Yeah, it is. But it's not minimal, so it's not interesting.coloin wrote:on looking manually,,,this may be another unavoidable set in grid2

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`36 (2) ..8..3.6....1..8...9368.1.4...8.74....5341.8...4..5.16..7526.4...2.3..7..897.4.3.`

Yeah, it is. But it's not minimal, so it's not interesting.coloin wrote:on looking manually,,,this may be another unavoidable set in grid2

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`36 (2) ..8..3.6....1..8...9368.1.4...8.74....5341.8...4..5.16..7526.4...2.3..7..897.4.3.`

- Red Ed
**Posts:**633**Joined:**06 June 2005

Cool, i like that (sorry to break the discussion). So you can choose any of the 60 cells, put in the number of the one or other solution and get 2 equivalent unique sudokus. From the solvers point of view that means, if you can solve the one, you can solve the other the same way.JPF wrote:Here is a puzzle with 2 solutions.

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`*-----------*`

|1..|2..|3..|

|...|.4.|.5.|

|...|..6|..7|

|---+---+---|

|.2.|.8.|...|

|9..|...|1..|

|..4|..7|...|

|---+---+---|

|...|9..|2..|

|.8.|.3.|.4.|

|..7|..1|..6|

*-----------*

The 2 solutions A and B are :

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`179258364632749851845316927726183495958462173314597682563974218281635749497821536`

146275389798143652352896417521689734973524168864317925415968273689732541237451896

They differ on 60 cells.

I tried e.g. to put in 5 in r5c2 and got an interesting sudoku. Its amazing that putting in 7 makes it the same.

[Added:] This also is a good solver test - most solvers do the puzzles stupidly from top to bottom, following there technique hierarchy.

Put 3, then 5 into r6c4 and look, how different the solution paths are.

- eleven
**Posts:**1971**Joined:**10 February 2008

Red Ed wrote:That's not quite what you mean. There are NO unavoidable sets in the empty cells. But there ARE unavoidable sets confined to those positions in either of the original solution grids.RW wrote:How many unavoidable sets are there in the 48 empty cells?

Actually, that was what I meant, though perhaps unavoidable set isn't the right term when speaking about empty cells...

The puzzle with 48 empty cells has 92 solutions. This means that if we examine the unavoidable sets in all 92 solutions, we should find a maximum of 91 different unavoidable sets (that all appear in at least two grids in two different permutations) that can be placed in the 48 empty cells without causing a contradiction with the 33 given clues.

91 is still such a small number that it should be possible to list. I just wonder if there is some faster way to find these than examining all possible solutions...

If we had a list of these unavoidable sets and a list that shows which unavoidable sets (and which permutations of these sets) appear in which solution, then it should be possible to determine the shortest path between the two grids.

RW

- RW
- 2010 Supporter
**Posts:**1000**Joined:**16 March 2006

Thanks Red Ed, my understanding has now been updated. In some ways it confirms that a clue can only be inserted if it doesnt appear in any of the minimal and non-minimal unavoidable sets remaining. In solving the puzzle with the other 2 [5] clues the 9 would then become superfluous.

OT

Is it worth updating the pseudo-puzzle / megaclue thread which outlines the extent of the unavoidables in every grid. [I suspect most people are completely oblivious]

Can an individual puzzles properties be pinpointed directly to the unavoidable sets that each individual clue or group of clues is responsible in "covering" ? I can see a few [trivial] clue patterns [4-clue][naked single] where a clue may be inserted. Any puzzle with an isomorphic example of this pattern will have this insertion.

In persuit of RW's line - we perhaps would need to find our most distant 2 grids with a max of 33 common clues [mc-grid and a.n.-othergrid] with the least [or most] ? <92 sol

Are ther really only 91 different minimal unavoidable sets with a different position pattern ? I am of the view that there will be at least 91

OT

Is it worth updating the pseudo-puzzle / megaclue thread which outlines the extent of the unavoidables in every grid. [I suspect most people are completely oblivious]

Can an individual puzzles properties be pinpointed directly to the unavoidable sets that each individual clue or group of clues is responsible in "covering" ? I can see a few [trivial] clue patterns [4-clue][naked single] where a clue may be inserted. Any puzzle with an isomorphic example of this pattern will have this insertion.

In persuit of RW's line - we perhaps would need to find our most distant 2 grids with a max of 33 common clues [mc-grid and a.n.-othergrid] with the least [or most] ? <92 sol

Are ther really only 91 different minimal unavoidable sets with a different position pattern ? I am of the view that there will be at least 91

- coloin
**Posts:**1743**Joined:**05 May 2005

JPF wrote:Here is a puzzle with 2 solutions.

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`*-----------*`

|1..|2..|3..|

|...|.4.|.5.|

|...|..6|..7|

|---+---+---|

|.2.|.8.|...|

|9..|...|1..|

|..4|..7|...|

|---+---+---|

|...|9..|2..|

|.8.|.3.|.4.|

|..7|..1|..6|

*-----------*

The 2 solutions A and B are :

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`179258364632749851845316927726183495958462173314597682563974218281635749497821536`

146275389798143652352896417521689734973524168864317925415968273689732541237451896

They differ on 60 cells.

This is an interesting phenomenon. And we may get more from it: the length of the path between these two isomorph grids in this case is 1 clue. However, if they are aligned properly at first, the length is 0, i.e. they are equivalent. If they were aligned in other ways, the length of the path could vary a lot. The key is how to find an optimized path from one grid to another.

- Addlan
**Posts:**62**Joined:**15 July 2005

coloin wrote:Are ther really only 91 different minimal unavoidable sets with a different position pattern ? I am of the view that there will be at least 91

Looking at those 48 cells in any solution out of the 92 possible solutions, each unavoidable set leads to at least one other possible solution. As a result N unavoidable sets will lead to a minimum of N+1 solutions. The puzzle has 92 solutions, therefore N<92.

RW

- RW
- 2010 Supporter
**Posts:**1000**Joined:**16 March 2006

Addlan wrote:RW wrote:Looking at those 48 cells in any solution out of the 92 possible solutions, each unavoidable set leads to at least one other possible solution. RW

Is it not that one unavoidable set leads to at least two solutions? If that is, N<=92/2=46

One unavoidable set leads to at least two solutions, yes. But each unavoidable after that leads to at least one more solution, not two more. For example two unavoidable sets with one common cell gives three solutions.

A new unavoidable set with no cells in common with previous sets doubles the amount of solutions. So two unavoidable sets with one cell in common and a third disjoint set gives six solutions.

RW

- RW
- 2010 Supporter
**Posts:**1000**Joined:**16 March 2006

Why didn't anyone just count the minimal unavoidables? There are 422, contrary to RW's claims.

An isolated sub-puzzle

Here's a nice small-scale counterexample. Consider the following isolated sub-puzzle:

This contains four minimal unavoidable sets, which I'll call "a", "A", "x" and "X".

"a" and "A" are dual min-unavs on digits 4 and 7; "x" and "X" are duals on digits 7, 8 and 9.

Solutions to the ISP

There are three solutions to this ISP.

The first contains min-unavs "a" and "x":

Permuting "a" to "A" gives the second solution, but kills "x" ...

... whereas permuting "x" to "X" gives the third solution but kills "a":

Conclusion

It's possible to have more minimal unavoidables than solutions.

An isolated sub-puzzle

Here's a nice small-scale counterexample. Consider the following isolated sub-puzzle:

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`. . . | . . . | . . .`

. . . | . . . | . . .

. 89 89 | . . . | . . .

-------------+-------------+------------

. . . | . . 47 | . . 47

. . 47 | . . 47 | . . .

. 79 479 | . . . | . . 47

-------------+-------------+------------

. . . | . . . | . . .

. . . | . . . | . . .

. 78 78 | . . . | . . .

This contains four minimal unavoidable sets, which I'll call "a", "A", "x" and "X".

"a" and "A" are dual min-unavs on digits 4 and 7; "x" and "X" are duals on digits 7, 8 and 9.

Solutions to the ISP

There are three solutions to this ISP.

The first contains min-unavs "a" and "x":

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`. . . | . . . | . . .`

. . . | . . . | . . .

. 8x 9x | . . . | . . .

-------------+-------------+------------

. . . | . . 4a | . . 7a

. . 4a | . . 7a | . . .

. 9x 7ax | . . . | . . 4a

-------------+-------------+------------

. . . | . . . | . . .

. . . | . . . | . . .

. 7x 8x | . . . | . . .

Permuting "a" to "A" gives the second solution, but kills "x" ...

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`. . . | . . . | . . .`

. . . | . . . | . . .

. 8 9 | . . . | . . .

-------------+-------------+------------

. . . | . . 7A | . . 4A

. . 7A | . . 4A | . . .

. 9 4A | . . . | . . 7A

-------------+-------------+------------

. . . | . . . | . . .

. . . | . . . | . . .

. 7 8 | . . . | . . .

... whereas permuting "x" to "X" gives the third solution but kills "a":

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`. . . | . . . | . . .`

. . . | . . . | . . .

. 9X 8X | . . . | . . .

-------------+-------------+------------

. . . | . . 4 | . . 7

. . 4 | . . 7 | . . .

. 7X 9X | . . . | . . 4

-------------+-------------+------------

. . . | . . . | . . .

. . . | . . . | . . .

. 8X 7X | . . . | . . .

Conclusion

It's possible to have more minimal unavoidables than solutions.

- Red Ed
**Posts:**633**Joined:**06 June 2005

Red Ed wrote:Why didn't anyone just count the minimal unavoidables? There are 422, contrary to RW's claims.

Here's a nice small-scale counterexample. Consider the following isolated sub-puzzle:

I think we are using terminology differently again... I said there'll be a maximum of 91 unavoidable sets that all appear in at least two grids in two different permutations. Swapping the digits in an unavoidable set doesn't create a new unavoidable set IMO, it's just a different permutation of the same set. So in your counterexample, I'd call "a" and "A" the same set, likewise "x" and "X".

If we consider all minimal sets that have the same digits and occupy the same cells as the same unavoidable set, there shouldn't be more than 91.

422 is quite a big number though... Are you sure you have filtered out all doubles?

RW

- RW
- 2010 Supporter
**Posts:**1000**Joined:**16 March 2006

RW, an unavoidable set consists of one value in each cell of the pattern. It was defined a long time ago and you can't change it now. In my previous post, "a" and "A" are different unavoidables.

Your concept appears to be that of an "ambiguous footprint". From that concept, I suppose a minimal ambiguous footprint (MAF) would be one that contains no smaller ambiguous footprint. If your claims pertain to those then I can't contradict them -- or rather I haven't tried to yet

Your concept appears to be that of an "ambiguous footprint". From that concept, I suppose a minimal ambiguous footprint (MAF) would be one that contains no smaller ambiguous footprint. If your claims pertain to those then I can't contradict them -- or rather I haven't tried to yet

- Red Ed
**Posts:**633**Joined:**06 June 2005

All bar two of the minimal unavoidables had valency 2, i.e. 2-1 = one other permutation with the same footprint. The other two had valency three. Grouping together min-unavs with the same footprint would give at least 210 groups (though how many exactly, I haven't checked).RW wrote:422 is quite a big number though... Are you sure you have filtered out all doubles?

Note that a permutation of a minimal unavoidable is not necessarily minimal (but of course is still an unavoidable).

- Red Ed
**Posts:**633**Joined:**06 June 2005

Sorry for the confusion...

Or perhaps we could rather use the more common term "deadly pattern". A minimal deadly pattern would be what we get if we remove all given clues from a minimal unavoidable set, right?

If this is true, then there must be some kind of "inner loops", where permutating different unavoidables eventually leads to the same new grid... Have to give it a thought...

Could you give an example of a minimal unavoidable whose permutation isn't a minimal unavoidable? I suppose this can happen only with unavoidbles that have valency >2... I am not very familiar with the properties of such unavoidables.

RW

Red Ed wrote:Your concept appears to be that of an "ambiguous footprint". From that concept, I suppose a minimal ambiguous footprint (MAF) would be one that contains no smaller ambiguous footprint. If your claims pertain to those then I can't contradict them -- or rather I haven't tried to yet

Or perhaps we could rather use the more common term "deadly pattern". A minimal deadly pattern would be what we get if we remove all given clues from a minimal unavoidable set, right?

Red Ed wrote:Grouping together min-unavs with the same footprint would give at least 210 groups (though how many exactly, I haven't checked).

If this is true, then there must be some kind of "inner loops", where permutating different unavoidables eventually leads to the same new grid... Have to give it a thought...

Red Ed wrote:Note that a permutation of a minimal unavoidable is not necessarily minimal (but of course is still an unavoidable).

Could you give an example of a minimal unavoidable whose permutation isn't a minimal unavoidable? I suppose this can happen only with unavoidbles that have valency >2... I am not very familiar with the properties of such unavoidables.

RW

- RW
- 2010 Supporter
**Posts:**1000**Joined:**16 March 2006

(For RW)

This trivalent minimal unavoidable ...

This trivalent minimal unavoidable ...

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`..3..6.8....8..1....813.4.6...7.48..9..583.1...5.91.64..2645.7...7.2..3.5.93.7.4.`

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`..3..6.8....8..1....813.4.6...7.48..9..583.1...5.91.64..2645.7...7.2..3.5.93.7.4.`

..8..3.6....1..8....368.1.4...8.74..5..391.8...9.45.16..5726.4...2.3..7.9.75.4.3.

..8..3.6....1..8....368.1.4...8.74..5..391.8...9.45.16..7526.4...2.3..7.9.57.4.3.

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`........................................................**................**.....`

- Red Ed
**Posts:**633**Joined:**06 June 2005

RW wrote:... minimal unavoidable whose permutation isn't a minimal unavoidable ... can happen only with unavoidables that have valency >2

Yes. I guess you know why, but for the benefit of others I'll explain. Imagine listing all the permutations (that have the same footprint) of a given unavoidable as I did in the previous post. An unavoidable in that list is minimal if & only if for each of its digits d@(r,c) it's the only unavoidable in the list that has d@(r,c).

If all the permuted unavoidables are minimal, I call them strongly minimal. Otherwise, the minimal ones (if any) are only weakly minimal. IIRC, most unavoidables are strongly minimal; only a few of the bigger (18+ cells) unavoidables are weakly minimal.

- Red Ed
**Posts:**633**Joined:**06 June 2005