The New Sudoku Players' Forum

[daj95376]

[Withdrawn: I keep trying to treat networks and chains differently, but there's an overlap of chains into networks that trips me up all too often.]

_

[David P Bird]

(7=56)r38c7 - (6)r3c4=(16)r3c4,r2c5 - (61=351)r6c5,r4c45 - (15=7)r4c3 => -7 r4c7 ; stte

Ronk, so you're resorting to using split nodes! But I think that it should be spelled out that listing gives the digits and cells they occupy in order.

However working though your chain I'm now questioning if the the links you've used are all sound and operate both forwards and backwards. It's also caused me to wonder if some of my applications of split nodes in the past might be suspect.

For example, is (61=351)r6c5,r4c45 a sound strong link? If (6)r6c5 is false the first term is false but it won't force the required distribution of digits in the second term. The same points arise if the link is taken in the reverse direction.

DPB

[DonM]

Simple pausing chain:
(3)r6c5=(3-8)r6c7=hp(89-6)r27c7=(6*-1)r2c5=r3c4-(1=5) => -56r6c5 stte

[SteveG48]

Thanks all. I expected and hoped for comments. I liked the logic of the solution, but was concerned with the notation. I like Ronk's re-write, but hope that David's comments lead to more discussion.

[daj95376]

[Withdrawn: I keep trying to treat networks and chains differently, but there's an overlap of chains into networks that trips me up all too often.]

_

[SteveG48]

An alternate perspective:

Code: Select all

 (57=6)r38c7 - (6=1)r3c4 - (1=5)r4c4 \
                         - (1=6)r2c5  - (56=3)r6c5 - r4c5 = 3r4c7  =>  -57 r4c7



_

OK, why not modify your alternative from (56=3)r6c5 - r4c5 = 3r4c7 => -57 r4c7 to (56=3)r6c5 - (3=15)r4c45 - (15=7)r4c3 etc. ?

[JC Van Hay]

Code: Select all

 *--------------------------------------------------------------------*
 | 6      59     8      | 3      7      4      | 1      59     2      |
 | 4      179    179    | 2     b16     5      | 6789   789    3      |
 | 3      2      157    |b16     9      8      |a567    4      57     |
 *----------------------+----------------------+----------------------|
 | 9      8     d157    |c15    c135    2      | 35-7   6      4      |
 | 17     1567   3      | 14568  1456   9      | 2      578    578    |
 | 2      4      56     | 568   c356    7      | 358    1      9      |
 *----------------------+----------------------+----------------------|
 | 17     1379   1479   | 45     2      6      | 5789   35789  578    |
 | 8      67     467    | 9      45     3      |a57     2      1      |
 | 5      39     2      | 7      8      1      | 4      39     6      |
 *--------------------------------------------------------------------*


(7=56)r38c7 - (6=16)b2p57 - (16=315)b5p128 - (15=7)r4c3 => -7 r4c7 ; stte

Or ... [(157)r4c345=3r4c5 - 3r6c5=XYWing(156)r34c4,r6c5 - (6=1)r2c5 - (1=6)r3c4 - (6=57)r38c6] - (57=3)r4c4; stte

[ronk]

(7=56)r38c7 - (6)r3c4=(16)r3c4,r2c5 - (61=351)r6c5,r4c45 - (15=7)r4c3 => -7 r4c7 ; stte

For example, is (61=351)r6c5,r4c45 a sound strong link? If (6)r6c5 is false the first term is false but it won't force the required distribution of digits in the second term. The same points arise if the link is taken in the reverse direction.

When you consider this term in the context of the doubly-linked expression, there is an obvious implied '&' between (6)r6c5 and (1)r4c4 for the l-to-r direction. If both are false it will "force the required distribution of digits in the second term." The right-to-left direction confuses me. I think it should be alright r-to-l, but really don't know.

Ronk, so you're resorting to using split nodes! But I think that it should be spelled out that listing gives the digits and cells they occupy in order.

I think split-nodes are virtually unavoidable when writing expressions for doubly-linked chains/networks.

[blue]

Code: Select all

+-----------------+-----------------+-------------------+
| 6   59    8     | 3      7      4 | 1      59     2   |
| 4   179   179   | 2      (16)   5 | 6789   789    3   |
| 3   2     157   | (16)   9      8 | (567)  4      57  |
+-----------------+-----------------+-------------------+
| 9   8     (157) | (15)   (135)  2 | 3-57   6      4   |
| 17  1567  3     | 14568  1456   9 | 2      578    578 |
| 2   4     56    | 568    (356)  7 | 358    1      9   |
+-----------------+-----------------+-------------------+
| 17  1379  1479  | 45     2      6 | 5789   35789  578 |
| 8   67    467   | 9      45     3 | (57)   2      1   |
| 5   39    2     | 7      8      1 | 4      39     6   |
+-----------------+-----------------+-------------------+

This is similar, but not identical to JC's alternate desctription of Steve's elimination:

(157=3)r4c345 - 3r6c5 = [XY-Wing: pivot=r6c5,pincers=r2c5,r3c4] - (1=6)r3c4 - (6=57)r38c7 => -57r4c7

It was produced from the network below, using the same SIS's and weak links that Steve used.
The bottom two streams end with the left/right reversal of the beginning of Steve's chain/net/whatever.
It's similar to the the left/right reversal of one of Danny's depictions as well.

Code: Select all

3r6c5 - (3=157)r4c345 ----------------------- 57r4c7
  ||
5r6c5 - (5=1)r4c4 - (1=6)r3c4 - (6=57)r38c7 - 57r4c7
  ||
6r6c5 - (6=1)r2c5 - (1=6)r3c4 - (6=57)r38c7 - 57r4c7

Merging the bottom two streams, gives:

Code: Select all

3r6c5 - (3=157)r4c345 ------------------------- 57r4c7
  ||
5r6c5 - (5=1)r4c4 --- (1=6)r3c4 - (6=57)r38c7 - 57r4c7
  ||                /
6r6c5 - (6=1)r2c5 -

The XY-Wing eliminating 1r3c4 stands out, and reversing the top stream to turn it into a chain, gave the result above.

(typos fixed)

[daj95376]

[Withdrawn: I keep trying to treat networks and chains differently, but there's an overlap of chains into networks that trips me up all too often.]

_

[daj95376]

[Withdrawn: somehow I ended up with an edited copy of my previous post.]

_

[SteveG48]

An alternate perspective:

Code: Select all

 (57=6)r38c7 - (6=1)r3c4 - (1=5)r4c4 \
                         - (1=6)r2c5  - (56=3)r6c5 - r4c5 = 3r4c7  =>  -57 r4c7



_

OK, why not modify your alternative from

(56=3)r6c5 - r4c5 = 3r4c7 => -57 r4c7

to

(56=3)r6c5 - (3=15)r4c45 - (15=7)r4c3 etc. ?

You know how the law says that you can't be tried twice for the same crime? Well, your scenario assigns <5> twice to the same cell.

It's already assigned by (1=5)r4c4, you can't assign it again by performing: (3=15)r4c45.

Why not? If it's established that r4c4 is 5 and that r4c5 is not 3, then doesn't it follow that (15) is true in r4c45? That's what the modified chain says to me.

[ronk]

Code: Select all

+-----------------+-----------------+-------------------+
| 6   59    8     | 3      7      4 | 1      59     2   |
| 4   179   179   | 2      (16)   5 | 6789   789    3   |
| 3   2     157   | (16)   9      8 | (567)  4      57  |
+-----------------+-----------------+-------------------+
| 9   8     (157) | (15)   (135)  2 | 3-57   6      4   |
| 17  1567  3     | 14568  1456   9 | 2      578    578 |
| 2   4     56    | 568    (356)  7 | 358    1      9   |
+-----------------+-----------------+-------------------+
| 17  1379  1479  | 45     2      6 | 5789   35789  578 |
| 8   67    467   | 9      45     3 | (57)   2      1   |
| 5   39    2     | 7      8      1 | 4      39     6   |
+-----------------+-----------------+-------------------+

This similar, but not identical to JC's alternate desctription of Steve's elimination:

(157=3)r4c345 - 3r6c5 = [XY-Wing: pivot=r6c5,pincers=r2c5,r3c4] - (1=6)r3c4 - (6=57)r38c7 => -57r4c7

I like it! Here's an alternative for the same target r4c7<>7.

r4c7 -7- r8c7 -5- r8c5 -4- r7c4 -5- r4c4 -1- r3c4 -6- als:r38c7 -57- r4c7

[David P Bird]

Ronk, I am bound up with some rather esoteric considerations which won't bother many here one iota, particularly those who are quite happy to use forcing chains and nets.

Firstly I'm adhering to the idea that every link in an AIC should be stand alone and independent of what went on earlier in the chain otherwise it ceases to be bidirectional and is forcing. Secondly I am only happy to accept AND logic (either declared or implied) when it is part of an accepted and recognisable pattern otherwise it represents branching. Keeping to these disciplines is a challenge, but that's the purpose of puzzles.

Although Steve's notation is wanting, his logic is valid and in my initial post I boxed off what I considered to be the ad hoc pattern he identified but in my view it really is a forcing net.

Playing with JC's and Blue's approach (well done!) I get this split node chain
(7=153)r3c345 - (3=56#1)r6c5 - (56=1#1)r4c4,r2c5 - (1=6)r3c4 - (6=57)r38c7 => r4c7 <> 7

The key point here is that in the blue node (1) only needs to be true or false in one of the cells to propagate the chain in either direction so I believe that this is a valid AIC.

BTW I've looked back over some of my previous split node chains and find they are mixed, some would pass and some would fail my present criteria.
PS I see you've posted an alternative chain while I've been writing, but you've cheated and used a completely different set of cells!

[Edit: typos highlighted by ronk corrected]

[ronk]

Playing with JC's and Blue's approach (well done!) I get this split node chain
(7=153)r3c345 - (3=56#1)r6c5 - (56=1#1)r4c4,r2c5 - (1=6)r3c4 - (6=57)r38c7 => r4c7 <> 7

The key point here is that in the blue node (1) only needs to be true or false in one of the cells to propagate the chain in either direction so I believe that this is a valid AIC.

"In either direction?" In the right-to-left direction, only r2c5=6 true or r4c4=5 true does not yield r6c5=3 because it is a tri-valued cell.

PS I see you've posted an alternative chain while I've been writing, but you've cheated and used a completely different set of cells!

Rather than cheating, I see my alternative chain as giving up, giving up on the effort to find an acceptable expression for a doubly-linked chain/network. While easy for me to conceptualize in SteveG's first post, a technically-correct chain/network expression is too complicated. That said, JC's, blue's, daj95376's approaches may be reduced versions of same.