- 2074 Bhadra · 8 marks
Dijkstra shortest path (2074 Bhadra)
Short answer
Dijkstra's algorithm yields a shortest path length of 17 from A to Z. The path is A–D–Z with weights 12 and 5.
Use Dijkstra's algorithm to find the length of shortest path from vertex A to vertex Z in the following weighted graph. Also highlight the shortest path/paths in the graph. [Figure: weighted graph on vertices A, B, C, D, E, Z with edges A-B 6, A-C 8, B-C 6, B-D 7, B-E 5, C-D 10, C-E 9, D-E 5, D-Z 5, E-Z 8, an outer arc of weight 12 joining A to D and an outer arc of weight 11 joining C to Z [some edge placements unclear in scan]]
Dijkstra's algorithm (non-negative weights) labels every vertex with the best distance known so far and fixes one vertex at a time:
- Give the source distance 0 and every other vertex ∞. All vertices are temporary.
- Make the temporary vertex with the smallest label permanent.
- For each temporary neighbour v of the new permanent vertex u: if d(u) + w(u,v) < d(v), set d(v) = d(u) + w(u,v) and record u as predecessor.
- Repeat steps 2 and 3 until the target (or all vertices) is permanent.
Graph as read
Reading: the outer arc A–D has weight 12 and the outer arc C–Z has weight 11; the other eight edges are as listed.
Edges: A–B 6, A–C 8, B–C 6, B–D 7, B–E 5, C–D 10, C–E 9, D–E 5, D–Z 5, E–Z 8, A–D 12, C–Z 11.
Trace
Each cell shows label = distance(predecessor). Bold marks the vertex made permanent in that step; ✓ means already permanent.
| Step | Fixed | A | B | C | D | E | Z |
|---|---|---|---|---|---|---|---|
| 1 | A | 0 | 6(A) | 8(A) | 12(A) | ∞ | ∞ |
| 2 | B | ✓ | 6(A) | 8(A) | 12(A) | 11(B) | ∞ |
| 3 | C | ✓ | ✓ | 8(A) | 12(A) | 11(B) | 19(C) |
| 4 | E | ✓ | ✓ | ✓ | 12(A) | 11(B) | 19(C) |
| 5 | D | ✓ | ✓ | ✓ | 12(A) | ✓ | 17(D) |
| 6 | Z | ✓ | ✓ | ✓ | ✓ | ✓ | 17(D) |
Result
Shortest distance from A to Z = 17.
Shortest path (highlight these edges): A → D → Z with edge weights 12 + 5 = 17.
Answer: length of the shortest path A to Z = 17, along A–D–Z (the direct edge A–D = 12, then D–Z = 5). The route A–B–E–Z costs 19 and A–C–Z costs 19.
✓ Answer checked by Claude Opus 5.5 (high effort)
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