The New Sudoku Players' Forum

[Havard]

its time to think more about the properties of low 20's clue puzzles, if any, that make good seeds
for {-2+1}xN derivations to 17's

Lars Petters (I will see if I can make him come and join us here) approach has been to make 100,000 random 20. Then do a -2+1 to get almost as many 19. Then again to get about 10,000 18's. Then repeated -1+1 on these 18's combined with -2+1 has so far produced around 50 new 17's, with a lot more to come, I expect.

I have a theory that the more you exhaust the +1-1 the better results you will get. It is almost like every 19 has a unique-ish potential for many 17's, but to find it you need to -1+1 a lot around it and its offspring (meaning the -2+1's)

Havard

[gsf]

Lars Petters (I will see if I can make him come and join us here) approach has been to make 100,000 random 20. Then do a -2+1 to get almost as many 19. Then again to get about 10,000 18's. Then repeated -1+1 on these 18's combined with -2+1 has so far produced around 50 new 17's, with a lot more to come, I expect.

I have a theory that the more you exhaust the +1-1 the better results you will get. It is almost like every 19 has a unique-ish potential for many 17's, but to find it you need to -1+1 a lot around it and its offspring (meaning the -2+1's)

Havard

thanks, so the general form for Lars' process is

Code: Select all

{-2+1}xA{-1+1}xB{-2+1}

where A reduces the input to 18 clues and B extends the 18 neighborhood
any observations on the range of B?

[JPF]

It turns out that the actual proposed algorithms end up by reducing a set of 18s to a set of 17s by a {-2+1}

Let's assume that the set G of actual 17s (the Gordon's list) is closed for {-1+1} and for {-2+2}, i.e :
if P is a puzzle of G, then the sets {-1+1}oP and {-2+2}oP are included in G.

Now, let Q be a 18 clues puzzle.
If Q includes one puzzle P of G, then there is no new 17 in Q.

JPF

[wintder]

Now, let Q be a 18 clues puzzle.
If Q includes one puzzle P of G, then there is no new 17 in Q.

JPF

Your post escapes me.

If I have an 18 I can 18 off and 17 on.

I will find all 17s.

I guess that there are no new 17s?

[re'born]

Now, let Q be a 18 clues puzzle.
If Q includes one puzzle P of G, then there is no new 17 in Q.

JPF

Your post escapes me.

If I have an 18 I can 18 off and 17 on.

I will find all 17s.

I guess that there are no new 17s?

IMO, the key part of JPF's statement is "no new 17 in Q".

[wintder]

The post was silly.

Take any 17. Recursively do 1 off, 1 on.

Eventually you will get them all.

Take any 18, do 1 off, 1 on, recursively, you will get all 18s.

I suspect that he meant something more distinct.

[JPF]

The post was silly.

Thanks for the compliment.

My statemement was not clear.

I should have say :

If Q includes one puzzle P of G, then there is no "new 17" by doing a {-2+1} on Q.

Q includes P means : Q= P +1 digit
new 17 means : not already in G.

The idea is to save time in screening the 18s.
There is no need to look at C(18,2)x65xp (1<=p<=9) puzzles if we can quickly say that one of the 18 17s is in G.

I don't know if it's practically true.

JPF

[wintder]

It turns out that the actual proposed algorithms end up by reducing a set of 18s to a set of 17s by a {-2+1}

This is the slow way of taking a 17 and doing a 1 off, 1 on until all are calculated.

Yes, I know that we don't currently have the processor power.

[wintder]

The post was silly.

Thanks for the compliment.

Well, the post was not very logical, and you usually are!

[wintder]

IMO, the key part of JPF's statement is "no new 17 in Q".

From any valid puzzle, all valid puzzles can be generated.

This is obvious, vapid.

It is true that there are no "new 17s", yet there are many 17s we have not found.

[JPF]

This is the slow way of taking a 17 and doing a 1 off, 1 on until all are calculated.

too bad...
most (at least many) of the 17s are isolated : doing a {-1+1} doesn't give any new 17.

JPF

[gsf]

asking for clarification is a lot more tasteful than vapidizing
clarification benefits all reader levels
vapidizing demeans the discourse ...

the {-n+m} notation used in this thread is derived from the implementation in my solver
where the puzzles derived from each {-n+m} round are required to be valid

puzzles derived by {-2+2} can be different from puzzles derived by {-1+1}x2

for any valid puzzle with C clues, {-1+1}xN will reach closure for some N,
and the closure will be much less than the set of all valid puzzles with C clues

so {-1+1}x* (apply {-1+1} until closure) can be used to partition entries in a catalog
and that partitioning can be used to prune the search for new entries to the catalog

I'll see about adding {n}* to denote {-n+n} until closure

there are probably some interesting relationships between say {1}* and {2}*:
{2}* for a given puzzle would have to contain all of the {1}* for the puzzle
and those {1}* would partition the {2}*

[wintder]

This is the slow way of taking a 17 and doing a 1 off, 1 on until all are calculated.

too bad...
most (at least many) of the 17s are isolated : doing a {-1+1} doesn't give any new 17.

JPF

Doing a single one off one on.

Yep.

[wintder]

gsf, JPF,

I respect you guys.

The attempt to make a rule is in trouble.

ANY valid puzzle can be used to create all 17s.

You need to constrain the number of 1 offs and 2 offs to make a rule.

[gsf]

ANY valid puzzle can be used to create all 17s.

You need to constrain the number of 1 offs and 2 offs to make a rule.

you misunderstand the {-n+m} notation which contains the constraint you seek
each round of {-n+m} is defined to produce only valid puzzles, where valid includes but is not limited to one solution

there exits 17 clue puzzles that have only 1 puzzle in the {-1+1} closure
e.g.

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sudoku -go{-1+1}x99 000405000003000090000008000040000000000060030500001000000000001800002500009030004

produces only the input puzzle