The New Sudoku Players' Forum

[denis_berthier]

Hi Paquita

-the braids for BxB, do they include bivalue chains and whips?

Both bivalue-chains and whips (and z-chains and t-whips) are special cases of braids of same length. Whether you decide to include them or not in a resolution path may change the path, but it can't change the final B rating.
Of course, the same applies to B-braids, i.e. braids with inner braids as right-linking objects. They may have inner whips, BC...

-there is some order as you indicated in PBC, what goes first, a longer bivalue or a shorter braid?

The priority order mentioned in [PBCS] is for a fixed length: bivalue-chain > z-chain > t-whip > whip
Between any two chains (or patterns) of different lengths (or sizes), the shorter one comes first.
This is the basis of the simplest-first strategy.

So what is the BxB number based on then?

Same answer as 1.
Of course, in the SHC, where speed is the goal, only braids are present.

-braids as oppposed to whips seem to have more than one assumption.

There is no assumption, neither in whips nor in braids. The word "assumption" was used by people who reasoned in terms of "inference steps"; all my approach is formulated in terms of patterns. Resolution isn't doing inferences; resolution is finding patterns and applying the theorems.
For whips or braids and for each fixed length, there is a pattern well defined in pure logic terms and there is a universal theorem (valid in any CSP) saying: if this pattern is present in the grid, then the target cannot be true.
The "continuity condition" in the braid[n] pattern is different from that in the whip[n] pattern, but that's all.

I wonder if that is a relation between the T&E level and the number of assumptions in a braid? It seems right that T&E(2) corresponds to 2 assumptions

Again, there are no assumptions in braids. But I've proven the following theorems long ago in [PBCS]:
- Solvable by braids (i.e. finite B rating) <==> in T&E(1)
- Solvable by Bp-braids (i.e. finite BpB rating) for some p <==> in T&E(2)
When you're reasoning in terms of T&E instead of braids, the T&E-depth is indeed the minimum number of simultaneous assumptions one must make at one point or another.
.

[coloin]

These 3 twin BxB6 puzzles are the only tridagon puzzles in my BxB6 file with gsf -q2 rating "M3"

Code: Select all

1.3...7...5678....7...........87.9.6.6.9.5.7..7..62.58...62...76.75.8.92..2.97... 97788 FNBP C33/M3.1351.393
1.3....8945..8.....8.........81.369....97..18..1..83.7...8...616.....87.81.6.79.3 97788 FNBP C33/M3.1351.393
1.3....8945..8.....8.........89.361....17..98..1..83.7...8...616.....87.81.6.79.3 97788 FNBP C33/M3.1351.393

Code: Select all

1.3...7...5678....7...........87.9.6.6.9.5.7..7..62.58...62...76.75.8.92..2.97... ED=11.7/10.5/4.0

+---+---+---+
|1.3|...|7..|
|.56|78.|...|
|7..|...|...|
+---+---+---+
|...|87.|9.6|
|.6.|9.5|.7.|
|.7.|.62|.58|
+---+---+---+
|...|62.|..7|
|6.7|5.8|.92|
|..2|.97|...|
+---+---+---+ 

The tridagon maybe doesnt look that complicated... but the M3 label { no double backdoor clues ?] might mean something ?

[denis_berthier]

These 3 twin BxB6 puzzles are the only tridagon puzzles in my BxB6 file with gsf -q2 rating "M3"

Code: Select all

1.3...7...5678....7...........87.9.6.6.9.5.7..7..62.58...62...76.75.8.92..2.97... ED=11.7/10.5/4.0
The tridagon maybe doesnt look that complicated...

There are 3 guardians, easily reducible to 2.
It's easily solved with digit replacement. Otherwise, it doesn't look so easy.

but the M3 label { no double backdoor clues ?] might mean something ?

No idea about that.

[coloin]

I guess it was just a hunch ... being the only puzzle with M3 ...and there are not that many with coresponding high ED

explained by eleven some time ago

Mladen has posted the first known puzzles with singles backdoor size 4, which means you have to guess (add) at least 4 correct numbers to make it a singles puzzle.

and from One flew over the backdoors

Backdoors and T&E are indeed very different.
As i understood it (without having checked it with Denis' definitions), T&E(1) corresponds to solving with singles chains, i.e. you can try each candidate - put it in - then solve all singles, and eliminate the candidate, if a contradiction arises (e.g. empty cell, missing digit for a unit). A puzzle is not solvable with singles chains, if you arrive at a grid, where no candidate can be eliminated that way.
T&E(2) then corresponds to nested singles chains (of depth 2), i.e. when you are stuck with the singles chain for a candidate, you can copy the grid, try to eliminate each candidate with singles chains - if so, eliminate it from the original grid (with the candidate tried), and then look if it leads to a contradiction there now. If so, you can eliminate the original candidate. If no progress can be achieved any more this way in the original puzzle, the candidate cannot be eliminated. Of course this is a very strong, but also elaborate method.
If a puzzle is not unique, this cannot be found with T&E - when no candidate can be eliminated, you don't know, if it has multiple solutions or the method is too weak.

If a solution is accepted, when it "happens" from trying a candidate, is a matter of taste. A solution is a solution. So it sounds strange, when in using T&E the solution is an Error But of course then T&E and pure guessing would be mixed.
I think that gsf's opinion, that (otherwise) extremely hard puzzles should be rated easier, if they have a single backdoor, has to be accepted (though i think different).
On the other side many solvers don't want to allow any T&E based solutions, only "pattern based" ones, where "pattern" refers to a predefined set of more or less easy-to-spot situations. What this set contains, differs between players, but those who allow chains too, are in a minority....

[denis_berthier]

.
No need to complicate something very simple: there's no relation between T&E and backdoors.
.