Lay a graph over the map: nodes are places the aircraft can be (waypoints or grid-cell centers), edges are legs it can fly, and each edge carries a non-negative weight — what that leg costs. A route's cost is the sum of its legs' weights, so the best route is the cheapest path from ingress to target.
Settle the frontier node with the smallest label \(g(u)\) — that label is now final — then relax every edge out of it: \(g(v) \leftarrow \min\big(g(v),\ g(u) + c_{uv}\big)\). Repeat until the target settles, then walk the predecessors back to read the route.
Settling the smallest label is safe only because every other way into that node starts at a label at least as large and adds weights that cannot be negative. A negative edge could undercut a node after it was settled, and the algorithm would return a wrong answer.
A tentative label is the cheapest cost-to-come found so far, and relaxation can still lower it when a cheaper way in turns up. A settled label is final. In the lesson's worked example, B is first reached at 5 directly and then drops to 4 through A before it settles.
graph and when digraph in MATLAB?graph(s, t, w) stores one weight per node pair, usable both ways — right when a leg costs the same in each direction. digraph stores one-way edges, so a leg that costs something different each way is two edges with two weights. shortestpath(G, s, t) runs Dijkstra on either.
Distance only finds the shortest route — straight through whatever threat is in the way. Exposure only has no reason to stop wandering: every unexposed leg is free, so any zero-exposure dogleg ties for best however long it is, and the planner can return a route no aircraft has the fuel to fly.
In \(c_e = w_d\,\ell_e + w_r\,r_e\) only the ratio matters, so set \(w_d = 1\): \(w\) is the number of km of extra flying you will accept to cut the exposure \(\sum r\) by one unit. At \(w = 200\) km the planner flies 2 km farther to cut the exposure by 0.01.
Survival multiplies along a route, \(P_S = \prod(1-p_e)\), but Dijkstra adds. Taking \(-\ln\) turns the product into a sum, \(-\ln P_S = \sum -\ln(1-p_e)\), and because the log is monotone the minimum-sum route is exactly the maximum-\(P_S\) route.
\(p \approx -\ln(1-p)\) only while \(p\) is small. With one heavy leg against several light ones the raw sum can pick the wrong route: one leg at 0.40 sums lower than two at 0.22 (0.40 against 0.44), yet its survival is 0.600 against 0.608.
A site takes a shot every \(\Delta s\) km, each killing with \(p_\text{shot}\), so its hazard is \(\lambda = -\ln(1-p_\text{shot})/\Delta s\) per km. An edge that spends \(\ell_{i,e}\) km inside site \(i\)'s lethal zone (detection range and MEZ both) carries \(r_e = \sum_i \lambda_i \ell_{i,e}\), and the route survives with \(P_S = e^{-\sum r_e}\).
Run Dijkstra once for each weight \(w\) in a sweep and plot each route's survival against its length. The front is the set of routes no other route beats on both at once; it shows what each increment of \(P_S\) costs in kilometers.
Apply it to the sweep's answers: fly the highest-\(P_S\) route whose length fits under \(L_\text{max}\), or call the mission infeasible if even the shortest route does not fit. A length budget inside Dijkstra would need a second label per node (length used), which breaks its one-label exactness — the hard resource-constrained shortest-path problem.
Cell centers become nodes, each linked to 8 neighbors by 10 km legs and \(10\sqrt{2} = 14.1\) km diagonals, so any map becomes a graph. The caveats: headings only every 45°, routes that zig-zag and run up to about 8% long, rings that snap to cells, and many equal-length ties.
A leg's heading sets the aspect each radar sees, \(\theta = \phi + 180^\circ - \psi\). Flown toward a radar the leg shows the nose; flown away it shows the tail. The RCS differs, so \(R_\text{det} \propto \sigma^{1/4}\), the lethal length and the risk differ, and one node pair carries two costs — a digraph.
With the worst-case \(\sigma\) it is always pessimistic: every ring is at its largest and it pays length to avoid threats it could have slipped past. With the nose-on \(\sigma\) it is optimistic wherever the aircraft is louder than its nose (it flies past radars beam-on) and pessimistic wherever it is quieter. Re-score the route with the real pattern.
Dijkstra is exact for the cost it is given, and that cost is a model's promise — a \(\sigma\), hazards, rings and a grid. In the demo, a nose-on plan promises \(P_S = 1.000\) through a gap that the aspect-aware model re-scores at 0.077. Re-score every plan under a better model than the one that chose it.