The New Sudoku Players' Forum

[David P Bird]

Nested ANSs

This puzzle from ttt demonstrates a type of opening that although probably not new, I've never been conscious of before. It exists when a house contains an ANS which can grow in steps by adding a cell at a time.

..1.7.......8..4..8.......31.3.2..8...45.37...5..1.3.22.......9..6..4.......9.1.. ttt 2009

Preliminary Steps: Show

(5)r4c7 = (5-4)r4c9 = (4)r9c9 - (4=3)r9c1 - (3=9)r8c1 - (9)r9c2 = (9)r4c2 => r4c7 <> 9
. . (9)BoxLine:b6c8 => r123c8 <> 9
(2)r3c6 - (2,9)r3c3,c2 = (9)r2c6 -[UR]- (2)r3c6 => r3c6 <> 2
(4)r6c4 = (4)r4c4 - (4)r4c9 = (4)r9c9 - (4)r9c1 = (4-5)r1c1 = (5-7)r2c1 = (7)r6c1 => r6c4 <> 7
(56=8)r47c7 - (8)r7c6 = (8)r9c6 - (38=4)r9c12 - (4)r9c9 = (4-5)r4c9 = (5)r4c7 => r138c7 <> 5
(7)r3c2 = (57-1)r3c68 = (1)r3c4 - (1)r8c4 = (1-9)r8c2 = (9)r4c2 => r4c2 <> 7
. . (7)Single r6c1

Code: Select all

*----------------------*----------------------*----------------------*
| 3456   346    1      | 2369   7      2569   | 2689   256    568    |
| 356    367f   29     | 8      356    12569  | 4      12567  1567a  |
| 8      67e    29     | 1269   4      12569  | 269    12567  3      |
*----------------------*----------------------*----------------------*
| 1      69d    3      | 4679   2      679    | 56     8      456    |
| 69c    2      4      | 5      8      3      | 7      169    16b    |
| 7      5      8      | 469    1      69     | 3      469    2      |
*----------------------*----------------------*----------------------*
| 2      1348   57     | 1367   356    15678  | 568    34567  9      |
| 39     1389   6      | 1237   35     4      | 28     2357   578    |
| 34     348    57     | 2367   9      25678  | 1      234567 45678  |
*----------------------*----------------------*----------------------*

In row2 cell r2c9 can be added to (3567)ANS:r2c125 to make (13567)ANS:r1259 as (1) is the only non-ANS digit in that lone cell.
The complementary AHS would then reduce from (129+x)AHS:r2c3689 to (29+x)AHS:r2c368.
The trick is to isolate the lone cell and use the smaller (3567)ANS:r2c125 with the smaller (29+x)AHS:r2c368.
If a chain from the non-ANS digit in the lone cell to any digit in the ANS can be found showing they can't be true together, then the other ANS digits can be eliminated from the smaller AHS.
(129)AHS:r2c368 = (1*)r2c9 - (1=6)r5c9 - (6)r5c1 = (6)r4c2 - (6=7)r3c2 - (7*=356)ANS:r2c125 => r2c68 <> 56

Row 2 actually provides a progression of nested ANSs and lone cell options:
(3567)r2c125 -> (1*3567)r2c1259* -> (12*3567)r2c1258*9 -> (1235679*)r2c1256*89
However the chain given seems to be the only one that produces eliminations.

A second example occurs in row 3 using (269)ANS:r3c37, (5+x)AHS:r3c68, (1269)LoneCell:r3c4
(15)r3c68 = (1*)r3c4 - (1)r8c4 = (1-8)r8c2 = (8)r8c79 - (8=56)r47c7 - (6*=29)r3c37 => r3c6 <> 29, r3c8 <> 2

In this case a segment of the external chain can be re-used to also eliminate an ANS digit from the lone cell
(2)r12c6 = (2-8)r9c6 = (8)r7c6 - (8=56)r47c7 - (6*=29)r3c37 => r3c4 <> 2

I feel that there is a tendency for solvers to ignore AHSs assuming that any eliminations they may provide will always be available through the complementary ANSs, but as these examples show, that can lead to opportunities being missed. Indeed, personally I find searching for AHSs easier.

(Although I was pleased to find these eliminations, I was disappointed to discover that I couldn't complete the puzzle using only linear methods.)

DPB
.
TAGdpbNestedANSs

[eleven]

David,

for me these eliminations can be found easier this way (see here):
If from an (naked) ALS 2 digits lead to a contradiction, the others must be in the ALS. (Sorry for using the bad word contradiction:) )

ALS 13567 r2c1259:
1r2c9->6r5c9
7r2c2->6r3c2->6r5c1
=> 356 must be in the ALS => r2c68 <> 56

So i also cannot see a reason, that it should not be found, if only ANS's are considered.
(356=17)r2c1259-(7=6)r3c2-r4c2=r5c1-(6=1)r5c9-(1=3567)r2c1259 => r2c368<>356
[edit:] corrected first link, see below

[David P Bird]

Eleven, pleased to hear from you!

My aim was to show that if a player finds different digits in a house containing nested ANSs are weakly linked there may be some bonus eliminations available. Consequently when identifying ANSs it's worth noting if they can be expanded or not. My presentation aimed to make the logic involved as simple as possible. Because I use colouring to show up the locked digits, I find AHSs easier to identify than ANSs (as I said), which slanted the way I wrote it, but I accept that you find ANSs easier to work with.

Regarding your contradiction route I think the purpose of a notation is to demonstrate the logical path, not to set mini-puzzles which is what I feel your links do (but then my Citroen 2CV synapses don't match your Ferrari Spider ones).

Actually on my personal elegance scale your forcing chains are more acceptable than branching but I prefer to avoid both. For someone with a good eye for openings FCs can be useful but otherwise I consider them a rather inefficient approach compared to AICs.

David
.

[eleven]

i guess, that the programmers will have included this stuff next week.
nice, that at least we can keep them working

[blue]

If from an (naked) ALS 2 digits lead to a contradiction, the others must be in the ALS. (Sorry for using the bad word contradiction:) )

ALS 13567 r2c1259:
1r2c9->6r5c9
7r2c2->6r3c2->6r5c1
=> 356 must be in the ALS => r2c68 <> 56

So i also cannot see a reason, that it should not be found, if only ANS's are considered.
(1356=7)r2c1259-(7=6)r3c2-r4c2=r5c1-(6=1)r5c9-(1=3567)r2c1259 => r2c368<>356

Hi eleven,

Your point is well taken, but your analysis seems to be ignoring the 7r2c9 posiblity.

For the contradiction argument, it isn't a big deal: 1r2c9 and 7r2c9 are ("already") contradictory [ and there are no other candidates for '1' ].
Said differently: there is really only one way to have both a '1' and a '7' in the ANS cells, and you've shown that it leads to a conflict/contradiction.

For the ANS chain, you need to change the first link, to give:
(356=7)r2c125-(7=6)r3c2-r4c2=r5c1-(6=1)r5c9-(1=3567)r2c1259 => r2c368<>356

At that point, it involves the two ANS that David was mentioning at the outset.
Also (and not by coincidence): changing the ending from "-(1=3567)r2c1259" to "-1r2c9=(HT: <129>r2c368)", gives the reversal of David's chain.

[eleven]

Thanks blue,

for the clarification and for pointing out my sloppy mistake in the first AIC link.

Yes, the only way to have 1 and 7 in the ALS is with 1r2c8 and 7rc2c2.
And the first ALS link of course has to be split differently. I changed it to (356=17)r2c1259.

[blue]

And the first ALS link of course has to be split differently. I changed it to (356=17)r2c1259.

Slick trick !
I love it ... it's filled with insight

[Added] Insight: What if there were multiple ways to have a <17> in the ANS, but they all lead to a (useful) common conclusion ?

[David P Bird]

Eleven, Blue,

Why not do the same at the other end of the chain and move the 3 over too? After all there aren't any 3s to eliminate.
(56=137)r2c1259 - (7=6)r3c2 - r4c2 = r5c1 - (6=1)r5c9 - (137=56)r2c1259 => r2c368 <> 56

Tricks like this are rather like riding a bike with no hands - yes it works, but should we be promoting the practice bearing in mind the earlier confusion over notating 1|7 and 1&7 ?

Blue, I don't think having more than one possible distribution of the 1s and 7s in the ANS makes any difference provided they are both in sub-groups that link to the external chain.