ACompressed recap
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MST: V − 1 edges; Kₙ has nⁿ⁻² spanning trees; cut property (lightest crossing edge safe), cycle property (strictly heaviest on a cycle excluded); distinct weights ⇒ unique MST; constant shift keeps MSTs. Kruskal: sort edges, union–find, O(E log E). Prim: one tree, cheapest edge out, Θ(V²) array or O(E log V) heap.
BMerged rule card
6.2.1 Minimum spanning trees: cut and cycle properties, uniqueness
- Spanning tree: V − 1 edges; Kₙ has nⁿ⁻² spanning trees.
- Cut property: lightest crossing edge is safe (unique ⇒ in every MST).
- Cycle property: strictly heaviest on a cycle ⇒ in no MST.
- Distinct weights ⇒ unique MST. Constant shift keeps MSTs, not shortest paths.
6.2.2 Kruskal's algorithm and union–find
- Kruskal: sort edges; add (u, v) if find(u) ≠ find(v); stop at V − 1 edges.
- O(E log E) = O(E log V); the partial result is a forest.
- Union by rank: height ≤ log n; plus path compression: ~O(α(n)) per operation.
6.2.3 Prim's algorithm; MST complexities
- Prim: grow one tree; add the outside vertex with the smallest key (cheapest edge in).
- Θ(V²) array (dense graphs); O(E log V) binary heap; O(E + V log V) Fibonacci heap.
- Prim one tree; Kruskal a forest; both via the cut property.
CInterleaved GATE-style questions
Q1NAT1 markMST
Number of spanning trees of K₄?
Numerical answer: ________
Q2MCQ1 markCycle property
In the six-town graph, which edge is strictly the heaviest on the cycle C–D–E?
- (A) none
- (B) C–D
- (C) C–E
- (D) D–E
Q3NAT2 marksKruskal
Kruskal on the six-town graph: what is the weight of the fifth (last) edge accepted?
Numerical answer: ________
Q4NAT2 marksMST count
A 4-cycle a–b–c–d–a with every edge of weight 1. How many MSTs?
Numerical answer: ________
Q5MCQ1 markKruskal
Kruskal's algorithm tests whether an edge creates a cycle using:
- (A) a union–find structure
- (B) DFS finish times
- (C) a max-heap
- (D) BFS from scratch every time
Q6NAT2 marksPrim
Prim from C on the six-town graph: weight of the second edge added?
Numerical answer: ________
Q7MCQ1 markPrim cost
For a sparse graph with E = Θ(V), the fastest of these Prim versions is:
- (A) neither works
- (B) both equal
- (C) array, Θ(V²)
- (D) binary heap, O(E log V) = O(V log V)
Q8MSQ2 marksMST facts
Which are true?
- (A) An MST is a shortest-path tree from every vertex.
- (B) Squaring all (positive) edge weights does not change the MSTs.
- (C) A graph with repeated weights may still have a unique MST.
- (D) An MST has V − 1 edges.
DAnswer Vault
Answer Vault
Try the questions first. Open when you are ready to check.
Q1 Answer: 161 mark
4².
Re-read: Lesson 6.2.1 — Minimum spanning trees: cut and cycle properties, uniqueness
Q2 Answer: (C)1 mark
Weights 8, 2, 10.
- (A) is wrong: C–E is strictly heaviest.
- (B) is wrong: 8 < 10.
- (D) is wrong: Lightest.
Re-read: Lesson 6.2.1 — Minimum spanning trees: cut and cycle properties, uniqueness
Q3 Answer: 52 marks
B–D.
Re-read: Lesson 6.2.2 — Kruskal's algorithm and union–find
Q4 Answer: 42 marks
Drop any one of the 4 edges.
Re-read: Lesson 6.2.2 — Kruskal's algorithm and union–find
Q5 Answer: (A)1 mark
find(u) = find(v) means a cycle.
- (B) is wrong: Not used.
- (C) is wrong: Not for cycle tests.
- (D) is wrong: Works but far slower.
Re-read: Lesson 6.2.2 — Kruskal's algorithm and union–find
Q6 Answer: 22 marks
C–B 1 first; from the tree {C, B} the cheapest leaving edge is C–A 2.
Re-read: Lesson 6.2.3 — Prim's algorithm; MST complexities
Q7 Answer: (D)1 mark
V log V beats V².
- (A) is wrong: Both work.
- (B) is wrong: No.
- (C) is wrong: Slower for sparse graphs.
Re-read: Lesson 6.2.3 — Prim's algorithm; MST complexities
Q8 Answer: (B) (C) (D)2 marks
MSTs and shortest-path trees differ in general.
- (A) is wrong: Triangle counterexample.
Re-read: Lesson 6.2.1 — Minimum spanning trees: cut and cycle properties, uniqueness