## Almost Locked Sets help

Post the puzzle or solving technique that's causing you trouble and someone will help

### Almost Locked Sets help

Could anyone suggest what I am doing wrong with Almost Locked Sets in chains.

This is the current state of my puzzle. I am concentrating on using ALS at this point.
.-----------------.---------------.--------------.
| 189 2 1478 | 3 147 5 | 79 78 6 |
| 5 14 478 | 17 6 479 | 379 378 2 |
| 39 6 37 | 2 8 79 | 5 4 1 |
:-----------------+---------------+--------------:
| 6 7 13 | 5 13 8 | 4 2 9 |
| 4 139 139 | 6 2 13 | 8 5 7 |
| 2 8 5 | 9 47 47 | 6 1 3 |
:-----------------+---------------+--------------:
| 13 5 6 | 4 9 2 | 137 37 8 |
| 7 139 289 | 18 5 13 | 123 6 4 |
| 138 134 12348 | 178 137 6 | 1237 9 5 |
'-----------------'---------------'--------------'

Now I thought I could include an ALS as a node in a Continuous Nice Loop.

-4[B2] -4[J2] +4[J3] -2[J3] +2[H3] -9[H3] +9[H2] -9[E2] +3[E2|E6] -3[H6] +3[J5] -7[J7] +7[J4] -7[B4] +1[B4] -1[B2]

It looks like a Nice Continuous Loop to me, with eliminations of 1, 3 and 8 at J3, 8 at H3 and 1 at E2 by virtue of the 1 eliminated by the ALS. However, the elimination of 3 at J3 is wrong, so can someone enlighten me what I have done wrong.
Wept

Wepwawet

Posts: 17
Joined: 19 November 2019

### Re: Almost Locked Sets help

Hi Wepwawet,

Wepwawet wrote:Could anyone suggest what I am doing wrong with Almost Locked Sets in chains.

Now I thought I could include an ALS as a node in a Continuous Nice Loop.

+4[B2] -4[J2] +4[J3] -2[J3] +2[H3] -9[H3] +9[H2] -9[E2] +3[E2|E6] -3[H6] +3[J5] -7[J5] +7[J4] -7[B4] +1[B4] -1[B2]

Your 3[E2|E6] doesn't have a weak link to 3[H6] because E2 doesn't see it, so the chain can't continue beyond that point. In other words:

Code: Select all
`.-------------------.----------------.--------------.| 189   2     1478  | 3    147   5   | 79    78   6 || 5    a14    478   | 17   6     479 | 379   378  2 || 39    6     37    | 2    8     79  | 5     4    1 |:-------------------+----------------+--------------:| 6     7     13    | 5    13    8   | 4     2    9 || 4   f(13)9  139   | 6    2   f(13) | 8     5    7 || 2     8     5     | 9    47    47  | 6     1    3 |:-------------------+----------------+--------------:| 13    5     6     | 4    9     2   | 137   37   8 || 7    e139  d289   | 18   5   !g13  | 123   6    4 || 138  b134  c12348 | 178  137   6   | 1237  9    5 |'-------------------'----------------'--------------'`

(1=4)r2c2 - r9c2 = (4-2)r9c3 = (2-9)r8c3 = r8c2 - (9=13)r5c26 // can't continue with a weak link to (3)r8c6
-SpAce-: Show
Code: Select all
`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

SpAce

Posts: 2559
Joined: 22 May 2017

### Re: Almost Locked Sets help

This is the answer to your other post here (because it was in the wrong place).

Wepwawet wrote:Trying to get my head around chained ALS's.

This is the stage I have arrived at in a puzzle and playing around with chains to familiarize myself with them better;

Code: Select all
`+---------------+----------------+---------------+| 5   2   378   | 1    4   9     | 678  36  78   || 346 48  9     | 3678 56  3578  | 178  2   14   || 346 1   378   | 3678 26  2378  | 9    34  5    |+---------------+----------------+---------------+| 7   3   6     | 4    28  1     | 28   5   9    || 14  45  15    | 9    3   28    | 2678 68  78   || 8   9   2     | 5    7   6     | 4    1   3    |+---------------+----------------+---------------+| 2   6   1458  | 378  18  378   | 15   9   14   || 9   7   14    | 68   156 58    | 3    48  2    || 13  58  138   | 2    9   4     | 158  7   6    |+---------------+----------------+---------------+`

I was trying this chain (hopefully, I have used the correct notation - first time I have used this style) ...
-2[C5] +2[C6] -2[E6] +8[E6] -8[E8] +6[E8] -6[A8] +3[A3|A8|A9] -3[J3] +5[J2|J3|J7] -5[G7] +1[G7] -1[G5] +8[G5] -8[D4] +2[D4] -2[C5]

You lost me in the red part. What were you trying to do? (Btw, you've missed a Hidden Pair 37 in box 8.)

Btw, our preferred notation is Eureka with the rYcX coordinates. That SudokuWiki notation is absolutely horrible.

Anyway, what puzzled me was J1 being made a 1 (though if I work it through starting from D5 going towards C5 I can see why) but I was led to believe the weak link of the 3 from A3 into the ALS {1358} at J237 would make it a locked set of 158 and that would eliminate the 1 from J1 (which doesn't work). Obviously, I am missing something here, but not quite sure what.
Could anyone suggest where my reasoning is wrong.

Hard to say. You should start by telling what you're trying to prove and how. If you're trying to eliminate 2[C5] then you should either start with +2[C5] and work from there to prove it can't be true, or otherwise start with -2[C6]. So, I think you probably have some gaps in your basic chaining skills. I'd concentrate on fixing those first before even trying ALS chains.
-SpAce-: Show
Code: Select all
`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

SpAce

Posts: 2559
Joined: 22 May 2017

### Re: Almost Locked Sets help

I am relatively new to using chain notations and there is a likelihood that there are errors there, but this is what I was trying to puzzle out.

Regarding the puzzle I posted on Nov 19th.
I had started off with c5 having a strong link on the 2 to c6 and went round the chain accordingly and I was initially puzzled by why r9c1 did not delete candidate 3. As I understand it, please correct me if I am wrong, that the internal links in an ALS are strong, which makes sense, as it is entered and exited by weak links, so the alternating inference nature of the chain is maintained. I believe this is where I have gone wrong and that I am missing the subtleties of the internal links of an ALS, and it cascaded from there, I know r1c8 cannot be a 6 (from the premise that c5 is not a 2) so it must be be either a 3 or 8, so if it is the 3 (which it is) it would make the internal link in the ALS to the 3 in r3 a weak link, which it cannot be. Is that correct?
You lost me in the red part. What were you trying to do?
It was just intended to just use the ALS of 3678 in row 1 as a convenient node in a chain (no deletions) to get around the board.

Btw, you've missed a Hidden Pair 37 in box 8
Yes, one of my Sudoku blindspots, hidden subsets.

Regarding the puzzle I posted on Nov 23rd.
It looks like I am misunderstanding some aspects of ALS's. Are you saying that all instances of the exit candidate that exist in an ALS must see the next node? Deducted from
Your 3[E2|E6] doesn't have a weak link to 3[H6] because E2 doesn't see it, so the chain can't continue beyond that point

If so, then I have missed that point about Almost Locked Sets, I cannot recollect seeing that point stated anywhere. It could explain why I could not fathom out why some of my reasoning failed me in some puzzles. That is the common theme in both of the puzzles, I am hoping that is the answer.

Edited to further explain.
Wept

Wepwawet

Posts: 17
Joined: 19 November 2019

### Re: Almost Locked Sets help

Further ALS explanation required.
I have gone to the Sudopedia Mirror web page http://sudopedia.enjoysudoku.com/ALS-XY-Chain.html and it states that ALS C and D have a restricted common of 1, yet ALS D has two 1's in it. Is this being regarded as a grouped node within the ALS (as they share the row), or is it a typo?
Wept

Wepwawet

Posts: 17
Joined: 19 November 2019

### Re: Almost Locked Sets help

Try the ALS intoduction here. It's an easy read and has worked examples.

The point is made that all instances of the Restricted Common Candidate (RCC) in successive ALS's in an ALS chain should see each other.

So, if any instance of the RCC in the first ALS is True it forces all of the instances of the RCC in the following ALS to be False, and so all of the remaining candidates must be True somewhere in the following ALS.

The Sudopedia web page you mentioned has a good example of this principle in action.

Leren
Leren

Posts: 3928
Joined: 03 June 2012

### Re: Almost Locked Sets help

Set a occupies a sector
Set b occupies a sector

Rc occupies a sector that intercepts a and b
All copies of N digit for a and b sets must be in this overlapping sector.
N cannot contain cellls that are shared by a and b (overlapping cella)
Some do, some teach, the rest look it up.

StrmCkr

Posts: 1170
Joined: 05 September 2006

### Re: Almost Locked Sets help

Thank you Loren, for the link, I had overlooked the prospect (for the RCC) having more than one candidate in either (or both ALS) as nearly all the examples only have single ones, and with the sentence from the HoDoKu page
Some explanations of ALS Chains (often called "ALS-XY-Chains") depend on the reversability of the chain
that kinda confirms the point I had missed about the exit candidate also seeing all occurrences of itself, something I suspected after reading SpAce reply. I thought I was quite happy with the usage of ALS now (well, the ones with single RCC's) until I read StrmCkr response. I am somewhat unsure of the terminology, what is the overlapping sector and what are N digits? I know that when ALS's overlap each other that the cells in the overlap area cannot contain a RC is that what you mean?

Edited to accomodate the post from StrmCkr (which I only noticed after Loren's)
Wept

Wepwawet

Posts: 17
Joined: 19 November 2019

### Re: Almost Locked Sets help

Removed.
Last edited by Leren on Wed Nov 27, 2019 9:09 pm, edited 3 times in total.
Leren

Posts: 3928
Joined: 03 June 2012

### Re: Almost Locked Sets help

Wepwawet wrote:
Hodoku wrote:Some explanations of ALS Chains (often called "ALS-XY-Chains") depend on the reversability of the chain

That sentence only shows that hobiwan probably didn't fully understand how ALS chains work when he wrote it. It wouldn't surprise me, because he also didn't fully grasp how group nodes work, which is why they're never used as end nodes in Hodoku chains (worked around with stupid DNLs). Nevertheless, I'm grateful that Hodoku still works as well as it does.

Reversibility has nothing to do with anything in normal chains. It's just a side effect of correctly formed AICs, but they would work just as well if they somehow were unidirectional (*). In ALS loops reversibility does matter a bit more because it shows clearly that the bystander digits get locked either way and can cause additional eliminations. Yet none of this matters because correctly constructed AICs or AIC-loops are always bidirectional, whether they have ALS terms or not.

Even more accurately, AICs (including AIC-like nets and krakens) have no concept of directionality at all, because they're not dynamic chains conceptually. Instead, they prove through static boolean logic that at least one of the end points must be true (because of a derived strong link, i.e. OR). Directionality only applies to implication chains which AICs are not natively, even though many people probably see them as such. Such a view point is possible because [a = b] <-> [a OR b] <-> [[~a -> b] & [~b -> a]], but such mental conversion is not necessary at all to read or write AICs. There's no need to assume anything or traverse the chain with AICs as long as your links are valid and alternating -- the static logic pattern itself proves what we need.

(*) Because we always have the intrinsic strong link (a OR not_a). Thus, if we have a unidirectional chain starting with not_a and implicating d somewhere, we have a derived strong link (a OR d) which is what we want (if there's something both a and d can eliminate):

a
||
~a -> b -> c -> d

...which is always reversible:

d
||
~d -> ~c -> ~b -> a

So, even if we think in unidirectional implication chains, AICs are always reversible and prove the same thing both ways (but we only need one).
Last edited by SpAce on Wed Nov 27, 2019 9:13 pm, edited 1 time in total.

SpAce

Posts: 2559
Joined: 22 May 2017

### Re: Almost Locked Sets help

Wepwawet wrote:As I understand it, please correct me if I am wrong, that the internal links in an ALS are strong, which makes sense, as it is entered and exited by weak links, so the alternating inference nature of the chain is maintained.

Yes, that is fully correct. In the ANS (almost naked subset) type of ALS the internal links are strong and the external links are weak. In an AHS (almost hidden subset) it's the other way around. ALS is usually used as a synonym for ANS, but it's a bit ambiguous, so I'll use ANS here.

An ANS has n cells with n+1 digits as candidates. If all candidates of any of those digits are removed, the other digits get locked as a naked subset (pair/triple/quad etc). So, any contained digit has a strong link with a naked subset formed of the other digits. I think that's the most intuitive way to see and use ANSs, and that's why I also prefer chains written like that: (1=234)r123c1. It means that either there's a digit 1 (and obviously two other digits, but we don't know or care which) or a naked triple 234 in those three cells. It also follows that any digit has a strong link with any other digit within the ANS, so it's technically correct to write (1=4)r123c1 as well, though I find it harder to read and compare to the grid.

As you said, since the ANS has an internal strong link, it must be surrounded by two weak links if used in a chain. They must obviously use different digits. In normal situations only one digit is used for either weak link, but it's also possible to use more on one or both sides for advanced linking. When the AIC is viewed as an implication chain and traversed in one direction, then one weak link enters and the other exits, as you said, but the roles obviously reverse if the chain is read the other way. Either way, what's important is that both weak links must see all instances of the linking digit(s) in both the ANS and the linked external nodes. That was the problem in your first example where only one of the locked 3s could see the intended linking target.

An example:

... - (5=23'1)r651c1 - (1=56'4)r1c568 - (4=89'3)r567c8 - ...

If read from left to right, the middle ANS (1456)r1c568 is entered with a weak link on 1 (the exit digit of the previous ANS; must be locked as a single in r1c1 to link to the second ANS), which locks the rest of the digits 456 as a naked triple. Any or all of them could be used for the exiting weak link, but in this case the 4 is used for linking. Again, it must be locked as a single in r1c8 to link to the third ANS in c8.

Btw, I've added the '-separator to highlight the right-linking candidates, though it's normally omitted (still the right-linking candidate(s) should be written last).

Regarding the puzzle I posted on Nov 23rd.
It looks like I am misunderstanding some aspects of ALS's. Are you saying that all instances of the exit candidate that exist in an ALS must see the next node? Deducted from
Your 3[E2|E6] doesn't have a weak link to 3[H6] because E2 doesn't see it, so the chain can't continue beyond that point

Exactly.

If so, then I have missed that point about Almost Locked Sets, I cannot recollect seeing that point stated anywhere. It could explain why I could not fathom out why some of my reasoning failed me in some puzzles. That is the common theme in both of the puzzles, I am hoping that is the answer.

Well, you do know how naked pairs/triples/quads etc work, right? That's what you should think about when considering how the "exit link" works. Once the incoming weak link "removes" one digit (completely) you have a naked subset with the rest. Look at that as a whole and think what it could eliminate in its surroundings if it were in fact a naked pair or triple or so on. Those are candidates with which you can link it. However, you should also look at where the individual digits are located within the locked set because they have different linking potentials (those that get locked as a single or a pointing pair are usually the best).

In your example with the ANS (139)r5c26, the incoming weak link removes the 9 so you're left with a naked pair 13. That far your chain was correct. However, since both digits have candidates in both cells, and they're not even in the same box, there's nowhere but the same row where either of those digits can (simply) link to. In fact, you could use the full pair to remove both 1 and 3 from r5c3 and get a 9r5c3 to continue the chain, but obviously you don't need the ANS for that (9r5c2 is directly strongly linked with 9r5c3).

Further ALS explanation required.
I have gone to the Sudopedia Mirror web page http://sudopedia.enjoysudoku.com/ALS-XY-Chain.html and it states that ALS C and D have a restricted common of 1, yet ALS D has two 1's in it. Is this being regarded as a grouped node within the ALS (as they share the row), or is it a typo?

It's not a typo. You can have several candidates of the weak-linking digit in both ANSs as long as they all see each other. That's why such a digit is called Restricted Common (RC) -- the restriction being that at most one of its candidates can be true in the combined cells of both ANSs (definition of a weak link). Btw, Hodoku unfortunately calls the same concept Restricted Common Candidate (RCC), which is a very confusing misnomer, because it obviously means digit as well:

Hodoku wrote:To combine two ALS they must share at least one candidate. "Share" means, that all instances of the candidate in ALS 1 see all instances of that candidate in ALS 2. Such a candidate is called a Restricted Common Candidate (or RCC for short).

Replace all instances of "candidate" with "digit" and that sounds about right. It's a totally different situation if two ALSs actually share candidates because it means they have overlapping cells (which can't contain RC candidates between those two ALSs, as Hodoku correctly mentions). Thus, I suggest sticking with Sudopedia (in this case) and "RC" which has no ambiguity.

Anyway, if that restricted digit is true in one ANS, it can't be true in the other, which locks the other digits in the latter. In other words, the presence of an RC between two ANSs guarantees that at least one of them gets locked with the other digits. That in turn means that the chain must propagate to at least one direction as long as both locked sets have a suitable weak link (for example through an RC with another ANS).

Here's my annotated version of the example on that page:

Code: Select all
`.----------------.--------------.-----------------------.| . a[4]6    .   |  .      .  . | d25  d258     .       || .   .      .   |  .      .  . |  .    .       .       || .   27-4   .   | c3'1    .  . |  .   d1'258  d1'28(4) |:----------------+--------------+-----------------------:| .   .      .   |  .      .  . |  .    .       .       || .   .     b2'6 | c6'73   .  . |  .    .       .       || .  a69'2   .   |  .      .  . |  .    .       .       |:----------------+--------------+-----------------------:| . a[4]69   .   |  .      .  . |  .    .       .       || .   .      .   | c73     .  . |  .    .       .       || .   .      .   |  .      .  . |  .    .       .       |'----------------'--------------'-----------------------'`

(4=69'2)r176c2 - (2=6)r5c3 - (6=73'1)r583c4 - (1=258'4)b3p1289 => -4 r3c2

What that AIC proves is that either we have a [4] in r17c2 (i.e. r1c2 or r7c2) or we have a (4) in r3c9 (or both). Either way there can't be a 4 in r3c2 which sees them all. The RCs are: a-2-b, b-6-c, c-1-d. The RC candidates in the last case are: 1r3c489.

We could also break the three-cell ANSs in c2 and c4 into smaller segments like this (note that the first weak link uses both digits 6 and 9):

(4=69)r17c2 - (6|9=2)r6c2 - (2=6)r5c3 - (6=7'3)r58c4 - (3=1)r3c4 - (1=258'4)b3p1289 => -4 r3c2

Now the RCs are: a-69-b, b-2-c, c-6-d, d-3-e, e-1-f.

Again, the '-separators are there just for educational purposes to highlight the RCs. Normally I only use them with loops to highlight bystander eliminations. So, my normal way to write the first chain would be:

(4=692)r176c2 - (2=6)r5c3 - (6=731)r583c4 - (1=2584)b3p1289 => -4 r3c2

...or if I wanted to highlight the eliminating digit at both ends (also correct but a bit harder to understand):

(4=692)r176c2 - (2=6)r5c3 - (6=731)r583c4 - (1258=4)b3p1289 => -4 r3c2

Note that it would also work (and the chain would be the same) if we had additional 4s in r3c8 and r6c2:

Code: Select all
`.------------------.--------------.--------------------------.| . a[4]6      .   |  .      .  . | d25  d258        .       || .   .        .   |  .      .  . |  .    .          .       || .   27-4     .   | c3'1    .  . |  .   d1'258(4)  d1'28(4) |:------------------+--------------+--------------------------:| .   .        .   |  .      .  . |  .    .          .       || .   .       b2'6 | c6'73   .  . |  .    .          .       || . a[4]69'2   .   |  .      .  . |  .    .          .       |:------------------+--------------+--------------------------:| . a[4]69     .   |  .      .  . |  .    .          .       || .   .        .   | c73     .  . |  .    .          .       || .   .        .   |  .      .  . |  .    .          .       |'------------------'--------------'--------------------------'`

But this wouldn't:

Code: Select all
`.----------------.--------------.-----------------------.| . a[4]6    .   |  .      .  . | d25  d258(4)  .       || .   .      .   |  .      .  . |  .    .       .       || .   247    .   | c3'1    .  . |  .   d1'258  d1'28(4) |:----------------+--------------+-----------------------:| .   .      .   |  .      .  . |  .    .       .       || .   .     b2'6 | c6'73   .  . |  .    .       .       || .  a69'2   .   |  .      .  . |  .    .       .       |:----------------+--------------+-----------------------:| . a[4]69   .   |  .      .  . |  .    .       .       || .   .      .   | c73     .  . |  .    .       .       || .   .      .   |  .      .  . |  .    .       .       |'----------------'--------------'-----------------------'`

The same chain is still valid but now it doesn't eliminate anything because 4r1c8 doesn't see r3c2:

(4=69'2)r176c2 - (2=6)r5c3 - (6=73'1)r583c4 - (1=258'4)b3p1289 => nothing

You'd have the same problem if you tried to continue the chain because now the quad 2458 is hard to weak-link with anything. All of its digits, including the 4s, are spread into multiple rows and columns within the box. Thus there's no simple way to link them as a unit with anything outside of the box (though it's possible but rather advanced).

Code: Select all
`.----------------.--------------.-----------------------.| . a[4]6    .   |  .      .  . | d25  d258     .       || .   .      .   |  .      .  . |  .    .       .       || .   247    .   | c3'1    .  . |  .   d1'258  d1'28(4) |:----------------+--------------+-----------------------:| .   .      .   |  .      .  . |  .    .       .       || .   .     b2'6 | c6'73   .  . |  .    .       .       || .  a69'2   .   |  .      .  . |  .    .       .       |:----------------+--------------+-----------------------:| . a[4]69   .   |  .      .  . |  .    .       .       || .   .      .   | c73'1   .  . |  .    .       .       || .   .      .   |  .      .  . |  .    .       .       |'----------------'--------------'-----------------------'`

Now the chain can't advance past this point:

(4=69'2)r176c2 - (2=6)r5c3 - (6=73'1)r583c4 // can't continue

...because 1r8c4 doesn't see the 1s in r3c89. There's no RC and thus no weak link between c and d.

Or even worse if we add 2r1c2:

Code: Select all
`.----------------.--------------.-----------------------.| . a[4]6'2  .   |  .      .  . | d25  d258     .       || .   .      .   |  .      .  . |  .    .       .       || .   247    .   | c3'1    .  . |  .   d1'258  d1'28(4) |:----------------+--------------+-----------------------:| .   .      .   |  .      .  . |  .    .       .       || .   .     b2'6 | c6'73   .  . |  .    .       .       || .  a69'2   .   |  .      .  . |  .    .       .       |:----------------+--------------+-----------------------:| . a[4]69   .   |  .      .  . |  .    .       .       || .   .      .   | c73'1   .  . |  .    .       .       || .   .      .   |  .      .  . |  .    .       .       |'----------------'--------------'-----------------------'`

Now the first ANS can't propagate anywhere, because there's no RC between a and b, so there's no chain at all beyond the internal strong link:

(4=69'2)r176c2 // can't continue

SpAce

Posts: 2559
Joined: 22 May 2017

### Re: Almost Locked Sets help

there is 27 sectors on a grid 9 for each box,row col

an almost locked set is a group of N cells each with at most N +1 digits that is contained with in a sector.

for example:
Code: Select all
`+------------+-------------+------------+| .  .     . | .  .      . | .  .     . || .  (19)  . | .  (789)  . | .  (89)  . || .  .     . | .  .      . | .  .     . |+------------+-------------+------------+| .  .     . | .  .      . | .  .     . || .  .     . | .  .      . | .  .     . || .  .     . | .  .      . | .  .     . |+------------+-------------+------------+| .  .     . | .  .      . | .  .     . || .  (17)  . | .  -7     . | .  .     . || .  .     . | .  .      . | .  .     . |+------------+-------------+------------+`

Code: Select all
`Almost Locked Set XZ-Rule: A=r2c258 {1789}, B=r8c2 {17}, X=1, Z=7 => r8c5<>7`

als- xz
combines two almost locked sets that have at least two digits in commonality

Set A=r2c258 {1789} {orange}
Set B=r8c2 {17} {blue}
Capture.PNG (6.96 KiB) Viewed 305 times

when both of the sets have at least one digit shared between the two sets we call this the restricted common candidate. {RCC}
as it can only exist in either set a or set b.

an RCC occurs in a sector that is overlapping both Set A & Set B marked in grey. {rcc are labeled x in the als move set}
Capture2.PNG (7.57 KiB) Viewed 305 times

the rcc is the digit{s} that is in both sets a & b, and can only be found in the grey area.
in this example its the digit "1"

because set a or b shares "1" digit we can conclude that the placement of this digits reduces either A or B by one digit
1 in A => B = N cells with N digits. { locked set}
1 in B => A = N cells with N digits. { locked set}

any Common Digit in A & B that is not the RCC {we call these the: non-restricted common candidate { labeled Z in the als move set}}
are potentially fixed in position by the placement of the RCC, any cells visible to these cells may be eliminated.

in this example its digit "7" is also shared by sets a & b and is not a RCC

that number is then excluded from any cell that is peer to all of cells containing this digit

ie 7 @ R8C5 can be removed as its visible to R2C5 & R8C2 which hold the digit "7"

I know that when ALS's overlap each other that the cells in the overlap area cannot contain a RC is that what you mean?

yes that's correct, to reiterate a RCC cannot be in an overlapping cells of set a and set b ,

since sets a & b are found in sectors the overlapping cells of sets A & B are a direct result of their sectors overlapping.

consider:
Set A = R28C1 (179) {orange}
Set B = R2C158 (1789) {blue}
overlapping Cells from A & B sector intersection is marked in Green.
Capture5.PNG (7.07 KiB) Viewed 305 times

the green cell cannot be used for the RCC

box 1 would be the RCC sector to use in this case however, we removed the green cell and there is no other cells left from a & B
so this is an invalid formation for the als-xz.

N digits

what is N digits: for my purpose.
N represents the possibility there is more then one RCC and more then one non-restricted Common candidate shared between both sets.

from here out its the same building blocks to form als-xy move by adding 1 extra sector, or adding N sectors {more then 1} the rules stay the same.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 1170
Joined: 05 September 2006

### Re: Almost Locked Sets help

Hi StrmCkr,

Very nice explanations and images. However, did you read what I said about "RCC" above? That term is clearly hobiwan's (rather surprising) mistake, as it makes zero sense to talk about a "candidate" when a "digit" is meant. It's not only wrong, it's also very confusing.

It should be just RC (Restricted Common) without the "candidate" part, just like Sudopedia and all the old threads here use it. Two connected ALSs can share one or more RCs (digits) but they can't share any RC candidates (instances of that digit), so it's actually a really significant difference. "RCC" should obviously mean the latter (candidate) but it actually means the former (digit), making it a completely useless term (because it can no longer be used for the latter meaning either without ambiguity).

Let's not spread such blatant errors, even if Hodoku does. Seems like a small thing, but such inaccuracies can cause endless amounts of confusion for newbies.

PS. I think the least confusing acronym would be RCD (Restricted Common Digit), because it would leave no doubt that we're talking about a digit and not a candidate.

SpAce

Posts: 2559
Joined: 22 May 2017

### Re: Almost Locked Sets help

These are the type of responses I needed, thanks guys. I haven't got time (at the moment) to digest the last few posts, but will look in depth at them sometime over the weekend, and no doubt will have more questions.
Wept

Wepwawet

Posts: 17
Joined: 19 November 2019

### Re: Almost Locked Sets help

http://forum.enjoysudoku.com/als-chains-a-tutorial-asi-3-t6443-30.html rcc predates hodoku and was used extensively on this forum, albeit a very overtly worded acronym.

As candidates represents the potential quantifiers for a space used by sudoku technique these are:
digits/numbers in the case of the Almost Naked set
Cells in the case of the Almost Hidden set.

So, yes it makes sense to a degree.
Even the pm grid was refereed to as potential Candidate markup shortened to pencil-marks or pencil-markup.

In layman's terms does it need the extra "c" to tell us exactly its restricting probably not since it applies to and is used on a sudoku grid within a specific technique branch.

History, the subject here in is discussed with lots of old links being referenced so i clearly defined those words.
Rcc can be upgraded or simplified to RC.
and may be either in old reference links.

There probably was a discussion dropping the extra "c" at some point as it could lead to ambiguity which you have brought up.

I can and am willing to edit my post to remove the extra "C" if you would prefer and shorten it to restricted common.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 1170
Joined: 05 September 2006

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