We present a deterministic exact algorithm for the problem $P_m\mid\mathrm{prec},p_j=1\mid C_{\max}$ with a fixed number of machines $m\ge 3$. For every constant $m$, the algorithm constructs an optimal schedule in polynomial time. Its running time is
$$ O(L) + m^{O(m^2)}\, n^{6m^2+13m+1}\, \log n $$
where $n\ge 2$ is the number of jobs, $L$ is the binary input length, and the word size is $O(m^2 \log n)$ bits. Using a fixed‑tape implementation, the number of steps becomes
$$ O(L) + m^{O(m^2)}\, n^{12m^2+26m+2}\, \log^3 n \,. $$
For three machines ($m=3$) the polynomial degrees are $94$ and $188$ respectively (ignoring logarithmic factors). This shows that, even with unit processing times and arbitrary precedence constraints, an optimal schedule can be found deterministically when the number of machines is constant.
Key techniques:
- Transform the scheduling problem into a finite‑state search, applying dynamic programming with aggressive pruning to keep the state space manageable.
- Perform a careful combinatorial counting argument to bound the number of states by $m^{O(m^2)}\, n^{6m^2+13m+1}$, yielding the overall time bound.
- Use the standard RAM model to count elementary operations, guaranteeing predictable resource usage in practice.
Implementation notes:
- Input is encoded in binary; its length $L$ captures the entire precedence graph.
- A word size of $O(m^2 \log n)$ suffices to store all intermediate counters and pointers without overflow.
- On a multi‑tape Turing machine, only $O(\log n)$ extra space is needed per step.
Significance: This algorithm closes the gap for exact deterministic solutions of precedence‑constrained scheduling with a fixed number of machines, providing a foundation for tackling more general scheduling models.
Review: The paper demonstrates that meticulous state compression and complexity analysis can yield deterministic polynomial‑time algorithms for otherwise hard combinatorial scheduling problems, marking a notable advance in theoretical scheduling research.