There is a lot of math you can apply to it. You can add up all the known blocks for a row/column and subtract that from the size of the row/column to get which blocks are known to be filled in there. (I am using O for filled in block and X for empty box). For 10x10 puzzle, If R1 = 5 4. That is 5 blocks+1space+4blocks=10 blocks. Every block is known= OOOOOXOOOO

If R1 = 5 3. That is 5 blocks+1space+3blocks=9 blocks out of 10 are known = ?OOOO??OO? The question marks reflect that because it uses 9 blocks it can fit at the front or end of the 10 block row with one space on each side. In between the blocks are 2 ? because each block can slide one block in each direction

If R1 = 3 4. That is 3 blocks+1space+4blocks= 8 blocks out of 10 are known = ??O???OO??

So then algebraically, for a row that is X long and where the rows and spaces add to up to Y, you will have (X-Y) unknown spots on each end, with each block shortened by (X-Y) due to uncertainty, and with spaces between blocks of (X-Y+1).

So for a 20 X 20 puzzle with R1 = 7 4 4 = 7+4+4+2(for spaces between the blocks) = 17. 20 - 17 = 3. So we will have 3 unknown blocks, then (7-3) = 4 filled boxes, followed by 4 unknown blocks, followed by (4-3) = 1 filled box, followed by 4 unknown blocks, followed by (4-3) = 1 filled box, followed by 3 unknown blocks at the end. = ???OOOO????O????O???

edge logic mentioned by someone else very helpful too.

I do all the nonogram daily apps but my favorite is nonogram999. It doesn’t have any daily apps but it has some large 30x30 puzzles that are challenging. I have to it on a tablet due to size of puzzle.

Let me know if any of that above needs clarification.