>dukuso wrote:

>

>

>in this one all pairs of symbols give chains of maximal length.

>It was not easy to find !

>

>(5,5,5 - 5,5,5)

>

>123456789

>897123456

>645789123

>312645978

>789312645

>564978312

>951267834

>438591267

>276834591

>

>I once tried to use it to find sudokus with few clues without success.

>Guenter.

>

>

>You're way ahead of me.

>

>But I don't understand what you mean by "all pairs of symbols give chains of maximal length"

>This grid has

>123......

>.........

>.........

>312......

>.............etc

>Is this not a chain of length 3?

>You probably mean something else.

yes, I think we had this earlier. You might have missed it.

Delete all symbols except 2 from a sudokugrid,

and consider the remaining 18 cells as vertices of a graph,

with two cells being joined by an edge, iff they are in the

same row,column or block. Then count the sizes of the

connected components.

If you do this in the grid above, you can always

reach each cell from each other cell, no matter which symbols

you choose or where you start.

No

1..2

2..1

or other things which are diconnected from the other 14 vertices.

I had uploaded a program here

http://magictour.free.fr/nppcs2.exewhich prints the coded cycle-structure of a sudokugrid.

This is invariant by the 9!*6^8*2 sudoku-transformations.

>--------------------------------

>Thank you for showing that the most canonical grid has no 18ers.

>So there may be some truth to the principle

>lots of symmetry <---> lots of clues needed

>but its probably of no use in finding a 16.

also the 1-gangster has 1728 compatible bands,

much more than the others. So even when the other 2 bands are

filled, there are still 1728 possibilities !

>I also don't understand the 1296 figure. I got 162.

>There are 9 3x3 Latin squares, each requiring 2 clues.

>For a given 3x3 Latin square, the first clue can be

>chosen in 9 ways, and the second clue can be chosen

>in 2 ways. (it can't be in the same row, column,

>or be the same digit)

the order of the clues doesn't matter, so divide this by 2.

>Since the 9 Latin squares partition the grid, there are 9x18=162

>ways of choosing 18 clues.

they are independent, so it's 9^9 , but

>What am I doing wrong?

we have another 9 latin squares vertically and these must match

with the clues too. See my other post, in reply to Gordon.

>---------------------------------------------------

>gfroyle said

>

>>Yes it does... if the complete grid is

>

>pointing out the size 6 unavoidable sets. Thank you!

>Those sets are subsets of the size 9 unavoidable sets,

>aka the 3x3 Latin squares.

>(Glad you like the term unavoidable )

>

>I was trying to formulate properties a good candidate grid

>for a 16 might have. And I was trying to make that particular

>grid fail them, since we know it requires at least 18.

>(and it has an 18 as dukuso found)

>

>I'm guessing so far at

>1) as little symmetry as possible

>2) no unavoidable sets of size 4

>3) 16 unavoidable sets of size 6 that cover all 81 cells

>

>you guys have probably been down that road. what conditions

>did you come up with? Maybe its not a profitable road to go down.

Some time ago Gordon tried a setcover method with all

the sets requiring a clue. Seems he gave up on this ?!