I see that Don has responded while I've been working on the (very) long post that follows.

He makes good points.

This is what I was working on.

I don't know if it's free of typos (and other errors) at this point, but I'll post it anyway ...

----------------

gurth wrote:(...) I have found the phobias about bifurcation, assumption, nets and T&E to be highly objectionable, irrational, idiosyncratic and, frankly, disgusting and ridiculous.

gurth wrote:Blue and Aran expose the difficulties arising from these stupid prejudices.

Such negativity,

gurth !

I'm shocked, and frankly I'm feeling a bit "nauseated".

The

absolute last thing I wanted to do with my post, was to ignite another Eureka-style firestorm.

I'm hoping, and indeed truly anticipating that after collecting his thoughts,

DonM will respond thoughtfully, respectfully, and with a full understanding that this can be an emotional topic for many people.

May I (respectfully and sincerely) suggest that your post would have been more effetive, if you had left the "stupid" out of "stupid prejudices", and replaced the first part with something along the lines of:

(...) I have found the misgivings about bifurcation, assumption, nets and T&E to be highly questionable.

---

aran wrote:blue

"assumptive"

It will be interesting to see what DonM may reply.

My sentiments exactly.

gurth wrote:Without "if" there can be no thought or logic in mathematics or in sudoku.

I'm pretty sure -- almost positive -- that I understand what you mean.

But ... and as a way, perhaps, of getting at what Don might mean by "less assumptive" ... let me cover three different situations.

1) A naked pair elimination: (NP: <12>r1c12) => -2r1c9

- One way of looking at it is this: If, in a solution to the puzzle, r1c9 was a '2', then neither of {r1c1,r1c2} could contain a '2'. In that case, the situation would be that the only value that could appear in {r1c1,r1c2}, would be '1'. If r1c1 contained a '1', there would be nothing left to place in r1c2, and vica-versa. Hence r1c9 cannot be a '2' in any solution to the puzzle.
- Another way of looking at it is this: In any solution to the puzzle, r1c1 must contain a '1' or a '2'. If it contained a '2', then r1c9 could not contain a '2'. The alternative to it containing a '2', would be for it to contain a '1', but in that case, r1c2 would need to contain a '2', and again r1c9 could not contain a '2'. Either way, r1c9 cannot contain a '2'.
- A third way of looking at it would be this: The only digits that can go in r1c1 and r2c2, are '1' and '2'. In any solution to the puzzle, one cell would contain a '1' and the, other a '2'. With that, r1c9 cannot contain a '2' in any solution to the puzle.

The three interpretations, seem to be ordered from "most" to "least" assumptive -- but I have no idea if that's what

DonM has in mind.

2) A skyscraper elimination. A "skyscraper" seems to sit right on the borderline between being a "chain elimination" and a "pattern elimination" -- some people see it one way, and some the other. For the sake of what follows, suppose that in the "chain view", it looks like this: 1r1c1 = 1r4c1 - 1r4c4 = 1r2c4 => -1r2c3. There are (at least) three way's of looking at the situation here, as well.

- One way of looking at it is this: If, in a solution to the puzzle, r2c3 contained a '1', then neither of {r1c1,r2c4} could contain a '1', and we would be faced with a situation where both r4c1 and r4c4 need to contain a '1'. According to the rules of Sudoku, though, they can't both contain a '1', and so it follows that r2c3 could not contain a '1' in any solution to the puzzle.
- Another way of looking at it is this: In c1, one of {r1c1,r4c1} must contain a '1' (in any solution to the puzzle). Ifr1c1 contained a '1', then r2c3 could not contain a '1'. The alternative is that r4c1 contains a '1', but that would mean that r4c4 cannot contain a '1', and since c4 must contain a '1', it would need to go in r2c4, and again r2c3 could not contain a '1'. In either case, r2c4 cannot contain a '1' in any solution to the puzzle.
- A third way of looking at it would be this: Since In any solution to the puzzle, r4c1 and r4c4 can't both be '1', and so one of {r4c1,r4c4} must be false. With that, and since 1's must appear in both columns, it follows that (in any solution to the puzzle) at least one of {r1c1,r2c4} would need to be a '1', and from that it follows that r2c3 cannot be a '1' in any solution to the puzzle.

In this case, the last option seems to be the "least assumptive" (in fact, "non-assumptive") ... and the first two, might seem to be "equally assumptive", since "

If" occurs only once in each. In the other hand, the first case amounts to a "trial and error" elimination, and the second case seems amount to an argument that "if r1c1 didn't contain a '1', then r2c4 would need to contain a '1'", and a statement of the "follow on" consequence.

3) A naked triple elimination, in which all three cells contain all three candidates: (NT: 123r1c123) => -1r1c9. Again, there are (at least) three different ways of looking at the situation:

- The first way, like it was above, would views the elimination as a "trial and error" elimination, where the argument is that in the end, 1r1c9 cannot be "true" in any solution to the puzzle, since if it was true, then (...). The rest of the argument, could be done in a few ways. Easiest, might be to argue that in that situation {r1c1,r1c2} would reduce to a naked pair ... "(NP: <23>r1c12)", which would make it impossible for r1c3 to contain a '2' or a '3', and to conclude that since (from the initial assumption) it also couldn't contain a '1', and no digit could be placed there, it follows that r1c9 cannot contain a '1' (in any solution to the puzzle).
- A second way, would be: First, note that in any solution to the puzzle, r1c1 would need to contain a '1', '2' or '3'. If it contained a '1' then r1c9 could not contain a '1'. (And) if it contained a '2', then r1c23 would be reduced to a naked pair, <13>r1c23, and again r1c9 could not contain a '1'. (And finally) the remaining alternative, would be for r1c1 to contain a '3', in which case r1c23 would be reduced to the naked pair, <12>r1c23, and still r1c9 could not contain a '1'. Since all three options lead to it being impossible for r1c9 to contain a '1', r1c9 cannot be a '1' in any solution to the puzzle.
- The third way, would be similar to the third way in the naked pair case: {r1c1,r1c2,r1c3} can only contain digits '1', '2' and '3', and in a solution to the puzzle, each number would need to appear in exactly one of those cells. Since that means that a '1' would need to appear in at least one of the cells, it follows that r1c9 cannot contain a '1' in any solution to the puzzle.

In this case, the 3rd option seems to be "least assumptive", and again "non-assumptive" in that the word "if", doesn't appear in the argument. The first two options include "if". In the first, it's only used once, and in the 2nd, it's used twice. On the other hand, the first two arguments appeal to known facts (read "additional arguments") regarding naked pairs ... and if you covered those in all of thier detail, I don't know but it seems like the 2nd argument would have more occurences of the word "if". On the other hand, the first argument, is a "T&E" argument, and many people prefer to avoid such arguments. Would

DonM think that the first one is "more assumptive" than the other, regardless of how many "if's" occur ? I really don't know.

Cheers,

Blue.