by **Leren** » Mon Oct 22, 2012 7:22 am

*-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