ALS Chains -A Tutorial ASI#3

Advanced methods and approaches for solving Sudoku puzzles

ALS Chains -A Tutorial ASI#3

ALS CHAINS -A TUTORIAL (further revised 03/25/2009)

This is substantial rewrite of a tutorial that was presented in the Eureka forum in July, 2007 by sirdave though it was to some extent a collaborative effort between the two of us. This tutorial has a direct application in what I call type A pattern-solving ie. pure visual pattern solving, but the concepts below can also be applied for use in type B pattern-solving (ie. the nice loop and AIC chains). For instance, those familiar with nice loops or AICs will immediately notice that some of the examples could be expressed as xy-chains or grouped nice loops/AICs.

Much of this tutorial has an application for newer solvers as well as more advanced solvers so some of the descriptions that follow will be taking that into account. However, in making the information accessible to less experienced solvers some of the descriptions will, by necessity, seem rather elementary to more experienced solvers. It is assumed that the reader understands what an ALS (almost locked set) is. The basic ALS (which provides the building blocks of ALS Chains) is a group of N cells in a single house (ie. row, column or box) with candidates for N+1 digit values. Removal of one of the digit values will result in a locked set (such as a naked pair or naked triple). In addition to the examples below, newer solvers might want to check out the excellent descriptions of the ALS XZ-rule at these sites:

http://www.sudocue.net/guide.php#ALS
http://www.sudopedia.org/wiki/ALS-XZ

First a short recap: ALS patterns were first placed on the map in an organized way by bennys in the ‘Almost Locked Sets (for now)’ thread on the Player’s forum back in December, 2005 ( http://forum.enjoysudoku.com/viewtopic.php?p=16334). Since then, ALSs have been generally described as fitting into 3 general categories: the ALS XZ-rule, ALS XY-wing and ALS XY-chains. In addition, Doubly-Linked ALSs have been described, but have received much less attention than the 3 main categories. There have been a number of ALS-related threads, but none have appeared to shed much light on how to use ALSs in a practical manner that is accessible to the average solver. On the contrary, the descriptions, except for the ALS XZ-rule and one other exception, have been somewhat obscure such that it’s not surprising that the power of ALSs has often not been taken advantage of in type A pattern-solving use. The other exception is a series of Eureka threads by Anne Morelot in May and June, 2006 who provided what were the first, if not the only, examples-in-graphics of virtually all forms of the ALS patterns mentioned above. Unfortunately, a change in that forum’s software has removed the graphics so a reference to that thread would not be of much use.

There are several reasons for this tutorial. First, it is to simplify the concept of the various ALS patterns from that of what is originally three, the ALS XZ-rule, ALS X-wing and ALS XY-chains to that of one: ALS Chains. There is no real good reason to distinguish between the three categories and keeping the separate names unnecessarily complicates and confuses the situation. After all, other than the number of individual ALSs (or sets) involved, they all have the same construct: A chain of ALSs whereby, if each is connected by a ‘restricted common’ digit, then if the flanking ALSs of the chain each has the same digit, each of which ‘sees’ a common digit, then that common digit can be eliminated.

Second, is to provide graphic-based examples of the use of ALS Chains and Doubly-Linked ALSs in type A pattern-solving. Doubly-Linked ALSs are not exactly the same as ALS Chains, but they do have a close similarity to a 2-set ALS Chain and thus, can be considered a subset of ALS Chains. Finally, is to show the potential power of this solving method to those who might have been unaware of it. The fact is that it can be used to solve some puzzles above the difficulty range or most newspaper puzzles (ie. such as the diabolicals). Many advanced solvers know about ALSs and use them occasionally in type A pattern-solving and probably more often in chains, but since there have been no in-depth, clear descriptions of their use, newer solvers may be less aware of them.

Above the concept of digits 'seeing' other digits was mentioned: This usually means that all the digits are in the same house or unit ie. the same row, column or box. However, digits can also 'see' each other if they are linked by what are called conjugate pairs as will be described in more detail later.

So, on to the examples. Initially, the traditional names of what have been classed as separate categories of ALSs will be given, but further on the patterns will be named based on the concept of ALS Chains. All of the examples use the same format. The first ALS (or set) in the chain is Green, the second Blue, and if present, the third Orange and the fourth Pink. The CEC (candidate elimination cell or cell in which the elimination occurs) will be coloured Yellow. Blue arrows denote the ‘restricted common’ links between ALSs, while Red arrows denote the line of site of the flanking ALSs to the digit or candidate for elimination.

2-Set ALS Chain (aka ALS XZ-Rule)

All ALS Chains are similarly constructed. Each ALS must be connected to the next ALS in the chain by what is known as a ‘restricted common’, a digit that cannot exist at the same time in both ALSs (ie. one of them will be lost as the puzzle is solved, thus exposing the underlying locked set). Because of this bond between the ALSs, any digit (other than the restricted common) that occurs in both flanking ALSs and that is in a cell that is in the ‘line-of-sight’ (same row, column or box) of that same digit in both ALSs can be eliminated. The 2-set ALS Chain (aka ALS xz-rule) is, of course, the simplest form of ALS Chain. In the one above, the Green ALS (or set as it is sometimes called) connects to the Blue ALS via the restricted common=9, but both ALSs also contain a digit 3 that ‘sees’ the 3 in the yellow cell so it can be eliminated. Note that it is always the ‘flanking’ ALSs (ie. the ALSs at each end of the chain) that must have the common digit that ‘sees’ the candidate for elimination. When this 2-set ALS Chain was first described the ‘restricted common’ digit was arbitrarily assigned the letter (or variable) ‘x’, while the candidate for elimination was given the letter ‘z’ hence the name XZ-rule.

3-Set ALS Chain (aka ALS XY-Wing Rule)

In this 3-Set ALS Chain, the Green ALS connects to the Blue ALS via the restricted common=2. Note that there are two 2s in the Blue ALS, so be warned that when there are more than one digit equalling the digit value of the restricted common in one or more of the connecting ALSs, all of the restricted common digits of one ALS must see all the restricted common digits in the other just as the 2 in the Green ALS above sees both 2s in the Blue ALS. The Blue ALS connects to the Orange ALS via the digit 6 (the single Blue 6 seeing both of the 2 Orange 6s). Now that the 3 ALSs are thus connected, the flanking Green and Orange ALSs ‘see’ a common digit 3 in the yellow cell, so it can be removed. You can see the obvious similarity to the 2-set ALS Chain. The only difference is that an extra set has been added. ALS Chains with greater than 2 sets or ALSs will have more than one restricted common (just as the above example has the two restricted commons, 2 and 6). Therefore, it is important to note that when searching out the components of ALS Chains of greater than 2 sets, the restricted commons in the chain must have different values for adjacent sets ie. a restricted common digit value can't be the same as that of the immediately preceding or immediately following restricted common in the chain. (Please see credits at the end of the tutorial regarding this rule.)

This is another example of a 3-set ALS Chain. The Green ALS connects to the Blue via the restricted common 4. The Blue ALS connects to the Orange ALS by the restricted common 7. The flanking Green and Orange ALSs converge on the 9 in the yellow cell so that it can be eliminated.

ALS Chains Greater Than 3 Sets (aka ALS XY-Chain)

Above is a 4-set ALS Chain. The Green, Blue, Orange and now Pink, ALSs are connected by the restricted commons, 4, 7, and 9, respectively and the flanking Green and Pink ALSs see the common digit 5 for elimination in the yellow cell. Note that 5 cannot be used as a restricted common linking the Green and Blue ALSs because all the 5s in the Green set do not see all the 5s in the Blue set.

Doubly-Linked ALSs

Doubly-linked ALSs are not chains in quite the same sense as those described above are, but they might be considered a form of the 2-Set chain and therefore a subset of ALS Chains in general.

Above is an example of Doubly-Linked ALSs. They are well-known to the most advanced Sudoku solvers, but again, perhaps due to the lack of accessible descriptions of them, they are unknown to less experienced solvers. This is unfortunate because they are especially powerful in that they have as many as four different avenues of eliminations. Doubly-Linked ALSs are so-called because they are linked by two restricted commons (though some people prefer to look on these as an interchangeable restricted common and common digit for elimination).

Two of the four possible sources of eliminations in Doubly-Linked ALSs (one for each ALS) derive from the fact that, if two ALSs are linked by two restricted commons, then any digits in cells (outside either ALS of course) that ‘see’ digits in either ALS that are not members of the restricted commons can be removed. In the above puzzle, the Green and Blue ALSs are linked by the restricted commons, 2 and 5. Therefore, since the 4 in the Green ALS sees the 4 in the yellow cell and is not one of the restricted commons, the 4 in the yellow cell can be removed. Likewise, by the same mechanism (but now using the Blue ALS), the 7 in the yellow-bordered cell can be removed.

The graphic also shows that the 2 in r3c4 can also be removed, but that is due to the principle that accounts for the other two means of eliminations which we'll get to in a moment.

The above puzzle introduces the concept of the use of the strong links of conjugate pairs to increase the reach and thus, the power of ALS Chains and Doubly-Linked ALSs. As was hinted at earlier, all that is necessary is that the solver understand that digits in cells connected by conjugate pair links ‘see’ each other just the same as if they were in the same row, column or box. These conjugate pair links allow both the connection of sets in a chain that are remote from each other and the elimination of digits in cells that are relatively far from one or more of the flanking ALSs. In the above puzzle image, the Green and Blue ALSs are doubly-linked by the restricted commons, 3 and 5, the latter through the conjugate pair indicated by the black-bordered cells. Therefore, the remaining, non-restricted common digit 1 in the Green ALS sees and thus, eliminates the 1 in the yellow cell.

Above, it was mentioned that Doubly-Linked ALSs had four potential sources of eliminations and an example of the first two was given. The third and fourth sources derive from the fact that any digit in a cell outside each ALS, but which sees a given restricted common in both ALSs can be removed. Since this situation can occur with both of the two restricted commons, you thus have the two addition sources of eliminations. Recall that in the first example of Double-Linked ALSs, the 2 in r3c4 was eliminated by this mechanism because it sees the restricted common 2s in r2c4 and r7c4.

The Doubly-Linked ALS example above consists of the Green Set A: (2,4,6) r38c1 and the Blue Set B: (4,6,7,9) r348c8. They are joined by the restricted commons, 4 and 6 as indicated by the blue lines. Looking at it from the point of view of the four possible avenues of eliminations, here is a summary of the possibilities:

1. Digits that 'see' the same digits in Set A, but which are not one of the restricted commons can be removed from cells outside of Set A, in this case: r2c1<>2

2. Digits that 'see' the same digits in Set B, but which are not one of the restricted commons can be removed from cells outside of Set B, in this case: r2c8<>7 and r9c8<>9

3. Digits that 'see' the first restricted common in both Sets can be removed from cells outside of both Sets, in this case there are no 6s in row 3 that can be removed.

4. Digits that 'see' the second restricted common in both Sets can be removed from cells outside of both Sets, in this case r8c3<>4 and r8c9<>4

Thus, this one pattern results in 5 eliminations! (Note: The above example replaces an example that was defective. It is the same example that is presented in a post further along in this thread. Since some posts refer to that particular post, I decided to not delete it even though it is largely reproduced above.)

Actual Puzzle Solution Example:

There is nothing like real-world solving examples to make potentially difficult concepts easier to grasp for newer solvers. Likewise, in this case the examples prove the perhaps unappreciated power of both ALS Chains and Doubly-Linked ALSs. The following is the solving sequence using only ALS Chains and Dual-Linked ALSs to solve an Extreme Diabolical level puzzle (SE rating above 7) starting from the point where most basic methods leave off . (However, this does not mean that this is the only way to solve this puzzle.) Note that RC below means restricted common.

A 3-set ALS Chain

Green ALS->RC 4->Blue ALS->RC 6->Orange ALS
Orange ALS->conjugate link->yellow cell common digit for elimination 2<-Green ALS
Thus yellow cell<>2

A 4-set ALS Chain

Green ALS->RC 2->Blue ALS->RC 4->Orange ALS ->RC 6->Pink ALS
Green ALS->yellow cell common digit for elimination 3<-Pink ALS
Thus yellow cell<>3

A 3-set ALS Chain

Green ALS->conjugate link->RC 6->Blue ALS->RC 3->Orange ALS
Green ALS->conjugate link->yellow cell common digit for elimination 9<-Orange ALS
Thus yellow cell<>9

A 4-set ALS Chain

Green ALS->conjugate link->RC 3->Blue ALS->RC 2->Orange ALS->RC6->Pink ALS
Green ALS->yellow cell common digit for elimination 4<-Pink ALS
Thus yellow cell<>4

A 3-set ALS Chain

Green ALS->conjugate link->RC 9->Blue ALS->RC 3->Orange ALS
Green ALS->yellow cell common digit for elimination 2<-Orange ALS
Thus yellow cell<>2

And the puzzle is solved.

Conclusion:
ALS Chains and Doubly-Linked ALSs are extremely powerful methods of type A pattern-solving especially if conjugate pair links are used where possible. They can take up where previous basic methods leave off and can be used to help solve very difficult puzzles. Adding ALS Chains and Doubly-Linked ALSs to one's solving arsenal is not all that difficult. Some hints: When looking at examples in tutorials do not make the mistake of trying to find chains in your puzzles that look like those in the examples. It’s far better to understand the underlying principles involved and you will find ALS Chains where you never thought possible. Practice a lot initially by looking for ALS Chains in all your puzzles. Do not try to work backwards. That is, do not try to find an ALS solution for a particular digit you want to remove. That rarely works and does nothing but waste your time and wear you out. Finally, the use of a solver that allows one to color cells, such as Angus Johnson’s Simple Sudoku or Ruud’s SudoCue, can be helpful in manually tracing an ALS Chain during the solving process.

By the way: The possibility for several eliminations from doubly-linked ALSs is closely related to the several eliminations possible from Sue De Coq, a tutorial for which is available here:

http://forum.enjoysudoku.com/viewtopic.php?t=6410

(Note: ASI stands for advanced solving illustrated. Credits: The final Doubly-Linked ALS graphic is based on an example first presented by Anne Morelot in mid-2006 on the Eureka forum. The warning about not using the same restricted common values in adjacent sets comes mainly from the recent work of Paul Isaacson. Prior to that the rules concerning re-use of restricted common values were less than clear.)
Last edited by DonM on Wed Mar 25, 2009 4:02 am, edited 4 times in total.
DonM
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Re: ALS Chains -A Tutorial ASI#3

DonM wrote:Doubly-Linked ALSs

Doubly-linked ALSs are not chains in quite the same sense as those described above are, but they might be considered a form of the 2-Set chain and therefore a subset of ALS Chains in general.

Above is an example of Doubly-Linked ALSs. They are well-known to the most advanced Sudoku solvers, but again, perhaps due to the lack of accessible descriptions of them, they are unknown to less experienced solvers. This is unfortunate because they are especially powerful in that they have as many as three different avenues of eliminations.

Au contraire, there are only two different avenues of eliminations. One is, as you show above, the elimination of a candidate which sees all occurrences of a digit locked in either ALS. The second is the elimination of a candidate which sees all occurrences of a restricted common digit in both ALSs, such as r3c4<>2 not shown above.

This is easier to see with the aid of a verbose nice loop:

als1:(r7c6,r7c7 =5|79|2= r7c4) -2- als2:(r2c4 =2|4|5= r2c6) -5- continuous loop

Digits 7 and 9 are locked in als1 and digit 4 is locked in als2. Weak inferences, possibly grouped, between the ALSs become conjugate inferences.
ronk
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DonM
Speaking as one of those people who is allergic to general ALS...and considering it a strategy that may be off-putting to many, I must first of all thank you for tackling in such clear visual format this thankless task

In that spirit, I therefore looked at each of your twelve examples to see what alternative I would have used.
In every case, hidden pairs or triples produce the same eliminations with for variety a deadly rectangle in n°3, and something else in n° 11

I'm sorry that I don't know how to copy your grids into this post, so here are the last three examples with the alternatives :

n° 12 (your last example) :
hidden triple (234) r7c357=>r7c2 <2>
not hidden triple=>8r7c5/c7-(8=9)r7c6-(9=4)r8c6-(4=2)r1c6-(2=6)r3c6-(6=2)r3c2=>r7c2 <2>

n° 11 : consider r46c5 : only possibilities are (2,4) or (8,2) : excluded (2,2).
4r6c5=>r7c5 <4>
8r4c5-(8=6)r4c6-(6=9)r4c9-(9=3)r4c1-(3=2)r4c3-(2=4)r7c4 =>r7c5 <4>

n°10 :
hidden triple (349) r148c9 => r6c9 <9>.
not hidden triple =>6r4c9-(6=8)r4c6-(8=9)r7c6-(9=4)r8c6-(4=9)r8c9 => r6c9 <9>

My point ultimately is that to anyone else like me who thinks that ALS is just too much like hard work, then there are simpler (well I think so) alternatives.
aran

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aran wrote:DonM
Speaking as one of those people who is allergic to general ALS...and considering it a strategy that may be off-putting to many, I must first of all thank you for tackling in such clear visual format this thankless task

In that spirit, I therefore looked at each of your twelve examples to see what alternative I would have used.
In every case, hidden pairs or triples produce the same eliminations with for variety a deadly rectangle in n°3, and something else in n° 11
....
My point ultimately is that to anyone else like me who thinks that ALS is just too much like hard work, then there are simpler (well I think so) alternatives.

Aran, thanks for your opening comment and post. These illustrated threads have more than one purpose. One of them is to try to clarify and/or make it easier to find and use patterns that some may have been confused about or have ignored. But another, is to promote discussions of manual solving related subjects which is my primary sudoku interest. The very good point you're making is one of those subjects and is very similar to the point some are making in the Sue De Coq threads. I find it fascinating how some of us are wired to look at patterns one way, while others find them another way. I try hard to be very careful about inferring that any method is 'the only way' (though sometimes I'm not above strongly promoting the method I prefer. ).
Last edited by DonM on Sat Nov 22, 2008 2:09 am, edited 1 time in total.
DonM
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This is a graphic based on a good example of a doubly-linked ALS first presented by Anne Morelot in mid-2006:

The pattern consists of the Green Set A: (2,4,6) r38c1 and the Blue Set B: (4,6,7,9) r348c8. They are joined by the restricted commons, 4 and 6 as indicated by the blue lines. Looking at it from the point of view of where one has to look for the eliminations (ie. where they are possible), there are four possibilities:

1. Digits that 'see' the same digits in Set A, but which are not one of the restricted commons can be removed from cells outside of Set A, in this case: r2c1<>2

2. Digits that 'see' the same digits in Set B, but which are not one of the restricted commons can be removed from cells outside of Set B, in this case: r2c8<>7 and r9c8<>9

3. Digits that 'see' the first restricted common in both Sets can be removed from cells outside of both Sets, in this case there are no 6s in row 3 that can be removed.

4. Digits that 'see' the second restricted common in both Sets can be removed from cells outside of both Sets, in this case r8c3<>4 and r8c9<>4

Thus, this one pattern results in 5 eliminations. The possibility for several eliminations from doubly-linked ALSs is closely related to the several eliminations possible from Sue De Coq.
DonM
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Don'M
These illustrated threads have more than one purpose. One of them is to try to clarify and/or make it easier to find and use patterns that some may have been confused about or have ignored

I never thought otherwise, and am reassured that you didn't take my post as a rebuff
But another, is to promote discussions of manual solving related subjects which is my primary sudoku interest

Mine too.

Just for interest here is the allergic path to the above eliminations :
consider r348c8 {4679}:
clearly there must be a 6 or 4 : otherwise=>(79)r348c8. Impossible
but there can't be a 6 and a 4 : otherwise=> 2r3c1+2r8c1. Impossible.
=>6 xor 4
=>necessarily pair 79 in r348c8 => r2c8 <7> r9c8 <9>
=>necessarily 2 in r38c1=> r2c1 <2>
=> strong link 4 r8c1/4r8c8 => r8c39 <4>.

No talk there of sets, restricted commons, digits that see restricted commons, digits that see digits that aren't restricted commons...
aran

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Re: ALS Chains -A Tutorial ASI#3

ronk wrote:Au contraire, there are only two different avenues of eliminations. One is, as you show above, the elimination of a candidate which sees all occurrences of a digit locked in either ALS. The second is the elimination of a candidate which sees all occurrences of a restricted common digit in both ALSs, such as r3c4<>2 not shown above.

I'm not following the 'Au contraire' part. Examples of both the categories (as you prefer to describe them), with essentially the same description, were shown in the tutorial, the second in the graphic just before the actual puzzle solution graphics. One can look on the categories as 2 broad ones as you correctly do so above. I prefer to look on them as 4 specific sources of eliminations as shown in the graphic I just added. Sirdave presented it as 2 specific + 1 broad categories.

In retrospect, I do think it would have been a good idea to point out that r3c4<>2 elimination & the 2nd (broad) category from the get-go using that graphic. Can't remember whether the plan was to present one point at a time or whether that elimination was simply missed. I provided many of the examples; sirdave did most of the graphics. If it was missed, then good eye!
DonM
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DonM
You could have been thrown by Ronk's inhabitual use of..."au contraire".
Is 2 the opposite of 3 ?
aran

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aran wrote:DonM
You could have been thrown by Ronk's inhabitual use of..."au contraire".
Is 2 the opposite of 3 ?

Perhaps we can suggest considering more frequent use of 'autrement'.
DonM
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Re: ALS Chains -A Tutorial ASI#3

DonM wrote:
ronk wrote:Au contraire, there are only two different avenues of eliminations. One is, as you show above, the elimination of a candidate which sees all occurrences of a digit locked in either ALS. The second is the elimination of a candidate which sees all occurrences of a restricted common digit in both ALSs, such as r3c4<>2 not shown above.

I'm not following the 'Au contraire' part.

While it may have been someone else's concept originally, I was objecting to your stated concept of three "avenues of eliminations." In the doubly-linked ALS xz-rule, there are two ALSs, two restricted commons, and two possible sets of locked candidates within the ALSs. That these even numbers should result in an odd number of avenues of elimination is illogical to me. However, I think one avenue (elimination rule) would be logical.

DonM wrote:Examples of both the categories (as you prefer to describe them), with essentially the same description, were shown in the tutorial, the second in the graphic just before the actual puzzle solution graphics.

Assuming that description actually led me to the correct graphic, I notice that ...
1) digit 2 is not a restricted common, and
2) the illustration is not even for doubly-linked ALSs.

[edit: added graphic below for clarity]

Aha, I see some trickery here with the naked pair between the two ALSs. I'll have to study that a bit more.
Last edited by ronk on Sat Nov 22, 2008 11:18 am, edited 1 time in total.
ronk
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Don, Your last example is a Disjoint Locked Set which uses two rows and two columns (and hence the similarity to Sue De Coqs which are also DLSs)

(24679)DLS:r38c2,r348[(6)r3,(4)r8,(2)c1,(79)c8] => r8c39<>4, r2c1<>2, r1c8<>6, r2c8<>7, r9c8<>9
(Five cells containing five digits, each of which is confined to a single house and so is restricted to one occurrence.)

Although an illustrative example, it's gross overkill for this puzzle, which can be solved by much simpler routes. This AIC with one ebedded ALS gives the same eliminations with the exception of (6)r1c8:

(4=2)r8c1 - (2=6)r3c1 - (6=79)ALS:r34c8 - (7|9=4)r8c8 - Loop => r2c1 <>2, r8c39 <> 4, r2c8<>7, r9c8<>9
(Being able to complete a closed loop of alternating inferences proves that they are all conjugate)

Aran, although you maintain there is no talk of sets in your logic chain, you're using set logic in your verbal description. If you understand the Eureka notation above, it amounts to your explanation which also omits the same (6) elimination. A rose by any other name…?
David P Bird
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David P Bird wrote:Although an illustrative example, it's gross overkill for this puzzle, which can be solved by much simpler routes.

It is about as clear an example of the pattern as one is going to get. The goal is to help people understand the patterns, not suggest them as the simplest routes.
DonM
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Don, If you put up illustrative examples, the first thing your readers will do is to ask themselves whether they should they try to master the pattern and add it to their repertoires. If they can see their existing methods would solve the given position, they are likely not to bother.

Now I fully accept that stripped down examples are excellent ways to show how a pattern works, but I think these need some pre-emptive words of explanation to prevent replies asking what the point is, and ideally a second example to illustrate the true power of the technique in a tough puzzle. Without these the chances are effort you put into the piece is likely to be wasted on many readers.

I'm disappointed that you ignored the rest of my post and picked on the one sentence that you took as an implied criticism, when my intention was to cover what I considered to be an omission on your part.
David P Bird
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David P Bird wrote:I'm disappointed that you ignored the rest of my post...

Okay, I'll respond to the rest of your post:
DPB: 'Don, Your last example is a Disjoint Locked Set which uses two rows and two columns (and hence the similarity to Sue De Coqs which are also DLSs)'

I considered that to be pretty much the same point I had already made at the bottom of the example: 'The possibility for several eliminations from doubly-linked ALSs is closely related to the several eliminations possible from Sue De Coq.'

DPB: 'Although an illustrative example, it's gross overkill for this puzzle, which can be solved by much simpler routes.'

Again, the idea was to provide a clear example, not sell it as the simplest route to the puzzle solution. However, in response to the solution of this very puzzle in which this very example was used, rather than pointing this out as overkill, a skilled sudoku solver replied: 'Anne, How many elves have you got in your production line? I can't keep up.' (David P. Bird June 8, 2006)

David P Bird wrote:... and picked on the one sentence that you took as an implied criticism, when my intention was to cover what I considered to be an omission on your part.

Have they changed the King/Queen's english over there? Over here, pointing out an omission is a criticism. I'm simply responding to it. And here's a more indepth response to the alleged omission:

Tutorials can be anything from primers that include only illustrations with brief descriptions to those that include simple examples to those that include complex examples. It all depends on the information one wants to transmit and the time one wishes to spend transmitting it.

In this case, except for the ALS xz-rule, there are practically no illustrated examples of these patterns anywhere. Those who will benefit from examples such as the one in question are those who have heard about these patterns, but can't understand them. Having now (hopefully) understood them, they will now be able to to try to apply them. For the most part, these people will typically not have existing methods to replace these methods. Those who do have existing methods will, by definition, be experienced enough to see the possibilities of these patterns...or not. As to 'Without these the chances are effort you put into the piece is likely to be wasted on many readers.', I haven't even remotely seen evidence of this. On the contrary, people are being given information here they haven't had available before; they are appreciative of the fact that someone is even providing it at all and are able to take the next step in applying the pattern. Omissions aren't really on their minds.
DonM
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Re: ALS Chains -A Tutorial ASI#3

DonM posted:

ronk wrote:Aha, I see some trickery here with the naked pair between the two ALSs. I'll have to study that a bit more.

Seven cover sets are required for the six cells. If it were a true ALS doubly-linked xz-rule, the number of base sets and cover sets would be equal. Since the ALSs are not doubly-linked, if follows that there is no corresponding "restricted common elimination" either.
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