---
title: "Dijkstra shortest path (2073 Magh)"
subject: "Discrete Structure"
chapter: 4
key: "semester_4/discrete_structure:dijkstra-2073-magh"
source: https://www.bctengineer.com/q/semester_4/discrete_structure/dijkstra-2073-magh
site: BCT Engineering
---

# Dijkstra shortest path (2073 Magh)

Discrete Structure · Chapter 4: Graph Theory

- Asked in: 2073 Magh, 6+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/discrete_structure/dijkstra-2073-magh
- Whole chapter: https://www.bctengineer.com/md/semester_4/discrete_structure/chapter-4

## Question

Find the shortest path from vertex a to vertex Z. Highlight the shortest paths in the graph. [Figure: weighted graph on vertices a, b, c, d, e, Z with edges a-b 4, a-c 1, a-d 4, b-c 3, b-Z 3, c-d 2, c-e 1 [?], d-e 2, e-Z 1 [exact placement of some weights unclear in scan]]

## Short answer

Using Dijkstra's algorithm, the shortest path from a to Z is a–c–e–Z with total length 3.

## Answer

Dijkstra's algorithm (non-negative weights) labels every vertex with the best distance known so far and fixes one vertex at a time:
1. Give the source distance 0 and every other vertex ∞. All vertices are temporary.
2. Make the temporary vertex with the smallest label permanent.
3. 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.
4. Repeat steps 2 and 3 until the target (or all vertices) is permanent.

#### Graph as read
Reading: the weight of edge c–e is taken as 1 (marked unclear in the scan); the other weights are as listed. Source a, target Z.

Edges: a–b 4, a–c 1, a–d 4, b–c 3, b–Z 3, c–d 2, c–e 1, d–e 2, e–Z 1.

#### 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** | 4(a) | 1(a) | 4(a) | ∞ | ∞ |
| 2 | c | ✓ | 4(a) | **1**(a) | 3(c) | 2(c) | ∞ |
| 3 | e | ✓ | 4(a) | ✓ | 3(c) | **2**(c) | 3(e) |
| 4 | d | ✓ | 4(a) | ✓ | **3**(c) | ✓ | 3(e) |
| 5 | Z | ✓ | 4(a) | ✓ | ✓ | ✓ | **3**(e) |
| 6 | b | ✓ | **4**(a) | ✓ | ✓ | ✓ | ✓ |

#### Result
Shortest distance from a to Z = **3**.

Shortest path (highlight these edges): **a → c → e → Z** with edge weights 1 + 1 + 1 = 3.

**Answer:** shortest path a to Z = a → c → e → Z, length 1 + 1 + 1 = 3. Other routes are longer (a–c–d–e–Z = 6, a–b–Z = 7).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. At each step Dijkstra's algorithm makes permanent:
   A. The vertex with the most neighbours
   B. The vertex with fewest edges to the target
   C. The temporary vertex with the smallest label
   D. A vertex chosen at random
   Answer: C. The smallest temporary label is final when all weights are non-negative. (Key point: Dijkstra step)

2. A neighbour v of the newly fixed vertex u gets a new label when:
   A. w(u,v) < d(u)
   B. d(v) is already permanent
   C. d(u) > d(v)
   D. d(u) + w(u,v) < d(v)
   Answer: D. The label is replaced only if the route through u is shorter. (Key point: Relaxation)

3. Shortest distance from a to Z:
   A. 3, along a-c-e-Z
   B. 4, along a-b-Z
   C. 6
   D. 7
   Answer: A. 1 + 1 + 1 = 3; a-b-Z costs 7. (Key point: Result)

4. After c is fixed (distance 1), the label of d becomes:
   A. 4, direct from a
   B. 3, from 1 + 2 via c
   C. 2
   D. 1
   Answer: B. The route through c (3) beats the direct edge a-d (4). (Key point: Improvement)

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