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Section A: Long Answer Questions

Attempt any TWO questions.

3 questions·10 marks each
1long10 marks

What are rules of inference? Show that the premises (p \rightarrow q), (q \rightarrow r) and (p) lead to the conclusion r using rules of inference.

Rules of Inference

Rules of inference are valid argument forms (templates) used to derive new true statements (conclusions) from a set of given true statements (premises). They form the basis of formal proofs in propositional logic. Common rules include:

RuleForm
Modus Ponensp,  pq    qp,\; p \rightarrow q \;\vdash\; q
Modus Tollens¬q,  pq    ¬p\neg q,\; p \rightarrow q \;\vdash\; \neg p
Hypothetical Syllogismpq,  qr    prp \rightarrow q,\; q \rightarrow r \;\vdash\; p \rightarrow r
Disjunctive Syllogismpq,  ¬p    qp \lor q,\; \neg p \;\vdash\; q
Additionp    pqp \;\vdash\; p \lor q
Simplificationpq    pp \land q \;\vdash\; p

Proof that pqp \rightarrow q, qrq \rightarrow r, pp lead to rr

StepStatementReason
1pqp \rightarrow qPremise
2qrq \rightarrow rPremise
3ppPremise
4qqModus Ponens on (1) and (3)
5rrModus Ponens on (2) and (4)

Hence the conclusion rr is derived. Alternatively, applying Hypothetical Syllogism to (1) and (2) gives prp \rightarrow r, and then Modus Ponens with premise pp again yields rr. \blacksquare

inferencelogic
2long10 marks

Define recurrence relation. Solve the recurrence relation (a_n = 5a_{n-1} - 6a_{n-2}) with initial conditions (a_0 = 1) and (a_1 = 0).

Recurrence Relation

A recurrence relation for a sequence {an}\{a_n\} is an equation that expresses each term ana_n in terms of one or more of the preceding terms an1,an2,a_{n-1}, a_{n-2}, \dots, for all nn greater than some fixed value, together with initial conditions.

Solving an=5an16an2a_n = 5a_{n-1} - 6a_{n-2}, a0=1a_0 = 1, a1=0a_1 = 0

Step 1 — Characteristic equation. Assume a solution an=xna_n = x^n:

x2=5x6    x25x+6=0.x^2 = 5x - 6 \;\Rightarrow\; x^2 - 5x + 6 = 0.

Step 2 — Find roots.

(x2)(x3)=0    x=2,  x=3.(x-2)(x-3) = 0 \;\Rightarrow\; x = 2,\; x = 3.

Step 3 — General solution (distinct roots):

an=A2n+B3n.a_n = A\cdot 2^n + B\cdot 3^n.

Step 4 — Apply initial conditions.

  • a0=1:A+B=1a_0 = 1:\quad A + B = 1
  • a1=0:2A+3B=0a_1 = 0:\quad 2A + 3B = 0

From the first, A=1BA = 1 - B. Substituting: 2(1B)+3B=02+B=0B=22(1-B) + 3B = 0 \Rightarrow 2 + B = 0 \Rightarrow B = -2, so A=3A = 3.

Step 5 — Final solution.

an=32n23n\boxed{a_n = 3\cdot 2^n - 2\cdot 3^n}

Check: a0=32=1a_0 = 3 - 2 = 1 ✓, a1=66=0a_1 = 6 - 6 = 0 ✓.

recurrencesolving
3long10 marks

Define tree and spanning tree. Explain Kruskal's algorithm to find a minimum spanning tree with a suitable example.

Tree

A tree is a connected, undirected graph with no simple circuits (cycles). Equivalently, a tree on nn vertices is a connected graph with exactly n1n-1 edges, in which there is a unique simple path between every pair of vertices.

Spanning Tree

A spanning tree of a connected graph GG is a subgraph that is a tree and contains all the vertices of GG. A minimum spanning tree (MST) of a weighted graph is a spanning tree whose sum of edge weights is the minimum possible.

Kruskal's Algorithm

Kruskal's algorithm builds an MST by repeatedly adding the smallest available edge that does not form a cycle.

Kruskal(G):
  Sort all edges in non-decreasing order of weight
  T = empty set (the MST)
  for each edge (u, v) in sorted order:
      if adding (u, v) to T does NOT form a cycle:
          add (u, v) to T          // use Union-Find to test
  until T has (n - 1) edges
  return T

Example

Consider a graph with vertices {A,B,C,D}\{A, B, C, D\} and weighted edges:

EdgeWeight
A–B1
B–C2
A–C3
C–D4
B–D5

Sorted edges: AB(1), BC(2), AC(3), CD(4), BD(5).

  1. Add A–B (1) → no cycle. ✓
  2. Add B–C (2) → no cycle. ✓
  3. Consider A–C (3) → forms cycle A–B–C–A. ✗ Skip.
  4. Add C–D (4) → no cycle. ✓ Now we have n1=3n-1 = 3 edges. Stop.

MST edges: {A–B, B–C, C–D} with total weight 1+2+4=71 + 2 + 4 = 7.

treespanning-tree
B

Section B: Short Answer Questions

Attempt any EIGHT questions.

9 questions·5 marks each
4short5 marks

Show that (p \rightarrow q) is logically equivalent to (\neg p \lor q).

We show pq¬pqp \rightarrow q \equiv \neg p \lor q using a truth table. Two propositions are logically equivalent if they have identical truth values for every assignment.

ppqq¬p\neg ppqp \rightarrow q¬pq\neg p \lor q
TTFTT
TFFFF
FTTTT
FFTTT

The columns for pqp \rightarrow q and ¬pq\neg p \lor q are identical in every row. Therefore:

pq¬pq.p \rightarrow q \equiv \neg p \lor q. \qquad \blacksquare
logic
5short5 marks

Define power set. Write the power set of {a, b, c}.

Power Set

The power set of a set SS, denoted P(S)P(S) or 2S2^S, is the set of all subsets of SS, including the empty set \emptyset and SS itself. If S=n|S| = n, then P(S)=2n|P(S)| = 2^n.

Power set of {a,b,c}\{a, b, c\}

Since n=3n = 3, there are 23=82^3 = 8 subsets:

P({a,b,c})={  ,  {a},  {b},  {c},  {a,b},  {a,c},  {b,c},  {a,b,c}  }P(\{a,b,c\}) = \{\;\emptyset,\; \{a\},\; \{b\},\; \{c\},\; \{a,b\},\; \{a,c\},\; \{b,c\},\; \{a,b,c\}\;\}
sets
6short5 marks

What is a partial order relation? Give an example.

Partial Order Relation

A relation RR on a set AA is a partial order (and (A,R)(A, R) is a poset) if it is:

  1. Reflexive: aRaa\,R\,a for all aAa \in A.
  2. Antisymmetric: if aRba\,R\,b and bRab\,R\,a, then a=ba = b.
  3. Transitive: if aRba\,R\,b and bRcb\,R\,c, then aRca\,R\,c.

Example

The "less than or equal to" relation \le on the set of integers Z\mathbb{Z} is a partial order, because:

  • aaa \le a (reflexive),
  • if aba \le b and bab \le a then a=ba = b (antisymmetric),
  • if aba \le b and bcb \le c then aca \le c (transitive).

Another common example is the divisibility relation aba \mid b on the positive integers, or the subset relation \subseteq on the power set of a set.

relations
7short5 marks

State the pigeonhole principle with an example.

Pigeonhole Principle

Statement: If nn objects (pigeons) are placed into kk boxes (pigeonholes) and n>kn > k, then at least one box contains two or more objects.

Generalized form: If nn objects are placed into kk boxes, then at least one box contains at least nk\left\lceil \dfrac{n}{k} \right\rceil objects.

Example

Among any 13 people, at least two must have been born in the same month. Here the 13 people are the objects and the 12 months are the boxes; since 13>1213 > 12, by the pigeonhole principle at least two people share a birth month.

counting
8short5 marks

Define complete graph and bipartite graph.

Complete Graph

A complete graph KnK_n is a simple undirected graph on nn vertices in which every pair of distinct vertices is joined by exactly one edge. It has n(n1)2\dfrac{n(n-1)}{2} edges, and every vertex has degree n1n-1. For example, K4K_4 has 4 vertices and 6 edges.

Bipartite Graph

A graph G=(V,E)G = (V, E) is bipartite if its vertex set VV can be partitioned into two disjoint sets V1V_1 and V2V_2 such that every edge connects a vertex in V1V_1 to a vertex in V2V_2 (no edge lies within V1V_1 or within V2V_2). A graph is bipartite if and only if it contains no odd-length cycle. The complete bipartite graph Km,nK_{m,n} joins every vertex of V1V_1 (size mm) to every vertex of V2V_2 (size nn).

graph
9short5 marks

Find the number of edges in a complete graph with 6 vertices.

The number of edges in a complete graph KnK_n is

(n2)=n(n1)2.\binom{n}{2} = \frac{n(n-1)}{2}.

For n=6n = 6:

6×52=302=15.\frac{6 \times 5}{2} = \frac{30}{2} = 15.

Therefore, K6K_6 has 15 edges (and each of the 6 vertices has degree 5).

graph
10short5 marks

Define a tree and state its basic properties.

Tree

A tree is a connected, undirected graph that contains no simple circuits (cycles). Equivalently, it is a connected graph in which any two vertices are joined by a unique simple path.

Basic Properties

For a tree with nn vertices:

  1. It has exactly n1n - 1 edges.
  2. It is connected, but removing any single edge disconnects it (every edge is a bridge).
  3. It is acyclic; adding any new edge between two existing vertices creates exactly one cycle.
  4. There is a unique simple path between every pair of vertices.
  5. A tree with n2n \ge 2 vertices has at least two leaves (vertices of degree 1).
  6. A connected graph is a tree if and only if it has n1n-1 edges and is acyclic (any two of: connected, acyclic, n1n-1 edges imply the third).
tree
11short5 marks

What is a tautology and a contradiction? Give one example of each.

Tautology

A tautology is a compound proposition that is always true, regardless of the truth values of its component propositions.

Example: p¬pp \lor \neg p (the law of excluded middle).

pp¬p\neg pp¬pp \lor \neg p
TFT
FTT

The result is always T.

Contradiction

A contradiction is a compound proposition that is always false, regardless of the truth values of its components.

Example: p¬pp \land \neg p.

pp¬p\neg pp¬pp \land \neg p
TFF
FTF

The result is always F.

(A proposition that is neither always true nor always false is called a contingency.)

logic
12short5 marks

Define the composition of two functions with an example.

Composition of Functions

If f:ABf : A \rightarrow B and g:BCg : B \rightarrow C are functions, their composition gf:ACg \circ f : A \rightarrow C is defined by

(gf)(x)=g(f(x)),for all xA.(g \circ f)(x) = g\big(f(x)\big), \quad \text{for all } x \in A.

That is, ff is applied first, then gg is applied to the result. (Note gfg \circ f is generally not the same as fgf \circ g.)

Example

Let f(x)=x+2f(x) = x + 2 and g(x)=x2g(x) = x^2, both functions on R\mathbb{R}.

  • (gf)(x)=g(f(x))=g(x+2)=(x+2)2=x2+4x+4.(g \circ f)(x) = g(f(x)) = g(x+2) = (x+2)^2 = x^2 + 4x + 4.
  • (fg)(x)=f(g(x))=f(x2)=x2+2.(f \circ g)(x) = f(g(x)) = f(x^2) = x^2 + 2.

For instance, (gf)(3)=(3+2)2=25(g \circ f)(3) = (3+2)^2 = 25, whereas (fg)(3)=32+2=11(f \circ g)(3) = 3^2 + 2 = 11, showing the two compositions differ.

functions

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