## Write Chain

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

### Write Chain

100036427460500103300100600021603074006410230034200816040720361270361040613000702

WriteChain.png (60.66 KiB) Viewed 251 times

Placing 8 OR 9 in r1c4 forces the same number into Row8 in Block9 via r3c2, r2c8, but the opposite number caused in r9c4 means r8c79 are 8,9
=> 5r9c8 stte. How would you write a chain which describes this, please?
Yogi
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Location: New Zealand

### Re: Write Chain

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`*----------------------------------------------*| 1    589 589 |c89  3     6    | 4   2    7   || 4    6   27  | 5  b789  b2789 | 1  a89   3   || 3    89  27  | 1   47    247  | 6   589  589 ||--------------+----------------+--------------|| 589  2   1   | 6   589   3    | 59  7    4   || 5789 589 6   | 4   1     5789 | 2   3    59  || 579  3   4   | 2   579   579  | 8   1    6   ||--------------+----------------+--------------|| 589  4   589 | 7   2     589  | 3   6    1   || 2    7   589 | 3   6     1    | 59  4    589 || 6    1   3   |d89  4589  4589 | 7   5-89 2   |*----------------------------------------------*`

A rather topical subject at the moment. See here. Double Kite (89) : Row 2, Column 4, Box3 => - 89 r9c8; stte

In Eureka style notation I'd write something like (89) r2c8 = r2c56 - r1c4 = (89) r9c4 => - 89 r9c8; stte

Leren

PS I suppose another way of looking at it is Remote Pairs (89). r1c4, r2c8 & r3c2 have one parity (A) and r9c4 has the other parity (B).

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`*---------------------------------------------*| 1     589 589 |A89 3    6    | 4   2    7   || 4     6   27  | 5  789  2789 | 1  A89   3   || 3    A89  27  | 1  47   247  | 6   589  589 ||---------------+--------------+--------------|| 589   2   1   | 6  589  3    | 59  7    4   || 5789  589 6   | 4  1    5789 | 2   3    59  || 579   3   4   | 2  579  579  | 8   1    6   ||---------------+--------------+--------------|| 589   4   589 | 7  2    589  | 3   6    1   || 2     7   589 | 3  6    1    | 59  4    589 || 6     1   3   |B89 4589 4589 | 7   5-89 2   |*---------------------------------------------*`

Since r8c9 can see both 89 parities => - (89 = 5) r9c8; stte. r3c2 has parity A but doesn't contribute to the eliminations in r9c8.

Leren
Leren

Posts: 4066
Joined: 03 June 2012

### Re: Write Chain

Hi Yogi,

Yogi wrote:Placing 8 OR 9 in r1c4 forces the same number into Row8 in Block9 via r3c2, r2c8, but the opposite number caused in r9c4 means r8c79 are 8,9
=> 5r9c8 stte. How would you write a chain which describes this, please?

The exact logic you're describing is hard to write correctly as a single chain. It's simplest with two grouped X-Chains:

(8)r9c4 = r1c4 - r2c56 = r2c8 - r9c8 = (8)r8c9 => -8 r9c8
(9)r9c4 = r1c4 - r2c56 = r2c8 - r9c8 = (9)r8c79 => -9 r9c8
I have sometimes proposed something like this as a way to combine them:

(8/9)r9c4 = r1c4 - r2c56 = r2c8 - r9c8 = (8/9)r8c79 => -89 r9c8
That, however, is not standard, and it should still be interpreted as two chains written as one. (Without the '/' it's completely incorrect no matter how enticing it might be.)

Fortunately your logic can be simplified by stopping the chains at r2c8. Then it can be written as a single loop (Double Grouped Kite Loop):

(8)r9c4 = r1c4 - r2c56 = (8-9)r2c8 = r2c56 - r1c4 = (9)r9c4 - loop => -89 r9c8, -89 r3c56
(Note that -89 r3c56 can also be eliminated with a naked triple first. That's why they're not present in Leren's pm.)

That pattern can also be called a Grouped Remote Pair. The shortest way to write it as a correct AIC:

(8,9,8,9 = 9,8,9,8)r9c4,r1c4,r2c56,r2c8 => -89 r9c8, -89 r3c56
This is not correct:

Leren wrote:In Eureka style notation I'd write something like (89) r2c8 = r2c56 - r1c4 = (89) r9c4 => - 89 r9c8; stte

Like I said, it needs the '/' or something to indicate that it's actually two X-Chains written as one. Otherwise it makes no sense.
-SpAce-: Show
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`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"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: 2672
Joined: 22 May 2017

### Re: Write Chain

Yogi wrote:100036427460500103300100600021603074006410230034200816040720361270361040613000702
Placing 8 OR 9 in r1c4 forces the same number into Row8 in Block9 via r3c2, r2c8, but the opposite number caused in r9c4 means r8c79 are 8,9
=> 5r9c8 stte. How would you write a chain which describes this, please?

What you describe is a forcing chain, but there's a much simpler solution (starting from the original puzzle):

Code: Select all
`**************************************************************************************************  SudoRules 20.1.s based on CSP-Rules 2.1.s, config = W+SFin***  Using CLIPS 6.32-r773***********************************************************************************************104 candidates, 512 csp-links and 512 links. Density = 9.56%whip[1]: r1n5{c3 .} ==> r3c3 ≠ 5, r3c2 ≠ 5hidden-pairs-in-a-column: c3{n2 n7}{r2 r3} ==> r3c3 ≠ 9, r3c3 ≠ 8, r2c3 ≠ 9, r2c3 ≠ 8;;; what's new wrt your PM:t-whip[2]: c4n8{r9 r1} - r2n8{c6 .} ==> r9c8 ≠ 8hidden-single-in-a-block ==> r8c9 = 8whip[1]: b7n8{r7c3 .} ==> r7c6 ≠ 8t-whip[2]: c4n9{r9 r1} - r2n9{c6 .} ==> r9c8 ≠ 9stte`
denis_berthier
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Location: Paris

### Re: Write Chain

ERI Pair
ERI Pair.png (23.32 KiB) Viewed 207 times
yzfwsf

Posts: 337
Joined: 16 April 2019

### Re: Write Chain

Thanx!
Yogi
2017 Supporter

Posts: 189
Joined: 05 December 2015
Location: New Zealand

### Re: Write Chain

OK, I get that attempting to cover both eliminations in one chain is confusing and I accept that my clumsy logic, asserting that “r8c79 had to be 8 & 9, therefore r9c8 had to be 5,” only muddied the waters. The two chains do it in a much more straightforward way to show that r9c8 can’t be 8 or 9. I don’t quite see it as a kite but I was mainly interested in the Eureka chaining language.
Here’s how I see it. Given that you have identified by logic or otherwise that the eliminations are in r9c8, you want to write the chain to explain it in such a way that the endpoints both see this cell. So we go from r9c4 to r2c8, and we will have an AIC if we can start and end with a strong link and alternate strong and weak links throughout the chain.
(8)r9c4=r1c4-r2c56=r2c8 => -8r9c8 and
(9)r9c4=r1c4-r2c56=r2c8 => -9r9c8 => 5r9c8 stte
The tricky part was seeing r1c4-r2c56 as a single weak link. I would not have thought of that as a single link but clearly it is valid. Are we seeing r2c56 as a single entity in the chain, a node, perhaps? Is the weak link here saying “When r1c4 is 8, NEITHER r2c5 nor r2c6 can be 8 and therefore r2c8 must be 8,” proving that r9c8 cannot be 8 because it can see both ends of the chain, at least one of which must be 8?
Finally, if everything works the way it should, does the AIC work in the opposite direction? Seems so.
If r2c8 is NOT 8, one of r2c5 or r2c6 must be 8, so r1c4 cannot be 8 and r9c4=8 => -8r9c8.
Works fine if r2c56 is a single point in the chain.
Yogi
2017 Supporter

Posts: 189
Joined: 05 December 2015
Location: New Zealand

### Re: Write Chain

Hi Yogi,

From what you wrote, I think your understanding of chains, group nodes, and the Eureka notation has taken tremendous steps forward! Great job!

Yogi wrote:The two chains do it in a much more straightforward way to show that r9c8 can’t be 8 or 9. I don’t quite see it as a kite but I was mainly interested in the Eureka chaining language.

Well, it's not a simple Kite so it's not as easy to see, but it's a (Dual) Grouped Kite. In a simple Kite you only have simple strong links, i.e. only two candidates of the digit in the relevant row and column, connected by a box. In a Grouped Kite you can have three candidates in one or both lines, as long as two of them are in the connecting box forming a group node (or two).

In this case the row has a group node in box 2 (forming a grouped strong link in that row), but the column strong link remains simple because it has only two candidates. The same logic would work if r3c4 had 8 and 9 as candidates, so you could have a group node (and a grouped strong link) in the column as well.

In a Kite the end-points of the chain must still be single candidates so that they can both see the elimination. On the other hand, in a Grouped Skyscraper one of the end-points is a group node (and it can have up to three candidates, thus a total of four in that line). It works because the elimination(s) is/are in the same box as the group node.

Here’s how I see it. Given that you have identified by logic or otherwise that the eliminations are in r9c8, you want to write the chain to explain it in such a way that the endpoints both see this cell. So we go from r9c4 to r2c8, and we will have an AIC if we can start and end with a strong link and alternate strong and weak links throughout the chain.
(8)r9c4=r1c4-r2c56=r2c8 => -8r9c8 and
(9)r9c4=r1c4-r2c56=r2c8 => -9r9c8 => 5r9c8 stte

Exactly. That's two Grouped Kites, both with a single group node (r2c56) in box 2/row 2. The r1c4-r2c56 is a grouped weak link and the r2c56=r2c8 is a grouped strong link.

The tricky part was seeing r1c4-r2c56 as a single weak link. I would not have thought of that as a single link but clearly it is valid. Are we seeing r2c56 as a single entity in the chain, a node, perhaps? Is the weak link here saying “When r1c4 is 8, NEITHER r2c5 nor r2c6 can be 8 and therefore r2c8 must be 8,” proving that r9c8 cannot be 8 because it can see both ends of the chain, at least one of which must be 8?

Exactly, on all accounts. The two candidates r2c56 form a group node because they're in the same box and row. Thus they can be used as a single entity within that box and row, but not in their columns (obviously, because they're in different ones).

Finally, if everything works the way it should, does the AIC work in the opposite direction?

Yes, as it should.

Seems so. If r2c8 is NOT 8, one of r2c5 or r2c6 must be 8, so r1c4 cannot be 8 and r9c4=8 => -8r9c8.
Works fine if r2c56 is a single point in the chain.

Exactly, again. When approaching from that direction, you can think of the three candidates in row 2 as an almost-claiming. If r2c8 is not 8, then the two remaining 8s in row 2 are both in box 2. That's a claiming, which would eliminate 8r1c4 (and any other 8 in that box that is not in row 2). A Grouped Skyscraper has an almost-claiming as well, but there the group node is an end-point. The same logic works for almost-pointing situations, which is how ERs (i.e. Grouped Cranes) work.

In general, you can think of simple bilocation strong links as almost-hidden-singles, while grouped strong links are either almost-claimings or almost-pointings.
-SpAce-: Show
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`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"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: 2672
Joined: 22 May 2017