The New Sudoku Players' Forum

[SteveG48]

I had something more in mind like:

Code: Select all


3r4c1 - (3=24)r4c47 - 4r5c5 = 4r9c5 - 4r9c1
      - 3r9c1

4r4c1 - r4c4 = 4r5c5 - (4=3)r9c5 - 3r9c1
      - 4r9c1



[Leren]

Fixed typo spotted by Marty in my other post. I think both chains are necessary, to prove the pattern. Once proved it becomes an accepted pattern and it's a one stepper remote pair move.

There was also a second way to prove the pattern:

Code: Select all

*--------------------------------------------------------------*
| 6     2348  7      | 258   123   358    | 134   9     234    |
| 139   239   39     | 6     123   4      | 7     8     5      |
| 1348  2348  5      | 278   9     378    | 134   6     234    |
|--------------------+--------------------+--------------------|
|a34    5     6      |b24    8     1      | 23    7     9      |
| 7     489   489    | 3     24    6      | 28    5     1      |
| 2     38    1      | 59    7     59     | 6     4     38     |
|--------------------+--------------------+--------------------|
| 5     1     348    |c478   6     378    | 9     2     48     |
| 489   6     489    | 1     5     2      | 48    3     7      |
| 8-34  7     2      |c489  d34    389    | 5     1     6      |
*--------------------------------------------------------------*

(3=4) r4c1 - r4c4 = r79c4 - (4=3) r9c5 and (3-4) r4c1 = r4c4 - r79c4 = (4-3) r9c5

Leren

[Sudtyro2]

Many thanks to Leren and Steve for the extended explanations and examples! My own thinking was that placements and/or eliminations result only from either single chains or from Kraken/SIS-based networks pointing to a common solution. JC's remote pair solution is new to me and is very interesting.

Much still to learn, as this "tyro" stage lingers on.

SteveC

[Marty R.]

Fixed typo spotted by Marty in my other post. I think both chains are necessary, to prove the pattern. Once proved it becomes an accepted pattern and it's a one stepper remote pair move.

There was also a second way to prove the pattern:

Code: Select all

*--------------------------------------------------------------*
| 6     2348  7      | 258   123   358    | 134   9     234    |
| 139   239   39     | 6     123   4      | 7     8     5      |
| 1348  2348  5      | 278   9     378    | 134   6     234    |
|--------------------+--------------------+--------------------|
|a34    5     6      |b24    8     1      | 23    7     9      |
| 7     489   489    | 3     24    6      | 28    5     1      |
| 2     38    1      | 59    7     59     | 6     4     38     |
|--------------------+--------------------+--------------------|
| 5     1     348    |c478   6     378    | 9     2     48     |
| 489   6     489    | 1     5     2      | 48    3     7      |
| 8-34  7     2      |c489  d34    389    | 5     1     6      |
*--------------------------------------------------------------*

(3=4) r4c1 - r4c4 = r79c4 - (4=3) r9c5 and (3-4) r4c1 = r4c4 - r79c4 = (4-3) r9c5

Leren

I'm hardly a notation expert, but It's hard to understand why the 1st chain doesn't suffice. I read it as if the premise is not 3 then the end IS 3 and if premise is 4 then end is NOT 4. If all four cells were bivalue 34, would the notation be (3=4)-(4=3)-(3=4)-(4=3), same start and end as this one? Would someone be calling for a 2nd chain?

[eleven]

I see it as a kite with a pair at the ends rather than a remote pair variation.
A nice, but rare pattern anyway.

[pjb]

Here's my 2 bob's worth:
if r4c1 is 3, r9c is 4 (via (4=3)r4c1 - (3=2)r4c7 - (2=4)r4c4 - r5c5 = r9c5))
if r4c1 is 4, r9c is 3 (via (3=4)r4c1 - r4c4 = r5c5 - (4=3)r9c5))
so either way r9c1 <> 34
Phil

[Leren]

In response to eleven's comment about the Kite interpretation:

1. The kite interpretation wouldn't work if there was an additional 4 in, say, r6c6, but the the Remote Pair construction would still work via the second method I detailed.

2. The Remote Pair link might be one of a number of links between other bi-value cells with the same 2 digits (in other puzzles).

So what I learn't from this puzzle is that two bi-value cells ab can have opposite parity if they don't share a Stack or Tier of boxes, but there are 3 Strong links on one of the digits (a or b) between them. I've never seen this mentioned in descriptions of Remote Pairs before.

Remote Pair solutions are pretty rare for Dan's one step wonder puzzles, but i did use it for the puzzle for November 17 2015. Can't remember any other occasion I've seen it used.

Leren

[David P Bird]

Leren, I would class JC's deduction as arising from an extended conjugate chain.

Code: Select all

 *-----------------*-----------------*-----------------*
 | 6    2348 <7>   | 258  123  358   | 134  9    234   |
 | 139  239  39    | <6>  123  <4>   | <7>  <8>  5     |
 | 1348 2348 <5>   | 278  <9>  378   | 134  6    234   |
 *-----------------*-----------------*-----------------*
 | 3'4" <5>  6     | 24'  <8>  1     | 23   7    <9>   |
 | <7>  489  489   | <3>  24"  <6>   | 28   5    <1>   |
 | <2>  3"8' 1     | 59   7    59    | 6    <4>  38    |
 *-----------------*-----------------*-----------------*
 | 5    <1>  348   | 478  <6>  378   | <9>  2    48    |
 | 489  <6>  489   | <1>  <5>  <2>   | 48   3    7     |
 | 8-34 7    2     | 489  3"4' 389   | <5>  1    6     |
 *-----------------*-----------------*-----------------*

(3'a ^ 4"b)r4c1 ^ (4')r4c4 ^ (4")r5c5 ^ (4'c ^ 3"d) => [ad] r9c1 <> 3, [bc] r9c1 <> 4

Either all the odd' or even" terms in a conjugate chain must be true, so any digit weakly linked to both odd and even terms must be false.

I would be reluctant to call this a remote pairs chain which <Andrew Stuart> says he originated. There you will see that all the cells in his chains contain the same pair which makes them easy to spot.

For me the essence of a pattern should be that it should be recognisable without needing to track truths, so I would also be reluctant to call this configuration a pattern without any further qualification. That would be the thin end of a wedge that would lead to classing chains of unlimited length as patterns.

David
.

[Marty R.]

So what I learn't from this puzzle is that two bi-value cells ab can have opposite parity if they don't share a Stack or Tier of boxes, but there are 3 Strong links on one of the digits (a or b) between them. I've never seen this mentioned in descriptions of Remote Pairs before.

Leren,

I think Keith's write-up covers that situation.

http://www.dailysudoku.co.uk/sudoku/for ... php?t=2143