Some 3-clue logic for PrismaPix
(solving help for PrismaPix, Fill-a-Pix, Minesweeper, etc)

Terminology: All squares start out Unmarked, and will eventually be marked as a Filled square or an Empty square.

So you think you've mastered the 2-Clue logic? Well that's better than I've done. (click HERE to go back to the 2-Clue page)

This is a work in progress. I welcome your input.

  Rule 1 - First Case 

         
         
    A B  
  B      
         
First of all, let's look at what we are working with. It's a little more complicated then before:

We have a clue A surrounded by some Other Clues, all labeled B

the sections we have are:
A-only (blue)
B-but-not-A (yellow)
A-and-one-B (green)
A-and-multiple-B's (orange)

The rule goes like this:
If the value of A is equal to the amount of A-only squares (if there are any) plus (the vaules of) all of the B's
Then the A-Only section (blue) will get Filled squares, the B-but-not-A and A-and-multiple-B sections will all get Empty squares.
(and the green A-and-one-B areas will get some filled squares, according to the value of each B)

Here's an example with an explanation
         
  1      
    4    
      1  
         
If we come across something like this, our 2-clue knowledge doesn't help.

but let's color it as in the explanation above.
         
  1      
    4    
      1  
         
A-only (blue)
B-but-not-A (yellow) (including overlapping clue areas, if any)
A-and-one-B (green)
A-and-multiple-B's (orange)

Now let's try something that, according to the rule, should be wrong . . .
         
  1      
    4   ?
      1 ?
    ? ? ?
If the lower 1 gets a Filled square anywhere in the yellow area, (the rule says it should be Empty) then all of the green area will need to be Empty.

The 4 needs all of its Filled squares.
We could put two in the blue squares, but any way we try to put the last two, both of them will be next to the upper 1.

Therefore the yellow section must be indeed be Empty.
         
  1      
    4    
      1  
         
According to the rule, the orange section should be empty also.
Let's test that . . .

If we put a Filled square there (in the orange area), then both 1's are satisfied and all of the yellow squares must be Empty.
That leaves only two more squares for the 4, for a total of three.

Therefore the orange section must be Empty.
         
  1      
    4    
      1  
         
Going back to the rule: 4 = 2 (blue A-only squares) + 1 + 1 (the other two clues)

So the A-Only section (blue squares) will get Filled squares,
the B-but-not-A (yellow squares)
and A-and-multiple-B sections (orange square) will all get Empty squares.

and the green sections will each get the amount of Filled squares that match the B clues (in this case, one each).

Here's a second example:
         
      1  
  2 4    
         
4 = 1 (A-only square) + 1 + 2 (the other two clues)
         
      1  
  2 4    
         
So the A-Only section (blue square) will get marked as a Filled square,
the B-but-not-A (yellow squares)
and A-and-multiple-B sections will all get Empty squares. (orange squares)

and the green sections will each get the amount of Filled squares that match the appropriate B clue

If you think about it, this rule applies to our 2-clue cases as well,
       
  4 1  
       
4 = 3 (A-only squares) + 1 (the other clue)

the A-Only section (blue squares) will get Filled squares,
the B-but-not-A (yellow squares) will all get Empty squares.
the A-and-multiple-B sections (all zero of them) will all get Empty squares.
and the green section will each get the amount of Filled squares that match the B clue (one)

  Rule 1 - Second Case 

The second case is basically the same thing, but instead of the clues being next to each other, we have reduced-area clues.

By that I mean that a clue usually refers to nine squares, like this
:
but if we have some of them marked (in this case 4 blue Filled and 2 grey Empty)
we can pretend we have a different kind of clue, like this:

that refers (in this case) to the 3 remaining unmarked squares.

Here is a simple example:
         
1        
         
  2      
      1  
In this contrived example where the grey cells are known to be Empty,
one of the 1's will have its Filled square in the blue area,
the other 1 will have its Filled square in the green area.

Those two Filled Squares will satisfy the 2, so all of the yellow squares will have to be Empty

Here is a slightly more complex situation taken from a PrismaPix puzzle
           
        8  
  2 4      
           
The blue squares have been Filled.
There were some other clues and some known Empty squares, but they don't matter right now.

So what can we tell here?
           
        8  
  2 4      
           
We know that the green area will have two more Filled squares than the yellow area (but will there be 2 or will there be 3?)

and we know that one of the two by the 8 will be Empty

This means that only two of the green squares will get Filled, and the lower one is one of them. and the three yellow will be Empty.
           
        8  
  2 4      
           
So we get this.

Not very much, but every little bit helps.
It's as if we changed the board to two regular (9-square) clues and a reduced (2-square) clue, like this:

and applied our rule
         
      1  
  2 4    
         
4 = 1 (A-only square) + 2 (regular clue) + 1 (reduced-area clue)

the A-Only section (blue square) will get a Filled square,
the B-but-not-A (yellow squares) will all get Empty squares.
the A-and-multiple-B sections (all zero of them) will all get Empty squares.
and the green sections (one light, one dark) will each get the amount of Filled squares that match the appropriate B clue.
and all of the white squares are outside of the scope of these clues.

  Rule 2 - AS_A 


           
  A S   A  
           
At first I thought this would rarely occur, but to my surprise, and trained eye, there were enough of this to warrant adding it to this "dissertation".

The rule is: In the case where both of the following are true:
  • the A's are (relatively) small numbers, and S is their Sum
  • The numbers are arranged A S (space) A (or vice versa), in a line, vertically or horizontally
Then all of the areas we have colored yellow, Must be marked as Empty squares
The yellow areas here are A-Only, and the green areas are S-and-one-A, there are no other areas. I have used two greens only for illustrative purposes.

           
  4 6   2  
           
Here is an example:

6 = 4 + 2

So, the yellow areas are all empty, the light green gets 4 Filled squares (we don't yet know where exactly) and the dark green gets 2 Filled squares (ditto)
     
  3  
     
  7  
  4  
     
Here is an another example:

7 = 4 + 3

So, the yellow areas are all empty, the light green gets 4 Filled squares (we don't yet know where exactly) and the dark green gets 3 Filled squares. This time we know exactly where. All of them!

(can you see that this is really Rule 1 in disguise?)
And remember, there may be additional clues (numbers), and some of the squares may already be known, but this rule still holds true.

That's all I have for now.
If you have any more rules or examples you'd like me to add here, please let me know.

keep thinkin'
Rammy

ya
mmo.c
y@ho
ra
m5om