The New Sudoku Players' Forum

[David P Bird]

Some interesting posts, but for now I'll concentrate on trying to answer eleven's question.

My view of these patterns is quite simple; for a SdC type pattern there are overlapping ANSs that combine to give a 2-sector Naked Set and for this pattern there are two overlapping AHSs that combine to give a 2-sector Hidden Set.

For both situations the combined digit and cell counts must obviously balance, and care must be taken to ensure that the counts for each digit in the combined pattern are forced. Usually this will mean each digit must occur once in the pattern, but there are other possibilities where individual digits or perhaps one or other of them must occur twice. I thought that the rules governing any second instances would be straightforward but as I was writing this, I realised I missed a case which makes them more complicated. That said however, when any potential 2-sector set is found it's easy enough to see how many member digits must be true in the combined cells.

I can't understand when everyone seems to be so comfortable with ANSs they don’t feel equally comfortable with AHSs. I certainly find them easier to spot*. When two AHSs overlap, each one needs one more digit to be restricted to the cell set to become fully locked, and when that digit must come from the locked members in the other AHS, there is a hit.

My approach is to look for AHSs in the boxes. For each one I identify I then check how the digits that aren't locked in that set are locked in the intersecting rows and columns. It doesn't take much practice to recognise the cases worth investigating.

As I don’t rely on a solver program I can't quickly run through a batch of puzzles looking for them, so my experiences are limited. So far it seems they often occur in conjunction with 2 sector ANS (as in the original puzzle in this thread), and the stacks and tiers worth investigating first are those that contain bivalue cells. They also seem to be rarer than 2 sector ANSs.

*To help me, as I switch between focus digits, my spreadsheet shows me the other digits that a) always, b) sometimes, and c) never occur with the focus digit in each house. It's not a comprehensive system, but it takes some of the drudgery out of finding ANS/AHS partitions.

[aran]

Since Xsudo was mentioned above in a discussion of constructs that are difficult to find manually eg. doubly-lined ALSs

Xsudo has been mentioned...but in particular by those who do not use it : eleven and myself.
You take the view that DL-ALS are too complicated for manual solving, and seem to regard DL-ALS with the sole exception of SDC as a re-engineeered product.
You even spoke of DL-ALS in the same breath as forcing chains...shocking !

I take the view that DL-ALS are simple enough to be understood and included in the manual solver's techniques.
What is an SDC ? a DL-ALS in which one ALS is a bivalue dual-linked to an "ordinary" ALS.
That's all it is.
When you think of the rather complicated definition given for SDC, there should be pause for thought.
In a "regular" DL-ALS the bivalue expands and the dual-links will often be located in two houses rather than one.

Once the solver can "see" an ALS rapidly, in a DL-ALS quest, he is looking for :
1.an ALS with an interesting distribution : specifically there must be a candidate x restricted to one box, and a candidate y restricted to one box (not necessarily the same box, unlike the SDC) eg
r1c1abcde r1c3 abce r1c5abd r1c6bcde
will be hopeless with its regular distribution, whereas
r1c1abcd r1c3abcd r1c5abce r1c6ace
is already more promising since candidates d and e are respectively restricted to b1 and b2.
2. now he looks to see if another ALS ties in : specifically an ALS with the same restricted candidates in the same boxes. So in other words, his search is heavily oriented.
Rather than deal with the general case, illustrate by reference to the above. For example :
r2c2pqd r2c3qrd r2c4pqse r2c5rse r2c8pqrs
ie an ALS pqrsde with d and e respectively restricted to b1 and b2
Hence {abcde} {edpqrs}
and then write out the eliminations at a stroke.

For anyone interested in the base cover version in a DL-ALS context
ALS 1 : m+1 candidates m cells
ALS 2 : n+1 candidates n cells
ALS 1 = (m-1+xy) where xy are the dual links
ALS 2 = (n-1+xy)
base = m+n cells
cover = h(m-1)+h(n-1)+hx+hy where h=r c b ie row column or box
=(m-1)+(n-1)+2=m+n
so rank = base-cover =(m+n)-(m+n) = 0.

Extending that to the "hidden" version
ALS 1 : m+1 cells m candidates
ALS 2 : n+1 cells n candidates
Reasoning "back to front" : assume this is rank 0 structure
Cover = h(m)+h(n) = m+n
Base is apparently m+1+n+1=m+n+2
But the base must be m+n.
Therefore there must be two overlapping cells.
eg as we saw in the David P Bird example with r7c1 and r7c2 being the overlaps.

[eleven]

Thanks David,

I can't understand when everyone seems to be so comfortable with ANSs they don’t feel equally comfortable with AHSs. I certainly find them easier to spot.

Yes, they are easy to spot, as long as i don't have pencilmarks. I see a lot of almost hidden sets, when looking for hidden sets, in any non trivial puzzle. I guess it would be the same in your example:

Code: Select all

+-------+-------+-------+
| . . 9 | 8 7 1 | . . . |
| . 8 2 | 6 . . | 1 . . |
| 1 . # | . .*2 | . 8 . |
+-------+-------+-------+
| . . . | . . . | . . 8 |
| 5 4 . | . 8*7 | . . . |
| 8 . . | . . . | 5 . . |
+-------+-------+-------+
| . . # | . 6 . | 8 . . |
| . . 8 | 3 . . | ° .*7 |
| . . 3 | .<1 8 | °<6<9 |
+-------+-------+-------+

The 45 pair in the #-ed cells is obvious (as well as 79 in box 3). Then there is the 153 in box 9.
Now it's natural to look for 247 in the 4 remaining cells of r7. You find 27 in c6 leaving the AHS.

But as soon as the candidates are placed in, i get blind for them (smiley with dark glasses).

[David P Bird]

Aran, I believe that your cover & base set balances don't properly allow for a variable number of cells being common to both Almost sets – or am I mistaken? Edit - scrub that, I now see where I was mistaken

Eleven, pleased to be of help. Obviously from the number of pieces of information your brain holds, your short term memory is better than mine.

[David P Bird]

Having identified another way of looking at rank 0 patterns that result in 2-sector locked sets, I consider them to be flashy rather than awe inspiring. If one can be spotted, it's a nice way of making multiple eliminations from a single step, but all of these would be available from AICs. Ronk's loop shows that with effort they may all be found at once, but more probably they'll be picked off in differerent steps, as shown by daj.

This then poses the question of where they should be placed in our hierarchy of techniques, particularly considering that they're not particularly frequent or easy to spot. Personally I think that they are worth a one-off, selective (rather than exhaustive), search when the PM grid looks promising ie with a reasonable number of cells containing pairs and triples. However, just being aware of the possible existence of these patterns should be of benefit in a regular chain hunt.

Turning to their notation I find that
It's quicker to identify all the eliminations from the MSLS than from the AIC loop.
The loop notation needs a more thorough understanding of its workings than the MSLS notation.
The MSLS notation is shorter.
The loop properly identifies all the cells involved whereas the MSLS doesn't.
The loop gives an indication of how the discovery was made while the MSLS doesn't
Considering all of these points, I personally favour the MSLS notation, but can accept that others won't.

[DonM]

Since Xsudo was mentioned above in a discussion of constructs that are difficult to find manually eg. doubly-lined ALSs

Xsudo has been mentioned...but in particular by those who do not use it : eleven and myself.
You take the view that DL-ALS are too complicated for manual solving, and seem to regard DL-ALS with the sole exception of SDC as a re-engineeered product.
You even spoke of DL-ALS in the same breath as forcing chains...shocking !

It was Eleven's comment that I was responding to: In my eyes the als, Sue de Coq, base/cover and ring interpretations are more of theoretical interest.All you have to spot is a simple forcing chain.

My response did not make a connection between DL-ALS and forcing chains. In my solving world, they have nothing to do with each other.

I take the view that DL-ALS are simple enough to be understood and included in the manual solver's techniques.
What is an SDC ? a DL-ALS in which one ALS is a bivalue dual-linked to an "ordinary" ALS.
That's all it is. When you think of the rather complicated definition given for SDC, there should be pause for thought.
In a "regular" DL-ALS the bivalue expands and the dual-links will often be located in two houses rather than one.

In my view, the definition of a DL-ALS is simple enough. Finding them is not so easy though I have done so in the past. On the other hand, IMO, while the definition of a SDC is complicated, finding them is easy.

While we can disagree on which is easier to find, actual manual solving history seems to prove my point: Having noted the solutions that were on Eureka and reviewing the solutions at the dailysudoku site and here, I can count on 2 or 3 fingers those that present a construct found as a DL-ALS. On the other, there has been quite a few SDCs and a number of almost-SDCs.

[eleven]

Having noted the solutions that were on Eureka and reviewing the solutions at the dailysudoku site and here, I can count on 2 or 3 fingers those that present a construct found as a DL-ALS. On the other, there has been quite a few SDCs and a number of almost-SDCs.

But note, that this relation is probably biased by the fact, that more/common solvers have implemented SDC, but not dl ALS (at least i don't know one for the latter).

[DonM]

Having noted the solutions that were on Eureka and reviewing the solutions at the dailysudoku site and here, I can count on 2 or 3 fingers those that present a construct found as a DL-ALS. On the other, there has been quite a few SDCs and a number of almost-SDCs.

But note, that this relation is probably biased by the fact, that more/common solvers have implemented SDC, but not dl ALS (at least i don't know one for the latter).

When I have seen the various SDC-based solutions I mentioned, it never occurred to me at the time that they were anything but manually found. I hope I wasn't wrong.

[eleven]

But note, that this relation is probably biased by the fact, that more/common solvers have implemented SDC, but not dl ALS (at least i don't know one for the latter).

Ah, hodoku can (i found it when looking for a SDC definition). And a lot more.
E.g. you can list all possible (implemented) steps for a grid ...

I now looked, what it has for the ER 7.3 of the current pattern game (where i gave up to find something nice). After some steps I saw a SDC and a DL-ALS.
Hard to decide now, if i should look for the SDC or the DL-ALS here:

Code: Select all

.------------------.-----------------------------.-------------------.
| 1     46789  479 | B579         2       5789   | 45689  3     5689 |
| 2689  3      279 | B579         589     4      | 1      289   5689 |
| 289   489    5   |  6           3       1      | 2489   7     89   |
:------------------+-----------------------------+-------------------:
| 2389  789    6   |  1-234-7-9   1489    2789   | 89     5     1789 |
| 4     5789   379 |  13-5-7-9    15689   56789  | 689    189   2    |
| 2589  1      279 | B2579        5689    256789 | 3      4     6789 |
:------------------+-----------------------------+-------------------:
| 569   4569   8   | B12459      -1-4569  3      | 7      129   159  |
| 7     2      14  |  8          A14      59     | 59     6     3    |
| 3569  569    139 | B1259        7       2569   | 2589   1289  4    |
'------------------'-----------------------------'-------------------'
Almost Locked Set XZ-Rule: A=r8c5 {14}, B=r12679c4 {124579}, X=1,4 =>
r7c5<>4, r7c5<>1, r4c4<>2, r5c4<>5, r45c4<>7, r45c4<>9
.------------------.-----------------------------.-------------------.
| 1     46789  479 | C579         2       5789   | 45689  3     5689 |
| 2689  3      279 | C579         589     4      | 1      289   5689 |
| 289   489    5   |  6           3       1      | 2489   7     89   |
:------------------+-----------------------------+-------------------:
| 2389  789    6   |  1-234-7-9   1489    2789   | 89     5     1789 |
| 4     5789   379 |  13-5-7-9    15689   56789  | 689    189   2    |
| 2589  1      279 | C2579        5689    256789 | 3      4     6789 |
:------------------+-----------------------------+-------------------:
| 569   4569   8   | A12459      -1-4569  3      | 7      129   159  |
| 7     2      14  |  8          B14      59     | 59     6     3    |
| 3569  569    139 | A1259        7       2569   | 2589   1289  4    |
'------------------'-----------------------------'-------------------'
Sue de Coq: r79c4 - {12459} (r126c4 - {2579}, r8c5 - {14}) =>
r7c5<>1, r7c5<>4, r4c4<>2, r45c4<>7, r45c4<>9, r5c4<>5

As soon as it is marked, both is easy to spot

But when looking at other samples, i don't think, that i will invest too much time for finding the one or other.

[David P Bird]

The MSLS equivalents are:
MSNS (14)b8,(2579)c4: 6 digits/constrained cells => r4c4 <> 279, r5c4 <> 579, r7c5 <> 14

MSHS (134)b5,(5689)c5: 7 digits/available cells => r4c4 <> 279, r5c4 <> 579, r7c5 <> 14

Corrected typos

[eleven]

MSHS (134)b5,(5689)c5: 7 digits/available cells => r4c4 <> 279, r5c4 <> 13, r7c5 <> 14

Ah thanks, i did not find it yesterday night. Of course its just the complement.

[ronk]

The MSLS equivalents are:
MSNS (14)b8,(2579)c4: 6 digits/constrained cells => r4c4 <> 279, r5c4 <> 13, r7c5 <> 14

MSHS (134)b5,(5689)c5: 7 digits/available cells => r4c4 <> 279, r5c4 <> 13, r7c5 <> 14

Assuming r5c4<>13 are typos, I don't understand why the 2nd isn't ...

(134)c4, (2569)b8: 7 digits/available cells => r4c4<>279, r5c4<>579, r7c5<>14

These are the strong inference set complements to the SDC cells r12679c4 and r8c5.

[eleven]

Yes, typos.
Another complement (as mentioned by David) is 134 in box 5 and 5689 in c5 (complementary cells to the DL-ALS in c45).
I guess you really could find it after spotting the 134 AHS, by checking the rows and columns 45 for a 2nd one (as David described).

Code: Select all

  +-----------+-----------+-----------+
  |  1  .  .  |  .  2  .  |  .  3  .  |
  |  .  3  .  |  .  b *4  |  1  .  .  |
  |  .  .  5  |  6  3 *1  |  .  7  .  |
  +-----------+-----------+-----------+
  |  .  .  6  |  a  C  .  |  .  5  .  |
  |  4  .  .  |  a  C  .  |  .  .  2  |
  |  . *1  .  |  .  b  .  | *3 *4  .  |
  +-----------+-----------+-----------+
  |  .  .  8  |  .  b *3  |  7  .  .  |
  |  7  2  X  | @8  X  Y  |  Y @6  3  |
  |  .  .  .  |  .  7  .  |  .  .  4  |
  +-----------+-----------+-----------+

X = 14, Y=59,
AHS 134 in aC (box 5).
Possible values in c5 14/5689.
5689 in bC (column 5) =>
134 in r45c4 -> r4c4 <> 279, r5c4 <> 579,
5689 in r267c5 -> r7c5 <> 14
134/5689 in r45c5 -> no eliminations

[David P Bird]

Typos confirmed and corrected.
As the ANS/AHS partitions in both b5 and b8 can be used, there are in fact 4 possible MSLS combinations that will make the same eliminations in 3 cells.

Code: Select all

1) MSNS:(14)b8,(2579)c4   6 digits/constrained cells  2) MSHS:(25679)b8,(134)c4 8 digits/available cs   
3) MSNS:(256789)b5,(14)c5 8 digits/constrained cells  4) MSHS:(134)b5.(5689)c5  7 digits/available cs

1) gives the smallest naked set and 4) gives the smallest hidden set.
As all eliminations are internal for hidden sets, 4) covers the fewest cells though.

This situation is a bit of a rarity as I've found that simple complements, where every digit appears once in the complementary sets, aren't always available with these two sector patterns.

I wrote this before I saw eleven's response which covers much the same ground.

[ronk]

Another complement (as mentioned by David) is 134 in box 5 and 5689 in c5 (complementary cells to the DL-ALS in c45).
I guess you really could find it after spotting the 134 AHS, by checking the rows and columns 45 for a 2nd one (as David described).

Yes, I recognized that "pattern" from David P Bird's description, but don't think it should be called a "complement." Complements traditionally occur in the same unit, i.e., row, column or box. A complementary pair is what I assumed his pair to be when ..

The MSLS equivalents are:
MSNS (14)b8,(2579)c4: 6 digits/constrained cells => r4c4 <> 279, r5c4 <> 579, r7c5 <> 14

MSHS (134)b5,(5689)c5: 7 digits/available cells => r4c4 <> 279, r5c4 <> 579, r7c5 <> 14

Apparently complementary pair was not his intent.

As the ANS/AHS partitions in both b5 and b8 can be used, there are in fact 4 possible MSLS combinations that will make the same eliminations in 3 cells.
1) MSNS:(14)b8,(2579)c4 6 digits/constrained cells 2) MSHS:(25679)b8,(134)c4 8 digits/available cs
...

Hmm, digit <7> is a clue or placement in b8 of eleven's pencilmarks.