ImplementCore implementation

Expert choice: capacity-constrained token routing

Implement `run(affinities, e, n, c)` returning the number of tokens k each expert selects in expert-choice routing, where each of the e experts picks its top-k tokens and k = ceil((n * c) / e). Round k UP to the nearest integer. n is the number of tokens, c the capacity factor (default 2).

python10 mindifficulty 2/4

From Mixture-of-Experts with Expert Choice Routing

Sample tests

run([[[0.1,0.9]],4,16,2])→8
run([[[0.5,0.5]],8,32,2])→8

Hints

Expert choice flips token choice: each expert independently selects k tokens instead of each token picking experts.
k = (n * c) / e with capacity factor c — the paper uses c = 2 to match GShard top-2's activation budget.
Use ceil so a non-integer ratio still fills every expert to full capacity.
Return k as a Python int; when c * n divides evenly by e, k is exact.

Hints

Code runs in an isolated sandbox on the OpenSandbox judge. No model key required.