Rule 1  First Case 

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: Aonly (blue) BbutnotA (yellow) AandoneB (green) AandmultipleB's (orange) 

If we come across something like this, our 2clue knowledge doesn't help. but let's color it as in the explanation above. 


Aonly (blue) BbutnotA (yellow) (including overlapping clue areas, if any) AandoneB (green) AandmultipleB's (orange) Now let's try something that, according to the rule, should be wrong . . . 


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. 


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. 


Going back to the rule: 4 = 2 (blue Aonly squares) + 1 + 1 (the other two clues)
and the green sections will each get the amount of Filled squares that match the B clues (in this case, one each). 

4 = 1 (Aonly square) + 1 + 2 (the other two clues)  

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

4 = 3 (Aonly squares) + 1 (the other clue) the AOnly section (blue squares) will get Filled squares, the BbutnotA (yellow squares) will all get Empty squares. the AandmultipleB 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 

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 

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? 


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. 


So we get this. Not very much, but every little bit helps. 

4 = 1 (Aonly square) + 2 (regular clue) + 1 (reducedarea clue) the AOnly section (blue square) will get a Filled square, the BbutnotA (yellow squares) will all get Empty squares. the AandmultipleB 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. 
Rule 2  AS_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 yellow areas here are AOnly, and the green areas are SandoneA, there are no other areas. I have used two greens only for illustrative purposes.  

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) 


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. 
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