On 8/9/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
... Extraordinarily long impasses are possible: the above's length-31 prefix 1121231123 1121231121 2311212311 2 may be followed by the foreign length-85 suffix 3121123121 1231213121 2131212131 2132112132 1121321132 1211321211 3212131211 3212113212 13121 yielding a non-billiard length-116 ternary word, all of whose projections are billiard words, but which cannot be further extended without invalidating some projection.
Claim unjustified, I'm afraid --- the actual back-up required there is much shorter than implied above. A more well-organised expedition to establish the shortest impasses prefixed by the dud 1121231123 uncovers length-26, 26, 29: 1121231123 1213121321 211321, 1121231123 1213211321 211321, 1121231123 1213211321 213211321. So while a (nontrivial) "pseudo-billiard" word --- one whose projections are all billiard words --- has in general some infinite pseudo-billiard extension, it is also conceivable that it has (always) another which terminates at an impasse within some boundable length. This would provide a combinatorial nec. & suff. criterion for a billiard word, independent of inequality analysis --- though the example above suggests that the bound may not be terrribly practical. [An efficient combinatorial criterion based on morphism inversion, and related to the Euclidean HCF algorithm, already exists when m = 2; it leads further to the explicit formula for f_2(n).] An alternative possibility: for given m, does there exist a finite set of words (like the length-10 dud above), such that every pseudo-billiard word has some factor in this set? Fred Lunnon