Advanced methods and approaches for solving Sudoku puzzles
ravel wrote:So my next question is, if this is a valid Braiding deduction too:
[...]
Lets take r12c1:
72 depletes r3c3.
78 gives a 2 in r3c3 and r7c2, then 6 in r4c2 and 5 in r4c1, therefore r8c1 depleted.
So r3c3 must be 7, solving the puzzle.

I suspect your too subtle point was that r12c1 was not even a traveling pair.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

I see your point, but sometimes chains and loops are not so easy to find as in this example...
Perhaps braids are easier to spot than chains, on occasion...
MrHamilton

Posts: 42
Joined: 14 March 2006

Code: Select all
`000001020300040500000600007001000006040080090500000300800002000050090400006700000 Ocean's New Year's present for ravel #2  *-----------* |...|..1|.2.| |3..|.4.|5..| |...|6..|..7| |---+---+---| |..1|...|..6| |.4.|.8.|.9.| |5..|...|3..| |---+---+---| |8..|..2|...| |.5.|.9.|4..| |..6|7..|...| *-----------*  *-----------------------------------------------------------------------------* | 4679    6789    45789   | 3589    357     1       | 689     2       3489    | | 3       126789  2789    | 289     4       789     | 5       168     189     | | 1249    1289    24589   | 6       235     3589    | 189     1348    7       | |-------------------------+-------------------------+-------------------------| | 279     23789   1       | 23459   2357    34579   | 278     4578    6       | | 267     4       237     | 1235    8       3567    | 127     9       125     | | 5       26789   2789    | 1249    1267    4679    | 3       1478    1248    | |-------------------------+-------------------------+-------------------------| | 8       1379    3479    | 1345    1356    2       | 1679    13567   1359    | | 127     5       237     | 138     9       368     | 4       13678   1238    | | 1249    1239    6       | 7       135     3458    | 1289    1358    123589  | *-----------------------------------------------------------------------------*`

I think we might agree, that if braid analysis can help with this puzzle, it is a technique to be reckoned with, as it has very elaborate chains in Sudoku Explainer program output.
First some basic eliminations:
Exclude 6 from r6c2, Multi-colors
Hidden single r5c1=6
"No hint available"

Code: Select all
`  *-----------------------------------------------------------------------------*  N  |  Z | 479     6789    45789   | 3589    357     1       | 689     2       34      |     | | 3       126789  2789    | 289*    4       789*    | 5       168*    189*    |     | | 1249    1289    24589   | 6       235     3589    | 189     34      7       |     | |-------------------------+-------------------------+-------------------------|     | | 279     23789   1       | 23459   2357    34579   | 278     4578    6       |  6  | | 6       4       237     | 1235    8       357     | 127     9       125     |     | | 5       2789    2789    | 1249    1267    4679    | 3       1478    1248    |     | |-------------------------+-------------------------+-------------------------|     | | 8       1379    3479    | 1345    1356    2       | 1679    13567   1359    |     | | 127     5       237     | 138     9       368     | 4       13678   1238    |     | | 1249    1239    6       | 7       135     3458    | 1289    1358    123589  |     | *-----------------------------------------------------------------------------*     | N         5 ------------------------------------------------------------------------------- Z                 6`

Tower 123 must be a braid, and the digit in r4c1 is a traveler with 5 or 6.
Not much else is obvious at this point.
Every strand in every tower and floor has four or five possible digits.

But, examine the strands in Floor 123:

N1 - 4789
Z1 - 5689

N2 - 23689
Z2 - 13789

N3 - 1589
Z3 - 2489

The 8 and 9 candidates appear on this list every time; but oddly enough they cannot be a traveling pair on this floor, since they don't occur together in any two cells of the critical boxrows r1c456, r1c789, r3c456, or r3c789, one of which would be required for travelers. As they do not travel together, by rule they can only occur together (intersect) in at most one boxrow in this floor, or perhaps in none, when both travel with other numbers. If 89 are present only in r2c23, and excluded from the marked cells, it would follow that exactly one of them travels; but this scenario would place a 6 in r7c7 and a 3 and 5 in boxcol r123c5, depleting r79c5. No suitable boxrow for both 8 and 9 is available that does not deplete our strand candidates somehow; both travel separately, and with 4 and 5; therefore they must appear in the marked cells; therefore exclude 8 and 9 from r2c23. Update the mapping:

N1 - 4789
Z1 - 5689

N2 - 236
Z2 - 137

N3 - 1589
Z3 - 2489

Now exclude 237 from r1367c3, Naked triples.
Since 9 travels in Floor 123, it cannot also do so in Tower 123.
Remove 9 from r4c1.
This sets up a "broken Z strand" in Tower 123 Z2, where at most one of the 9's marked below can be true.
Code: Select all
`  *-----------------------------------------------------------------------------* | 479     6789*   4589    | 3589    357     1       | 689     2       34      | | 3       1267    27      | 289     4       789     | 5       168     189     | | 1249    1289*   4589    | 6       235     3589    | 189     34      7       | |-------------------------+-------------------------+-------------------------| | 27      23789   1       | 23459   2357    34579   | 278     4578    6       | | 6       4       237     | 1235    8       357     | 127     9       125     | | 5       2789    89      | 1249    1267    4679    | 3       1478    1248    | |-------------------------+-------------------------+-------------------------| | 8       1379    49*     | 1345    1356    2       | 1679    13567   1359    | | 127     5       237     | 138     9       368     | 4       13678   1238    | | 1249    1239    6       | 7       135     3458    | 1289    1358    123589  | *-----------------------------------------------------------------------------*`

These broken strands with three digits can be useful when you can establish that one of the three digits is true, leading to further exclusions; but that is not immediately the case here.

(As Simple Sudoku is stumped again and Sudoku Explainer now sees nothing easier than 2 Dynamic Forcing Chains of 24 and 25 steps, proving that r1c1 isn't 7 because then r6c5 has to be 6 AND not be 6, just maybe there is an easier Braiding deduction somewhere?)

To be continued...
MrHamilton

Posts: 42
Joined: 14 March 2006

I use a basic form of this method. I don't write anything down, but I do observe which numbers are moving together within a chute. One thing that I have noticed, though, is that I rarely see a chute where numbers move in triplets. So when I do, it always seems somewhat inelegant to me. Has anyone else noticed this?
Lardarse

Posts: 106
Joined: 01 July 2005

Hey.......a lot of work was done on this here Canonical form

We looked at the minlex representation of all essentially different grids.

The repeating minirow pattern is found in a band in approx 29% of all grids !

In the min lex representation there are two choices for the clue at r2c3

All grids can be reduced to this :
Code: Select all
`123|456|789 45?|...|... ...|...|... ---+---+--- 2..|...|... ...|...|... ...|...|... ---+---+--- ...|...|... ...|...|... ...|...|...`
If there is a repeating minrow r2c3 will be a 6 !

The 6 always comes before a 7 in min lex order

This happened 29% of the time - for any one of the 6 bands, therefore it only occurs 5% of the time in a band !

Incidentaly it happened 54/6 % of the time if the grid had a 17puzzle [which makes sense]

So you will be right if you guess NOT a repeating minirow 95% of the time.

Does this concur ?

C
coloin

Posts: 1743
Joined: 05 May 2005

I don't assume that it doesn't repeat. I just notice when it does...
Lardarse

Posts: 106
Joined: 01 July 2005

Not sure if I pointed this out already but--
since side notation is fully adequate to describe all valid grids, i.e. no two different valid grids have the same side notation and no valid instance of side notation represents more than one valid grid, I would think it could be demonstrated that the number of valid unique grids must be equal to (no greater than) the number of valid unique side notation descriptions.
To be valid, a grid must meet the usual constraints AND not allow the same traveling pair in a tower and floor simultaneously. That's the tricky constraint, otherwise you could just count the permutations of 1-9 in towers and floors along NZ lines. Triples limit your choices a lot. You could also account for rotational symmetry I suppose...
Fastest might be to calculate the number of contradictory side notations which imply one or more pairs in both tower and floor, as they're sure to be invalid grids, and then subtract that from the total number of side notations.
If I were attempting this, I'd break it down by first counting grids of six triples, then five, four, three, two, one, and none (the most frequent case).
MrHamilton

Posts: 42
Joined: 14 March 2006

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