The New Sudoku Players' Forum

by **ArkieTech** » Sat May 16, 2015 11:15 pm

Code: Select all: *-----------* |...|...|...| |...|4..|.52| |74.|65.|1..| |---+---+---| |1..|..3|..7| |.3.|8.5|.6.| |8..|9..|..3| |---+---+---| |..3|.79|.24| |21.|..4|...| |...|...|...| *-----------*

Play/Print this puzzle online

by **SteveG48** » Sun May 17, 2015 12:12 am

Code: Select all: *--------------------------------------------------------------------* | 35 2589 12589 | 7 39 128 | 4689 489 689 | | 36 689 1689 | 4 39 18 | 7 5 2 | | 7 4 289 | 6 5 28 | 1 3 89 | *----------------------+----------------------+----------------------| | 1 c569 4569 | 2 c46 3 | 4589 489 7 | |b49 3 7 | 8 b14 5 | 2 6 b19 | | 8 256 2456 | 9 146 7 | 45 14 3 | *----------------------+----------------------+----------------------| |a5-6 d568 3 | 1 7 9 |d68 2 4 | | 2 1 a69 | 35 8 4 | 369 7 569 | |a49 7 489 | 35 2 6 | 389 189 1589 | *--------------------------------------------------------------------*

(5=694)b7p167 - (4=194)r5c159 - (94=65)r4c25 - (5=86)r7c27 => -6 r7c1 ; stte

by **Leren** » Sun May 17, 2015 12:36 am

Code: Select all: *--------------------------------------------------------------* |b35 2589 12589 | 7 a39 128 | 468-9 48-9 68-9 | | 36 689 1689 | 4 39 18 | 7 5 2 | | 7 4 e289 | 6 5 28 | 1 3 f89 | |--------------------+--------------------+--------------------| | 1 569 4569 | 2 46 3 | 4589 489 7 | | 49 3 7 | 8 14 5 | 2 6 19 | | 8 256 2456 | 9 146 7 | 45 14 3 | |--------------------+--------------------+--------------------| |c56 568 3 | 1 7 9 | 68 2 4 | | 2 1 d69 | 35 8 4 | 369 7 569 | | 49 7 489 | 35 2 6 | 389 189 1589 | *--------------------------------------------------------------*

(9=3) r1c5 - (3=5) r1c1 - (5=6) r7c1 - (6=9) r8c3 - r3c3 = r3c9 => - 9 r1c789; stte, or

Code: Select all: *--------------------------------------------------------------* |b35 2589 1258-9 | 7 39 128 | 4689 489 689 | |c36 c689 c1689 | 4 39 c18 | 7 5 2 | | 7 4 28-9 | 6 5 28 | 1 3 89 | |--------------------+--------------------+--------------------| | 1 569 4569 | 2 46 3 | 4589 489 7 | | 49 3 7 | 8 14 5 | 2 6 19 | | 8 256 2456 | 9 146 7 | 45 14 3 | |--------------------+--------------------+--------------------| |a56 568 3 | 1 7 9 | 68 2 4 | | 2 1 a69 | 35 8 4 | 369 7 569 | | 49 7 489 | 35 2 6 | 389 189 1589 | *--------------------------------------------------------------*

ALS XY Wing: (9=5) r7c1, r8c3 - (5=3) r1c1 - (3=9) r2c1236 => - 9 r13c3; stte

Leren

by **pjb** » Sun May 17, 2015 1:00 am

Code: Select all: 35 2589 12589 | 7 39 128 | 4689 489 689 36 689 1689 | 4 39 18 | 7 5 2 7 4 28-9 | 6 5 28 | 1 3 89 ------------------------+----------------------+--------------------- 1 569 4569 | 2 46 3 | 4589 489 7 49 3 7 | 8 14 5 | 2 6 19 8 256 2456 | 9 146 7 | 45 14 3 ------------------------+----------------------+--------------------- 56 568 3 | 1 7 9 | 68 2 4 2 1 69 | 35 8 4 | 369 7 569 49 7 489 | 35 2 6 | 389 189 1589

(6-8)r7c7 = (8-5)r7c2* = (5-6)r7c1 = r2c1 - (68=9)r2c2* - r3c3
(6)r8c7 - (6=9)r8c3 - r3c3
(6)r8c9 - (6=9)r8c3 - r3c3 => -9 r3c3; stte

Phil

by **Sudtyro2** » Tue May 19, 2015 6:28 pm

SteveG48 wrote:
Code: Select all
*--------------------------------------------------------------------* | 35 2589 12589 | 7 39 128 | 4689 489 689 | | 36 689 1689 | 4 39 18 | 7 5 2 | | 7 4 289 | 6 5 28 | 1 3 89 | *----------------------+----------------------+----------------------| | 1 c569 4569 | 2 c46 3 | 4589 489 7 | |b49 3 7 | 8 b14 5 | 2 6 b19 | | 8 256 2456 | 9 146 7 | 45 14 3 | *----------------------+----------------------+----------------------| |a5-6 d568 3 | 1 7 9 |d68 2 4 | | 2 1 a69 | 35 8 4 | 369 7 569 | |a49 7 489 | 35 2 6 | 389 189 1589 | *--------------------------------------------------------------------*
(5=694)b7p167 - (4=194)r5c159 - (94=65)r4c25 - (5=86)r7c27 => -6 r7c1 ; stte

Steve, your 4-node AIC above is very interesting, but I can't seem to fully decipher the logic. Nodes 1 and 4 are clearly ALS, and node 3 appears to be an AALS with the 94-digits logically OR'd. Node 2 and how it links to node 3 remain a mystery.

The best I could do using only your cells and the same ending digits is the following network. It treats the bivalue cell (46)r4c5 as a starting SIS.

Code: Select all: 4r4c5 – (4=19)r5c59 – (9=4)r5c1 – (4=695)b7p167 – 6r7c1 || / / || 9r4c2 ------ / || || / 6r4c5 – 6r4c2 / || / 5r4c2 – (5=86)r7c27 ----------------

But, I can't see how to morph this network into a single AIC. Any suggestions welcomed!

SteveC

by **daj95376** » Tue May 19, 2015 7:00 pm

Sudtyro2 wrote:But, I can't see how to morph this network into a single AIC. Any suggestions welcomed!

SteveG48's compressed notation is not an AIC, it's a network misrepresented as an AIC.

Code: Select all: +-----------------------------------------------------------------------+ | 35 2589 12589 | 7 39 128 | 4689 489 689 | | 36 689 1689 | 4 39 18 | 7 5 2 | | 7 4 289 | 6 5 28 | 1 3 89 | |-----------------------+-----------------------+-----------------------| | 1 h569 4569 | 2 g46 3 | 4589 489 7 | | d49 3 7 | 8 f14 5 | 2 6 e19 | | 8 256 2456 | 9 146 7 | 45 14 3 | |-----------------------+-----------------------+-----------------------| | a56 68-5 3 | 1 7 9 | 68 2 4 | | 2 1 b69 | 35 8 4 | 369 7 569 | | c49 7 489 | 35 2 6 | 389 189 1589 | +-----------------------------------------------------------------------+ # 74 eliminations remain (5=6)r7c1 - (6=9)r8c3 - (9=4)r9c1 - (4=9*)r5c1 - (9=1)r5c9 - (1=4)r5c5 - (4=6)r4c5 - (*96=5)r4c2 => -5 r7c2

This leaves a Hidden Single for 5r7c1 and results in -6r7c1.

A simpler representation of SteveG48's logic is the SIN:

Code: Select all: 6r7c1 9r8c3 4r9c1 9r5c1 1r5c9 4r5c5 6r4c5 5r4c2 [r7]-5 => -6 r7c1

_

by **JC Van Hay** » Tue May 19, 2015 8:00 pm

Sudtyro2 wrote:The best I could do using only your cells and the same ending digits is the following network. It treats the bivalue cell (46)r4c5 as a starting SIS.
Code: Select all
4r4c5 – (4=19)r5c59 – (9=4)r5c1 – (4=695)b7p167 – 6r7c1 || / / || 9r4c2 ------ / || || / 6r4c5 – 6r4c2 / || / 5r4c2 – (5=86)r7c27 ----------------

But, I can't see how to morph this network into a single AIC. Any suggestions welcomed!

SteveC

As a kraken cell (569)r4c2 or AAIC :

Code: Select all: +--------------------+---------------+-----------------+ | 35 2589 12589 | 7 39 128 | 4689 489 689 | | 36 689 1689 | 4 39 18 | 7 5 2 | | 7 4 289 | 6 5 28 | 1 3 89 | +--------------------+---------------+-----------------+ | 1 (569) 4569 | 2 (46) 3 | 4589 489 7 | | 4(9) 3 7 | 8 (14) 5 | 2 6 (19) | | 8 256 2456 | 9 146 7 | 45 14 3 | +--------------------+---------------+-----------------+ | 5-6 (568) 3 | 1 7 9 | (68) 2 4 | | 2 1 (69) | 35 8 4 | 369 7 569 | | 4(9) 7 489 | 35 2 6 | 389 189 1589 | +--------------------+---------------+-----------------+

(68=5)r7c72 - 5r4c2=*[(9=*64)r4c25 - (4=19)r5c59] - 9r5c1=(96)r9c1,r8c3 :=> -6r7c1
or
(68=5)r7c72 - 5r4c2=*[(9=*64)r4c25 - (4=19)r5c59] - (9=4)r5c1 - (4=965)r9c1,r8c3,r7c1 :=> -6r7c1

by **SteveG48** » Tue May 19, 2015 11:46 pm

daj95376 wrote:
Sudtyro2 wrote:But, I can't see how to morph this network into a single AIC. Any suggestions welcomed!

SteveG48's compressed notation is not an AIC, it's a network misrepresented as an AIC.

Code: Select all
+-----------------------------------------------------------------------+ | 35 2589 12589 | 7 39 128 | 4689 489 689 | | 36 689 1689 | 4 39 18 | 7 5 2 | | 7 4 289 | 6 5 28 | 1 3 89 | |-----------------------+-----------------------+-----------------------| | 1 h569 4569 | 2 g46 3 | 4589 489 7 | | d49 3 7 | 8 f14 5 | 2 6 e19 | | 8 256 2456 | 9 146 7 | 45 14 3 | |-----------------------+-----------------------+-----------------------| | a56 68-5 3 | 1 7 9 | 68 2 4 | | 2 1 b69 | 35 8 4 | 369 7 569 | | c49 7 489 | 35 2 6 | 389 189 1589 | +-----------------------------------------------------------------------+ # 74 eliminations remain (5=6)r7c1 - (6=9)r8c3 - (9=4)r9c1 - (4=9*)r5c1 - (9=1)r5c9 - (1=4)r5c5 - (4=6)r4c5 - (*96=5)r4c2 => -5 r7c2

This leaves a Hidden Single for 5r7c1 and results in -6r7c1.

Hmm. Interesting discussion. First, let's be clear about the intended overall logic. Going back to my original chain:

(5=694)b7p167 - (4=194)r5c159 - (94=65)r4c25 - (5=86)r7c27 => -6 r7c1,

The idea is that (5)b7p1 (r7c1 if you prefer) strong links to the 68 pair at r7c27. This eliminates 6 at r7c1. Hopefully that's clear.

As for the chain itself, I don't know why it wouldn't be considered a proper AIC. The first term uses the set [4569]b7p167. Eliminating 5 locks the set with all candidates in known positions. In particular, r9c1 is a 4.

The second term uses the set [194]r5c159. This set is already locked, but with the candidates in unknown positions. The weak link from the first term eliminates 4 at r5c1, making the position of all candidates known. In particular, r5c1 is a 9 and r5c5 is a 4.

The third term uses the unlocked set [4569]r4c25. Separate weak links from the second term eliminate 9 from r4c2 and 4 from r4c5, locking the set with r4c2 at 5. (I think of this as a parallel move, for want of a better description). The weak link from there to the fourth term locks the 68 set, establishing the result.

I'm sure that everyone here understands the logic. What I don't understand is why this is a misrepresentation of an AIC. It seems like a perfectly conventional AIC using locked sets to me.

by **daj95376** » Wed May 20, 2015 1:16 am

SteveG48 wrote:
daj95376 wrote:SteveG48's compressed notation is not an AIC, it's a network misrepresented as an AIC.

Code: Select all
+-----------------------------------------------------------------------+ | 35 2589 12589 | 7 39 128 | 4689 489 689 | | 36 689 1689 | 4 39 18 | 7 5 2 | | 7 4 289 | 6 5 28 | 1 3 89 | |-----------------------+-----------------------+-----------------------| | 1 h569 4569 | 2 g46 3 | 4589 489 7 | | d49 3 7 | 8 f14 5 | 2 6 e19 | | 8 256 2456 | 9 146 7 | 45 14 3 | |-----------------------+-----------------------+-----------------------| | a56 68-5 3 | 1 7 9 | 68 2 4 | | 2 1 b69 | 35 8 4 | 369 7 569 | | c49 7 489 | 35 2 6 | 389 189 1589 | +-----------------------------------------------------------------------+ # 74 eliminations remain (5=6)r7c1 - (6=9)r8c3 - (9=4)r9c1 - (4=9*)r5c1 - (9=1)r5c9 - (1=4)r5c5 - (4=6)r4c5 - (*96=5)r4c2 => -5 r7c2

This leaves a Hidden Single for 5r7c1 and results in -6r7c1.

Hmm. Interesting discussion. First, let's be clear about the intended overall logic. Going back to my original chain:

(5=694)b7p167 - (4=194)r5c159 - (94=65)r4c25 - (5=86)r7c27 => -6 r7c1,

The idea is that (5)b7p1 (r7c1 if you prefer) strong links to the 68 pair at r7c27. This eliminates 6 at r7c1. Hopefully that's clear.

As for the chain itself, I don't know why it wouldn't be considered a proper AIC. The first term uses the set [4569]b7p167. Eliminating 5 locks the set with all candidates in known positions. In particular, r9c1 is a 4.

The second term uses the set [194]r5c159. This set is already locked, but with the candidates in unknown positions. The weak link from the first term eliminates 4 at r5c1, making the position of all candidates known. In particular, r5c1 is a 9 and r5c5 is a 4.

The third term uses the unlocked set [4569]r4c25. Separate weak links from the second term eliminate 9 from r4c2 and 4 from r4c5, locking the set with r4c2 at 5. (I think of this as a parallel move, for want of a better description). The weak link from there to the fourth term locks the 68 set, establishing the result.

I'm sure that everyone here understands the logic. What I don't understand is why this is a misrepresentation of an AIC. It seems like a perfectly conventional AIC using locked sets to me.

Hmmm! Where to begin.

Let's start with your second term: (4=194)r5c159. This translates into ... if all of the 4s in r5c159 are false, then 194 must be true in r5c159. Really????? At best, your logic might read:

(5=694)b7p167 - (4=9*)r5c1 - (9=4)r5c95 - (*94=65)r4c25 - (5=86)r7c27 => -6 r7c1

This is network logic in chain-style notation.

Now you know why I split your ALS terms into sequences of bivalue terms in my reply. It demonstrates that the "9" elimination in r4c2 comes from a different term than the term producing the "4" elimination in r4c5.

As for what your chain is saying. The first strong link says that "6" is true in r7c1 ... and your last strong link says that "68" is true in r7c27. Do you really want to hang your hat on this logic? Maybe your notation could be shortened to:

(6=94)b7p67 - (4=9*)r5c1 - (9=4)r5c95 - (*94=65)r4c25 - (5=86)r7c27 => -6 r7c1

Now, moving on to "separate weak links from the second term eliminate 9 from r4c2 and 4 from r4c5". I'm pretty sure there's nothing in Ruud's AIC definition that allows for two, concurrent weak links. Yes, I'm aware that Leren used two weak links from an ALS to get -8r1c8 and -5r3c2 in another puzzle. The double-linked ALS' were the main topic ... and I didn't want to muddy it by arguing that -8r1c8 prevented the chain from being an AIC.

_

by **SteveG48** » Wed May 20, 2015 1:43 am

daj95376 wrote:Hmmm! Where to begin.

Let's start with your second term: (4=194)r5c159. This translates into ... if all of the 4s in r5c159 are false, then 194 must be true in r5c159. Really????? At best, your logic might read:

(5=694)b7p167 - (4=9*)r5c1 - (9=4)r5c95 - (*94=65)r4c25 - (5=86)r7c27 => -6 r7c1

This is network logic in chain-style notation.

OK, I'll buy that. I didn't really like the idea of (4=4..). And I see why you called my notation "compressed". I've been trying to avoid that after earlier discussions.

Now you know why I split your ALS terms into sequences of bivalue terms in my reply. It demonstrates that the "9" elimination in r4c2 comes from a different term than the term producing the "4" elimination in r4c5.

As for what your chain is saying. The first strong link says that "6" is true in r7c1 ... and your last strong link says that "68" is true in r7c27. Do you really want to hang your hat on this logic? Maybe your notation could be shortened to:

(6=94)b7p67 - (4=9*)r5c1 - (9=4)r5c95 - (*94=65)r4c25 - (5=86)r7c27 => -6 r7c1

Now wait. Suppose the chain had said something like (5)r7c1 .....= (6)r7c2 => -6 r7c1. Surely you wouldn't object to that. You taught me yourself that (5)r7c1 = (6)r7c2 would imply -6 r7c1 and -5 r7c2, since the strong link works both ways. Why, then, object to (5)r7c1 = (68)r7c27 => -6 r7c1? (I do, however, like the shortening of the first term. I should have seen that.)

Now, moving on to "separate weak links from the second term eliminate 9 from r4c2 and 4 from r4c5". I'm pretty sure there's nothing in Ruud's AIC definition that allows for two, concurrent weak links. Yes, I'm aware that Leren used two weak links from an ALS to get -8r1c8 and -5r3c2 in another puzzle. The double-linked ALS' were the main topic ... and I didn't want to muddy it by arguing that -8r1c8 prevented the chain from being an AIC.

_

OK, I can't argue that point, not being familiar enough with the formal definition for an AIC. I don't see why two concurrent links wouldn't be allowed, but if they aren't, they aren't. I like the "parallel move", but I won't call it an AIC

.

Thanks, Danny. Your thoughts are always helpful.

by **daj95376** » Wed May 20, 2015 7:16 am

SteveG48 wrote:
Now you know why I split your ALS terms into sequences of bivalue terms in my reply. It demonstrates that the "9" elimination in r4c2 comes from a different term than the term producing the "4" elimination in r4c5.

As for what your chain is saying. The first strong link says that "6" is true in r7c1 ... and your last strong link says that "68" is true in r7c27. Do you really want to hang your hat on this logic? Maybe your notation could be shortened to:

(6=94)b7p67 - (4=9*)r5c1 - (9=4)r5c95 - (*94=65)r4c25 - (5=86)r7c27 => -6 r7c1

Now wait. Suppose the chain had said something like (5)r7c1 .....= (6)r7c2 => -6 r7c1. Surely you wouldn't object to that. You taught me yourself that (5)r7c1 = (6)r7c2 would imply -6 r7c1 and -5 r7c2, since the strong link works both ways. Why, then, object to (5)r7c1 = (68)r7c27 => -6 r7c1? (I do, however, like the shortening of the first term. I should have seen that.)

Yes, I stretched my point too far. Sorry! I tend to "over-apply" the possible network secondary eliminations when I encounter chain-style notation for a network. Thus, to try and recreate your logic, I treated all of your assignments as "real" up to =5r4c2 ... and then came to the conclusion that (68) couldn't exist in r7c27. I don't have this problem when a diagram is presented or asterisk (*) is incorporated in the notation.

Note: I'll be away for a few days, but I'll check on this thread when I return.

_

by **SteveG48** » Thu May 21, 2015 1:55 pm

daj95376 wrote:Note: I'll be away for a few days, but I'll check on this thread when I return.

Actually, I think we're about done, but the discussion is fun so let's summarize. I started with:

(5=694)b7p167 - (4=194)r5c159 - (94=65)r4c25 - (5=86)r7c27 => -6 r7c1

After discussion I would change it to this:

(6=94)b7p67 - (4=9*)r5c1 - (9=14)r5c59 - (4*9=65)r4c25 - (5=86)r7c27 => -6 r7c1

The only point remaining is the highlighted term, the "parallel move". I seems to me that if both 4 and 9 are true in r5c15, then they must both be false in r4c25, so while I called it 2 separate weak links, it's still a proper weak link between two sets, albeit two houses, b4 and b5, are involved. (Naturally, the fact that the 9 is in b4 and the 4 is in b5 is critical, but there's nothing unusual about using the known locations of candidates in a set.) Would we agree, then, that the chain as modified is a proper AIC, given the star notation?

by **Sudtyro2** » Thu May 21, 2015 5:10 pm

Code: Select all: *--------------------------------------------------------------* | 35 2589 12589 | 7 39 128 | 4689 489 689 | | 36 689 1689 | 4 39 18 | 7 5 2 | | 7 4 289 | 6 5 28 | 1 3 89 | |--------------------+--------------------+--------------------| | 1 d569 4569 | 2 c46 3 | 4589 489 7 | |b49c 3 7 | 8 c14 5 | 2 6 b19 | | 8 256 2456 | 9 146 7 | 45 14 3 | |--------------------+--------------------+--------------------| | 5-6 e568 3 | 1 7 9 |e68 2 4 | | 2 1 a69 | 35 8 4 | 369 7 569 | |a49 7 489 | 35 2 6 | 389 189 1589 | *--------------------------------------------------------------*

The “parallel move” issue reminds me of Myth Jellies' CoALS rule which can be applied to the two critical overlapping ALS, (149)r5c15 and (146)r45c5. The ALS overlap digits are the 4s and the 1, while the non-overlap digits are only 6 and 9.

All digit pairs in the chain below are AND'd except as noted in r4c2. The resulting non-network AIC seems “clean.” The only hitch may be that cell r5c1 is used twice in the chain, but there doesn't appear to be any conflict with the CoALS rule.

(6=94)b7p67 - (4=91)r5c19 – (14=69)r45c5,r5c1 – (6|9=5)r4c2 - (5=86)r7c27 => -6 r7c1

SteveC

by **SteveG48** » Thu May 21, 2015 6:06 pm

Sudtyro2 wrote:
Code: Select all
*--------------------------------------------------------------* | 35 2589 12589 | 7 39 128 | 4689 489 689 | | 36 689 1689 | 4 39 18 | 7 5 2 | | 7 4 289 | 6 5 28 | 1 3 89 | |--------------------+--------------------+--------------------| | 1 d569 4569 | 2 c46 3 | 4589 489 7 | |b49c 3 7 | 8 c14 5 | 2 6 b19 | | 8 256 2456 | 9 146 7 | 45 14 3 | |--------------------+--------------------+--------------------| | 5-6 e568 3 | 1 7 9 |e68 2 4 | | 2 1 a69 | 35 8 4 | 369 7 569 | |a49 7 489 | 35 2 6 | 389 189 1589 | *--------------------------------------------------------------*

The “parallel move” issue reminds me of Myth Jellies' CoALS rule which can be applied to the two critical overlapping ALS, (149)r5c15 and (146)r45c5. The ALS overlap digits are the 4s and the 1, while the non-overlap digits are only 6 and 9.

All digit pairs in the chain below are AND'd except as noted in r4c2. The resulting non-network AIC seems “clean.” The only hitch may be that cell r5c1 is used twice in the chain, but there doesn't appear to be any conflict with the CoALS rule.

(6=94)b7p67 - (4=91)r5c19 – (14=69)r45c5,r5c1 – (6|9=5)r4c2 - (5=86)r7c27 => -6 r7c1

SteveC

Nice. Should the 4 in the third term be on the other side of the = sign?

I like how it works, but if the parallel move doesn't qualify as a proper AIC, I don't see why the CoALS rule would. The third term involves ALSs in two different houses, which isn't normally allowed. I think I would write it:

(6=94)b7p67 - (4=91)r5c19 – (1=4)r5c5 - [4=(9)r5c1&(6)r4c5] – (69=5)r4c2 - (5=86)r7c27 => -6 r7c1

When I write it this way, it certainly doesn't look "non-network", because of the fourth term.

by **Sudtyro2** » Thu May 21, 2015 7:16 pm

SteveG48 wrote:...Should the 4 in the third term be on the other side of the = sign?

As I understand it, the digit placement is correct as shown. The CoALS rule says there is a strong link between the AND'd digits in the overlap cell(s) and the AND'd digits in the non-overlap cells.

SteveG48 wrote:I like how it works, but if the parallel move doesn't qualify as a proper AIC, I don't see why the CoALS rule would. The third term involves ALSs in two different houses, which isn't normally allowed.

In the above link Myth carefully defines his CoALS structure as the logical union of two regular ALS to form a Combined ALS. He then proves his rule and provides some good examples of how it can be used in a regular AIC. I think Myth was pretty careful in most everything he did. Plus, I know there's a long history of topics dealing with other aspects of overlapping ALS. Most of that, unfortunately, is pretty much over my head. IOW, even though I've learned a lot in this thread, I'll have to defer to the experts to help out with the hard questions.

SteveC

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