The New Sudoku Players' Forum

[JPF]

Although an optimistic Havard probably has a chance at a John Buchan [39].....at least more chance than a Ringo Star [16]

Am I missing some british joke ?

On that line - quite possibly with all the 36s and 37s found we might have a minimal 2-stepper ? ]

There are no singles in the ravel’s lists

JPF

[gsf]

Although an optimistic Havard probably has a chance at a John Buchan [39].....at least more chance than a Ringo Star [16]

Am I missing some british joke ?

I had to resort to wiki for: John Buchan wrote The Thirty-Nine Steps
and knew Ringo sang You're Sixteen ...

[coloin]

Apologies for the irreverance.......its to do with "bingo" calling - [although it probably is "sweet" 16] - but you are correct gsf

I dont think that we will get a 39 easily though.......

Having said that, sometimes when you make assertions in sudokupuzzleland - along comes an exception.......however I do now think the maximum count of any clue in a minimal puzzle is 6......you always need 3 missing it seems.

To have a 7 clues filled you need to have a B1B2B3B4B7 equivalent filled - which of course leaves the clue in B1 superfluous !

Code: Select all

+---+---+---+
|1..|...|...|
|...|1..|...|
|...|...|1..|
+---+---+---+
|...|...|...|
|.1.|...|...|
|...|...|...|
+---+---+---+
|..1|...|...|
|...|...|...|
|...|...|...|
+---+---+---+


This reduces the number of 48-clue regions in a grid somewhat....but there are still a vast amount .

C

[Havard]

the search for a 39 is proving to be quite hard...

I have found it (on average) to be about 1 38 per 250 37. (1off2on)

Who knows what that ratio is going from 38 to 39?

I would at least need 1000 38 to have a fair go, and so far I have only found 31.

I am not giving up quite yet though...

Havard

[ravel]

Congratulations on finding the 38's, Havard. I am glad, that someone continues this search. With my capacities it would take years until i would have a fair chance for a 39. Good luck!

[Mauricio]

I had some fun with ravel's 38, and found something remarkable/interesting (I dont know where else to post it, and certainly it doesn't deserve a new topic):

Code: Select all

Ravel's 38, minlex form. Minimal.
+-------+-------+-------+
| . . . | . . . | . . . |
| . . 1 | . 2 3 | . 4 5 |
| . 2 4 | 6 1 5 | 3 7 . |
+-------+-------+-------+
| . . . | . 8 2 | 7 . . |
| . 1 8 | 5 . 7 | . 6 . |
| 2 . 7 | 1 . 6 | 8 5 . |
+-------+-------+-------+
| . 6 2 | 3 5 1 | 4 8 . |
| . 8 . | . . . | . 3 . |
| 1 . . | . . 8 | . . 6 |
+-------+-------+-------+


Code: Select all

Ravel's 38 -14 clues +1 clue. Minimal. Same solution
+-------+-------+-------+
| . . . | . . . | . . . |
| . . . | . . 3 | . 4 . |
| . 2 4 | . . 5 | 3 7 . |
+-------+-------+-------+
| . . . | . . 2 | 7 . . |
| . 1 8 | . . 7 | . 6 . |
| 2 . . | 1 . 6 | . 5 . |
+-------+-------+-------+
| . 6 . | . . . | 4 8 . |
| . 8 . | . . . | . 3 . |
| 1 . . | 9 . . | . . 6 |
+-------+-------+-------+


[JPF]

Nice find.

See here a posting I made a while ago, more or less on the same subject.

JPF

[coloin]

Indeed - We have one clue doing the job of 14 clues......

Possibly I could open a new thread line in The Megaclue ! ....done

25 clue minimal puzzle - 9 @ r9c4 = 32213 grid solutions !

A good one to set the ball rolling for us competitive types !

Actually very good.....

The best I could do on first try was this

Code: Select all

+---+---+---+
|.2.|...|...|
|...|...|.6.|
|687|..9|.34|
+---+---+---+
|..3|9..|621|
|.7.|..2|...|
|...|5.8|..7|
+---+---+---+
|..2|.1.|..3|
|...|..3|8..|
|.4.|...|..6|
+---+---+---+  omitting the 1 at r7c5 gives 6040 grid solutions

C

[ravel]

From the other thread:

By doing a "reverse" 2off1on on the list of 38's three times, I get a massive list of 35's that I then can exclude from all the 35's I am making. This then prunes the search a lot, allowing me to already at the 35 stage get rid of a lot of puzzles that are not needed.

I dont quite understand. From the 35 you arrived with 3 2off1on's, you obviously can get back to the same 38 by doing 3 1off2on's. But why shouldn't you find a new 38 also, theoretically with a distance of 9off9on from the first one ?

[Havard]

From the other thread:

By doing a "reverse" 2off1on on the list of 38's three times, I get a massive list of 35's that I then can exclude from all the 35's I am making. This then prunes the search a lot, allowing me to already at the 35 stage get rid of a lot of puzzles that are not needed.

I dont quite understand. From the 35 you arrived with 3 2off1on's, you obviously can get back to the same 38 by doing 3 1off2on's. But why shouldn't you find a new 38 also, theoretically with a distance of 9off9on from the first one ?

I thought I would too, but no new 38 came out of doing the process backwards. Do you think it is possible?

Havard

[gsf]

From the other thread:

By doing a "reverse" 2off1on on the list of 38's three times, I get a massive list of 35's that I then can exclude from all the 35's I am making. This then prunes the search a lot, allowing me to already at the 35 stage get rid of a lot of puzzles that are not needed.

I dont quite understand. From the 35 you arrived with 3 2off1on's, you obviously can get back to the same 38 by doing 3 1off2on's. But why shouldn't you find a new 38 also, theoretically with a distance of 9off9on from the first one ?

I thought I would too, but no new 38 came out of doing the process backwards. Do you think it is possible?

this question came up on another thread
it depends on the definition of the {-off+on} operator
I have implemented it as a sequence of {-n+m} (n-off m-on), where each {-n+m} produces only valid puzzles
so that {-1+1}x2 may produce different output than {-2+2} because the {-2+2} can "leap"
over some 1-off invalid puzzles that would have been pruned by {-1+1}

[ravel]

I thought I would too, but no new 38 came out of doing the process backwards. Do you think it is possible?

I dont have any concrete samples. What i made, was a 2off/1on for 37's, then 1off/2on (including 1off/1on) on the 36's (until closure). This was a relatively fast way to get new 36's (and 37's) compared to a 2off/2on for 36's.

Mauricio,
i forgot to mention yesterday, that your find is another etraordinary example for the sudoku mysteries - 1 clue for 14. Great.

[Havard]

I thought I would too, but no new 38 came out of doing the process backwards. Do you think it is possible?

I dont have any concrete samples. What i made, was a 2off/1on for 37's, then 1off/2on (including 1off/1on) on the 36's (until closure). This was a relatively fast way to get new 36's (and 37's) compared to a 2off/2on for 36's.

Interesting. I am finding that with a large enough supply of 36, the 1off1on will increase the amount for every run, so the supplies of 36 are hence "endless". However, I have only been getting about 6.000 37 from 100.000 36, so the most effective thing I have found to be 2off2on on the collection I have made of 37s. As far as I can tell, it will be able to keep going for a while, much in the same way as 1off1on on 36 are. So far I have 40.000 37, but the goal is over 100.000. Might still not be enough 38's though...

Havard

[ravel]

Ah, i had this feeling at the end (with some 1000's 36's) too, that 1off/1on would never end for the 36's. But at the beginning the closure was reached for several times and i had to use 2off/2on to continue.
Now have you tried for the 38's, what i described for 37's above ?

[Havard]

Ah, i had this feeling at the end (with some 1000's 36's) too, that 1off/1on would never end for the 36's. But at the beginning the closure was reached for several times and i had to use 2off/2on to continue.
Now have you tried for the 38's, what i described for 37's above ?

About the 36, it seems to me that the collection needs a certain size before it will start expanding like that. Critical mass of sudoku if you like...

And I did try to go 2off/1on on the 38 (have 240 of them now), but this made no new 37.

While the 1off/1on on 36 can be exponential, the 2off/2on on the 37 delivers a steady 2000 new 37 every 4 hours.

Havard