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So, down to a limit, the fewer the starting values given in a grid, determine how difficult it is to complete?

That seems logical to me.

With any given number of starting values (let's say 23 for instance), does the complexity of solving the grid vary with the initial position of these values, or would all grids starting with 23 values need the same solving techniques? IE. Be the same complexity.

(I can't follow the math shown above).

Or am I way off?

Way off.

First, the math above, and the problem being posed, only deals with representing the final grid of values and doesn't address the starting clues.

The difficulty of solving a puzzle, given a set of initial clues, is dependent on several factors.

For a given grid of symbols, there are many sets of initial clues that are locally minimal. A local minimum is a set of clues that is sufficient such that there is exactly one unique grid that is consistent with them and such that if you were to remove any one of those clues there would be multiple valid solutions. But not all of those sets are the same size. So, if I give you two different sets of starting clues for the same grid, one that has, say, twenty symbols and one that has twenty-three, it's entirely possible that the one with twenty-three symbols is a local minima while the one with twenty actually has three more than is necessary to solve the puzzle and may, therefore, be easier to solve.

Another way to view this is to consider a puzzle with a given starting set of clues. Now I want to add an additional clue to make it easier to solve. Are all of the possible symbols I could add equally valuable as a clue? Nope. Some might make it considerably easier, while others might have virtually no impact. Think of most puzzles that you work -- don't you find that there is almost always some low hanging fruit? Spaces that you almost immediately can fill in with very little effort? Would adding those symbols to the clue set therefore make the puzzle much easier to solve? Not much at all, since you would have filled in that symbol almost immediately anyway. But, again thinking of most of the puzzles you work, particularly the more challenging ones where you get to a point where you really have to work to figure out another symbol. Don't you find yourself saying that if you could only figure out the value of this or that symbol, it would all break open? Well, what if that symbol were added to the initial clue set? That would make the puzzle much easier to solve.

 

For a given grid of symbols, there are many sets of initial clues that are locally minimal. A local minimum is a set of clues that is sufficient such that there is exactly one unique grid that is consistent with them and such that if you were to remove any one of those clues there would be multiple valid solutions. But not all of those sets are the same size. So, if I give you two different sets of starting clues for the same grid, one that has, say, twenty symbols and one that has twenty-three, it's entirely possible that the one with twenty-three symbols is a local minima while the one with twenty actually has three more than is necessary to solve the puzzle and may, therefore, be easier to solve.

If these are the starting values for the same grid and 23 is the 'minima', by only giving 20 of these values, how can that grid possibly be resolved? Let alone be easier?

I can see, if there is a minima starting set and some values are added, it may or may not be easier to solve, but just have fewer values to figure out.

 

If these are the starting values for the same grid and 23 is the 'minima', by only giving 20 of these values, how can that grid possibly be resolved? Let alone be easier?

I can see, if there is a minima starting set and some values are added, it may or may not be easier to solve, but just have fewer values to figure out.

23 might be the minima for a particular set of starting values.

But there might be a different set of starting values for that same grid that only needs 20.

Imagine that I begin with the set of 23 that is minimal. That only means that if I remove one of those 23, the puzzle becomes unsolvable (i.e., has multiple valid solutions).

But what if I, instead, add an additional symbol? That symbol may add enough information that I can now remove four (not any four, but a specific four) of the symbols from the original set before I once again have a minimal set.

But, more likely, the minimal set of 23 and the minimal set of 20 may have few, if any, members in common.

For most solution grids, there are millions, and sometimes billions, of distinct minimal clue sets. The number of clues in each set range from 17 to 40.