The New Sudoku Players' Forum

[champagne]

But as we could see for Golden Nugget, use of ABI loop is partial in exocet with 4 digits

What means "partial?"

just a piece of the post

what to do when an exocet has been found

The second example is the start of Golden Nugget
Golden Nugget has no potential for elimination in floors 1247 out of the exocet direct effect.

Golden nugget

[ronk]

But as we could see for Golden Nugget, use of ABI loop is partial in exocet with 4 digits

What means "partial?"

Golden nugget has no potential for elimination in floors 1247 out of the exocet direct effect.

In your post and your link, the "partial" term is not used even once, so I'm as mystified as before. Are you referring to the "abi-loop" elimination being r1c7<>7 but not including r2c7<>7?

[edit: Questions about "exocet direct effect" deleted.]

[champagne]

In your post and your link, the "partial" term is not used even once, so I'm as mystified as before. Are you referring to the "abi-loop" elimination being r1c7<>7 but not including r2c7<>7?

On my side I am amazed that simple facts can not be understood.

In fata morgana, the ABI loop leads to the conclusion pair 36r5c46 valid

In Golden Nugget, Only 2 of the 6 possibilities in the base can be cleared through the ABI loop
which is "partial" compared to "fata morgana"

To cover completely the point we have in row 7 of the reduced PM

7+ 1+ X |12+ X 4+ |1247+ X 1247+

digit '1' appears twice outside the stack 3.
this does not prevent the ABI loop to be there for part of the pairs.

So, in my stats, as I noticed, the lot 4 has surely some potential for use of the ABI loop.

I guess this is true as well for lot 0, but in that case, we have small chances to come, as in "fata morgana" to the validation of one of the pairs in the base.

Note : I am sorry but I did not look in detail each of the 3.5 million puzzles.


EDIT just reminding that post of may 23

may be on more remark about the ABI loop;

when we have an exocet with 3 digits, we have seen that, usually, the ABI loop will give the right pair to put in the base.

With 4 digits, the best we can hope in the UR columns is to have the 4 digits equally shared in 2 boxes.
In that case, the ABI loop should kill four of the 6 possibilities for the base and keep as possible 2 complementary pairs.

In Golden Nugget, we can only clear 2 possibilities out of the 6.

[ronk]

In your post and your link, the "partial" term is not used even once, so I'm as mystified as before. Are you referring to the "abi-loop" elimination being r1c7<>7 but not including r2c7<>7?

On my side I am amazed that simple facts can not be understood.

]
On my side I am amazed that you think the definition of a term need not include that term somewhere!

In fata morgana, the ABI loop leads to the conclusion pair 36r5c46 valid

Most likely there won't be even one elimination in the two base cells when each holds the same four candidate values. Therefore, when there are four in one and three in the other, just treat it like an "almost abi loop", i.e., treat the oddball 4th digit like a fish fin.

fin false ==> two 3-digit non-partial "abi loops" leading to one candidate false in both base cells,
fin true ==> all other candidates false in the base cell with four candidates,
therefore, one candidate eliminated from the base cell with four candidates

EDIT just reminding that post of may 23

may be on more remark about the ABI loop;

when we have an exocet with 3 digits, we have seen that, usually, the ABI loop will give the right pair to put in the base.

With 4 digits, the best we can hope in the UR columns is to have the 4 digits equally shared in 2 boxes.
In that case, the ABI loop should kill four of the 6 possibilities for the base and keep as possible 2 complementary pairs.

In Golden Nugget, we can only clear 2 possibilities out of the 6.

I'm sorry, I don't see the "partial" term here either.

[champagne]

I extracted that puzzle form lot0
Let's call it unnamed1

9876..5..4...8..3.........72...4..1....1.5.........6..1.....3.8.943...2..2..1....

That puzzle has a 3 digits exocet
r7c2 r7c3 r8c5 r9c8 567
in that position (I'll check the SE rating before an after exocet exploration)

Code: Select all

9     8     7     |6     23    123   |5     4    12   
4     156   1256  |2579  8     1279  |129   3    1269 
356   1356  12356 |2459  259   1249  |1289  689  7     
----------------------------------------------------
2     3567  35689 |789   4     36789 |789   1    359   
3678  3467  3689  |1     23679 5     |24789 789  2349 
3578  13457 13589 |2789  2379  23789 |6     5789 23459
----------------------------------------------------
1     567   56    |24579 25679 24679 |3     5679 8     
5678  9     4     |3     567   678   |17    2    156   
35678 2     3568  |5789  1     6789  |479   5679 4569 


The pattern is not the preferred pattern of David, but we have a partial use of the ABI loop.

Code: Select all

X    X    7g   |6g   X    X    |5g   X    X   
X    56+  56+  |57+  X    7+   |X    X    6+   
56+  56+  56+  |5+   5+   X    |X    6+   7g   

X    567+ 56+  |7+   X    67+  |7+   X    5+   
67+  67+  6+   |X    67+  5g   |7+   7+   X   
57+  57+  5+   |7+   7+   7+   |6g   57+  5+   

X    567  56   |57+  567+ 67+  |X    567+ X   
567+ X    X    |X    567  67+  |7+   X    56+ 
567+ X    56+  |57+  X    67+  |7+   567+ 56+ 

the pattern is perfect for the use of ABI loop for 56r7c23
so 56r7c23 is not valid
r7c2=7 and due to the exocet property r8c7=1

At that point, it's clear that we have an Xwing on digit '7'
either 7r56c1r56c5
or 7r56c1r56c8

the potential of the floors 567 shows that it should not be to tough to clear 5r7c3,
but I was to short in time to do it.

This is a first example of a puzzle with a non optimal 3 digits exocet.

It was not so easy to catch it.

Most of the lot0 puzzles are downgraded forms of the exocet.

[champagne]

.....
I'm sorry, I don't see the "partial" term here either.

I have to confess that on my side you have for long a nickname

guess which one it is


Sneezy
Sleepy
Dopey
Doc
Happy
Bashful
Grumpy

[ronk]

I have to confess that on my side you have for long a nickname ... guess which one it is ...
Sneezy ... Sleepy ... Dopey ... Doc ... Happy ... Bashful ... Grumpy

Doc, of course! Thanks but, since this is not a chat room, we should probably cease this exchange of pleasantries.

[David P Bird]

champagneThanks for your efforts – I'm sorry that I haven't matched them and have yet to make use of your new collections.

The pattern is not the preferred pattern of David, but we have a partial use of the ABI loop.

The puzzle is a perfectly good JExocet.
Digit (7) offers no UR threat in columns 2 & 3 because (78)r1c23 is already established.
(5)r9c8 = (5)r6c8 – (5)r4c9 = (5#2)r4c23
(6)r9c8 = (6)r3c8 – (6)r2c9 = (6#2)r2c23
So these digits are incompatible => r7c23 <> 56
giving r7c2 = 7, r8c7 <> 7, and r8c6 <> 8 (which you didn't include.)

Your use of "partial Exocet" seems to describe the case when an Exocet is located which can't be fully reduced to a pair of digits in the base cells using ABI loops.

If that's right, then using ABI loops can give 3 possible outcomes:
(1) ABI Irreducible - no conflicts identified
(2) ABI Partially Reducible - some conflicts identified
(3) ABI Reducible - conflicts identified for all but one combination of digits.

These outcomes are not really Exocet properties, but belong to the ABI loop method.

- 0 – 0 – 0 – 0 -

I haven't been completely idle though and have been looking at puzzles with Exocets and SK loops.

..4.....9.1...3.7.6.....2.......8....5..3..8....1.5.....2.....4.7.3...1.9...7.6.. tarx0037,tarek 7.0 SK *BB r46c5 r2c4 r8c6

After SK Loop eliminations:

Code: Select all

 *----------------------*----------------------*----------------------*
 | 3578   238    4      | 25678  158    1267   | 1358   356    9      |
>| 258    1      589    | 2469   2469   3      | 458    7      568    |<
 | 6      389    3578   | 45789  158    1479   | 2      345    1358   |
 *----------------------*----------------------*----------------------*
 | 12347  2469   13679  | 24679  2469 # 8      | 13459  2469   123567 |
>| 1247   5      1679   | 24679  3      24679  | 149    8      1267   |<
 | 23478  2469   36789  | 1      2469 # 5      | 349    2469   2367   |
 *----------------------*----------------------*----------------------*
 | 1358   368    2      | 5689   158    169    | 7      359    4      |
>| 458    7      568    | 3      2469   2469   | 589    1      258    |<
 | 9      348    1358   | 2458   7      124    | 6      235    358    |
 *----------------------*----------------------*----------------------*
   24     -      69                              49     -      26

Compatibility checks show that the base cells for (2469)Jexocet:r46c5,r2c4,r8c6 must hold either (29) or (46)

Now r46c8 (one side of a UR to be avoided) will hold just one instance of a base digit, so the second one must be true in r13c8 or r79c8. In turn this means that (35)r13c8 and (35)r79c8 must hold 1 truth each.

This then holds for the rest of the SK loop which I like to notate as a multi-sector locked set:
SK Loop: (38)c2,(46)b7,(58)r8,(29)b9,(35)c8,(46)b3,(58)r2,(29)b1 16 digits, 16 intersection cells.

Each box/line intersection will hold one truth from the digits shown for the line and one from those shown for the boxes.

This check is much quicker to perform than working through the complete SK loop looking for cross-loop conflicts in the cells in the other boxes – which I don't think works here anyway.

[David P Bird]

champagne now I have more time I'm picking through your posts from yesterday.

I thought you had put your new puzzle collections onto your web site, but when I checked they weren't there. Will you be publishing them?

EDIT :
I forgot to mention that in that code, I added a new constraint for Jexocets,
All given of the band/stack + digits of the base = 9 digits
With that constraint, I lost only one Jexocet in the file of potential hardest in the following puzzle
....5..8...67..1..7....3..42..9............1.96....2...3..4.....7...8.5...21..6..;554;elev;864

I don’t understand this as there is this: (3458)JExocet:r5c12,r6c4,r4c8 => r4c8 <> 67
The compatibility checks show that the base digits are either (35) or (48)

Your wording doesn't make your new condition very clear, but whatever it is, it doesn't seem to be a watertight test.

[champagne]

champagne now I have more time I'm picking through your posts from yesterday.

I thought you had put your new puzzle collections onto your web site, but when I checked they weren't there. Will you be publishing them?

EDIT :
I forgot to mention that in that code, I added a new constraint for Jexocets,
All given of the band/stack + digits of the base = 9 digits
With that constraint, I lost only one Jexocet in the file of potential hardest in the following puzzle
....5..8...67..1..7....3..42..9............1.96....2...3..4.....7...8.5...21..6..;554;elev;864

I don’t understand this as there is this: (3458)JExocet:r5c12,r6c4,r4c8 => r4c8 <> 67
The compatibility checks show that the base digits are either (35) or (48)

Your wording doesn't make your new condition very clear, but whatever it is, it doesn't seem to be a watertight test.

I as well during the day go back to previous posts.

2 points here

1) The constraint I added is more "to see". I'll withdraw it later.
I cleared all puzzles where "given" in the band/stack plus digits of the base are not equal to 9
In that puzzle, we have 4 digits given in the stack plus 4 digits in the base = 8
What is surprising is that this is the only puzzle I found with that specificity.

2) I am surely thinking of making my data base of seeds public, but I have to find the right way.

It can not be on my website, the file is too big
One possibility is to upload an appropriate zip file on the skfr project in Google.
I have first to define more precisely what to load.
my data base is just a collection in maxtext canonical form of all puzzles I have used as seeds to search potential hardest puzzles.
it contains all puzzles of my data base of potential hardest, but this is a very small part of it.
the test I run is on a file where all puzzles of the data base of potential hardest have been cleared.

Some comments on what would be of interest can help.

I have also to do more checking on that first run done with some fresh code.

[champagne]

Your use of "partial Exocet" seems to describe the case when an Exocet is located which can't be fully reduced to a pair of digits in the base cells using ABI loops.

let me answer first to that.

I am sure I used the word partial in at least to expressions:

a) partial exocet
b) partial use of abi loop
=======================================

a) In my head, "partial exocet" is a precise concept and here is the wording I have in the text in preparation to come in the "definitions" post

Partial Exocets
We have a partial exocet if the exocet digit per digit rule is verified for 2 or more digits.
In a partial exocet, we have no direct elimination, but the scenario study can be applied to the pairs of digits verifying the rule.

I think I'll add one example as that one from the post "what to do after an exocet has been seen"

The third example is the complete solution of a puzzle (number 12177 in the data base) with a large potential for elimination in the floors of a “nearly” exocet.
That example shows how, using the exocets properties (even partial as in that example) side eliminations can be done.
I did not find use of the "abi" loop in that example

puzzle 12177

as you can see, I used here the expression "nearly exocet" and not "partial exocet". I'll change the text

==================================

b) partial use of abi loop

The definition here is not as clear.
I used that expression in reference to the situation in fata morgana where 2 abi loops lead to the validation of one pair in the base

By difference, I am using the expression partial use of abi loop for any other situaton (including the use of the abi loop in a partial exocet when it comes.

As in a 4 digits exocet the maximum of pair eliminations is 4 out of 6, I feel that when this is true, the situation should be seen as a full use of abi loops in a 4 digit exocet

Nothing here of importance

[David P Bird]

champagne Thanks for your response which I understand OK.

I've put your findings into a table which make the interesting results easier to to digest. I'd like you to check it please as I've corrected a typo regarding lot 6.

Code: Select all

*------------*-------------*-----------*------*---------*
| Collection | Base Digits | Reducible | Twin |  Count  |
*------------*-------------*-----------*------*---------*
|     0      |     3       |    -      |  -   |  182343 | 
|     1      |     3       |   Yes     |  -   |  175836 |
|     2      |     3       |    -      | Yes  |     300 |
|     3      |     3       |   Yes     | Yes  |    4451 |
*------------*-------------*-----------*------*---------*
|            |             |           |      |  362930 |
*------------*-------------*-----------*------*---------*
|     4      |     4       |    -      |  -   | 2870053 |
|     5      |     4       |   Yes     |  -   |  120083 |
|     6      |     4       |    -      | Yes  |    5099 |
|     7      |     4       |   Yes     | Yes  |     413 |
*------------*-------------*-----------*------*---------*
|            |             |           |      | 2995648 |
*------------*-------------*-----------*------*---------*
Reducible = Candidate distributions allow ABI loops to be used 
Twin      = Target and companion cells contain a hidden pair so may be in either order

How would you categorise the puzzles in Lots 24 and 42?

As I work manually I'd be happy for a sample of 500 for each of these collections which would allow me to randomly select ones to explore. Those such as daj with batch solvers would probably want a sample size of 5000 though.

I'm finding that with the most difficult puzzles although I can now make more starting eliminations using patterns, the only way to finish solving them is to split them into cases, which is not my aim. Your new collection for "difficult" rather than "extreme" puzzles will therefore be very helpful to me.

What I'll study first will be lots 0 and 1 to compare ABI loops with my alternative system and to explore other possibilities.

I'm quite happy for them to be presented in your maxtext canonical form.

Thanks again,

DPB

[champagne]

champagne Thanks for your response which I understand OK.

I've put your findings into a table which make the interesting results easier to to digest. I'd like you to check it please as I've corrected a typo regarding lot 6.

I corrected the typo

champagne Thanks for your response which I understand OK.


Reducible = Candidate distributions allow ABI loops to be used
Twin = Target and companion cells contain a hidden pair so may be in either order[/code]

a slightly different expectation for "reducible"
this lot should contain many puzzles allowing the maximum possible use of the ABI loop
finding the valid pair with 3 digits
finding the 2 complementary valid pairs with 4 digits

All is possible with lots 0;4 but i expect to have at least a partial use of the ABI loop

Twin is "could be twin".
The selection can deliver 1+2 and 2+2 (from the code)
As I mentioned, it must still be checked whether we have an AHS
The most interesting is clearly twin + reducible

How would you categorise the puzzles in Lots 24 and 42?

It's the same pattern just depending on the order in which "exocets" have been found

'4' means clearly a 4 digits exocet has been found
'2' means we have also a pattern that could be a twin 3 digits exocet

As I work manually I'd be happy for a sample of 500 for each of these collections which would allow me to randomly select ones to explore.

no problem to do that to-day or to-morrow.
I'll pick up one every nnn as sample to avoid as much as possible to have similar puzzles

I'm finding that with the most difficult puzzles although I can now make more starting eliminations using patterns, the only way to finish solving them is to split them into cases, which is not my aim. Your new collection for "difficult" rather than "extreme" puzzles will therefore be very helpful to me.

In the list of hardest, we had (except 2) only exocets with 4 digits.
with such exocets, as I said, the best you can expect as easy logic (ABI loop) is to reduce the number of possible pairs to 2.
In general, you can only reduce the count by 2 as in Golden Nugget. (this should be the situation in lot 4, by far the biggest )
IMO, most of the 3 digits exocets should be solvable by a manual player.

(my next post will be the end of the example I started yesterday)

[champagne]

The puzzle is a perfectly good JExocet.
Digit (7) offers no UR threat in columns 2 & 3 because (78)r1c23 is already established.
(5)r9c8 = (5)r6c8 – (5)r4c9 = (5#2)r4c23
(6)r9c8 = (6)r3c8 – (6)r2c9 = (6#2)r2c23
So these digits are incompatible => r7c23 <> 56
giving r7c2 = 7, r8c7 <> 7, and r8c6 <> 8 (which you didn't include.)

I have no evidence of r8c6 <> 8 so I suspect this is a typo for <9> r9c8.

I just restart that puzzle where I stopped, telling <5>r7c23 should not be to hard to establish

9876..5..4...8..3.........72...4..1....1.5.........6..1.....3.8.943...2..2..1....

That puzzle has a 3 digits exocet
r7c2 r7c3 r8c5 r9c8 567
let's restart with 7r7c1 1r8c7 established

Code: Select all

X    X    7g   |6g   X    X    |5g   X    X   
X    56+  56+  |57+  X    7+   |X    X    6+   
56+  56+  56+  |5+   5+   X    |X    6+   7g   

X    56+  56+  |7+   X    67+  |7+   X    5+   
67+  6+   6+   |X    67+  5g   |7+   7+   X   
57+  5+   5+   |7+   7+   7+   |6g   57+  5+   

X    7    56   |5+   56+  6+   |X    56+  X   
56+  X    X    |X    567  67+  |x    X    56   
56+  X    56+  |5+   X    6+   |7+   567  56+ 

<5>r7c3 comes easily

5r9c8 + 7r8c5 leaves no '5' in box 8
7r9c8 + 5r8c5 =>5r6c8 =>5r3c1=>5r2c4=>7r2c6 no '7' in box 7
so the final solution for the exocet is 67
and we are quickly in that position

Code: Select all

9     8     7     |6     23    123   |5     4    12   
4     156   125   |2579  8     1279  |29    3    1269 
356   1356  1235  |2459  259   1249  |289   68   7     
----------------------------------------------------
2     356   3589  |789   4     36789 |789   1    39   
367   346   389   |1     23679 5     |2489  78   2349 
37    134   1389  |289   2379  2389  |6     5    2349
----------------------------------------------------
1     7     6     |245   25    24    |3     9    8     
8     9     4     |3     67    67    |1     2    5   
35    2     35    |89    1     89    |47    67   46   

more '6' can be cleared through the now classical loop
6r8c5 - 6r9c8 = 6r3c8 - 6r3c1 = 6r5c1 - 6r5c5 = 6r8c5

at that point, the puzzle is rated 7.3 by Sudoku explainer

[ronk]

you can have a "partial exocet". I posted recently that example puzzle 12177

A couple of quotes from this link:

We now study 23r12c3 due to digit 2, we look in row 3



23r12c3 + 2r3c9 + 3r3c5 conflict r4c2

23r12c3 + 2r3c4 + 3r3c5

3r4c2 3r6c6 2r5c1 2r7c6 6r9c6 2r4c9

RI <3>r12c7 3r9c7 3r7c1 6r7c3 1r7c8

HP12r12c7 6r6c6 =>6r4c3 conflict

I had to use (23)AUR:r12c37 here, probably because I got lost at the "RI". BTW what means "RI"?

13r12c3 {3r7c1 1r4c2 1r3c4 3r3c9 3r4c5 3r6c2 1r7c8 1r8c1}



23r12c3 {3r7c1 2|6r4c2 2r3c4 3r3c9 3r9c7 2r7c6 6r7c3 1r7c8 6r9c6 3r4c5}



26r12c3 {6r7c1 6r3c8 2r3c4 6r4c2 6r6c7 2r4c9 2r1c7 2r5c1 2r7c6 6r9c6 2r8c2}



36r12c3 {3r7c1 6r4c2 6r3c8 3r3c9 6r9c1 6r7c6 3r6c2 3r4c5 3r1c6}



We see in that summary that

<13+>r7c1 (3r7c1 3r7c1 6r7c1 3r7c1)

<3+>r4c2 (1r4c2 2|6r4c2 6r4c2 6r4c2)

<1>1r3c8 (1r3c4 1r7c8 6r3c8 6r3c8)

<2>2r3c9 (3r3c9 2r3c4 2r3c4 3r3c9)

How do these four cases justify r7c1<>3?

BTW cut & paste from your link produces two line breaks where there should be one.