The New Sudoku Players' Forum

[DonM]

Code: Select all

 +--------------------------------------------------------------------------------+
 |  6       147     1237    |  13      479     5       |  379     8       279     |
 |  23457   147     123578  |  13      4789    789     |  35679   579     2679    |
 |  357     9       3578    |  2       6       78      |  357     4       1       |
 |--------------------------+--------------------------+--------------------------|
 |  8       16      1269    |  7       5       29      |  19      3       4       |
 |  79      5       4       |  8       1       3       |  2       6       79      |
 |  279     3       127     |  4       29      6       |  8       179     5       |
 |--------------------------+--------------------------+--------------------------|
 |  1       8       5679    |  56      3       4       |  5679    2       679     |
 |  34579   467     35679   |  56      27      127     |  145679  1579    8       |
 |  457     2       567     |  9       78      178     |  14567   157     3       |
 +--------------------------------------------------------------------------------+
 # 105 eliminations remain


Please forgive me for interrupting, but does the following work? (I don't know how to write it as a chain.)

Code: Select all

 [r4c2]=1 [r4c7]<>1 r89c7=14 [r9c7]<>6 [r9c3]=6 [r4c3]<>6 [r4c2]=6 => [r4c2]<>1


This should work (if I haven't made any typos ):

NL: r4c2 -1- r4c7 =1= hp(14)r89c7 -6- r9c7 =6= r9c3 -6- r4c3 =6= r4c2

AIC: (1)r4c7 = hp(14)r89c7 - (6)r9c7 = (6)r9c3 - (6)r4c3 = (6)r4c2

=> r4c2<>1

Incidentally, this is a reason why I prefer the AIC notation. Except for the fact that traditionally you start and end and AIC with a strong link and so the chain doesn't start with (1)r4c2 - (1)r4c7 etc., the AIC (IMO at least) more closely mirrors daj95376's printed logic.

[ronk]

This should work (if I haven't made any typos ):

NL: r4c2 -1- r4c7 =1= hp(14)r89c7 -6- r9c7 =6= r9c3 -6- r4c3 =6= r4c2

AIC: (1)r4c7 = hp(14)r89c7 - (6)r9c7 = (6)r9c3 - (6)r4c3 = (6)r4c2

=> r4c2<>1

The problem with that notation is ... there isn't a hidden pair in r89c7. There may be ... if digit <1> is removed from r4c7.

OTOH the ALS in r12347c7 doesn't have that baggage, it is. Likewise the AHS in r489c7 is.

[hobiwan]

NL: r4c2 -1- r4c7 =1= hp(14)r89c7 -6- r9c7 =6= r9c3 -6- r4c3 =6= r4c2

OTOH the ALS in r12347c7 doesn't have that baggage, it is. Likewise the AHS in r489c7 is.

So, where does that take us?

Code: Select all

r4c2 -1- AHS:r489c7 =6= r9c3 -6- r4c3 =6= r4c2

Is this correct?

[DonM]

This should work (if I haven't made any typos ):

NL: r4c2 -1- r4c7 =1= hp(14)r89c7 -6- r9c7 =6= r9c3 -6- r4c3 =6= r4c2

AIC: (1)r4c7 = hp(14)r89c7 - (6)r9c7 = (6)r9c3 - (6)r4c3 = (6)r4c2

=> r4c2<>1

The problem with that notation is ... there isn't a hidden pair in r89c7. There may be ... if digit <1> is removed from r4c7.

OTOH the ALS in r12347c7 doesn't have that baggage, it is. Likewise the AHS in r489c7 is.

I don't see the problem. The idea of notation is to convey the logic the solver is following. In this case I think it does: If r4c7<>1 then it is not an ahs and hp(14)r89c7 is true. One may also look on that pattern as an ahs until proven otherwise and legitimately prefer using the AHS and that's just fine, but afaik either way is purely discretionary.

[ronk]

NL: r4c2 -1- r4c7 =1= hp(14)r89c7 -6- r9c7 =6= r9c3 -6- r4c3 =6= r4c2

OTOH the ALS in r12347c7 doesn't have that baggage, it is. Likewise the AHS in r489c7 is.

So, where does that take us?

Code: Select all

r4c2 -1- AHS:r489c7 =6= r9c3 -6- r4c3 =6= r4c2

Is this correct?

No, because with different labels on two sides of a node (cells, group of cells), the inferences must be the same. For an AHS, I don't know a better answer than what the AHS folks are doing. That's one of the reasons I've always used the complementary ALS.

[Luke]

Code: Select all

 +--------------------------------------------------------------------------------+
 |  6       147     1237    |  13      479     5       |  379     8       279     |
 |  23457   147     123578  |  13      4789    789     |  35679   579     2679    |
 |  357     9       3578    |  2       6       78      |  357     4       1       |
 |--------------------------+--------------------------+--------------------------|
 |  8       16      1269    |  7       5       29      |  19      3       4       |
 |  79      5       4       |  8       1       3       |  2       6       79      |
 |  279     3       127     |  4       29      6       |  8       179     5       |
 |--------------------------+--------------------------+--------------------------|
 |  1       8       5679    |  56      3       4       |  5679    2       679     |
 |  34579   467     35679   |  56      27      127     |  145679  1579    8       |
 |  457     2       567     |  9       78      178     |  14567   157     3       |
 +--------------------------------------------------------------------------------+
 # 105 eliminations remain


Please forgive me for interrupting, but does the following work? (I don't know how to write it as a chain.)

Code: Select all

 [r4c2]=1 [r4c7]<>1 r89c7=14 [r9c7]<>6 [r9c3]=6 [r4c3]<>6 [r4c2]=6 => [r4c2]<>1


Nice. You have a knack for taking the complicated and confusing and making it simple and clear. Thanks.

[Pat]

Code: Select all

 6       147     1237    |  13      479     5       |  379     8       279
 23457   147     123578  |  13      4789    789     |  35679   579     2679
 357     9       3578    |  2       6       78      |  357     4       1
 ------------------------+--------------------------+----------------------
 8       16      1269    |  7       5       29      |  19      3       4
 79      5       4       |  8       1       3       |  2       6       79
 279     3       127     |  4       29      6       |  8       179     5
 ------------------------+--------------------------+----------------------
 1       8       5679    |  56      3       4       |  5679    2       679
 34579   467     35679   |  56      27      127     |  145679  1579    8
 457     2       567     |  9       78      178     |  14567   157     3



does the following work?
(I don't know how to write it as a chain.)

Code: Select all

[r4c2]=1 [r4c7]<>1 r89c7=14 [r9c7]<>6 [r9c3]=6 [r4c3]<>6 [r4c2]=6
=> [r4c2]<>1


or going in the other direction
i don't need the duo
but then i'm using a "net" rather than a "chain"

Code: Select all

r4c2=1 --> r4c3=6 --> r9c7=6 --> r8c7=4 --\
       \                     \------------|--> no 1 in c7
        \---------------------------------/



[daj95376]

or going in the other direction
i don't need the duo
but then i'm using a "net" rather than a "chain"

Code: Select all

r4c2=1 --> r4c3=6 --> r9c7=6 --> r8c7=4 --\
       \                     \------------|--> no 1 in c7
        \---------------------------------/


Excellent Pat

You caught me! I took a SIN and rewrote it to look like a chain. In fact, your net matches my SIN.

Now, all I have to do is see if I can make sense of any of the notation examples supplied for it.

[Luke]

If anyone is so inclined to check the validity of this one, I'd be grateful. My uncertainty arises as the final link actually ends up back in the starting set. I'm trying to use this idea in the puzzle below:

Code: Select all

 *--------------------------------------------------------------------*
 | 1      378    378    | 9      2      6      | 348    48     5      |
 | 568    2      4      | 3      58     1      | 7      689    69     |
 | 356    9      358    | 48     458    7      | 1      268    236    |
 |----------------------+----------------------+----------------------|
 | 7      1      589    | 468    34689  345    | 2      469    349    |
 | 2      345    359    | 7      3469   345    | 3469   1      8      |
 | 38     348    6      | 148    13489  2      | 49     5      7      |
 |----------------------+----------------------+----------------------|
 | 58     568    1      | 2      346    34     | 4689   7      469    |
 | 9      67     27     | 146    146    8      | 5      3      246    |
 | 4      368    238    | 5      7      9      | 68     268    1      |
 *--------------------------------------------------------------------*

(38)r6c12=(4)r6c2-(4=9)r6c7-(9)r4c8=(9)r2c8-(9=6)r2c9-(6)r2c1=(6-3)r3c1=(3)r6c1 => r5c23<>3, r6c5<>3.

[aran]

If anyone is so inclined to check the validity of this one, I'd be grateful. My uncertainty arises as the final link actually ends up back in the starting set. I'm trying to use this idea in the puzzle below:

Code: Select all

 *--------------------------------------------------------------------*
 | 1      378    378    | 9      2      6      | 348    48     5      |
 | 568    2      4      | 3      58     1      | 7      689    69     |
 | 356    9      358    | 48     458    7      | 1      268    236    |
 |----------------------+----------------------+----------------------|
 | 7      1      589    | 468    34689  345    | 2      469    349    |
 | 2      345    359    | 7      3469   345    | 3469   1      8      |
 | 38     348    6      | 148    13489  2      | 49     5      7      |
 |----------------------+----------------------+----------------------|
 | 58     568    1      | 2      346    34     | 4689   7      469    |
 | 9      67     27     | 146    146    8      | 5      3      246    |
 | 4      368    238    | 5      7      9      | 68     268    1      |
 *--------------------------------------------------------------------*

(38)r6c12=(4)r6c2-(4=9)r6c7-(9)r4c8=(9)r2c8-(9=6)r2c9-(6)r2c1=(6-3)r3c1=(3)r6c1 => r5c23<>3, r6c5<>3.

Luke : it's as simple as this
the chain establishes a strong link between the pair 38r6c12 and 3r6c1.
=> at least one of those is true (and indeed both could be true, to illustrate a recent point made elsewhere : 3r6c1, 8r6c2)
=> necessarily 3 somewhere in r6c12.
It could be r6c1 or r6c2.
So the elimination logic must take that into account : ie eliminate anything seen by both r6c1 and r6c2 : <3>r5c23 <3>r6c5.
Note that 3r9c2 cannot be eliminated since though seen by r6c2, it remains invisible to r6c1. Ditto 3r1c2. Nor 3r3c1, within the ocular scope of r6c1 alone.

[Luke]

Thanks for verifying my chain. I have a way of getting things wrong the first time I try something new.

=> at least one of those is true (and indeed both could be true, to illustrate a recent point made elsewhere : 3r6c1, 8r6c2)

There you go. Interesting that this would provide an example of your earlier point.

[Pat]

Code: Select all

 | 1      378    378    | 9      2      6      | 348    48     5      |
 | 568    2      4      | 3      58     1      | 7      689    69     |
 | 356    9      358    | 48     458    7      | 1      268    236    |
 |----------------------+----------------------+----------------------|
 | 7      1      589    | 468    34689  345    | 2      469    349    |
 | 2      345    359    | 7      3469   345    | 3469   1      8      |
 | 38     348    6      | 148    13489  2      | 49     5      7      |
 |----------------------+----------------------+----------------------|
 | 58     568    1      | 2      346    34     | 4689   7      469    |
 | 9      67     27     | 146    146    8      | 5      3      246    |
 | 4      368    238    | 5      7      9      | 68     268    1      |



(38)r6c12=(4)r6c2-(4=9)r6c7-(9)r4c8=(9)r2c8-(9=6)r2c9-(6)r2c1=(6-3)r3c1=(3)r6c1
=> r5c23<>3, r6c5<>3

i don't need the duo
but then i'm using a "net" rather than a "chain"

Code: Select all

       /--------------------------------------------------------\
r6c5=3 --> r3c1=3 --> r2c1=6 --> r2c9=9 --> r4c8=9 --> r6c7=4 --|--> r6c2 no possibility
       \-> r6c1=8 ----------------------------------------------/

________________________________________________________________________________________



there are some out there who would wish to deny to the solving public (via the pejorative taunt of "net") this oh so simple bit of logic.

"net" is not pejorative.
my above example may look messy --
only because i took the trouble to write it all out.

[aran]

Code: Select all

 | 1      378    378    | 9      2      6      | 348    48     5      |
 | 568    2      4      | 3      58     1      | 7      689    69     |
 | 356    9      358    | 48     458    7      | 1      268    236    |
 |----------------------+----------------------+----------------------|
 | 7      1      589    | 468    34689  345    | 2      469    349    |
 | 2      345    359    | 7      3469   345    | 3469   1      8      |
 | 38     348    6      | 148    13489  2      | 49     5      7      |
 |----------------------+----------------------+----------------------|
 | 58     568    1      | 2      346    34     | 4689   7      469    |
 | 9      67     27     | 146    146    8      | 5      3      246    |
 | 4      368    238    | 5      7      9      | 68     268    1      |



(38)r6c12=(4)r6c2-(4=9)r6c7-(9)r4c8=(9)r2c8-(9=6)r2c9-(6)r2c1=(6-3)r3c1=(3)r6c1
=> r5c23<>3, r6c5<>3

i don't need the duo
but then i'm using a "net" rather than a "chain"

Code: Select all

       /--------------------------------------------------------\
r6c5=3 --> r3c1=3 --> r2c1=6 --> r2c9=9 --> r4c8=9 --> r6c7=4 --|--> r6c2 no possibility
       \-> r6c1=8 ----------------------------------------------/

________________________________________________________________________________________



there are some out there who would wish to deny to the solving public (via the pejorative taunt of "net") this oh so simple bit of logic.

"net" is not pejorative.
my above example may look messy --
only because i took the trouble to write it all out.
Pat
I am happy to agree that yes that was a net !
Presentation clear.
As to "net" not being a pejorative term, well I do seem to have acquired a contrary impression on my travels...