The New Sudoku Players' Forum

[tso]

I got stuck here:

Code: Select all

 6 3 4 | 7 . . | . 1 .
 9 5 1 | . 6 . | . 7 .
 8 2 7 | 1 . . | . . 4
-------+-------+------
 . 1 . | 6 . 7 | . 5 9
 . 6 . | 2 . 9 | . . .
 7 9 . | 8 . . | . 2 6
-------+-------+------
 1 4 . | . . 6 | . 8 .
 . 8 . | . 7 1 | . 9 2
 . 7 . | . . . | 3 . 1

Code: Select all

  6     3     4     | 7     2589  258   | 2589  1     58   
  9     5     1     | 34    6     2348  | 28    7     38   
  8     2     7     | 1     359   35    | 569   36    4     
 -------------------+-------------------+-------------------
  234   1     238   | 6     34    7     | 48    5     9     
  45    6     58    | 2     145   9     | 1478  34    378   
  7     9     35    | 8     1345  345   | 14    2     6     
 -------------------+-------------------+-------------------
  1     4     239   | 39    23    6     | 57    8     57   
  35    8     356   | 345   7     1     | 46    9     2     
  25    7     2569  | 459   248   248   | 3     46    1     


No simple forcing chain exists using only the 2-candidate cells -- right?

I tried to form a chain between the first two 2-candidate cells in the puzzle using one cell that had more than 2 candidates:

r2c4=3 -> r3c6=5
r2c4=4 -> r8c5<>4 -> r8c7=4 -> r9c8=6 -> r3c8=3 -> r3c6=5
Therefore, r3c6=5

Changing the last link in the chain makes another elimination:
r2c4=3 -> r2c9=8
r2c4=4 -> r8c5<>4 -> r8c7=4 -> r9c8=6 -> r3c8=3 -> r2c9=8
Therefore, r2c9=8

(The second pair of chains alone all that is needed to reduce the rest of the solution to naked singles.)

Here are just the cells involved for clarity:

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  .    .    .  |  .    .    .  |  .    .    . 
  .    .    .  | 34    .    .  |  .    .   38 
  .    .    .  |  .    .   35  |  .   36    . 
---------------+---------------+---------------
  .    .    .  |  .    .    .  |  .    .    . 
  .    .    .  |  .    .    .  |  .    .    . 
  .    .    .  |  .    .    .  |  .    .    . 
---------------+---------------+---------------
  .    .    .  |  .    .    .  |  .    .    . 
  .    .    .  | 345   .    .  | 46    .    .  <-- only two cells can contain a '4' in this row.
  .    .    .  |  .    .    .  |  .   46    . 

My questions:

1) Do we have a special name for these types of mixed forcing chain?

2) Could labeling edges be used somehow to find this type of chain?

3) Anyone see a better "next step"?

[stuartn]

tso wrote:

No simple forcing chain exists using only the 2-candidate cells -- right?

Either a 3 or 4 in R2C4 force a 5 in R7C7 - (and a 4 in R8C7) - then it all falls apart. The chain from the 3 is somewhat circuitous but quite clear - the 4 chain is very obvious but I'll post if anybody wants.

stuartn

[tso]

tso wrote:

No simple forcing chain exists using only the 2-candidate cells -- right?

Either a 3 or 4 in R2C4 force a 5 in R7C7 - (and a 4 in R8C7) - then it all falls apart. The chain from the 3 is somewhat circuitous but quite clear - the 4 chain is very obvious but I'll post if anybody wants.

stuartn

I don't see a link from r3c4=4 that is a 2-candidate cell.
What do you have?

[stuartn]

What's the basis (apart from a purely academic one - which I totally respect) of insisting that a forcing chain has to rely on 2 candidate cells?

IMHO, in identifying a forcing chain one needs to follow exactly the same eliminatory rules as always. Obviously using 2 candidate cells only is simple and clear - but (as with this grid) eliminating a candidate from two 3 candidate cells leaves us with a double - which is one of the most useful strategies ever.

stuartn

[tso]

What's the basis (apart from a purely academic one - which I totally respect) of insisting that a forcing chain has to rely on 2 candidate cells?

IMHO, in identifying a forcing chain one needs to follow exactly the same eliminatory rules as always. Obviously using 2 candidate cells only is simple and clear - but (as with this grid) eliminating a candidate from two 3 candidate cells leaves us with a double - which is one of the most useful strategies ever.

stuartn

No, I have no reason to insist on a chain of only 2-candidate cells. The best chain is the one that I can see most easily. A shortish chain that includes one or more multiple candidate cells that is easier to spot than a longish chain of 2-candidate cells is preferable to me.

However, if a pure xy-chain type forcing chain exists:

a) it will require only 2-candidate cells (bivalue); one may not even need to fill in all the candidates in cells requiring more than two. In some cases, these can be very easy to spot even when they many cells long, which can be subjectively gratifying to identify.
b) can be found by drawing labeled edges and looking for a loop with a single repitition. In other words, there is at least one sure-fire method of either finding an existing xy-chain, or showing that none exist.

(This is *supposed* to apply as well to bilocation graphs -- but in my experience, they're always far to tedious and complicated to be of any use.)

The chain I found was mixed -- not pure xy-chain, not pure bilocation links. I didn't find it by an identifiable method, but by inspection and intuition. I was just wondering if the labeled edge method used for finding xy-chains could be extended some way to be useful for finding this type of chain. I believe this type of chain is much more common, and since xy-chains are a subset of these, any method that would work for a mixed chain would work for simple xy-chains.


What do you mean by "eliminating a candidate from two 3 candidate cells leaves us with a double - which is one of the most useful strategies ever."

[stuartn]

With a 3 in R2C4 both R1C5 and R1C6 become 28 - which helps to open up the chain. I'm sure you'll agree that finding pairs like this is one of the commonest and most useful strategies - most of the Times fiendish's are solvable through this method.... as you'll no doubt be aware.

stuartn

[tso]

With a 3 in R2C4 both R1C5 and R1C6 become 28 - which helps to open up the chain. I'm sure you'll agree that finding pairs like this is one of the commonest and most useful strategies - most of the Times fiendish's are solvable through this method.... as you'll no doubt be aware.

stuartn

... but (as with this grid) eliminating a candidate from two 3 candidate cells leaves us with a double - which is one of the most useful strategies ever.

I still don't understand what your getting at.

Here's what I have at the moment before the forcing chains are used. R1c4 has 4 candidates, not 3. Though r2c4=4 does *eventually* lead to r1c45=[28][28] -- that's not something that I would do in my head as part of a simple forcing chain. How would you use that as part of a simpler forcing chain than what I had in my first post?

Code: Select all

  6     3     4     | 7     2589  258   | 2589  1     58   
  9     5     1     | 34    6     2348  | 28    7     38   
  8     2     7     | 1     359   35    | 569   36    4     
 -------------------+-------------------+-------------------
  234   1     238   | 6     34    7     | 48    5     9     
  45    6     58    | 2     145   9     | 1478  34    378   
  7     9     35    | 8     1345  345   | 14    2     6     
 -------------------+-------------------+-------------------
  1     4     239   | 39    23    6     | 57    8     57   
  35    8     356   | 345   7     1     | 46    9     2     
  25    7     2569  | 459   248   248   | 3     46    1   


Could you post the complete implication chain or net you found? It would be easier for me to keep track of what you're refering to.


I *think* the consensus is something like this: In a simple forcing chain, each step implies the next, by itself. If a step needs more than one previous step to be inferred, then you have a forcing NET. In my opinion a NET is generally more difficult to follow in one's head, more difficult to find, more difficult to write down.

[stuartn]

Here goes:

R2C4=3 =>r3C6=5 and R3C5=9 and R3C7=6. As R3C5=9 => R1C5 /C6=28 =>R1C7=9 =>R3C7=6=>R8C7=45. As R3C7=6 => R3C8 =3 => R5C8 =4 and R6C7 = 1 and R4C7=8 => R5C7=7=> R7C7=5 and R8C7 = 4.

R2C4 = 4 => R8C4=35 =>R8C7 = 4

[Nick70]

r2c4=4 => r8c4<>4 => r8c7=4 => r9c8<>4 => r5c8=4
r2c4=3 => r2c9<>3 => r5c9=3 => r5c8<>3 => r5c8=4

[stuartn]

Yep - thanks Nick70 - another one that makes it all fall apart. (the chain - not you!). Thanks - I thought I was all alone there for a bit.

stuartn

[tso]

r2c4=4 => r8c4<>4 => r8c7=4 => r9c8<>4 => r5c8=4
r2c4=3 => r2c9<>3 => r5c9=3 => r5c8<>3 => r5c8=4

This chain uses 8 cells, two of which have 3 candidates. One of the original chains I posted ...

r2c4=3 -> r2c9=8
r2c4=4 -> r8c5<>4 -> r8c7=4 -> r9c8=6 -> r3c8=3 -> r2c9=8

... has only 7, only one of which has 3 candidates -- only naked singles are required after this.

Is labeling edges useless for finding mixed forcing chains like this? How did you go about finding your chain?

[Scott H]

Is labeling edges useless for finding mixed forcing chains like this? How did you go about finding your chain?

I came up with the labelled edges view of things precisely for finding general ("mixed") forcing chains. I haven't given many examples, but Jeff has developed the technique and given several nice examples.

When you label the edges in a forcing chain, adjacent edges must change either sign (+/-) or absolute value, but not both. This gives all forcing chains.

The three "pure" subclasses of forcing chains are:

1. bivalue or xy-chains. These have all negative (weak) edges.
The strong implications in xy-chains are all when switching values in 2-valued nodes.

2. bilocation chains. These have all positive (strong) labelled edges. The strong implications in these chains appear as edges between cells, and the weak implications are when switching values within cells.

3. number chains (my term, are there others?). All edges use a single +/- number and alternate signs. These are also the chains findable by coloring one number and using the full range of multiple color deductions.

When searching for general forcing chains, I mark candidates (not cells) that could be part of a strong link. I also try to mark which unit(s), i.e. row/col/box/cell, the strong link(s) could lie in. It's a challenge to mark this much info without getting too confusing and then search it to find forcing chains. I don't a final pat solution, but Jeff and I (and probably others) have had success with this approach in finding general forcing chains.