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Both conventions are valid. You call it binary when you view it as a rooted tree, or ternary if you view it just as a graph.

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nyrikki 5 days ago | parent [-]

But it _all_ triples?

> I sketch how the stereographic projection of the Stern–Brocot tree forms an ordered binary tree of Pythagorean triples, which can be used to compute best approximations of turn angles of points on the circle and finally trigonometric functions

The permutation and stack problem in the page seem to indicate this is a potential method for approximations, but insufficient for _all_

That said I am reading this on mobile and may have missed something.

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not2b 5 days ago | parent [-]

The ternary tree contains all primitive triples (where the GCD of the terms is 1), where a<b<c. So it contains (3,4,5) but not (6,8,10) or (4,3,5).

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nyrikki 5 days ago | parent [-]

Yes, but the binary projection does not according to the link.

345 and 435 would require two binary trees.

	▲	AnotherGoodName 5 days ago \| parent \| next [-]
		I think skipping transposed values is fine though. You could just mirror the output at 45degrees for that if you wanted it. It does hit all distinct triples including the multiples of triples so it’s more inclusive of everything than the ternary tree.
	▲	hakmem 5 days ago \| parent \| prev [-]
		You can see both triples are contained in one binary tree using the big diagram in section 3. The triple [3 4 5] has the "path" RR. The triple [4 3 5] the path R.