## Vanhegan extreme December 18, 2012

Post puzzles for others to solve here.

### Re: Vanhegan extreme December 18, 2012

tlanglet wrote:
Marty R. wrote:I went for the DP as well and don't know if the notation I used is acceptable.

(9r8c8=2r9c8)-(1235=9)r5c78,r6c7,r4c7=>r7c7<>9

Marty,

Your notation is fine by me except you do need to indicate the basis of the first strong link, which in this case is the potential DP. Note that David had one format and I used an alternate format to provide that detail. I am not aware of a "standard" notation to handle these types of circumstances.

Ted

Ted, thanks, you are 100% correct. I got lazy because the DP had been previously discussed. Had I been the first, I would have posted a basis.
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### Re: Vanhegan extreme December 18, 2012

Leren wrote:
daj 95376 Wrote: Hmmm! I don't see it. For example, I don't see B=1, if it existed, creating a conflict in your six cells.

The way this move works is as follows:

If A is 1 and B is 3 then d = nothing! = > conflict.

If A is 1 and B is 2 then c is 6, b is 8 and a is ... nothing! = > conflict.

Thus whatever the value of B, A cannot be 1 = > A cannot be 1.

Leren

Hi, Leren. Did I ever mention I hate the I-word? If you speak my lingo....

Code: Select all
`*--------------------------------------------------------------------------------*| 8       1       7        | 4       3       5        | 2       6       9        || 6       2       3        | 7       9       1        | 4       8       5        || 5       4       9        | 268     268     28       | 7       1       3        ||--------------------------+--------------------------+--------------------------|| 2       356     48       | 368     1       378      | 359     3479    67       || 7       356    *18       | 9      *268     4        | 35-1   *23     *26       || 9       36      14       | 236     5       237      |*13      2347    8        ||--------------------------+--------------------------+--------------------------|| 1       7       6        | 238     28      2389     | 39      5       4        || 34      8       2        | 5       47      39       | 6       379     1        || 34      9       5        | 1       47      6        | 8       237     27       |*--------------------------------------------------------------------------------*`

To me, it's another (suspiciously common around here) ALS-xz, which can always be expressed as a very tidy AIC (complement David's ANS.)
ronk-style:
(1=2368)als:r5c3589-(3=1)r6c7 ==>r5c7<>1

If one's into the ALS-xz approach rather than the AIC, there's an established way of expressing that as well (identify your sets and the restricted common/other common.)

BTW, you used the term "ALS Pair Exclusion," which ain't in my lexicon. Aligned Pair Exclusion? Maybe, but that's considered subsumed by other methods.

Luke
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### Re: Vanhegan extreme December 18, 2012

Leren wrote:
daj 95376 Wrote: Hmmm! I don't see it. For example, I don't see B=1, if it existed, creating a conflict in your six cells.

The way this move works is as follows:

If A is 1 and B is 3 then d = nothing! = > conflict.

If A is 1 and B is 2 then c is 6, b is 8 and a is ... nothing! = > conflict.

Thus whatever the value of B, A cannot be 1 = > A cannot be 1.

Okay, but I think you have the cart before the horse. If we just consider the candidates in r5c8, then forcing chain logic is essentially equivalent to your results.

Code: Select all
`r5c8=2 , r5c9=6 , r5c5=8 , r5c3=1 => r5c7<>1r5c8=3 , r6c7=1                   => r5c7<>1`

As far as I'm concerned, the use of "als" in a chain format is usually redundant. So, I would write:

(1=8)r5c3 - (8=62)r5c59 - (2=3)r5c8 - (3=1)r6c7 => r5c7<>1

- or compress it to -

(1=8623)r5c3598 - (3=1)r6c7 => r5c7<>1

As Luke mentioned, this is equivalent to an ALS-xz using your abBcd cells.
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### Re: Vanhegan extreme December 18, 2012

Without any preconceptions, an Almost Locked Set can be appreciated as being one which would become fully locked if one digit in these cells were resolved one way or another. Depending on whether the deciding candidate is resolved to be true or false there are two possibilities:

An Almost Naked Set is N cells that contain N+1 candidates. If one of these candidates is false all the others become locked in these cells, and there is a strong internal link between this candidate and the other unlocked candidates.

An Almost Hidden Set is N cells containing N-1 locked candidates. If one of the unlocked candidates is true all the others are false and there is a weak internal link between them.

Originally cells sets notated as ALSs were being used with both weak and strong internal links, but it caused confusion as the validity checks required are different for the two types. Nowadays to avoid such confusion I don't describe sets as being ALSs but use either ANS or AHS, depending on the type of the internal link that’s involved.

Code: Select all
`*--------------------------------------------------------------------------------*| 8       1       7        | 4       3       5        | 2       6       9        || 6       2       3        | 7       9       1        | 4       8       5        || 5       4       9        | 268     268     28       | 7       1       3        ||--------------------------+--------------------------+--------------------------|| 2       356     48       | 368     1       378      | 359     3479    67       || 7       356    *18       | 9      *268     4        | 35-1   *23     *26       || 9       36      14       | 236     5       237      |*13      2347    8        ||--------------------------+--------------------------+--------------------------|| 1       7       6        | 238     28      2389     | 39      5       4        || 34      8       2        | 5       47      39       | 6       379     1        || 34      9       5        | 1       47      6        | 8       237     27       |*--------------------------------------------------------------------------------*`

ANSs and AHSs form complementary cell sets either of which can be used in a chain. We've seen the ANS version

(1=3)r6c7 - (2368=1)ANS:r5c3589 => r5c7 <> 1

But using the complementary AHS is neater:

(1=3)r6c7 – (3)r5c8 = (35)AHS:r5c27 => r5c7 <> 1

It's become common to omit the ANS and AHS abbreviations, but just because they haven't been included, it doesn't mean that ALS logic hasn't been employed Danny.
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### Re: Vanhegan extreme December 18, 2012

myself wrote:As far as I'm concerned, the use of "als" in a chain format is usually redundant.

David P Bird wrote:ANSs and AHSs form complementary cell sets either of which can be used in a chain. We've seen the ANS version

(1=3)r6c7 - (2368=1)ANS:r5c3589 => r5c7 <> 1

But using the complementary AHS is neater:

(1=3)r6c7 – (3)r5c8 = (35)AHS:r5c27 => r5c7 <> 1

It's become common to omit the ANS and AHS abbreviations, but just because they haven't been included, it doesn't mean that ALS logic hasn't been employed Danny.

Okay I see how my meaning could be interpreted other than as I meant. What I meant was that the inclusion of the ALS/ANS/AHS qualifiers in a chain were often redundant -- not the logic. If you can provide another interpretation for the following chains, then I will support including the qualifiers.

Code: Select all
`(1=3)r6c7 - (2368=1)r5c3589 => r5c7 <> 1(1=3)r6c7 – (3)r5c8 = (35)r5c27 => r5c7 <> 1 `

In fact, the first chain should be:

Code: Select all
`(1=3)r6c7 - (3268=2681)r5c3589 => r5c7 <> 1`

Because one quadtuple being false forces another to be true. However, I'm tempted to just use:

Code: Select all
`(1=3)r6c7 - (3=1)r5c8953 => r5c7 <> 1`

And let the use of four cells for a strong inference be self-evident of an ANS.

What I find interesting is that no one proposed:

Code: Select all
`(1=3)r6c7 - (3=1)r5c8953 => r6c3 <> 1`
daj95376
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### Re: Vanhegan extreme December 18, 2012

My oft-repeated statement on labels: :

Isn't the whole purpose of notation to transmit the logic of our chains to others, ideally as clearly as possible? With that in mind, don't we really use labels such as 'als' and 'ahs' as a favor to others? In the case of an ALS, particularly in a longer AIC, the strong link of the ALS, as distinguished from all the other 'regular' strong links, may not jump out at you right away, even if you are an experienced solver, but the 'als' label signals its presence making the reading of the chain easier.

BTW, speaking of terminology, I've seen the term 'nice loop' being used here to describe an AIC notated with Eureka notation. Am I mistaken or has it not always been true that the term 'nice loop' has always been associated only with chains using 'nice loop' notation? Thus, if we are using Eureka AIC notation, we can have a continuous (sometimes notated with the label: 'loop') or discontinuous AIC, but we don't call them 'nice loops'.
DonM
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### Re: Vanhegan extreme December 18, 2012

Danny, I actually agree with much of what you say, and would prefer the ALSs designations to be omitted altogether rather than being used in a confusing way for newcomers. The problem in this thread is that the contributors here have arrived from different directions and don't share the same AIC/Eureka knowledge base. If you like, we are a mixed ability class!

Many ALS patterns in the puzzles we see here are simultaneously both ANSs and AHSs because they contain 2 unlocked candidates. These are the ones that throw learners in my experience.

Your alternative ways of combining different cells hadn't hit me when I wrote before and are interesting. However I disagree with you on some individual points

You say the notation in full should be:

(1=3)r6c7 - (3268=2681)r5c3589 => r5c7 <> 1

but there is no need to give the full digit set for the two ANS options as if we split them into two terms

(2368)r5c3589 = (1)r5c3589

it makes sense as each node is a Boolean and they can't both be false.

Then you suggest

(1=3)r6c7 - (3=1)r5c8953 => r6c3 <> 1

but this creates an extra burden on readers as they must trawl through the cells (which aren't listed in ascending order) counting the passenger digits. If you make a mistake it's then difficult for us to work out whether you've miscounted or mistyped!

Don, As I said, when the term ALS was first introduced it covered both ANSs and AHSs but later it came to mean only ANSs which resulted in having to use AHS for the missing case. IMO the complement to an AHS is an ANS and it should be called just that for the sake of clarity. I believe that we share common ground in wanting to make our findings as accessible as possible for newcomers as it is the best way to attract new blood. We should therefore not only avoid creating new artificial obstacles, but also should try to slowly pull the old ones down.

You make a good point about Nice Loop terminology being inappropriate when Eureka style Boolean chains are being used. The two rival systems ran side by side for quite a while, but Nice Loop notations couldn’t handle the addition of patterns into the mix so well and eventually lost out.
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### Re: Vanhegan extreme December 18, 2012

DonM, I understand your position on trying to make a solution as clear as possible. I'll give it further consideration.

David P Bird wrote:You say the notation in full should be:

(1=3)r6c7 - (3268=2681)r5c3589 => r5c7 <> 1

but there is no need to give the full digit set for the two ANS options as if we split them into two terms

(2368)r5c3589 = (1)r5c3589

it makes sense as each node is a Boolean and they can't both be false.

For a bivalue cell, we can write (3=2)r3c6 ... or we can write (3)r3c6 = (2)r3c6 without loss of context. I'm saying that the quad strong inference in r5c3579 should be written (3268=2681)r5c3589 -- and that it can be interpreted as (3268)r5c3589 = (2681)r5c3589. In the case of (2368=1)r5c3589, the expression can't be split and still make sense.

David P Bird wrote:Then you suggest

(1=3)r6c7 - (3=1)r5c8953 => r6c3 <> 1

but this creates an extra burden on readers as they must trawl through the cells (which aren't listed in ascending order) counting the passenger digits. If you make a mistake it's then difficult for us to work out whether you've miscounted or mistyped!

Actually, my (condensed) expression comes as close as any other for being accurate. I've actually taken the burden off of the reader because I've listed the cells in the order the ANS network would decompose in the relationship.

Code: Select all
`r5c8<>3 , r5c8=2 , r5c9=6 , r5c5=8 , r5c3=1     ==>>     (3=1)r5c8953`

The reader now knows (from my chain) that an assumption of r6c7<>1 leads to r5c3=1. Since r6c3 sees both cells, the elimination r6c3<>1 follows.

As for my making a mistake, anyone who presents a chain with a mistake will place some level of burden on the reader. Do you reallllllly want to get into what levels of possible mistakes are acceptable and not acceptable?

Note: Sometimes multiple cells are needed for intermediate pairs/triples. As long as these cells are listed next to each other, that should be sufficiently clear to determine their relationship. Consider the ANS reading right-to-left:

Code: Select all
`r5c3<>1 , r5c3=8 , r5c5<>8 , r5c59=26 , r5c8<>2 , r5c8=3     ==>>     (1=3)r5c3598`
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### Re: Vanhegan extreme December 18, 2012

daj95376 wrote: In the case of (2368=1)r5c3589, the expression can't be split and still make sense.

Danny, < here > you wrote

M-Ring B (6=1)r7c8 - r23c8 = (1-6)r3c9 = (6)r789c9 - loop => r3c9<>5; r89c8<>6

So you can make sense of the red Boolean when you write it but not when I do!
In words: (these cells contain 2368) and (1 is contained in one of these cells) are both True/False conditions.

On the rest of your post, from past experience I've learnt that once you get an idée fixe nothing will shift it, so after this last shot I won’t try further.

1. Every node in an AIC must be a Boolean and the alternating inferences must be shown.

2. The Eureka notation was a massive compromise reached after some debate. For example it would be far more efficient to use an A1 cell reference rather than r1c1 but people weren't prepared to change. Eventually everyone adopted it even though it wasn't to individual tastes simply so we could understand one another. However, ever since then every new recruit has thought that the rest of the regiment should fall in step with him.
David P Bird
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### Re: Vanhegan extreme December 18, 2012

David P Bird wrote:
daj95376 wrote: In the case of (2368=1)r5c3589, the expression can't be split and still make sense.

Danny, < here > you wrote

M-Ring B (6=1)r7c8 - r23c8 = (1-6)r3c9 = (6)r789c9 - loop => r3c9<>5; r89c8<>6

So you can make sense of the red Boolean when you write it but not when I do!
In words: (these cells contain 2368) and (1 is contained in one of these cells) are both True/False conditions.

In my post that you reference, <6> is a candidate in all three cells of r789c9 as part of a grouped strong inference. When you can demonstrate that <1> exists in all four cells of r5c3589, then I'll take your use of (2368=1)r5c3589 seriously.

My use of (3=1)r5c8953 is every bit as improper as (2368=1)r5c3589. The only difference is that I'm willing to admit that my expression is a condensation of a more complete term. DPB feels that his expression is a complete term.
Last edited by daj95376 on Thu Dec 20, 2012 4:36 am, edited 1 time in total.
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### Re: Vanhegan extreme December 18, 2012

daj95376 wrote:

What I find interesting is that no one proposed:

Code: Select all
`(1=3)r6c7 - (3=1)r5c8953 => r6c3 <> 1`

However, earlier in this thread I posted a solution which included this elimination:
interestingly, same cells, same result for the 1-4 ALS: A: 12368 at r5c3589; B: 13 at r6c7; X=3; Z=1; => -1r5c7, -1r6c3; stte
pjb
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### Re: Vanhegan extreme December 18, 2012

pjb wrote:daj95376 wrote:

What I find interesting is that no one proposed:

Code: Select all
`(1=3)r6c7 - (3=1)r5c8953 => r6c3 <> 1`

However, earlier in this thread I posted a solution which included this elimination:
interestingly, same cells, same result for the 1-4 ALS: A: 12368 at r5c3589; B: 13 at r6c7; X=3; Z=1; => -1r5c7, -1r6c3; stte

pjb,

My apologies. I was concentrating on chains and missed your solution as being equivalent.
daj95376
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### Re: Vanhegan extreme December 18, 2012

daj95376 wrote:
Code: Select all
`(1=3)r6c7 - (3=1)r5c8953 => r5c7 <> 1(1=3)r6c7 - (3=1)r5c8953 => r6c3 <> 1`

Thanks Danny -- a learning moment.
dan

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### Re: Vanhegan extreme December 18, 2012

Luke451 wrote:To me, it's another (suspiciously common around here) ALS-xz, which can always be expressed as a very tidy AIC (complement David's ANS.)
ronk-style:
(1=2368)als:r5c3589-(3=1)r6c7 ==>r5c7<>1

You are apparently referring to the "als:" label. In entirety, my preferences would be ...
(1=3)als:r5c3589-(3=1)r6c7 ==> r5c7<>1
... or ...
(1268=2683)als:r5c3589-(3=1)r6c7 ==> r5c7<>1
... with or without the label.
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### Re: Vanhegan extreme December 18, 2012

David P Bird wrote:
daj95376 wrote: In the case of (2368=1)r5c3589, the expression can't be split and still make sense.

Danny, < here > you wrote

M-Ring B (6=1)r7c8 - r23c8 = (1-6)r3c9 = (6)r789c9 - loop => r3c9<>5; r89c8<>6

So you can make sense of the red Boolean when you write it but not when I do!
In words: (these cells contain 2368) and (1 is contained in one of these cells) are both True/False conditions.

In reply daj95376 wrote:In my post that you reference, <6> is a candidate in all three cells of r789c9 as part of a grouped strong inference. When you can demonstrate that <1> exists in all four cells of r5c3589, then I'll take your use of (2368=1)r5c3589 seriously.

My use of (3=1)r5c8953 is every bit as improper as (2368=1)r5c3589. The only difference is that I'm willing to admit that my expression is a condensation of a more complete term. DPB feels that his expression is a complete term.

Let's analyse this: daj is saying that (1)r5c3589 is an invalid node because (1) can only possibly occupy one of these cells, namely r5c3. I'm saying that the node is valid because it will be true if (1) occupies any of the cells (or in other words if it exists in the ANS) and false otherwise which is all that is required.

Now daj has to get himself out of a hole he dug for himself, so he goes onto say that he accepts his alternative fails to meet his unnecessary requirement, but he seems to grant himself a special licence because he's pointed that fact out. However, presumably he refuses to grant me the same licence on the grounds that I don’t accept that his condition is required.

If we run with that, then how does he propose that in a few months time anyone will know why one form is accepted and the other form isn't?

I've been contributing to this sub-forum in an attempt to help contributors understand the disciplines involved in notating AICs, and furthermore to write them in a reader-friendly way - such as with spaces and enough information to make them easy to follow and validate. One of my hopes was that this would increase the number of participants. However I consider my efforts have largely failed, and that my posts have probably been considered to be unwelcome intrusions by most. In the light of that and also this inane argument, I've decided to stop intruding.
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