The New Sudoku Players' Forum

[David P Bird]

I should have said grouped kites (finned mutant x-wings). The first one is near the start:

2 Truths = {6R4 6C4}
3 Links = {6r9 6c9 6b5}
1 Elimination --> r9c9<>6

Thanks, when I didn't find a simple one, not knowing your methods, I wrongly assumed you had used branched chains somewhere.
In Eureka Notation:
(6)r4c9 = (6)r4c56 - (6)r56c4 = (6)r9c4 => r9c9 <> 6
and later
(4)r9c1 = (4)r7c3 - (4)r4c3 = (4)r4c7 => r9c7 <> 4

For me Xsudo is a fine research tool, but is somewhat lacking in explaining the way the eliminations can be located. Your use of virtual sets in the JExocet analysis is definitely a step in the right direction I feel. It equates more directly to my mental model.

A plea though; when you post examples of XSudo grids, I would be helpful if you a) notated the puzzle string and b) the link and truth sets, it would be a great time saver for me, and I believe many of the rest of us.

BTW in checking my previous post I found a rather critical omission so I've had to edit it.

[ronk]

So you have now 3 of the 4 digits that can not occupy both AAHS

The fourth digit (2) must be in one of the AAHS, may be in both.

I must be missing something basic here. Unless 1 of the 3 need not occupy an AAHS, how can the 4th be true in both AAHSs?

For me Xsudo is a fine research tool, but is somewhat lacking in explaining the way the eliminations can be located. Your use of virtual sets in the JExocet analysis is definitely a step in the right direction I feel. It equates more directly to my mental model.

I don't think sultan vinegar used an XSudo virtual set in his post, so to what do you refer?

[champagne]

So you have now 3 of the 4 digits that can not occupy both AAHS

The fourth digit (2) must be in one of the AAHS, may be in both.

I must be missing something basic here. Unless 1 of the 3 need not occupy an AAHS, how can the 4th be true in both AAHSs?

This can be a problem of wording. With my poor English,"may be" is not more than a possibility.

In fact, without more information, we only know that the fourth digit is there, so we can clear all candidates of that digit seeing both AAHS.

[sultan vinegar]

For me Xsudo is a fine research tool, but is somewhat lacking in explaining the way the eliminations can be located. Your use of virtual sets in the JExocet analysis is definitely a step in the right direction I feel. It equates more directly to my mental model.

I agree that Xsudo's explanations could be better. IMO, the triplet idea is great for the cases where there is only one (or maybe two) triplets, as you can understand a complicated net of logic just by understanding one (or two) triplet(s). But, these exotic patterns have many triplets in general, and so the triplet idea is not so useful anymore.

A plea though; when you post examples of XSudo grids, I would be helpful if you a) notated the puzzle string and b) the link and truth sets, it would be a great time saver for me, and I believe many of the rest of us.

No worries. I've edited my post with the Guardians/AADJE solution, as well as the post with the alternative explanation for the DJE with this information.

[David P Bird]

A walk through on the false ADJE pattern I produced:

Requirements
1) An Almost Double JExocet with 4 digits
2) Only one base digit should eventually be capable of being true three times in the partial fish cells.
3) One of the mini-lines along the JE band must contain a target cell and two cells that can't contain a base digit.

Code: Select all

 *----------------*-------------------*-----------------*
 | abcd abcd /    | .     .      .    | .     .    .    | abcd = base digits 
 | .    .    .    | .     /      .    | abcdx /    /    | x = any other digit
 | .    .    .    | .     abcdx  .    | /     .    .    | / = no base digit
 *----------------*-------------------*-----------------*

If r3c5 holds a base digit, the same digit must be true in r2c7 because r2c89 are unavailable - which makes the JE false.
It's therefore impossible for the two target cells to hold different base digits.
So the second true digit in the base cells must be true in r2b2 and r3b3.
With an ADJE, this requires this digit to be true in both sets of base cells, and therefore false in all four target cells.
In turn this requires this digit to be true 3 times in the partial fish cells and be the single spoiler digit.

Taking (a) to the single spoiler digit, the following pattern is now forced.
It results because (b) (c) & (d) can only be true in two partial fish cells, and must each occur at least once in b1c3,b2c5,b3c7

Code: Select all

 *----------------*-------------------*-----------------*
 | ab   ab   /    | .     dx     .    | cx    .    .    | abcd = base digits 
 | .    .    dx   | ac    /      ac   | bx    .    .    | x = any other digit
 | .    .    cx   | .     bx     .    | /     ad   ad   | / = no base digit
 *----------------*-------------------*-----------------*

Inferences & Eliminations:
1) In three mini-rows (a) occurs with a different base digit in each box together with a non-base digit.
Therefore any non-base digit in the cells shown with (a) as a candidate can be eliminated

2) In six mini-rows one of the other base digits occurs with two non-base digits.
If one cell must contain a base digit
a) all base digits can be eliminated from the other two cells
b) any base digit eliminated from that cell can be eliminated from two other mini-rows on the repeating diagonal

3) The cell pairs containing (bx), (cx) and (dx) must each contain one instance of a base digit
If a non-base digit must be true in one of them,
a) any non-base digits can be eliminated from the other.
b) the partial fish cells for the base digit must now contain two truths which may allow other eliminations

4) The spoiler digit (a) must be true 3 times in the partial fish cells, and so may be restricted to certain base digits.
If one of the cells in the pairs (ab), (ac) or (ad) contains a digit that can't be the spoiler, the other cell can only hold a potential spoiler.

There will also be inferences available for the non-base digits in the JE band as three of them form repeating pairs with different members of (bcd), but generally these will be trivial.

[David P Bird]

To quickly catch up on some outstanding replies:

Leren, for your tough example
98.7..6..7...6......6....5.4..3.9.....8.573.....8......2......1..9...5....7.839..;975059;GP;13_03;21;1234 ;3;34

I can only find these eliminations using un-branched chains
(9-7)r7c5 = (7-124)r8c5 = DJE - (9)r3c5 = Loop => r7c5 <> 4 (first weak link is conjugate)
(3)r1c5 = (3-9)r3c5 = (9-7) = (7-124)r8r5 = DJE => r1c3 <> 3 (DJE excludes 1234 r1c3)
(4-9)r7c4 = (9-7)r7c5 = (7-124)r8c5 = DJE - (4)r7c468 => r7c4 <> 4 (Spot cells)
(9)r7c4 = (9-7)r7c5 = (7-124)r8c5 = DJE - (124=56)r67c6 => r7c4 <> 5 (Spot cells)

After these I then have to admit defeat and start using derived inferences/branching.

Champagne,
Quickly examining the two puzzles you posted
9..8..7...8..7..6...5..4...8....53...2.1.........8...94....8..3.1.2..8......6..7.;11.10;1.20;1.20;GP;Kz1 b;15666
98.7..6..5...9..4...3..2...81......5.3...5.....5...2....8..3..1...6..9......4..7.;11.30;11.30;9.40;GP;Kz1 b;17065

I find nothing pattern-like in the first
For the second there is an almost SK loop for 4679 which should give something

Ronk,
As you know I have only a passing acquaintance with XSudo. I would like to see an example where virtual sets gathered from complex patterns are used which I believe would do much to simplify them. If the reader is familiar with the theorem that produces the virtual set, I feel there is no need for XSudo to re-prove it. I don't know what Xsudo's capabilities are in that direction though, so my wording wasn't the best.

Sultan Vinegar
Thanks for your reply and responding to my plea.
For me there is a general problem with calling a Kraken anything a pattern as the supporting chain is external to the defined elements, is unlimited in length, and can take any shape.

Sorry for the delays but I've got a lot on at the moment, and have to steal time to keep up with this discussion.

DPB

[JC Van Hay]

9..8..7...8..7..6...5..4...8....53...2.1.........8...94....8..3.1.2..8......6..7.;11.10;1.20;1.20;GP;Kz1 b;15666
98.7..6..5...9..4...3..2...81......5.3...5.....5...2....8..3..1...6..9......4..7.;11.30;11.30;9.40;GP;Kz1 b;17065

these 2 puzzles have complementary AAHS;

I found no identified exotic pattern and could not prove the complementary property with my code (voluntarily limited in that situation).

The two puzzles can be solved in the same way :

The first puzzle :

1. Potential Complementary AAHS : (1245) in B36 as there are no given for the digits 1245 in stack 3.
2. It is easy to prove r3c9,r5c8=8.
2. If r12c9=ab->r46c8=cd where {a,b,c,d}={1,2,4,5}, then -abcdr13c8(=39),r45c9(=67). Furthermore, r56c7=ab, r23c7=cd :=> -abcdr79c7(=69)
3. r79c7=69->3 Singles; HP(67)r45c9; "Jellyfish"(2B5R39C8) :=> LC(2r46c8) :=> r7c3=2; ste
4. All the other solutions of r79c7 lead to easily proven contradictions.

The second puzzle :

1. Potential Complementary AAHS : (2467) in B14 as there are no given for the digits 2467 in stack 1.
2. If r36c1=ab->r23c2=cd where {a,b,c,d}={2,4,6,7}, then -abcdr3c1(=1),r6c2(=9). Furthermore, r12c3=ab, r45c3=cd :=> -abcdr89c3(=19)
3. r89c3=19->6 Singles; FXWing(46C36) :=> -46r6c1(=7); 14 Singles; NP(48)r68c9; ste
4. All the other solutions of r89c3 lead to easily proven contradictions.

Conclusion : if the Complementary AAHS is too hard to establish, its potential eliminations can be less hard to prove.

[sultan vinegar]

for your tough example
98.7..6..7...6......6....5.4..3.9.....8.573.....8......2......1..9...5....7.839..;975059;GP;13_03;21;1234 ;3;34

I can only find these eliminations using un-branched chains
(9-7)r7c5 = (7-124)r8c5 = DJE - (9)r3c5 = Loop => r7c5 <> 4 (first weak link is conjugate)
(3)r1c5 = (3-9)r3c5 = (9-7) = (7-124)r8r5 = DJE => r1c3 <> 3 (DJE excludes 1234 r1c3)
(4-9)r7c4 = (9-7)r7c5 = (7-124)r8c5 = DJE - (4)r7c468 => r7c4 <> 4 (Spot cells)
(9)r7c4 = (9-7)r7c5 = (7-124)r8c5 = DJE - (124=56)r67c6 => r7c4 <> 5 (Spot cells)

After these I then have to admit defeat and start using derived inferences/branching.

Great chains. I'd like to add one to the party (excuse my notation, first time I've typed it up):

I show a strong inference connecting 9r3c45, so 9r2c4 and 9r3c9 can be eliminated:

9r3c5 == 9r7c5 -- 7r7c5 == 7r8c5 -- 124r8c5 == DJE -- 124r67c6 == 5r7c6 -- 5r12c6 == 5r2c4 -- 9r2c4 == 9r3c45 => r2c4<>9, r3c9 <> 9.

I should clarify the AAALS part to be clear; if the DJE is true, then 124r67c6 are all false, leaving a locked set in r67c6 with only one instance of candidate 5 in r7c6. This is the same inference used in DPBs 4th chain, but I notate it differently.

This gives us a nice little set of eliminations. I'm still struggling to understand the eliminations in rows 4 and 6, as well as the elimination of 35r2c3 in the following Xsudo grid (hybrid work of Ronk, DPB and SV):

Hidden Text: Show

98.7..6..7...6......6....5.4..3.9.....8.573.....8......2......1..9...5....7.839..

20 Truths = {9R7 124C3 12347C5 124C7 3N1 3N2 67N6 1N8 1N9 59B2}
35 Links = {1234r1 12349r3 12r4 124r6 4r7 5c46 6c6 9c4 2n3 27n4 13678n5 26n7 124b1 124b3}
15 Eliminations --> r6c689<>4, r4c28<>1, r4c89<>2, r7c45<>4, r12c3<>3, r2c3<>5, r2c4<>9,
r3c9<>9, r7c4<>5.

[champagne]

9..8..7...8..7..6...5..4...8....53...2.1.........8...94....8..3.1.2..8......6..7.;11.10;1.20;1.20;GP;Kz1 b;15666

The first puzzle :
2. It is easy to prove r3c9,r5c8=8.

Hi JC

I had a look to the PM and I ask my solver to confirm my impression.

With my classical set of rules, this is not easy at all.

Can you tell us what is "easy"

[JC Van Hay]

9..8..7...8..7..6...5..4...8....53...2.1.........8...94....8..3.1.2..8......6..7.;11.10;1.20;1.20;GP;Kz1 b;15666

The first puzzle :
2. It is easy to prove r3c9,r5c8=8.

Can you tell us what is "easy"

Code: Select all

#1. r9c3=8; UP23
#2. r3c8=8=r5c9->UP29
                 "Swordfish"(6C167) : r1c6=XWing(r56c67)-r56c1=r3c1 :=> r1c6=6; UP31
                 Chain[4] : (2=1)r1c3-1r4c3=(1-2)r4c8=2r4c5-2r6c6=2r2c6 :=> -2r1c5, NP(13)r2c13; UP34
                 LC(1r13c9,2r23c9) :=> r3c7=9
                 C(S) :=> r3c9=8=r5c8; UP26

Analysis of the puzzle from r79c7 :

#3. r8c9=6->C(sk-basics) :=> r7c7=6; UP27; HP(67)r45c9

r8c9=6->"Swordfish"(6C167) : r1c6=XWing(r56c67)-r56c1=r3c1 :=> r1c6=6; UP28
        HP(67)r3c12
        "Jellyfish"(2R367C1) : r2c1=r9c1-r7c3="Swordfish"(r367c78,r3c5,r6c6) -> r2c1=r6c6=r3c5 :=> -2r2c6; UP29
        r4c9=7->C(S) :=> r5c9=7; UP31
        NT(359)r237c4; UP33
        NT(359)r8c13,r9c2; UP34
        NP(67)r37c3 :=> NP(49)r4c25 :=> NP(12)r4c89 :=> r6c8=5
        C(S)

#4. "Jellyfish"(2R4C369) :=> r4c8=r4c5-r6c6="Swordfish"(r12c369,r7c3,r9c9) -> r4c8=r7c3=r9c9 :=> -2r7c8; r7c3=2; UP28
#5. r7c8=9->NP(45)r8c89; C(S) :=> r38c8=9
#6a. r8c8=9+r2c7=9->Loop[3] : (4=5)r8c9-5r12c9=(5-4)r1c8=4r12c9 @ :=> r1c8=45,r9c9=12; UP31
                    "Jellyfish"(5R258C8) : r2c4=r2c9-r1c8,r8c9=*[r7c8=*r6c8-r5c8=r5c1-r8c1=*r8c5] :=> r2c4=r7c8=r8c5 :=> -5r7c4
                    XYWing(5-79)r37c4,r3c2 :=> r7c4=7; UP33
                    NP(45)r8c59 :=> Skyscraper(5R28) :=> r2c4=5; UP41
                    "Swordfish"(4R258) :=> -4r146c259
                    NT(167)r167; C(S)
#6b. r8c8=9+r3c7=9->"Swordfish"(1r347c58,r3c1,r4c3) :=> -1r13c3,r6c1; NT(346)r1c23,r2c3; UP34
                    HP(67)r56c6; UP35
                    XYWing(14-5)r47c8,r5c7 :=> -5r6c8,r9c6
                    NP(45)r8c59 :=> Kite(5R8C8) :=> r2c4=5; C(S)
#6c. Conclusion : r3c8=9; UP81

[JC Van Hay]

A solution of 98.7..6..5...9..4...3..2...81......5.3...5.....5...2....8..3..1...6..9......4..7.;11.30;11.30;9.40;GP;Kz1 b;17065

Code: Select all

+----------------------+----------------------+---------------------+
| 9      8       124   | 7      135    14     | 6     1235   23     |
| 5      267     1267  | 138    9      168    | 1378  4      2378   |
| 1467   467     3     | 1458   1568   2      | 1578  1589   789    |
+----------------------+----------------------+---------------------+
| 8      1       24679 | 2349   2367   4679   | 347   369    5      |
| 2467   3       24679 | 12489  12678  5      | 1478  1689   46789  |
| 467    4679    5     | 13489  13678  146789 | 2     13689  346789 |
+----------------------+----------------------+---------------------+
| 2467   245679  8     | 259    257    3      | 45    256    1      |
| 12347  2457    1247  | 6      12578  178    | 9     2358   2348   |
| 1236   2569    1269  | 12589  4      189    | 358   7      2368   |
+----------------------+----------------------+---------------------+

1. Potential Complementary AAHS : (2467) in B14 as there are no given for the digits 2467 in stack 1.
2. If r36c1=ab->r23c2=cd where {a,b,c,d}={2,4,6,7}, then -abcdr3c1(=1),r6c2(=9). Furthermore, r12c3=ab, r45c3=cd :=> -abcdr89c3(=19)
3. Analysis of the puzzle from r89c3 :

Hidden Text: Show

Code: Select all

r8c3=1+r9c3=2->UP40
               LC(8r8c56) :=> NP(23)r18c9; UP42
               LC(8r56c4) :=> NT(467)r6c126; C(S)   
r8c3=1+r9c3=6->UP27
             * r1c5=1or3->C(S) :=> r1c5=5; UP29
               LC(1r12c6) :=> r9c4=1; UP34
               LC(8r8c56) :=> NP(23)r18c9; C(S)
r8c3=1+r9c3=9->UP29
               FXWing(46C36) :=> -46r6c1(=7); UP43
               NP(48)r68c9; UP81

r8c3=2+r9c3=1->UP33
               LC(4r6c12) :=> r4c6=6; UP34
               FXWing(6C36) :=> LC(6)r45c3; UP37
               LC(7r6c12) + LC(5r13c5) :=> r8c6=7, r7c5=2; C(S)
r8c3=2+r9c3=6->UP39
               LC(8r3c45) :=> -8r3c789
               "Swordfish"(7C9B14) : r23c9=r6c9-r6c12=r45c3-r2c3=r3c12 :=> -7r3c7
               Chain[4] : (1=5)r3c7-(5=4)r7c7-(4=8)r8c9-8r2c9=8r2c7 :=> -1r2c7; C(S)
r8c3=2+r9c3=9->UP30
               FXWing(4C36) :=> LC(4r45c6); UP32
               HP(13)r89c1; FXWing(6C36) :=> r6c1=7; UP37
               HP(26)r7c8,r9c9
               Skyscraper(5R17) :=> -5r3c7,r8c8
               XYWing(4-56)r7c17,r9c2 :=> r7c7=5; C(S)

r8c3=4+r9c3=1->C(S)
r8c3=4+r9c3=2->UP35
               XWing(5R17) :=> -5r38c58; C(S)
r8c3=4+r9c3=6->UP29
               NT(238)r189c9; UP31
               LC(2r1c89) :=> r1c3=1; C(S)
r8c3=4+r9c3=9->UP32
               HP(13)r89c1
               Skyscraper(2C14) :=> -2r9c2,r7c5
               XWing(5C47) :=> -5r39c258; UP36
               NP(18)r5c7,r6c8; C(S)

r8c3=7+r9c3=1->UP27
             * r1c5=1or3->C(S) :=> r1c5=5; UP29
               LC(1r12c6) :=> r8c5=1; UP33
               LC(8r3c45) :=> NT(579)r3c789; UP34
               LC(2r12c3) :=> r5c1=2; UP39
               NP(46)r6c26; LC(8r56c8); C(S)
r8c3=7+r9c3=2->UP28
               LC(7r6c12) :=> r4c6=7
               FXWing(4C36) :=> LC(4r45c3); C(S)
r8c3=7+r9c3=6->UP27
               HP(13)r89c1
             * r7c7=4->C(S) :=> r7c7=5; UP29
             * r1c5=1or3->C(S) :=> r1c5=5; UP35
               "Swordfish"(4C367) :=> NP(29)r47c4
               Skyscraper(2C14) :=> -2r4c3,r5c5
               LC(6r4c56) :=> NP(18)r5c45; C(S)
r8c3=7+r9c3=9->C(S)

[champagne]

2. It is easy to prove r3c9,r5c8=8.

Can you tell us what is "easy"

Code: Select all

#1. r9c3=8; UP23
#2. r3c8=8=r5c9->UP29
                 "Swordfish"(6C167) : r1c6=XWing(r56c67)-r56c1=r3c1 :=> r1c6=6; UP31
                 Chain[4] : (2=1)r1c3-1r4c3=(1-2)r4c8=2r4c5-2r6c6=2r2c6 :=> -2r1c5, NP(13)r2c13; UP34
                 LC(1r13c9,2r23c9) :=> r3c7=9
                 C(S) :=> r3c9=8=r5c8; UP26

[/quote]

Hi JC,

Le me react at that point, before your update and your complete solution.

my first reaction was

surely, esaier would be a move for beginners

More seriously several remarks.

Technically, this would classified above the highest difficulty for a move in serate mode (nested chain in dynamic plus mode for the qualification of 6r1c6).

But what bothers me is the very late show up of the contradiction.

Regarding the difficulty, I think for long that for skill players, when they have a reason to fight against a candidate, they don't care about the kind of move they uses.

Congratulations and thanks for that complete solutions of the 2 puzzles.

I'll continue to decipher your paths.

I am however wondering if we don't have a simpler attack and have a look in that sense to the solver proposal for the first puzzle in a version not passing the first level of nested chains, but using more than serate in dynamic mode.

[champagne]

9..8..7...8..7..6...5..4...8....53...2.1.........8...94....8..3.1.2..8......6..7.;11.10;1.20;1.20;GP;Kz1 b;15666

Code: Select all

#2. r3c8=8=r5c9->UP29
                 "Swordfish"(6C167) : r1c6=XWing(r56c67)-r56c1=r3c1 :=> r1c6=6; UP31
                 Chain[4] : (2=1)r1c3-1r4c3=(1-2)r4c8=2r4c5-2r6c6=2r2c6 :=> -2r1c5, NP(13)r2c13; UP34
                 LC(1r13c9,2r23c9) :=> r3c7=9
                 C(S) :=> r3c9=8=r5c8; UP26

r8c9=6->"Swordfish"(6C167) : r1c6=XWing(r56c67)-r56c1=r3c1 :=> r1c6=6; UP28
        HP(67)r3c12
        "Jellyfish"(2R367C1) : r2c1=r9c1-r7c3="Swordfish"(r367c78,r3c5,r6c6) -> r2c1=r6c6=r3c5 :=> -2r2c6; UP29
        r4c9=7->C(S) :=> r5c9=7; UP31
        NT(359)r237c4; UP33
        NT(359)r8c13,r9c2; UP34
        NP(67)r37c3 :=> NP(49)r4c25 :=> NP(12)r4c89 :=> r6c8=5
        C(S)


One interesting point in JC's solution for puzzle 1 is that start with 2 bi values.

JC looks for a contradiction in that case using all possible rules, and mainly, as usual for him, all "fish eliminations" including the hardest.

After r3c9=8 and r7c7=6, skfr rates that puzzle 9.0/7.8/3.4, nothing to block a skilled player.

This can be applied in many hardest puzzles, and fits with manual players skills.
In hardest puzzles, the number of bi_values is relatively limited.

For the second puzzle, the choice of r89c3 as start for a scenario analysis is somehow "tailor made".

[champagne]

The last update of the data base of potential hardest contains 1096206 puzzles.


I made in several steps a complete check for exocets.
The risk of error is high, but should not change much the distribution.

The clues distribution in the range 20_26 is the following

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20   21    22     23      24      25     26
262  10525 74134  247220  390692  333130 40164


I found an exocet pattern in 924298 puzzles, 84.3% of the total

The split is the following

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894431 puzzles have a standard JE appearing quickly
   197 puzzles have a JE coming later
 26749 puzzles have a pattern corresponding to the extended JE
  2921 puzzles have another exocet pattern, not recognised as JE

The so called "extended JE" pattern fits with my code, but a deeper analysis would be required to see whether David would still qualify them as JE.

The split of the total per number of clues is the following

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20   45   262   17.18%
21   1763   10525   16.75%
22   42323   74134   57.09%
23   213777   247220   86.47%
24   353664   390692   90.52%
25   284314   333130   85.35%
26   28412   40164   70.74%
         
   924298   1096127   84.32%

I was expecting a lower frequency for number of clues>25, it seems correct.

I'll load in the next days the corresponding files in my google drive

[sultan vinegar]

Great work, I look forward to analysing your files. Can you clarify exactly what patterns are covered by the "extended JE", as so many have been mentioned in multiple posts in multiple threads that it's hard to keep track of them all. Also, where does DJE fit in these results?

If it would help, I volunteer to write and maintain an "Ultimate Exocetidae Guide", collecting all the definitions, patterns, extensions, secondary eliminations etc in the one post for easy reference. Champagne could then periodically copy and paste the post into one of the reserved posts at the very beginning of this thread whenever I update it. Yes/No/Better Ideas???