||Wed 2014-03-12 14:14|
|For s(4,9) my method gives a bound of s<=9/5|
Hmm, does it? I tried to boil down your method into something I could state in a general way, and then applied that, and for s(4,9) I got what I think is actually the right bound.
If every piece is at least s and at most m-s, then every stick of length n is cut into at least n/(m-s) and at most n/s pieces, hence at least ⌈n/(m-s)⌉ and at most ⌊n/s⌋, hence the total number of pieces P in the dissection must satisfy m⌈n/(m-s)⌉ ≤ P ≤ m⌊n/s⌋. By similarly considering the number of pieces that a stick of length m is cut into, we derive a second inequality n⌈m/(m-s)⌉ ≤ P ≤ n⌊m/s⌋ (identical to the first except that n,m are swapped everywhere they appear except in the divisor m-s). So when s becomes big enough that either of those inequalities is self-inconsistent (with LHS > RHS) or when the two are incompatible (because they define ranges for P that do not meet), there can be no dissection with s that large.
It is true that s=9/5 is a threshold beyond which something goes wrong, namely that the first inequality for P is self-inconsistent (with s > 9/5 we would need 20 ≤ P ≤ 16). But something else has gone wrong before that – at s=7/4 the two intervals stop overlapping (because the first inequality switches from 16 ≤ P ≤ 20 to 20 ≤ P ≤ 20, while the second was at 18 ≤ P ≤ 18 all along). And the computer search data I have says that 7/4 really is the best solution.
I agree with you, on the other hand, that the method doesn't get the exact bound for s(7,8), and hence is not reliable in all circumstances.