Editing Nonogram (section)

===Mathematical approach===
It is possible to get a start to a puzzle using a mathematical technique to fill in blocks for rows/columns independent of other rows/columns. This is a good "first step" and is a mathematical shortcut to techniques described above. The process is as follows:
[[File:Mathematical approach.png|right]]
# Add the clues together, plus 1 for each "space" in between. For example, if the clue is 6 2 3, this step produces the sum 6 + 1 + 2 + 1 + 3 = 13.
# Subtract this number from the total available in the row (usually the width or height of the puzzle). For example, if the clue in step 1 is in a row 15 cells wide, the difference is 15 - 13 = 2.  Note: If spaces can be used on the left or right (top or bottom) borders, this "shrinks" the available area. If it is known that the rightmost cell is a space, the difference is 14 - 13 = 1.
# Any clues that are greater than the number in step 2 will have some blocks filled in. In the example, this applies to the clues 6 and 3, but not 2.
# For each clue in step 3, subtract the number in step 2 to determine the number of blocks that can be filled in. For example, the 6 clue will have (6 - 2 =) 4 blocks filled in and the 3 clue will have (3 - 2 =) 1. Note: Applying the same procedure to a clue that "failed" step 3 will produce a non-positive number, indicating that no blocks will be filled in for this clue. The clue 2 produces the number (2 - 2 =) 0; if there were a 1 clue, it would produce the number (1 - 2 =) -1.
# To fill in the blocks, assume the blocks are all pushed to one side, count from that side "through" the blocks, and backfill the appropriate number of blocks. This can be done from either direction. For example, the 6 clue can be done either of two ways as follows:
## From the left: Since the 6 is the first number, count 6 blocks from the left edge, ending in the 6th block.  Now "backfill" 4 blocks (the number obtained in step 4), so that cells 3 through 6 are filled.
## From the right: Starting from the right, the clues that are to the right of the 6 clue must be accounted for. Starting from cell 15, count 3 cells for the 3 clue (to cell 13), then a space (12), then the 2 clue (10), then a space (9), then the 6 clue (3). From the 3rd cell, "backfill" 4 blocks, filling cells 3 through 6. The results are the same as doing it from the left in the step above.
# Repeat step 5 for all clues identified in step 3.

In the illustration, row 1 shows the cells that are filled under this procedure, rows 2 and 4 show how the blocks are pushed to one side in step 5, and rows 3 and 5 show the cells backfilled in step 5.

Using this technique for all rows and columns at the start of the puzzle produces a good head start into completing it.  Note: Some rows/columns won't yield any results initially. For example, a row of 20 cells with a clue of 1 4 2 5 will yield 1 + 1 + 4 + 1 + 2 + 1 + 5 = 15. 20 - 15 = 5. None of the clues are greater than 5. Also, this technique can be used on a smaller scale. If there are available spaces in the center or either side, even if certain clues are already discovered, this method can be used with the remaining clues and available spaces.