A shortest route and a changed edge
Choose your method, connect the parts and explain your conclusions. These original questions include written reasoning and sketches as well as exact answer checks.
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Before you start
Shortest paths with Dijkstra’s algorithm. Read the relevant method, then decide which parts you can solve without hints.
Check a misconception first · Try a connected problem
Read an original example question
An undirected network has positive edge weights. Find a route from A to F.
| Edge | Weight |
|---|---|
| AB | 2 |
| AC | 5 |
| BC | 1 |
| BD | 4 |
| CD | 1 |
| CE | 4 |
| DE | 1 |
| DF | 4 |
| EF | 1 |
Part a
Use Dijkstra’s algorithm to find the shortest distance.
Part b
State a shortest route and show your labelled-node working.
Part c
The edge BC now has weight 5; all others are unchanged. Find the new shortest distance.
Part d
Explain why Dijkstra’s usual settled-node rule depends on nonnegative weights.
Model solution and review criteria
Part a
Permanent distances are A:0, B:2, C:3, D:4, E:5, F:6. Record tentative updates and predecessors as each node is settled.
- Start at A and update adjacent tentative distances.
- Settle the least tentative node each time.
- Record predecessors as well as lengths.
Part b
A–B–C–D–E–F follows the predecessor chain. A route length alone does not demonstrate the algorithm.
- Give the full route.
- Show settled and tentative labels in a table.
Part c
A–C–D–E–F has length 8; A–B–D–E–F also has length 8. Recompute labels: the old route is no longer shortest.
- Recompute rather than changing only the old route length.
- Accept either tied shortest route in the explanation.
Part d
A later path through a negative edge could improve a supposedly final label. Positive weights justify the settled-node guarantee; negative-edge graphs need a suitable different method.
- Explain the guarantee that would fail.
Compare your own reasoning. No automatic examination marks are awarded.