4..6....4..3.2.....1...4.6.9..5......4..3......2..1;1819;GP;H364


My solver finds here 2 rank 0 SLGs tightly linked to the exocet

SLG rank 0
16 Truths = {1R1247 2R1247 3R1247 4R127 3N3 }
16 Links = {1c368 2c359 3c359 4c68 1n7 2n24 4n1 7n1 }
elims
1r3c6 1r3c8 1r8c3 1r8c6
2r3c9 2r6c3 2r8c3 2r8c9
3r3c9 3r6c3 3r9c3 3r9c5
4r3c6 4r3c8 4r9c8
5r2c2 58r4c1 8r7c1
19 elims done


SLG rank 0
16 Truths = {1C1247 2C1247 3C1247 4C127 3N3 }
16 Links = {1r358 2r368 3r369 4r39 1n7 2n24 4n1 7n1 }

Same eliminations


EDIT
I continued to study that puzzle to see whether the rank 1 logic could be applied after the rank 0 eliminations

1) it works
2) then, using the rank1 property, it is easy to show 24 false in the base,
knowing that 12 and 23 are false due to the "abi loop", "2" can not be in the base (nor in the target)

[Leren]

Champagne wrote:

I had a look at the first "not cleared" scenario of the puzzle 7654. Seen by me, clearing would require a "Nishio" elimination, what is done by JC Van Hay, but not in my limited set of rules.
Do you apply such techniques on your side?

(you could also post your clearings, but I am not sure you can do it easily)

Champagne,

I do use Nishio as a last resort for the SK/V Loop scenario analysis and can confirm that none of the 9 puzzles achieves full SK/V Loop clearence without it.

In addition to basic techniques, Skyscrapers and Kites I use Type 1 URs, basic Fish and XY Wings (plus Nishio).

Hidden Text: Show

1 r2c7 = 4 r3c8 = 8 r7c8 = 1 r8c9 = 5 r8c3 = 4 r7c2 = 8 r1c2 = 1 r2c3 = 5 TRUE TRUE
2 r2c7 = 4 r3c8 = 8 r7c8 = 1 r8c9 = 5 r8c3 = 4 r7c2 = 8 r3c2 = 5 r2c1 = 1 TRUE TRUE
3 r2c7 = 4 r3c8 = 8 r7c8 = 1 r8c9 = 5 r8c1 = 8 r9c2 = 4 r1c2 = 1 r2c3 = 5 FALSE FALSE
4 r2c7 = 4 r3c8 = 8 r7c8 = 1 r8c9 = 5 r8c1 = 8 r9c2 = 4 r3c2 = 5 r2c1 = 1 TRUE FALSE
5 r2c7 = 4 r3c8 = 8 r9c8 = 5 r8c7 = 1 r8c3 = 4 r7c2 = 8 r1c2 = 1 r2c3 = 5 FALSE FALSE
6 r2c7 = 4 r3c8 = 8 r9c8 = 5 r8c7 = 1 r8c3 = 4 r7c2 = 8 r3c2 = 5 r2c1 = 1 TRUE TRUE
7 r2c7 = 4 r3c8 = 8 r9c8 = 5 r8c7 = 1 r8c1 = 8 r9c2 = 4 r1c2 = 1 r2c3 = 5 TRUE TRUE
8 r2c7 = 4 r3c8 = 8 r9c8 = 5 r8c7 = 1 r8c1 = 8 r9c2 = 4 r3c2 = 5 r2c1 = 1 TRUE TRUE
9 r2c9 = 8 r1c8 = 4 r7c8 = 1 r8c9 = 5 r8c3 = 4 r7c2 = 8 r1c2 = 1 r2c3 = 5 TRUE TRUE
10 r2c9 = 8 r1c8 = 4 r7c8 = 1 r8c9 = 5 r8c3 = 4 r7c2 = 8 r3c2 = 5 r2c1 = 1 FALSE FALSE
11 r2c9 = 8 r1c8 = 4 r7c8 = 1 r8c9 = 5 r8c1 = 8 r9c2 = 4 r1c2 = 1 r2c3 = 5 TRUE TRUE
12 r2c9 = 8 r1c8 = 4 r7c8 = 1 r8c9 = 5 r8c1 = 8 r9c2 = 4 r3c2 = 5 r2c1 = 1 FALSE FALSE
13 r2c9 = 8 r1c8 = 4 r9c8 = 5 r8c7 = 1 r8c3 = 4 r7c2 = 8 r1c2 = 1 r2c3 = 5 TRUE TRUE
14 r2c9 = 8 r1c8 = 4 r9c8 = 5 r8c7 = 1 r8c3 = 4 r7c2 = 8 r3c2 = 5 r2c1 = 1 FALSE FALSE
15 r2c9 = 8 r1c8 = 4 r9c8 = 5 r8c7 = 1 r8c1 = 8 r9c2 = 4 r1c2 = 1 r2c3 = 5 TRUE TRUE
16 r2c9 = 8 r1c8 = 4 r9c8 = 5 r8c7 = 1 r8c1 = 8 r9c2 = 4 r3c2 = 5 r2c1 = 1 TRUE TRUE

TRUE = contradiction found FALSE = contradiction not found

The table shows my scenario clearences for puzzle 7654 with Nishio removed. The first clearance column uses my
full set of checks. The second clearance column is with the fish and XY Wing checks also removed.

Champagne wrote:

I continued to study that puzzle to see whether the rank 1 logic could be applied after the rank 0 eliminations

1) it works
2) then, using the rank1 property, it is easy to show 24 false in the base,
knowing that 12 and 23 are false due to the "abi loop", "2" can not be in the base (nor in the target)

Champagne,

Could you provide more details of your use of the Rank 1 property here, I don't understand what you are saying.

Leren

[champagne]

Hi Leren,

I'll answer in a separate post for the puzzle 1819.

I compared my results to yours (with no use of nishio).
You have still 11 possibilities at the end, I have six.

My set of rules for the scenario clearance is basically serate's dynamic plus expansion with a start of 8 "true", using
pointing,claiming,pairs,Xwing as in serate
plus
kites and sky scrappers
UR type 1 and UR having left one digit in excess (including diagonal)

In fact, considering the way expansion is done, it would not be easy in my program to add chains and likely not so easy to add XY wings.
I should add in the near future triplets and swordfish, but I guess it would not change the results here

An interesting result however is that only this small lot was not solved using my reduced (compared to you) set of rules