The New Sudoku Players' Forum

[denis_berthier]

The least number of independently redundant clues I find is three:

Now reduced to just one:

Code: Select all

 . . . 4 8 1 5 . .
 . . 4 . . . . 6 .
 . 8 . . 5 3 . . 2
 . 1 . 2 . . 3 . 8
 . 9 . . . . 2 . 7
 . . 7 . . 8 . . 6
 8 . . 7 4 . . 3 .
 . 4 . . . . 7 . .
 . . 1 3 2 5 . . .    r9c4  is redundant

That's a nice result.

[Mauriès Robert]

On the other hand, to make me do this from time to time, I can tell you that for a manual solver, the search for short chains of contradiction is more complicated than the exploitation of conjuguated tracks. It is complicated for the manual solver to find the Z candidates which allow you to develop chains of contradiction (your chains), whereas it is easy for your programme.

The problem is probably you haven't yet understood that the start of a chain is not a target, but a partial-whip[1] pattern - which is not more complex than intersections - and that the continuity condition plays a major role in simplifying the extensions.

and further on you write to me

And, once more, a manual solver doesn't have to use the simplest-first strategy. If 69 candidates out of 146 lead to a contradiction, then the manual solver has a high chance of finding one by starting from anywhere. He has still a much larger chance if he starts from a partial-whip[1].

No doubt that I did not understand this notion of the partial whip that allows you to find the whip. Perhaps you could explain it to me by taking an example, for example the whip [4] of your solution of this puzzle.
Robert

[denis_berthier]

On the other hand, to make me do this from time to time, I can tell you that for a manual solver, the search for short chains of contradiction is more complicated than the exploitation of conjuguated tracks. It is complicated for the manual solver to find the Z candidates which allow you to develop chains of contradiction (your chains), whereas it is easy for your programme.

The problem is probably you haven't yet understood that the start of a chain is not a target, but a partial-whip[1] pattern - which is not more complex than intersections - and that the continuity condition plays a major role in simplifying the extensions.

and further on you write to me

And, once more, a manual solver doesn't have to use the simplest-first strategy. If 69 candidates out of 146 lead to a contradiction, then the manual solver has a high chance of finding one by starting from anywhere. He has still a much larger chance if he starts from a partial-whip[1].

No doubt that I did not understand this notion of the partial whip that allows you to find the whip. Perhaps you could explain it to me by taking an example, for example the whip [4] of your solution of this puzzle.

I already told you to read the Basic User Manual for a graphical description of whips, partial-whips and t-whips.

[DEFISE]

Hi Denis, Robert, Paolo,

Sterile debate!
Denis's program "Simplest First" is made not only to solve the grid but to estimate its level.
It's almost impossible by hand !

[Mauriès Robert]

I already told you to read the Basic User Manual for a graphical description of whips, partial-whips and t-whips.

OK, I don't have that manual. Where can I find it?
Robert

[JPF]

Probably off-topic, here is a minimal spiral:

Code: Select all

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


Minimal ; SER=8.4

JPF

[SpAce]

I already told you to read the Basic User Manual for a graphical description of whips, partial-whips and t-whips.

OK, I don't have that manual. Where can I find it?

Here.

The problem is probably you haven't yet understood that the start of a chain is not a target, but a partial-whip[1] pattern - which is not more complex than intersections - and that the continuity condition plays a major role in simplifying the extensions.

For Robert. As far as I understand, in the sudoku context a partial-whip[1] is basically a "Remote-Finned Cyclopsfish", i.e. an Almost-Hidden-Single or an Almost-Claiming or an Almost-Pointing. Instead of being an actual whip[1] the pattern has a spoiler (or several), which becomes the first right-linking-candidate (or group in case of a partial-g-whip).

Questions for Denis:

If 69 candidates out of 146 lead to a contradiction, then the manual solver has a high chance of finding one by starting from anywhere. He has still a much larger chance if he starts from a partial-whip[1].

Why? This is the part I don't understand. The possible targets are still fixed and very limited, except with t-whips that allow growing the chain both ways. I don't see how that possibility depends on starting with a partial-whip[1].

The problem is probably you haven't yet understood that the start of a chain is not a target, but a partial-whip[1] pattern - which is not more complex than intersections - and that the continuity condition plays a major role in simplifying the extensions.

Do you always start looking for whips with a partial-whip[1]? What about whips that start with a bivalue cell? Those obviously exist so how were they started? Were they originally t-whips that were left-extended?

Or do you count bivalue cells as partial-whips[1], too? That would be logical (*) but extremely confusing, considering that you always tell that whip[1] means intersections, which (I bet) everyone interprets as either claiming or pointing in the sudoku context. If whip[1] (and consequently partial-whip[1]) actually has a wider meaning, it would be nice to clarify that.

(*) Logically a naked single should be a whip[1], and thus a bivalue cell should be a partial-whip[1].

(This is also my response to this post. Sorry I didn't have time before.)

[denis_berthier]

Probably off-topic, here is a minimal spiral:

Code: Select all

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


Minimal ; SER=8.4

It's nice to have an example not too far from the original pattern, so that the spiral shape is still recognisable.

[denis_berthier]

If 69 candidates out of 146 lead to a contradiction, then the manual solver has a high chance of finding one by starting from anywhere. He has still a much larger chance if he starts from a partial-whip[1].

Why? This is the part I don't understand. The possible targets are still fixed and very limited

They are obviously more limited with the partial-whip[1] condition than with no condition at all. Thus, the manual user has fewer chances of trying a candidate that leads nowhere.

except with t-whips that allow growing the chain both ways. I don't see how that possibility depends on starting with a partial-whip[1].

In order to build a t-whip[n], you must already have a partial-whip[n-1] (A full t-whip[1] is a whip[1].)

Do you always start looking for whips with a partial-whip[1]?

Yes.

What about whips that start with a bivalue cell?

It's not a problem. It's just the particular case when the first partial-whip[1] has no t-candidate.

Or do you count bivalue cells as partial-whips[1], too?

I now guess you mean rc-bivalue. First, not the bivalue cell itself; the target z must also be specified. But, with these conditions, yes, of course, they also make partial-whips[1].

but extremely confusing, considering that you always tell that whip[1] means intersections

I don't say that a whip[1] is an intersection in the sense that it'd be the mathematical intersection of a row (or columnn) and a block. It is an intersection in terms of a resolution rule, including a target.
What may confuse you is, you have to realise that there is no whip[1] consisting of a target plus an rc-bivalue cell. But that may make a partial-whip[1].

More generally, it's interesting to notice that, in any CSP, one may have whips[2 or more] without having whips[1] - said otherwise, no whips[1] but partial-whips[1]. This is the case in Latin Squares. In such CSPs, there are no g-whips.

(*) Logically a naked single should be a whip[1]

Logically, there'd be no difference between Naked or Hidden Singles.
Singles are not included in the definition of whips. The definition requires a precise pattern.

[Ajò Dimonios]

Hi Denis

I would like to ask a question. From your demonstration indicating that the Braids produce the same result as the T&E method, I would like to understand if by T&E method we mean singles to the contradiction or basics to the contradiction?

Paolo

[denis_berthier]

From your demonstration indicating that the Braids produce the same result as the T&E method, I would like to understand if by T&E method we mean singles to the contradiction or basics to the contradiction?

Hi Paolo
For any resolution theory T with the confluence property, one can define a T&E(T, 1) procedure [and more generally a T&E(T, n) procedure].
Here, the confluence property is essential for the procedure to have a non-ambiguous outcome.

What I call T&E is a shorthand for T&E(BRT, 1), where BRT is Singles+ECP (ECP = elementary Constraints Propagation), i.e. the simplest resolution theory one can imagine (BRT = Basic Resolution Theory, upon which all the other resolution theories are built).

In PBCS, I have indeed stated and proven several theorems relating various T&E procedures with various types of extended braids:

- A puzzle is solvable by T&E=T&E(BRT, 1) iff it is solvable by (ordinary) braids
- A puzzle is solvable by gT&E=T&E(W1, 1) iff it is solvable by g-braids
- A puzzle is solvable by T&E(S, 1) iff it is solvable by S-braids (where S is the set of rules for Whips[1] plus (Naked, Hidden and Super-Hidden) Subsets
- A puzzle is solvable by T&E(Bp, 1) iff it is solvable by Bp-braids (where Bp-braids are relatively complex braids including braids of lengths ≤ p as right-linking elements.
And
- A puzzle is in T&E(2)=T&E(BRT, 2) iff it is in T&E(B, 1) iff it is solvable by B-braids of any length with inner braids of any length.

Notice that any of these theorems can also be formulated in a "local" version, i.e. restricted to a particular elimination.
Notice also that these theorems can easily be restricted to 2D versions.
And in the case you call Basics, i.e. BRT plus Naked+Hidden Subsets, it's also easy to have corresponding braids without inner Fish - but this notion of Basics is in violation of my super-symmetric view of Sudoku.

[m_b_metcalf]

It's nice to have an example not too far from the original pattern, so that the spiral shape is still recognisable.

Here's another, with the other end missing:

Code: Select all

 . . . 1 2 3 4 . .
 . . 5 . . . . 6 .
 . 2 . . 6 5 . . 7
 . 8 . . . . 5 . 3
 . 4 . . . . 7 . 6
 . . 7 . . 4 . . 1
 7 . . 4 8 . . 1 .   Minimal, 29 clues
 . 6 . . . . 8 . .
 . . 8 2 5 6 . . .   ED=7.3/1.2/1.2

[denis_berthier]

It's nice to have an example not too far from the original pattern, so that the spiral shape is still recognisable.

Here's another, with the other end missing:

Code: Select all

 . . . 1 2 3 4 . .
 . . 5 . . . . 6 .
 . 2 . . 6 5 . . 7
 . 8 . . . . 5 . 3
 . 4 . . . . 7 . 6
 . . 7 . . 4 . . 1
 7 . . 4 8 . . 1 .   Minimal, 29 clues
 . 6 . . . . 8 . .
 . . 8 2 5 6 . . .   ED=7.3/1.2/1.2

Nice example also.
It has a simple solution using only a few bivalue chains:

Hidden Text: Show

***********************************************************************************************
*** SudoRules 20.1.s based on CSP-Rules 2.1.s, config = BC+SFin
*** Using CLIPS 6.32-r773
***********************************************************************************************
singles and whips[1]
biv-chain[4]: r6c7{n9 n2} - b3n2{r2c7 r2c9} - r7c9{n2 n5} - c2n5{r7 r6} ==> r6c2 ≠ 9
naked-pairs-in-a-column: c2{r6 r7}{n3 n5} ==> r2c2 ≠ 3
hidden-single-in-a-block ==> r2c1 = 3
biv-chain[3]: r2n1{c2 c7} - r3c7{n1 n9} - b2n9{r3c4 r2c4} ==> r2c2 ≠ 9
singles ==> r2c2 = 1, r9c2 = 9, r9c9 = 4,> r9c1 = 1, r3c7 = 1
biv-chain[5]: r6c5{n3 n9} - r6c7{n9 n2} - b3n2{r2c7 r2c9} - r7n2{c9 c3} - b7n3{r7c3 r7c2} ==> r6c2 ≠ 3
stte

[m_b_metcalf]

It has a simple solution using only a few bivalue chains:

It was the best I could do. Others:

Code: Select all

...1234....4....5..6..54..7.2....8.3.7....9.5..9..8..19..87..1..5....7....7245...   ED=2.0/1.2/1.2   
...1234....4....5..2..54..6.7....8.5.8....6.3..6..8..16..87..1..5....7....7245...   ED=7.2/1.2/1.2   

[denis_berthier]

It has a simple solution using only a few bivalue chains:

It was the best I could do.

I've noticed this also: some patterns easily give hard puzzles and some patterns give only easy to medium ones.