Unavoidables, clue generation and minimum number of solutions.

A new clue is generated when it cant be in an unavoidable set which isnt already covered by definite clues. This is done by "covering " the 8 other spaces which would be in the same unavoidable set as this clue. This is the common pattern of 3 clues in a box in many of Gfroyles 17s.

- Code: Select all
`+---+---+---+`

|...|24.|7..|

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

|.1.|...|...|

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

|...|8..|..6|

|7..|...|...|

|4..|...|2..|

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

|3..|.7.|...|

|...|...|..2|

|...|..6|.18|

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

The 7 is insereted in r8c8 - the 2,1& 8 in box 9 would be in the set. The other unknown clues in box 9 cant make an unavoidable set with the 7 in box 9 because of the 7s in c7 and r7 preclude it.

This is a good way ---- but not nessesarily the best way to achieve clue separation and therefore maximum set coverage.

What we need is 16 of the most "diverse" clues in set coverage - not nessessarily clue generators. Of course if ALL the sets are covered all the other superfluos clues will be inserted on insertion of the final [16th ?]clue. Perhaps this means using all nine clues in all the boxes..... and perhaps a maximum of 2 of any particular clue number.

In any completable [minimal] puzzle ALL the sets are covered by all of the clues.

The 17 numbers cover All the sets and then it is solved uniquely.

If there are 17 disjointed sets then it cant be done in 16 clues.

The Sf apparently cant be done in 16 - therefore there must exist extra set[s] which cant be covered by any combination of 16 clues.

Sets have to be at least 4 clues in them. Next size up is 6 clues.

The most [completely] disjointed sets a grid can have is 20 - this is very unlikely to happen.[You need 20 different 4 sets of numbers][4*20=80]

Does this mean that all sudoku grids can be reduced to at least 20 clues - or 19 ?

15 different 4s and 3 different 6s give 18 sets possible - this cant be done in 17 clues.

With 18 and 19 clues the chances of hitting all the sets are much improved - so much so that I think all grids can be done in 19 - apart from the grid which has 20 - which I believe wont happen !

with no 4 set unavoidables the maximum sets possible will be 13 [78] [This is the SFB]

Only 2 4s in the grid means that there can only be 12 6s give a total of 14 completly disjointed sets possible. [80][this is the SF - EDIT which actually has only 9]. Gfroyle did allude to this months ago.

With 5 "4 set" unavoidables the maximum disjointed sets possible are 15.

This would explain the wide range of MCN in all the 17s. If you cover all the sets then it is solved.

Of course you are more likely to cover them all if you have less to cover as in the SF and the SFB [perhaps].

With 11 "4 set" the maximum number of sets is 17 [80] [This is Mosch's grid - which actually has 14.] I think someone got many 19s in this grid - which surprised many.

I have a new method which attempts to separate the clues to try for more efficient clue coverage [if it is possible] of all the unavoidable sets....await results ! If I get a new 17 then there will be mileage in this method.