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# Row, Column and 3x3 Resulting Insufficiency Messages Explained

Most puzzles typically labeled as "diabolical" or "extremely difficult" in newspapers and magazines do not require indirect argumentation of the sort described in this section to solve. In practice, the simple constraints mentioned in earlier sections suffice for all but a few.

However, in principle, eliminating a possible value for a given cell may require some indirect reasoning, i.e. the chaining of reasoning steps. Some examples of these kind of cases from the sample puzzles supplied with the program are given below.

## Example 1

This is from an example puzzle that I've listed as difficult.

Here is how to read the color-coded board indicators in conjunction with the two messages shown.

 [Click here for a larger version of the diagram.] Messages: ```(3,7) = 3 results in inconsistency (8,5) = 6 = (9,5) ... in 3x3 at (77): 3 occurs uniquely, (9,8) = [7,9,5,3] now 3 ``` Explanation: Assuming (3,7) is 3 (highlighted in red with a circle) results in an inconsistent board. If that is true, (3,7) cannot be 3 and the offending possibility can be removed. Making the assumption (3,7)=3 results in (8,5) and (9,5) both having the value 6, which, of course, is inconsistent, as the two sixes are in the same column, namely 5. (This is highlighted by the pair of red 6s.) Let's take each case separately. (In what follows, red is used for contradicting values, blue for pivotal values in the chain of inference, and orange for values that will be eliminated along the way.)

Subparts of the reasoning are as follows:

 Establishing (8,5)=6 (8,5) must be 6 (shown in red) if it cannot be 1. [6,1] are the only two possibilities at this stage for the cell. (1 is highlighted in orange.) (8,5) cannot be 1 if the cell on the same row, namely (8,2), must be 1. (This pivotal value 1 is highlighted in blue). (8,2) must be 1 if the other possibility for the cell, namely 6, cannot be true. (6 for (8,2) is highlighted in orange.) (8,2) cannot be 6 if the cell on the same column, namely (3,2), must be 6. (3,2) must be 6 (highlighted in blue) if it cannot be 3. (3 for (3,2) is highlighted in orange.) But (3,2) cannot be 3 based on the original assumption that (3,7), a cell on the same row as (3,2), was 3 to begin with. Establishing (9,5)=6 (9,5) must be 6 (shown in red) if it cannot be 3. [6,3] are the only two possibilities at this stage for the cell. (3 is highlighted in orange.) (9,5) cannot be 3 if the cell on the same row, namely (9,8), must be 3. (9,8) must be 3 due as indicated in the 2nd message. [Note: the indentation indicated by the dots (ellipsis) indicates it is a subpart of the larger proof.] In the 3x3 area anchored at (7,7) (highlighted in brown/orange), 3 is uniquely mentioned in cell (9,8). In other words, it cannot take any of the values [5,7,9] because then 3 would be "lost", i.e. not assigned in the 3x3 area. (The values [5,7,9] in (9,8) eliminated by the uniqueness requirement (Heuristic 4) are highlighted in orange.) 3 is uniquely mentioned in (9,8) with respect to the 3x3 area at (7,7) if the only other mention in (7,7) can be eliminated. (3 in (7,7) is highlighted in orange.) But (7,7) cannot be 3 based on the original assumption that (3,7), a cell on the same column as (7,7), was 3 to begin with.

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