DAA - Unit 3 | ankushraj.dev

🗺️Chapter 1 — Graph Algorithms

🗺️ GRAPH ALGORITHMS — MIND MAP

Representation → Adjacency Matrix O(V²) space | Adjacency List O(V+E) space

DFS → Stack/Recursion | O(V+E) | Explores depth-first | Applications: cycle detection, topo sort

BFS → Queue | O(V+E) | Explores level-by-level | Applications: shortest path (unweighted), level order

Topological Sort → DFS-based or Kahn's (in-degree) | Only for DAGs (Directed Acyclic Graphs)

Dijkstra → Greedy | Single Source Shortest Path | No negative weights | O(V²) or O(E log V)

Bellman-Ford → DP | Single Source | Handles negative weights | Detects negative cycles | O(VE)

Floyd-Warshall → DP | All-Pairs Shortest Path | O(V³) | D[i][j] = min(D[i][j], D[i][k]+D[k][j])

Sollin's (Borůvka's) → MST | Each component picks cheapest outgoing edge | O(E log V)

1. Representation of Graphs

Graph G = (V, E) where V = set of vertices, E = set of edges.
Types: Directed/Undirected, Weighted/Unweighted, Cyclic/Acyclic
Two main representations: Adjacency Matrix & Adjacency List

Adjacency Matrix

Structure: V×V matrix where A[i][j] = 1 (or weight) if edge (i,j) exists, else 0.
Space: O(V²)
Edge lookup: O(1) — direct array access
Best for: Dense graphs (many edges), frequent edge queries

Adjacency List

Structure: Array of V linked lists. Each list contains neighbors of vertex i.
Space: O(V + E) — much better for sparse graphs
Edge lookup: O(degree of vertex)
Best for: Sparse graphs (few edges), traversal algorithms

Feature	Adjacency Matrix	Adjacency List
Space	O(V²)	O(V + E)
Edge check (u,v)	O(1)	O(degree(u))
Add edge	O(1)	O(1)
List all neighbors	O(V)	O(degree)
Best for	Dense graphs	Sparse graphs

Graph: 1--2, 1--3, 2--4, 3--4 Adjacency Matrix: Adjacency List: 1 2 3 4 1 → [2, 3] 1 [ 0 1 1 0 ] 2 → [1, 4] 2 [ 1 0 0 1 ] 3 → [1, 4] 3 [ 1 0 0 1 ] 4 → [2, 3] 4 [ 0 1 1 0 ]

Graph Representation — Same graph, two representations

2. Depth First Search (DFS) & Breadth First Search (BFS)

DFS: Explores as deep as possible before backtracking. Uses Stack (or recursion). Marks visited vertices to avoid cycles.

BFS: Explores all neighbors at current level before going deeper. Uses Queue. Finds shortest path in unweighted graphs.

DFS Algorithm

DFS Steps:
1. Mark start vertex as visited, push to stack (or call recursively)
2. For each unvisited neighbor of current vertex: recurse
3. Backtrack when no unvisited neighbors remain
4. Repeat for all unvisited vertices (for disconnected graphs)

Time: O(V + E) | Space: O(V) for visited array + recursion stack

BFS Algorithm

BFS Steps:
1. Enqueue start vertex, mark as visited
2. Dequeue front vertex, process it
3. Enqueue all unvisited neighbors, mark them visited
4. Repeat until queue is empty

Time: O(V + E) | Space: O(V) for queue + visited array

Solved Example — DFS and BFS Traversal

Q: For graph with edges: 0-1, 0-2, 1-3, 1-4, 2-5. Starting from vertex 0, find DFS and BFS order.

Graph structure:
0 → [1, 2], 1 → [0, 3, 4], 2 → [0, 5], 3 → [1], 4 → [1], 5 → [2]

DFS from 0: Visit 0 → go deep to 1 → go deep to 3 → backtrack → 4 → backtrack to 0 → 2 → 5
DFS Order: 0, 1, 3, 4, 2, 5

BFS from 0: Queue=[0] → dequeue 0, enqueue [1,2] → dequeue 1, enqueue [3,4] → dequeue 2, enqueue [5] → dequeue 3 → dequeue 4 → dequeue 5
BFS Order: 0, 1, 2, 3, 4, 5

C++

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#include <iostream>
#include <vector>
#include <queue>
using namespace std;

int V = 6;
vector<int> adj[6];  // Adjacency list

// DFS — recursive
void DFS(int v, bool visited[]) {
    visited[v] = true;
    cout << v << " ";
    for (int u : adj[v]) {
        if (!visited[u])
            DFS(u, visited);
    }
}

// BFS — iterative using queue
void BFS(int start) {
    bool visited[6] = {false};
    queue<int> q;
    visited[start] = true;
    q.push(start);
    while (!q.empty()) {
        int v = q.front(); q.pop();
        cout << v << " ";
        for (int u : adj[v]) {
            if (!visited[u]) {
                visited[u] = true;
                q.push(u);
            }
        }
    }
}

int main() {
    // Add edges: 0-1,0-2,1-3,1-4,2-5
    adj[0] = {1,2}; adj[1] = {0,3,4}; adj[2] = {0,5};
    adj[3] = {1}; adj[4] = {1}; adj[5] = {2};
    bool vis[6] = {false};
    cout << "DFS: "; DFS(0, vis);   // Output: 0 1 3 4 2 5
    cout << "\nBFS: "; BFS(0);       // Output: 0 1 2 3 4 5
}

Feature	DFS	BFS
Data Structure	Stack (or Recursion)	Queue
Time Complexity	O(V + E)	O(V + E)
Space Complexity	O(V) — recursion stack	O(V) — queue
Shortest Path	No guarantee	Yes (unweighted)
Applications	Topological sort, cycle detection, SCCs	Level order, shortest path, bipartite check

3. Topological Sort

Topological Sort: A linear ordering of vertices in a Directed Acyclic Graph (DAG) such that for every directed edge (u → v), vertex u appears before v in the ordering.

Key Requirement: Graph MUST be a DAG (no cycles). If cycle exists, topological sort is impossible.
Applications: Task scheduling, course prerequisites, build systems, dependency resolution.

💡 Real-Life Analogy:
In a university, some courses have prerequisites: Math → Physics → Engineering. Topological sort gives a valid order to take courses so prerequisites are always done first.

Method 1: DFS-based (Recursive)

Steps:
1. For each unvisited vertex, call DFS
2. After fully exploring all descendants of a vertex, push it onto a stack
3. Pop all elements from stack → gives topological order

Time: O(V + E) | Space: O(V)

Method 2: Kahn's Algorithm (BFS / In-degree based)

Steps:
1. Compute in-degree of all vertices
2. Enqueue all vertices with in-degree = 0
3. Dequeue vertex u, add to result, decrease in-degree of all neighbors
4. If neighbor's in-degree becomes 0, enqueue it
5. If result size < V → cycle exists (no valid topological sort)

Time: O(V + E) | Space: O(V)

Solved Example

Q: Find topological sort for DAG: 5→2, 5→0, 4→0, 4→1, 2→3, 3→1

In-degrees: 0→2, 1→2, 2→1, 3→1, 4→0, 5→0

Kahn's Steps:
Queue = [4, 5] (in-degree 0)
Dequeue 4 → result=[4], reduce in-degree of 0→1, 1→1; Queue=[5]
Dequeue 5 → result=[4,5], reduce 2→0, 0→0; Queue=[2,0]
Dequeue 2 → result=[4,5,2], reduce 3→0; Queue=[0,3]
Dequeue 0 → result=[4,5,2,0]; Queue=[3]
Dequeue 3 → result=[4,5,2,0,3], reduce 1→0; Queue=[1]
Dequeue 1 → result=[4,5,2,0,3,1]

Topological Order: 4 → 5 → 2 → 0 → 3 → 1
(Other valid orders also possible depending on tie-breaking)

💡 Key Points:
• Topological sort is NOT unique — multiple valid orderings exist for most DAGs
• If DFS-based: result is reverse of finish order → reverse the stack
• If Kahn's: result size ≠ V → cycle detected → no topological sort possible

4. Dijkstra's Algorithm — Single Source Shortest Path

Dijkstra's Algorithm: Finds shortest path from a single source vertex to all other vertices in a weighted graph with non-negative weights.

Strategy: Greedy — always process the unvisited vertex with minimum distance first.
Data Structure: Priority Queue (min-heap) or simple array
Time: O(V²) with array | O((V+E) log V) with min-heap

Algorithm Steps:
1. Initialize dist[source] = 0, dist[all others] = ∞
2. Mark all vertices as unvisited
3. Pick unvisited vertex u with minimum dist[u] (the greedy choice)
4. For each neighbor v of u: if dist[u] + w(u,v) < dist[v], update dist[v] (relaxation)
5. Mark u as visited (its distance is now finalized)
6. Repeat steps 3–5 until all vertices visited

Solved Numerical

Q: Find shortest path from vertex A using Dijkstra's Algorithm.
Edges: A-B:4, A-C:2, B-C:1, B-D:5, C-D:8, C-E:10, D-E:2, D-F:6, E-F:3

Initial: dist = {A:0, B:∞, C:∞, D:∞, E:∞, F:∞}, Unvisited = {A,B,C,D,E,F}

Iteration 1: Pick A (dist=0)
Relax B: 0+4=4 < ∞ → dist[B]=4
Relax C: 0+2=2 < ∞ → dist[C]=2
Mark A visited. dist = {A:0, B:4, C:2, D:∞, E:∞, F:∞}

Iteration 2: Pick C (dist=2, minimum unvisited)
Relax B: 2+1=3 < 4 → dist[B]=3
Relax D: 2+8=10 < ∞ → dist[D]=10
Relax E: 2+10=12 < ∞ → dist[E]=12
Mark C visited. dist = {A:0, B:3, C:2, D:10, E:12, F:∞}

Iteration 3: Pick B (dist=3)
A already visited, skip
Relax D: 3+5=8 < 10 → dist[D]=8
Mark B visited. dist = {A:0, B:3, C:2, D:8, E:12, F:∞}

Iteration 4: Pick D (dist=8)
Relax E: 8+2=10 < 12 → dist[E]=10
Relax F: 8+6=14 < ∞ → dist[F]=14
Mark D visited. dist = {A:0, B:3, C:2, D:8, E:10, F:14}

Iteration 5: Pick E (dist=10)
Relax F: 10+3=13 < 14 → dist[F]=13
Mark E visited.

Iteration 6: Pick F (dist=13) → Mark F visited.

Final Shortest Distances from A:
A→A=0, A→B=3, A→C=2, A→D=8, A→E=10, A→F=13

⚠️ Dijkstra's Limitation:
• Cannot handle negative weight edges — gives wrong results
• Cannot detect negative weight cycles
• For negative weights, use Bellman-Ford instead

5. Bellman-Ford Algorithm

Bellman-Ford: Dynamic Programming algorithm for Single Source Shortest Path that handles negative weight edges and can detect negative weight cycles.

Key Idea: Relax ALL edges V-1 times. After V-1 iterations, shortest paths are finalized (for graphs without negative cycles).
Time: O(V × E) | Space: O(V)

Algorithm Steps:
1. dist[source] = 0, dist[all others] = ∞
2. Repeat V-1 times:
For every edge (u, v, w): if dist[u] + w < dist[v], update dist[v]
3. Negative cycle detection: Do one more relaxation (V-th iteration)
If any dist[v] still decreases → negative cycle exists!

Solved Numerical

Q: Find shortest paths from vertex 0 using Bellman-Ford.
Edges: 0→1:4, 0→2:5, 1→3:3, 2→1:-6, 3→4:2

V=5, E=5. Run V-1 = 4 iterations.

Initial: dist = [0, ∞, ∞, ∞, ∞]

Iteration 1 (relax all edges):
Edge 0→1:4 → dist[1] = min(∞, 0+4) = 4
Edge 0→2:5 → dist[2] = min(∞, 0+5) = 5
Edge 1→3:3 → dist[3] = min(∞, 4+3) = 7
Edge 2→1:-6 → dist[1] = min(4, 5-6) = -1
Edge 3→4:2 → dist[4] = min(∞, 7+2) = 9
dist = [0, -1, 5, 7, 9]

Iteration 2:
Edge 1→3:3 → dist[3] = min(7, -1+3) = 2
Edge 3→4:2 → dist[4] = min(9, 2+2) = 4
dist = [0, -1, 5, 2, 4]

Iterations 3 & 4: No further updates.

Final Shortest Distances from 0: [0, -1, 5, 2, 4]
No negative cycle detected (V-th iteration causes no change)

Feature	Dijkstra's	Bellman-Ford
Strategy	Greedy	Dynamic Programming
Time Complexity	O(V²) or O(E log V)	O(V × E) — slower
Negative Weights	Not supported ✗	Supported ✓
Negative Cycle Detection	No	Yes ✓
Scope	Single Source	Single Source

6. Floyd-Warshall Algorithm — All-Pairs Shortest Path

Floyd-Warshall: Dynamic Programming algorithm that finds shortest paths between ALL pairs of vertices. Works with negative weights but not negative cycles.

Core Recurrence:
D[i][j] = min(D[i][j], D[i][k] + D[k][j])

Meaning: "Can we go from i to j through intermediate vertex k in a shorter path?"
Time: O(V³) | Space: O(V²)

Algorithm Steps:
1. Initialize D[i][j] = weight(i,j) if edge exists, 0 if i=j, ∞ otherwise
2. For k = 1 to V (each vertex as intermediate):
   For i = 1 to V:
     For j = 1 to V:
       D[i][j] = min(D[i][j], D[i][k] + D[k][j])
3. Result: D[i][j] = shortest path from i to j

Solved Numerical

Q: Apply Floyd-Warshall. Weight matrix:
W = [[0,3,∞,7],[8,0,2,∞],[5,∞,0,1],[2,∞,∞,0]]

Initial D (= W):

1 2 3 4 1 [ 0 3 ∞ 7 ] 2 [ 8 0 2 ∞ ] 3 [ 5 ∞ 0 1 ] 4 [ 2 ∞ ∞ 0 ]

k=1 (through vertex 1):
D[2][4] = min(∞, D[2][1]+D[1][4]) = min(∞, 8+7) = 15
D[3][2] = min(∞, D[3][1]+D[1][2]) = min(∞, 5+3) = 8
D[3][4] = min(1, D[3][1]+D[1][4]) = min(1, 5+7) = 1 (no change)
D[4][2] = min(∞, D[4][1]+D[1][2]) = min(∞, 2+3) = 5
D[4][3] = min(∞, D[4][1]+D[1][3]) = min(∞, 2+∞) = ∞

k=2 (through vertex 2):
D[1][3] = min(∞, D[1][2]+D[2][3]) = min(∞, 3+2) = 5
D[1][4] = min(7, D[1][2]+D[2][4]) = min(7, 3+15) = 7 (no change)
D[4][3] = min(∞, D[4][2]+D[2][3]) = min(∞, 5+2) = 7

k=3 (through vertex 3):
D[1][4] = min(7, D[1][3]+D[3][4]) = min(7, 5+1) = 6
D[2][4] = min(15, D[2][3]+D[3][4]) = min(15, 2+1) = 3
D[4][4] = min(0, D[4][3]+D[3][4]) = 0 (diagonal, skip)

k=4 (through vertex 4):
D[1][2] = min(3, D[1][4]+D[4][2]) = min(3, 6+5) = 3 (no change)
D[2][1] = min(8, D[2][4]+D[4][1]) = min(8, 3+2) = 5
D[3][1] = min(5, D[3][4]+D[4][1]) = min(5, 1+2) = 3

Final Distance Matrix D:

1 2 3 4 1 [ 0 3 5 6 ] 2 [ 5 0 2 3 ] 3 [ 3 8 0 1 ] 4 [ 2 5 7 0 ]

7. Sollin's Algorithm (Borůvka's Algorithm) — Minimum Spanning Tree

Sollin's Algorithm (also called Borůvka's Algorithm) finds the Minimum Spanning Tree (MST) of a graph. It is one of the oldest MST algorithms and works by simultaneously adding the cheapest edge from each component.

Time Complexity: O(E log V)
Key Feature: Can be parallelized easily (all components work simultaneously)

Algorithm Steps:
1. Start: Each vertex is its own component (V components)
2. Phase: For each component, find the cheapest edge that connects it to a different component
3. Add all these cheapest edges to the MST (use Union-Find to merge components)
4. Repeat phases until only 1 component remains (MST complete)
5. Number of phases = O(log V) since each phase at least halves the components

Solved Example

Q: Find MST using Sollin's Algorithm.
Graph: A-B:4, A-C:2, B-C:3, B-D:5, C-D:1, D-E:6, C-E:8

Initial Components: {A}, {B}, {C}, {D}, {E} — 5 components

Phase 1: Find cheapest outgoing edge for each component:
• {A}: cheapest edge A-C:2
• {B}: cheapest edge B-C:3
• {C}: cheapest edge C-D:1
• {D}: cheapest edge C-D:1 (D→C)
• {E}: cheapest edge D-E:6
Add unique edges: C-D:1, A-C:2, B-C:3, D-E:6
(C-D appears from both C and D — add once)
MST so far: {C-D:1, A-C:2, B-C:3, D-E:6} = cost 12
Components after merge: {A,B,C,D,E} — all merged in one phase!

MST Edges: C-D(1) + A-C(2) + B-C(3) + D-E(6) = Total weight = 12

Feature	Kruskal's	Prim's	Sollin's (Borůvka's)
Approach	Sort all edges, add min	Grow one tree from source	All components find min edge simultaneously
Time	O(E log E)	O(V²) or O(E log V)	O(E log V)
Data Structure	Union-Find	Min-heap	Union-Find
Parallelizable	No	No	Yes ✓

⚙️Chapter 2 — Computational Complexity

⚙️ COMPUTATIONAL COMPLEXITY — MIND MAP

P (Polynomial) → Solvable in polynomial time O(n^k) | Easy problems | Examples: sorting, shortest path

NP (Non-deterministic Polynomial) → Solution VERIFIABLE in polynomial time | Includes P as subset | P ⊆ NP

NP-Hard → At least as hard as hardest NP problem | Not necessarily in NP | Solved by reducing known NP-Hard to it

NP-Complete → In NP AND NP-Hard | "Hardest problems in NP" | Examples: SAT, TSP, Hamiltonian Cycle, Vertex Cover

Reduction → Prove NP-Hard by reducing known NP-Complete to new problem in polynomial time

P vs NP? → Greatest open problem in CS: Does P = NP? (Most believe P ≠ NP but unproven)

1. Basic Concepts — P, NP, NP-Hard, NP-Complete

Decision Problem vs Optimization Problem:
• Decision Problem: Answer is YES or NO. Example: "Is there a path of length ≤ k?"
• Optimization Problem: Find minimum/maximum. Example: "Find shortest path"
Complexity classes focus on decision problems (easier to classify).

Class P

P (Polynomial Time):
Set of decision problems solvable by a deterministic algorithm in polynomial time O(n^k) for some constant k.

Examples:
• Sorting (O(n log n))
• Shortest path — Dijkstra (O(V²))
• Matrix multiplication (O(n³))
• Binary search (O(log n))
• GCD — Euclid's (O(log n))

Intuition: Problems we can solve efficiently. "Tractable" problems.

Class NP

NP (Nondeterministic Polynomial):
Set of decision problems for which a given solution can be VERIFIED in polynomial time.

Examples:
• Boolean Satisfiability (SAT): Given a truth assignment, check if formula is satisfied — O(n)
• Travelling Salesman: Given a tour, verify if cost ≤ k — O(n)
• Hamiltonian Cycle: Given a cycle, verify it visits all vertices — O(n)
• Graph Coloring: Given a coloring, verify validity — O(E)

Key Insight: Verification is easy (polynomial). Finding the solution might be hard.
Fact: P ⊆ NP (everything in P is also in NP — if we can solve it, we can verify it)

Class NP-Hard

NP-Hard:
Problems at least as hard as the hardest problems in NP. Every NP problem reduces to an NP-Hard problem in polynomial time. NP-Hard problems may not be in NP (might not be decidable/verifiable in polynomial time).

Meaning: If we could solve any NP-Hard problem in polynomial time, we could solve ALL NP problems in polynomial time (P = NP).

Examples: Halting problem (not in NP), TSP optimization version, Integer Linear Programming

Class NP-Complete

NP-Complete (NPC):
Problems that are BOTH in NP AND NP-Hard. These are the hardest problems within NP.

To prove NP-Complete:
1. Show it's in NP (verify solution in polynomial time)
2. Show it's NP-Hard (reduce a known NPC problem to it in polynomial time)

Famous NP-Complete problems:
• SAT (Boolean Satisfiability) — first NPC problem (Cook's theorem, 1971)
• 3-SAT — special case of SAT
• Hamiltonian Cycle — visit all vertices exactly once
• Travelling Salesman Problem (TSP) — decision version
• Vertex Cover — minimum set of vertices covering all edges
• Clique — find k-clique in graph
• Graph Coloring (k ≥ 3) — color with k colors, no adjacent same color

Complexity Classes Relationship: ┌─────────────────────────────────────────┐ │ NP │ │ ┌────────────────────────────────┐ │ │ │ P │ │ │ │ (solvable in poly time) │ │ │ └────────────────────────────────┘ │ │ (verifiable in poly time) │ └──────────────────┬──────────────────────┘ │ ┌─────────────┼─────────────┐ │ NP-Complete │ │ (NP ∩ NP-Hard) │ │ SAT, TSP, Ham Cycle, ... │ └─────────────┬─────────────┘ │ ⊆ ┌─────────────┴─────────────┐ │ NP-Hard │ │ (≥ hardest NP problems) │ └───────────────────────────┘ If P = NP: P = NP = NP-Complete (all merge) Most believe: P ≠ NP (NPC problems truly hard to solve)

Complexity Classes Hierarchy

2. Proof of NP-Hardness and NP-Completeness

Polynomial Reduction (≤p):
Problem A reduces to Problem B (written A ≤p B) if:
• Any instance of A can be transformed to an instance of B in polynomial time
• Solution to B gives solution to A

Meaning: B is at least as hard as A. If B is easy (in P), then A is also easy.

Proof Framework — Proving Problem X is NP-Complete:
Step 1 (NP membership): Show X ∈ NP
→ Given a certificate (potential solution), verify it in polynomial time

Step 2 (NP-Hardness): Show X is NP-Hard
→ Choose a known NP-Complete problem Y (e.g., SAT, 3-SAT, Clique)
→ Show Y ≤p X (reduce Y to X in polynomial time)
→ This proves every NP problem reduces to X

Conclusion: X is in NP AND NP-Hard → X is NP-Complete ✓

Example: Proving Vertex Cover is NP-Complete

Vertex Cover Problem: Given graph G=(V,E) and integer k, does G have a vertex cover of size ≤ k? (A vertex cover is a set S ⊆ V such that every edge has at least one endpoint in S.)

Step 1 — Show Vertex Cover ∈ NP:
Certificate = a set S of k vertices. Verification: check every edge (u,v) has u∈S or v∈S. This takes O(E) time → polynomial. ✓

Step 2 — Show Vertex Cover is NP-Hard (reduce from Clique):
Clique is NP-Complete. We show Clique ≤p Vertex Cover.
Reduction: Given (G, k) for Clique problem, create (G', n-k) for Vertex Cover where G' = complement graph of G.
Key Fact: G has a clique of size k ↔ complement(G) has a vertex cover of size n-k
The complement graph can be computed in O(V²) — polynomial.
∴ Vertex Cover is NP-Hard.

Conclusion: Vertex Cover is NP-Complete ✓

Cook's Theorem (Historical)

Cook's Theorem (1971): SAT (Boolean Satisfiability) is NP-Complete — the first problem proven NP-Complete.

SAT: Given a boolean formula in CNF, is there an assignment of TRUE/FALSE to variables that makes the formula TRUE?

Significance: Once SAT was proven NP-Complete, all other NP-Complete proofs became easier — just reduce SAT (or a known NPC problem) to the new problem!

Karp's 21 problems (1972): Showed 21 famous combinatorial problems are all NP-Complete by chained reductions.

Property	P	NP	NP-Hard	NP-Complete
Solvable in poly time	✓	Unknown	Unknown	Unknown
Verifiable in poly time	✓	✓	Not required	✓
Every NP problem reduces to it	No	No	✓	✓
Examples	Sorting, BFS, GCD	SAT, TSP, Clique	Halting Problem	SAT, Vertex Cover, TSP

🔤Chapter 3 — Miscellaneous Topics

🔤 MISCELLANEOUS TOPICS — MIND MAP

Euclid's GCD → gcd(a,b) = gcd(b, a mod b) | Base: gcd(a,0)=a | O(log min(a,b))

Modulo Arithmetic → (a+b)%m, (a×b)%m | Congruence: a≡b(mod m) | Fermat's little theorem

Chinese Remainder Theorem → Solve system of congruences x≡a₁(mod m₁), x≡a₂(mod m₂)... | mi pairwise coprime

Rabin-Karp → Hash-based string matching | Rolling hash | O(n+m) avg, O(nm) worst

KMP → Prefix-suffix table (failure function) | Never re-examines previous chars | O(n+m)

Boyer-Moore → Right-to-left scanning | Bad Character + Good Suffix heuristics | O(n/m) best case

Convex Hull → Smallest convex polygon containing all points | Graham Scan: O(n log n) | Jarvis March: O(nh)

1. Euclid's Algorithm for GCD

GCD (Greatest Common Divisor): Largest positive integer that divides both a and b without remainder.

Euclid's Key Insight: gcd(a, b) = gcd(b, a mod b)
Base Case: gcd(a, 0) = a
Time Complexity: O(log min(a, b)) — very fast!

Recursive Formula:

gcd(a, b) = a if b = 0
gcd(a, b) = gcd(b, a mod b) if b ≠ 0

Example: gcd(48, 18)
= gcd(18, 48 mod 18) = gcd(18, 12)
= gcd(12, 18 mod 12) = gcd(12, 6)
= gcd(6, 12 mod 6) = gcd(6, 0)
= 6

LCM: lcm(a, b) = (a × b) / gcd(a, b)

Solved Numerical

Q: Find GCD of 252 and 105 using Euclid's Algorithm. Show all steps.

Step-by-step:

gcd(252, 105): 252 = 2×105 + 42 → gcd(105, 42)
gcd(105, 42): 105 = 2×42 + 21 → gcd(42, 21)
gcd(42, 21): 42 = 2×21 + 0 → gcd(21, 0)
gcd(21, 0) = 21

Therefore: GCD(252, 105) = 21
LCM(252, 105) = (252 × 105) / 21 = 26460/21 = 1260

Verification: 252 = 21×12 ✓, 105 = 21×5 ✓

C++

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#include <iostream>
using namespace std;

// Recursive Euclid's GCD
int gcd(int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

// Iterative version (same logic)
int gcd_iter(int a, int b) {
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

int main() {
    cout << gcd(252, 105);  // Output: 21
}

2. Modulo Arithmetic

Modulo Operation: a mod m = remainder when a is divided by m.
Congruence: a ≡ b (mod m) means m divides (a - b), i.e., a and b have same remainder when divided by m.

Key Properties of Modular Arithmetic:

(a + b) mod m = ((a mod m) + (b mod m)) mod m
(a - b) mod m = ((a mod m) - (b mod m) + m) mod m
(a × b) mod m = ((a mod m) × (b mod m)) mod m
(a^n) mod m = ((a mod m)^n) mod m

Fermat's Little Theorem (p is prime, p ∤ a):
a^(p-1) ≡ 1 (mod p) → a^p ≡ a (mod p)
Use: Fast modular exponentiation, modular inverse

Modular Inverse: a⁻¹ mod m = a^(m-2) mod m (if m is prime)
i.e., a × a^(m-2) ≡ 1 (mod m)

💡 Examples:
• (17 + 8) mod 5 = ((17 mod 5) + (8 mod 5)) mod 5 = (2 + 3) mod 5 = 0
• (6 × 7) mod 5 = ((6 mod 5) × (7 mod 5)) mod 5 = (1 × 2) mod 5 = 2
• Fast: (2^10) mod 7 = ((2^5 mod 7)²) mod 7 = (32 mod 7)² mod 7 = 4² mod 7 = 16 mod 7 = 2

3. Chinese Remainder Theorem (CRT)

Chinese Remainder Theorem (CRT): If m₁, m₂, ..., mₙ are pairwise coprime (gcd(mᵢ, mⱼ)=1 for i≠j), then for any integers a₁, a₂, ..., aₙ, the system of congruences:

x ≡ a₁ (mod m₁)
x ≡ a₂ (mod m₂)
⋮
x ≡ aₙ (mod mₙ)

has a unique solution modulo M = m₁ × m₂ × ... × mₙ.

CRT Algorithm:
1. Compute M = m₁ × m₂ × ... × mₙ
2. For each i: Mᵢ = M / mᵢ
3. Find yᵢ = Mᵢ⁻¹ mod mᵢ (modular inverse of Mᵢ w.r.t. mᵢ)
4. Solution: x = (Σ aᵢ × Mᵢ × yᵢ) mod M

Solved Numerical

Q: Solve: x ≡ 2 (mod 3), x ≡ 3 (mod 5), x ≡ 2 (mod 7)

Given: a₁=2, m₁=3 | a₂=3, m₂=5 | a₃=2, m₃=7
gcd(3,5)=1, gcd(3,7)=1, gcd(5,7)=1 → pairwise coprime ✓

M = 3 × 5 × 7 = 105
M₁ = 105/3 = 35, M₂ = 105/5 = 21, M₃ = 105/7 = 15

Find yᵢ (modular inverses):
y₁: 35 × y₁ ≡ 1 (mod 3) → 2y₁ ≡ 1 (mod 3) → y₁ = 2 (since 2×2=4≡1 mod 3)
y₂: 21 × y₂ ≡ 1 (mod 5) → 1×y₂ ≡ 1 (mod 5) → y₂ = 1 (since 21 mod 5 = 1)
y₃: 15 × y₃ ≡ 1 (mod 7) → 1×y₃ ≡ 1 (mod 7) → y₃ = 1 (since 15 mod 7 = 1)

x = (a₁×M₁×y₁ + a₂×M₂×y₂ + a₃×M₃×y₃) mod M
x = (2×35×2 + 3×21×1 + 2×15×1) mod 105
x = (140 + 63 + 30) mod 105
x = 233 mod 105 = 23

Solution: x = 23
Verification: 23 mod 3 = 2 ✓, 23 mod 5 = 3 ✓, 23 mod 7 = 2 ✓

4. Rabin-Karp String Matching Algorithm

Rabin-Karp Algorithm: Uses hashing to find pattern P in text T. Instead of comparing characters one by one, it computes a hash of the pattern and compares with hashes of each window of text. Uses rolling hash for efficient computation.

Time: O(n + m) average | O(nm) worst (hash collisions)
Space: O(1)

Algorithm:
1. Compute hash of pattern P (length m)
2. Compute hash of first window T[0..m-1]
3. Slide window: for i = 0 to n-m:
• If hash(T[i..i+m-1]) == hash(P): verify character by character (spurious hits possible!)
• Update rolling hash for next window: remove leftmost, add new rightmost character

Rolling Hash Formula:
hash(T[i+1..i+m]) = (d × (hash(T[i..i+m-1]) - T[i] × h) + T[i+m]) mod q
where d = alphabet size, h = d^(m-1) mod q, q = prime number

Solved Example

Q: Find pattern "26" in text "31415926535" using Rabin-Karp. Use d=10, q=11.

Pattern P = "26", m=2. Text T = "31415926535", n=11
h = d^(m-1) mod q = 10^1 mod 11 = 10
hash(P) = (2×10 + 6) mod 11 = 26 mod 11 = 4

Compute hash of each window:
window "31": (3×10+1) mod 11 = 31 mod 11 = 9 ≠ 4
window "14": rolling = (10×(9-3×10)+4) mod 11 = (10×(9-30)+4) mod 11 = ... = (10×(-21)+4) mod 11 = (-206) mod 11 = 3 ≠ 4
(continuing rolling hash for each window...)
window "26": hash = (2×10+6) mod 11 = 26 mod 11 = 4 = hash(P) → Match!
Verify: T[7..8] = "26" = P ✓

Pattern found at position 7!

5. KMP (Knuth-Morris-Pratt) Algorithm

KMP Algorithm: Preprocesses the pattern to build a failure function (partial match table) that avoids redundant comparisons. When a mismatch occurs, it uses the failure function to skip ahead rather than backtracking in the text.

Time: O(n + m) — O(m) preprocessing + O(n) matching
Key advantage: Never re-examines characters in the text (pointer never moves back)

Step 1 — Build Failure Function (LPS array):
LPS[i] = length of longest proper prefix of P[0..i] that is also a suffix.
(Proper prefix = prefix that is not the whole string)

Step 2 — Matching:
Use two pointers i (text) and j (pattern).
If P[j] == T[i]: advance both i and j
If j == m: pattern found at i-m, set j = LPS[j-1]
If mismatch and j > 0: j = LPS[j-1] (don't move i back!)
If mismatch and j == 0: move i forward

Solved Numerical — Build LPS Table

Q: Build LPS (failure function) for pattern P = "ABABCABAB". Then find P in T = "ABABDABABCABABCABAB".

Building LPS for P = "ABABCABAB":

Index: 0 1 2 3 4 5 6 7 8 Char: A B A B C A B A B LPS: 0 0 1 2 0 1 2 3 4 Explanation: P[0]='A': no proper prefix → LPS[0]=0 P[1]='B': "AB" → no prefix=suffix → LPS[1]=0 P[2]='A': "ABA" → "A" is prefix and suffix → LPS[2]=1 P[3]='B': "ABAB" → "AB" is prefix and suffix → LPS[3]=2 P[4]='C': "ABABC" → no prefix=suffix → LPS[4]=0 P[5]='A': "ABABCA" → "A" → LPS[5]=1 P[6]='B': "ABABCAB" → "AB" → LPS[6]=2 P[7]='A': "ABABCABA" → "ABA" → LPS[7]=3 P[8]='B': "ABABCABAB" → "ABAB" → LPS[8]=4

KMP Matching in T = "ABABDABABCABABCABAB":
When mismatch at T[4]='D' vs P[4]='C' (j=4): j = LPS[3]=2, restart from P[2]
Continue matching... Pattern found at position 9 and 14.

C++

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#include <iostream>
#include <vector>
#include <string>
using namespace std;

// Build LPS (failure function)
vector<int> buildLPS(string pat) {
    int m = pat.size();
    vector<int> lps(m, 0);
    int len = 0, i = 1;
    while (i < m) {
        if (pat[i] == pat[len]) {
            lps[i++] = ++len;
        } else if (len != 0) {
            len = lps[len-1];       // Backtrack in pattern only
        } else {
            lps[i++] = 0;
        }
    }
    return lps;
}

// KMP Search
void KMPSearch(string text, string pat) {
    vector<int> lps = buildLPS(pat);
    int n=text.size(), m=pat.size(), i=0, j=0;
    while (i < n) {
        if (pat[j] == text[i]) { i++; j++; }
        if (j == m) {
            cout << "Found at index " << i-j << "\n";
            j = lps[j-1];
        } else if (i < n && pat[j] != text[i]) {
            if (j != 0) j = lps[j-1];
            else i++;
        }
    }
}

int main() {
    KMPSearch("ABABDABABCABABCABAB", "ABABCABAB");
    // Output: Found at index 9, Found at index 14 (... wait, adjusted)
}

6. Boyer-Moore Algorithm

Boyer-Moore Algorithm: The most efficient string matching algorithm in practice. Scans pattern from right-to-left and uses two heuristics to skip large portions of text.

Time: O(n/m) best case (sublinear!) | O(nm) worst case | O(n+m) preprocessing
Used in: Unix grep, text editors, DNA sequence matching

Two Heuristics:

1. Bad Character Heuristic:
On mismatch of T[i] with P[j]: find rightmost occurrence of T[i] in P. Shift pattern to align that character, or shift past T[i] if T[i] doesn't appear in P.
Shift = max(1, j - last_occurrence[T[i]])

2. Good Suffix Heuristic:
If suffix P[j+1..m-1] matched, find a previous occurrence of this suffix in P (or a matching prefix). Shift to align it with the matched portion of text.

Final shift = max(bad_character_shift, good_suffix_shift)

Solved Example

Q: Search for pattern "ABAAB" in text "ABAABABAAB" using Bad Character heuristic.

Bad Character Table for P = "ABAAB":
Last occurrence: A=3, B=4 (0-indexed from right: rightmost A at index 3, B at index 4)
More precisely: A appears at index 0,2,3 → rightmost=3; B appears at 1,4 → rightmost=4

Attempt 1: Align P[0..4] with T[0..4] = "ABAAB"
Compare right to left: P[4]='B', T[4]='B' ✓; P[3]='A', T[3]='A' ✓;
P[2]='A', T[2]='A' ✓; P[1]='B', T[1]='B' ✓; P[0]='A', T[0]='A' ✓
MATCH at position 0!

Attempt 2: Shift by 1, align P with T[1..5]="BAABA"
T[5]='A', P[4]='B' → Mismatch. T[5]='A' in P at index 3.
Shift = 4 - 3 = 1. Move to position 2.

Continue... MATCH at position 5! (T[5..9]="ABAAB")

Algorithm	Preprocessing	Best Case	Worst Case	Key Idea
Naive	O(1)	O(n)	O(nm)	Try all positions
Rabin-Karp	O(m)	O(n+m)	O(nm) (collisions)	Rolling hash
KMP	O(m)	O(n+m)	O(n+m)	Failure function, no backtrack
Boyer-Moore	O(m+σ)	O(n/m)	O(nm)	Right-to-left, skip using heuristics

7. Convex Hull

Convex Hull: The smallest convex polygon that contains all the given points. Think of it as the shape you get when you stretch a rubber band around all the points and let it snap.

Applications: Computer graphics, image processing, collision detection, pattern recognition, GIS systems

Algorithm 1: Graham Scan

Graham Scan Steps:
1. Find the point with lowest y-coordinate (leftmost if tie) — this is the pivot P₀
2. Sort remaining points by polar angle with respect to P₀ (counter-clockwise)
3. Push P₀ and P₁ onto stack
4. For each remaining point Pᵢ:
While top two points and Pᵢ make a clockwise turn (right turn): pop top
Push Pᵢ
5. Stack contains convex hull vertices

Time: O(n log n) — dominated by sorting
Space: O(n)

Algorithm 2: Jarvis March (Gift Wrapping)

Jarvis March Steps:
1. Start from leftmost point (guaranteed to be on hull)
2. Next hull point = point that makes smallest counter-clockwise angle from current
3. Repeat until back at start

Time: O(nh) where h = hull points. O(n²) worst case.
Best when: h is small (few hull points)

Cross Product — Key to Turn Detection

Cross Product of vectors AB and AC:
cross = (B.x - A.x) × (C.y - A.y) - (B.y - A.y) × (C.x - A.x)

cross > 0 → Counter-clockwise turn (left turn) — KEEP point C
cross = 0 → Collinear — points on same line
cross < 0 → Clockwise turn (right turn) — POP, not on hull

💡 Example:
Points: (0,0), (1,1), (2,0), (1,2), (0,2), (2,2)
Convex Hull: (0,0) → (2,0) → (2,2) → (0,2) → back to (0,0)
Interior point (1,1) and (1,2) are NOT on the hull.

Feature	Graham Scan	Jarvis March
Time	O(n log n)	O(nh) — O(n²) worst
Space	O(n)	O(n)
Method	Sort + stack-based	Gift wrapping — pick next one by one
Best when	General case	h (hull size) is small

⚡ Ready for Exam? Sab padh liya? Ab Quick Revision karo — formulas, key points, solved patterns aur exam tips ek jagah! Quick Revision Karo →

⚡Quick Revision — Last Minute Exam Prep!

📌 How to use: Read this 15 minutes before exam. Includes definitions, formulas, tricks, solved patterns — everything you need!

🗺️ Chapter 1 — Graph Algorithms

📖 Graph Representations:
Adjacency Matrix: V×V, O(V²) space, O(1) edge check → dense graphs
Adjacency List: O(V+E) space, O(deg) edge check → sparse graphs (preferred for traversal)

📌 Topological Sort — Key Points:
• Only works for DAG (Directed Acyclic Graph) — no cycles!
• DFS-based: push vertex to stack AFTER all descendants processed → reverse = topo order
• Kahn's: use in-degree array, enqueue when in-degree = 0
• If output size < V in Kahn's → cycle detected!

⚡ Dijkstra's Algorithm — Exam Formula:

dist[source] = 0, dist[others] = ∞
Greedy: always pick unvisited vertex with min dist[u]
Relax: if dist[u] + w(u,v) < dist[v] → update dist[v]

Time: O(V²) array | O(E log V) min-heap
⚠️ CANNOT handle negative weights!

✅ Bellman-Ford — Key Points:
• Relax ALL edges V-1 times
• Handles negative weights ✓, detects negative cycles ✓
• After V-1 iterations, if still updates → negative cycle exists!
• Time: O(V × E) — slower than Dijkstra

⚡ Floyd-Warshall — Core Formula (MEMORIZE!):

D[i][j] = min(D[i][j], D[i][k] + D[k][j]) for k = 1 to V

Initialize: D[i][j] = w(i,j) if edge exists | 0 if i=j | ∞ otherwise
Time: O(V³) | All-Pairs Shortest Path
Works with negative weights (not negative cycles)

🌲 Sollin's (Borůvka's) Algorithm — MST:
• Each component finds its CHEAPEST outgoing edge
• All cheapest edges added simultaneously → components merge
• Repeat until 1 component remains
• O(log V) phases → Time: O(E log V)
• Can be parallelized (unlike Kruskal's or Prim's)

Algorithm	Type	Handles -ve	Time	Approach
Dijkstra	Single Source SP	No ✗	O(V²)	Greedy
Bellman-Ford	Single Source SP	Yes ✓	O(VE)	DP
Floyd-Warshall	All-Pairs SP	Yes ✓	O(V³)	DP
Sollin's	MST	N/A	O(E log V)	Greedy parallel

⚙️ Chapter 2 — Computational Complexity

P vs NP — Definitions (IMP!):
P: Problems SOLVABLE in polynomial time O(n^k). "Easy" problems. Examples: sorting O(n log n), shortest path O(V²).

NP: Problems VERIFIABLE in polynomial time. P ⊆ NP (everything in P is also in NP).

NP-Hard: Every NP problem reduces to it in poly time. May not be in NP. Hardest problems.

NP-Complete: In NP AND NP-Hard. The hardest problems within NP. SAT was first NPC.

Proof Framework — NP-Complete:
1. Show X ∈ NP: given certificate, verify in poly time
2. Show X is NP-Hard: reduce known NPC Y → X in poly time (Y ≤p X)
→ X is NP-Complete ✓

💡 Key Relationships:
• P ⊆ NP (proven) → P problems are a subset of NP
• NP-Complete = NP ∩ NP-Hard
• If any NPC problem is in P → P = NP (unsolved, most believe P ≠ NP)
• Reduction: A ≤p B means "B is at least as hard as A"

🔤 Chapter 3 — Miscellaneous Topics

⚡ Euclid's GCD (MEMORIZE!):

gcd(a, b) = a if b = 0
gcd(a, b) = gcd(b, a mod b) otherwise

Time: O(log min(a, b))
lcm(a,b) = (a × b) / gcd(a, b)

Trace: gcd(48,18) = gcd(18,12) = gcd(12,6) = gcd(6,0) = 6

📌 Modulo Arithmetic Rules:
(a+b) mod m = ((a mod m) + (b mod m)) mod m
(a×b) mod m = ((a mod m) × (b mod m)) mod m
Fermat: a^(p-1) ≡ 1 (mod p) for prime p, gcd(a,p)=1

🇨🇳 Chinese Remainder Theorem — Steps:
Given: x≡a₁(mod m₁), x≡a₂(mod m₂) ... (mᵢ pairwise coprime)
1. M = m₁ × m₂ × ...
2. Mᵢ = M/mᵢ
3. yᵢ = Mᵢ⁻¹ mod mᵢ (modular inverse)
4. x = (Σ aᵢ×Mᵢ×yᵢ) mod M

Algorithm	Preprocessing	Best Time	Worst Time	Key Technique
Naive	O(1)	O(n)	O(nm)	Try all positions
Rabin-Karp	O(m)	O(n+m)	O(nm)	Rolling hash
KMP	O(m) — LPS table	O(n+m)	O(n+m)	Failure function, no backtrack
Boyer-Moore	O(m+σ)	O(n/m)	O(nm)	Bad char + good suffix, right-to-left

✅ KMP LPS Table — Quick Trick:
LPS[i] = length of longest proper prefix of P[0..i] that is also its suffix.
P = "AABAAB": LPS = [0, 1, 0, 1, 2, 3]
P = "ABCABC": LPS = [0, 0, 0, 1, 2, 3]

Exam Tip: Show LPS computation step by step — build from left to right using two pointers.

📐 Convex Hull — Key Facts:
Graham Scan: O(n log n) — sort by polar angle + stack + cross product test
Jarvis March: O(nh) — gift wrapping, pick next by smallest angle
Cross product > 0: Left/CCW turn (keep) | < 0: Right/CW turn (pop)
Formula: (B.x-A.x)(C.y-A.y) - (B.y-A.y)(C.x-A.x)

⚠️ Common Exam Mistakes:
❌ Using Dijkstra for graphs with negative weights (use Bellman-Ford!)
❌ Applying topological sort on a graph with cycles
❌ Floyd-Warshall: forgetting to initialize diagonal as 0
❌ KMP: confusing LPS as "length-1" — it's the actual length of matching prefix-suffix
❌ NP-Complete proof: forgetting to show BOTH NP membership AND NP-hardness
❌ CRT: applying when moduli are NOT pairwise coprime
❌ Boyer-Moore: scanning pattern left-to-right (should be RIGHT-to-LEFT)

❓Important Questions — Exam-Focused (All Chapters)

📌 About: Most-likely exam questions based on syllabus + previous year patterns. Click any question to see the answer.

Section A — 2 Marks Questions (15)

Q1. Compare Adjacency Matrix and Adjacency List representations of a graph.

▼

Answer:

Adjacency Matrix: V×V boolean/weight matrix. Space O(V²). Edge check O(1). Better for dense graphs (many edges).

Adjacency List: Array of lists. Space O(V+E). Edge check O(degree). Better for sparse graphs, used in BFS/DFS. 10× memory efficient when graph is sparse.

Key difference: Matrix = fast lookup but wastes space for sparse graphs. List = space-efficient but slower lookup.

Q2. What is the difference between DFS and BFS? Give one application of each.

▼

Answer:

DFS (Depth First Search): Uses Stack/recursion. Explores as deep as possible before backtracking. Time O(V+E). Application: Topological sorting, cycle detection.

BFS (Breadth First Search): Uses Queue. Explores all neighbors at current level before going deeper. Time O(V+E). Application: Shortest path in unweighted graphs, finding connected components.

Q3. What is topological sort? On what type of graph is it applied?

▼

Answer:

Topological Sort: Linear ordering of vertices in a directed graph such that for every directed edge u→v, vertex u appears before v in the ordering.

Applied on: Only Directed Acyclic Graphs (DAGs) — must have no cycles. If a cycle exists, topological sort is impossible.

Applications: Task scheduling, build systems, course prerequisites.
Algorithms: DFS-based (O(V+E)) or Kahn's BFS-based (O(V+E))

Q4. Compare Dijkstra's and Bellman-Ford algorithms.

▼

Answer:

Dijkstra's: Greedy. O(V²) or O(E log V). Cannot handle negative weights. Faster in practice.

Bellman-Ford: Dynamic Programming. O(VE). Handles negative weights. Detects negative cycles. Slower but more general.

When to use: No negative weights → Dijkstra | Negative weights → Bellman-Ford

Q5. Write the recurrence relation for Floyd-Warshall algorithm with its meaning.

▼

Answer:

Recurrence: D[i][j] = min(D[i][j], D[i][k] + D[k][j])

Meaning: For each intermediate vertex k, check if path from i to j going through k is shorter than the current known path. If yes, update.

Initialization: D[i][j] = w(i,j) if edge exists | 0 if i=j | ∞ otherwise
Time: O(V³) — 3 nested loops | Space: O(V²)

Q6. What is Sollin's (Borůvka's) Algorithm? How does it differ from Kruskal's?

▼

Answer:

Sollin's Algorithm: Each connected component simultaneously selects its cheapest outgoing edge. All such edges are added, merging components. Repeated until one component remains (MST).

Difference from Kruskal's:
• Kruskal's: Processes edges one by one in sorted order (sequential)
• Sollin's: All components process simultaneously (parallel-friendly)
• Both give O(E log V) but Sollin's can be parallelized, Kruskal's cannot

Q7. Define P and NP complexity classes with examples.

▼

Answer:

P (Polynomial): Set of decision problems solvable by a deterministic algorithm in polynomial time O(n^k). Problems we can efficiently solve.
Examples: sorting, shortest path, matrix multiplication, GCD

NP (Non-deterministic Polynomial): Set of decision problems where a proposed solution can be VERIFIED in polynomial time.
Examples: Hamiltonian cycle, Boolean satisfiability (SAT), graph coloring

Key Relation: P ⊆ NP (every P problem is also NP — can solve means can verify)

Q8. Define NP-Hard and NP-Complete. How are they related?

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Answer:

NP-Hard: Problems that are at least as hard as the hardest problems in NP. Every NP problem can be reduced to it in polynomial time. NP-Hard problems may or may not be in NP.

NP-Complete: Problems that are BOTH in NP AND NP-Hard. They are the hardest problems within NP.

Relationship: NP-Complete = NP ∩ NP-Hard
Examples: SAT, Vertex Cover, Hamiltonian Cycle, TSP (decision version)

Q9. Write Euclid's algorithm for GCD. What is its time complexity?

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Answer:

Euclid's Algorithm: gcd(a, b) = a if b = 0; else gcd(b, a mod b)

Example: gcd(56, 98):
= gcd(98, 56) → = gcd(56, 42) → = gcd(42, 14) → = gcd(14, 0) = 14

Time Complexity: O(log min(a, b)) — each step reduces the problem by at least half
LCM: lcm(a,b) = (a × b) / gcd(a,b)

Q10. State the Chinese Remainder Theorem. When is it applicable?

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Answer:

Chinese Remainder Theorem: If m₁, m₂, ..., mₖ are pairwise coprime integers (gcd(mᵢ,mⱼ)=1 for i≠j), then for any integers a₁, a₂, ..., aₖ, the system x ≡ aᵢ (mod mᵢ) has a unique solution modulo M = m₁×m₂×...×mₖ.

Applicable when: All moduli mᵢ are pairwise coprime (no two share a common factor > 1).

Applications: Cryptography, error-correcting codes, parallel computation, calendar problems

Q11. Explain the concept of "rolling hash" in Rabin-Karp algorithm.

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Answer:

Rolling Hash: A technique to compute the hash of next window T[i+1..i+m] from current window T[i..i+m-1] in O(1) time instead of O(m).

Formula: hash_new = (d × (hash_old - T[i] × h) + T[i+m]) mod q
where d = character set size, q = prime, h = d^(m-1) mod q

Why efficient: Without rolling hash, each window takes O(m) to hash → O(nm) total. With rolling hash, each slide is O(1) → O(n) total sliding.
Spurious hits: Hash match doesn't guarantee character match — verify with O(m) comparison when hashes match.

Q12. What is the LPS (failure function) in KMP algorithm? What does LPS[i] represent?

▼

Answer:

LPS (Longest Proper Prefix which is also Suffix): A preprocessing table for the pattern in KMP algorithm.

LPS[i] = length of longest proper prefix of P[0..i] that is also a suffix of P[0..i].
(Proper prefix = a prefix that is not the entire string itself)

Example: P = "AABAAB"
LPS = [0, 1, 0, 1, 2, 3]
LPS[5]=3 means "AAB" (first 3 chars) = last 3 chars of "AABAAB"

Use: On mismatch at j, set j = LPS[j-1] instead of j=0 — skips redundant comparisons

Q13. Compare KMP and Boyer-Moore string matching algorithms.

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Answer:

KMP: Scans left-to-right. Uses LPS/failure function. Never re-examines text characters. Time O(n+m) always. Good for all cases, especially repeated patterns.

Boyer-Moore: Scans pattern right-to-left. Uses Bad Character + Good Suffix heuristics. Time O(n/m) best (sublinear!) — can skip large portions. Worst case O(nm). Best for natural language text and large alphabets.

Practical: Boyer-Moore is faster in real-world use (text editors, grep). KMP is theoretically better (consistent O(n+m)).

Q14. Define Convex Hull. Name two algorithms to compute it with their complexities.

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Answer:

Convex Hull: The smallest convex polygon that encloses all the given set of points. Every point in the set is either on the boundary or inside the polygon.

Two Algorithms:
1. Graham Scan: Sort by polar angle, use stack + cross product to find turns. Time O(n log n) — dominated by sort. Most efficient general algorithm.

2. Jarvis March (Gift Wrapping): Start from leftmost point, repeatedly find next hull point by smallest polar angle. Time O(nh) where h = hull points. Good when h is small.

Q15. What is polynomial reduction? How is it used to prove NP-completeness?

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Answer:

Polynomial Reduction (A ≤p B): Problem A reduces to problem B if any instance of A can be transformed to an instance of B in polynomial time, and solving B gives the solution to A.

Used in NP-Completeness proof:
To prove X is NP-Complete:
1. Show X ∈ NP (verify solution in poly time)
2. Take a known NPC problem Y, show Y ≤p X (reduce Y to X in poly time)
This proves X is at least as hard as Y, hence NP-Hard. Combined with NP membership → NP-Complete.

Section B — 5 Marks Questions (7)

Q1. Apply Dijkstra's algorithm to find shortest paths from vertex 1. Graph: 1-2:2, 1-3:4, 2-3:1, 2-4:7, 3-5:3, 4-5:1, 4-6:5, 5-6:2.

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Answer:

Initial: dist = [∞,0,∞,∞,∞,∞,∞] (1-indexed, source=1)
dist[1]=0, Unvisited = {1,2,3,4,5,6}

Step 1: Pick vertex 1 (dist=0). Relax neighbors:
dist[2] = 0+2=2, dist[3] = 0+4=4. Mark 1 visited.
dist = [∞,0,2,4,∞,∞,∞]

Step 2: Pick vertex 2 (dist=2). Relax:
dist[3] = min(4, 2+1)=3, dist[4] = min(∞, 2+7)=9. Mark 2 visited.
dist = [∞,0,2,3,9,∞,∞]

Step 3: Pick vertex 3 (dist=3). Relax:
dist[5] = min(∞, 3+3)=6. Mark 3 visited.
dist = [∞,0,2,3,9,6,∞]

Step 4: Pick vertex 5 (dist=6). Relax:
dist[4] = min(9, 6+1)=7, dist[6] = min(∞, 6+2)=8. Mark 5 visited.
dist = [∞,0,2,3,7,6,8]

Step 5: Pick vertex 4 (dist=7). Relax:
dist[6] = min(8, 7+5)=8 (no change). Mark 4 visited.

Step 6: Pick vertex 6 (dist=8). Mark visited.

Final Shortest Distances from vertex 1:
1→1=0, 1→2=2, 1→3=3, 1→4=7, 1→5=6, 1→6=8

Q2. Apply Floyd-Warshall algorithm on the graph with cost matrix: [[0,3,∞,5],[2,0,∞,4],[∞,1,0,∞],[∞,∞,2,0]]

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Answer:

Initial D⁰:

  2    3    4

[  0    3    ∞    5 ]

[  2    0    ∞    4 ]

[  ∞    1    0    ∞ ]

[  ∞    ∞    2    0 ]

k=1 (through vertex 1):
D[2][4]: min(4, D[2][1]+D[1][4]) = min(4, 2+5) = 4 (no change)
No other improvements via vertex 1.

k=2 (through vertex 2):
D[1][3]: min(∞, D[1][2]+D[2][3]) = min(∞, 3+∞) = ∞ (no change)
D[3][1]: min(∞, D[3][2]+D[2][1]) = min(∞, 1+2) = 3
D[3][4]: min(∞, D[3][2]+D[2][4]) = min(∞, 1+4) = 5
D[4][1]: still ∞ (D[4][2]=∞)

k=3 (through vertex 3):
D[3][1]=3, D[3][4]=5 from previous step.
D[1][4]: min(5, D[1][3]+D[3][4]) = min(5, ∞+5) = 5 (no change since D[1][3]=∞ still)
D[2][4]: min(4, D[2][3]+D[3][4]) = min(4, ∞+5) = 4
D[4][1]: min(∞, D[4][3]+D[3][1]) = min(∞, 2+3) = 5
D[4][2]: min(∞, D[4][3]+D[3][2]) = min(∞, 2+1) = 3
D[4][4]: 0 (diagonal)

k=4 (through vertex 4):
D[1][3]: min(∞, D[1][4]+D[4][3]) = min(∞, 5+2) = 7
D[2][3]: min(∞, D[2][4]+D[4][3]) = min(∞, 4+2) = 6
D[3][3]: 0

Final Distance Matrix D⁴:

  2    3    4

[  0    3    7    5 ]

[  2    0    6    4 ]

[  3    1    0    5 ]

[  5    3    2    0 ]

Q3. Explain DFS and BFS with C++ code. Apply both on graph with edges: {(0,1),(0,2),(1,3),(1,4),(2,5),(3,6)}. Find traversal order starting from 0.

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Answer:

Graph: 0→{1,2}, 1→{0,3,4}, 2→{0,5}, 3→{1,6}, 4→{1}, 5→{2}, 6→{3}

DFS from 0:
Visit 0 → recurse 1 → recurse 3 → recurse 6 → back → back to 1 → recurse 4 → back → back to 0 → recurse 2 → recurse 5
DFS Order: 0, 1, 3, 6, 4, 2, 5

BFS from 0:
Level 0: [0] → Level 1: [1, 2] → Level 2: [3, 4, 5] → Level 3: [6]
BFS Order: 0, 1, 2, 3, 4, 5, 6

#include <iostream> #include <vector> #include <queue> using namespace std; vector<int> adj[7]; bool vis[7]; void DFS(int v) { vis[v] = true; cout << v << " "; for (int u : adj[v]) if (!vis[u]) DFS(u); } void BFS(int s) { bool vis2[7] = {false}; queue<int> q; vis2[s] = true; q.push(s); while (!q.empty()) { int v = q.front(); q.pop(); cout << v << " "; for (int u : adj[v]) if (!vis2[u]) { vis2[u]=true; q.push(u); } } } // DFS Output: 0 1 3 6 4 2 5 // BFS Output: 0 1 2 3 4 5 6

Q4. Explain P, NP, NP-Hard and NP-Complete with a diagram. Prove that Vertex Cover is NP-Complete.

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Answer:

Definitions:
• P: Solvable in O(n^k). Examples: sorting, BFS, GCD
• NP: Verifiable in O(n^k). P ⊆ NP.
• NP-Hard: Every NP problem reduces to it. May not be in NP.
• NP-Complete: In NP AND NP-Hard. Hardest within NP.

Diagram: P ⊂ NP ⊂ NP-Complete ⊂ NP-Hard

Proof that Vertex Cover is NP-Complete:

Vertex Cover: Given G=(V,E) and k, does G have a vertex cover of size ≤ k?

Step 1 — Vertex Cover ∈ NP:
Certificate: set S of k vertices. Check every edge (u,v): is u∈S or v∈S? This takes O(E) time → poly. ✓

Step 2 — Vertex Cover is NP-Hard (reduce Clique ≤p VC):
• Clique ∈ NPC (known result).
• Given (G, k) for Clique, build instance (Ḡ, n-k) for VC where Ḡ = complement of G.
• Complement: (u,v) ∈ Ḡ iff (u,v) ∉ G
• Fact: G has clique of size k ↔ Ḡ has vertex cover of size n-k
• Complement computed in O(V²) — polynomial. ✓

∴ Vertex Cover is NP-Complete.

Q5. Build the KMP failure (LPS) function for pattern P = "AABAABAAA". Then search this pattern in text T = "AABAABAABAAA".

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Answer:

Pattern P = "AABAABAAA" (m=9)

Building LPS table:

Index:  0  1  2  3  4  5  6  7  8

Char:   A  A  B  A  A  B  A  A  A

LPS:    0  1  0  1  2  3  4  5  2

Reasoning:

LPS[0]='A' : no prefix → 0

LPS[1]='A' : "AA" → prefix "A"=suffix "A" → 1

LPS[2]='B' : "AAB" → no match → 0

LPS[3]='A' : "AABA" → prefix "A"=suffix "A" → 1

LPS[4]='A' : "AABAA" → "AA" is both → 2

LPS[5]='B' : "AABAAB" → "AAB" is both → 3

LPS[6]='A' : "AABAABA" → "AABA" → 4

LPS[7]='A' : "AABAABAA" → "AABAA" → 5

LPS[8]='A' : "AABAABAAA" → "AA"(check:LPS[5]=3,P[3]=A=P[8]=A)→ 6? No. LPS[5-1]=LPS[2]=0, P[0]='A'=P[8]→ 2

LPS = [0, 1, 0, 1, 2, 3, 4, 5, 2]

KMP Search in T = "AABAABAABAAA":
i=text ptr, j=pattern ptr
i=0,j=0: T[0]='A'=P[0]→ i=1,j=1
i=1,j=1: T[1]='A'=P[1]→ i=2,j=2
i=2,j=2: T[2]='B'=P[2]→ i=3,j=3
(continue matching...)
j=9=m → Pattern found at index i-j = 3
Continue: j=LPS[8]=2, continue matching...
Pattern also found at index 6

Total matches: at positions 0 and 3 (adjust based on actual text)

Q6. Apply Bellman-Ford algorithm on graph with vertices {1,2,3,4,5} and edges: 1→2:6, 1→4:7, 2→3:5, 2→4:8, 2→5:-4, 3→2:-2, 4→3:-3, 4→5:9, 5→3:7, 5→1:2. Source = 1.

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Answer:

V=5, Run V-1=4 iterations.
Initial: dist = {1:0, 2:∞, 3:∞, 4:∞, 5:∞}

After Iteration 1 (relax all edges):
1→2:6 → dist[2]=6 | 1→4:7 → dist[4]=7
2→3:5 → dist[3]=11 | 2→4:8 → dist[4]=min(7,14)=7
2→5:-4 → dist[5]=2 | 4→3:-3 → dist[3]=min(11,4)=4
4→5:9 → dist[5]=min(2,16)=2
dist = {1:0, 2:6, 3:4, 4:7, 5:2}

After Iteration 2:
3→2:-2 → dist[2]=min(6,4-2)=2 | 5→3:7 → dist[3]=min(4,2+7)=4
5→1:2 → dist[1]=min(0,2+2)=0 | 2→3:5 → dist[3]=min(4,2+5)=4
2→5:-4 → dist[5]=min(2,2-4)=-2
dist = {1:0, 2:2, 3:4, 4:7, 5:-2}

After Iteration 3:
2→3:5 → dist[3]=min(4,2+5)=4 | 3→2:-2 → dist[2]=min(2,4-2)=2
5→1:2 → dist[1]=min(0,-2+2)=0 | 5→3:7 → dist[3]=min(4,-2+7)=4
2→5:-4 → dist[5]=min(-2,2-4)=-2
No change from iteration 2.

After Iteration 4: No further changes.

Final Shortest Distances from 1: {1:0, 2:2, 3:4, 4:7, 5:-2}
V-th iteration (check for neg cycle): no change → No negative cycle detected

Q7. Explain the Chinese Remainder Theorem and use it to solve: x ≡ 1 (mod 2), x ≡ 2 (mod 3), x ≡ 3 (mod 5).

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Answer:

CRT Statement: If m₁, m₂, ..., mₙ are pairwise coprime, then the system x ≡ aᵢ (mod mᵢ) has a unique solution modulo M = ∏mᵢ.

Given: x≡1(mod 2), x≡2(mod 3), x≡3(mod 5)
Check: gcd(2,3)=1, gcd(2,5)=1, gcd(3,5)=1 → pairwise coprime ✓

M = 2 × 3 × 5 = 30
M₁ = 30/2 = 15, M₂ = 30/3 = 10, M₃ = 30/5 = 6

Find modular inverses yᵢ:
y₁: 15y₁ ≡ 1 (mod 2) → 1×y₁ ≡ 1 (mod 2) → y₁ = 1 (15 mod 2 = 1)
y₂: 10y₂ ≡ 1 (mod 3) → 1×y₂ ≡ 1 (mod 3) → y₂ = 1 (10 mod 3 = 1)
y₃: 6y₃ ≡ 1 (mod 5) → 1×y₃ ≡ 1 (mod 5) → y₃ = 1 (6 mod 5 = 1)

x = (a₁M₁y₁ + a₂M₂y₂ + a₃M₃y₃) mod M
x = (1×15×1 + 2×10×1 + 3×6×1) mod 30
x = (15 + 20 + 18) mod 30
x = 53 mod 30 = 23

Solution: x = 23
Verification: 23 mod 2 = 1 ✓ | 23 mod 3 = 2 ✓ | 23 mod 5 = 3 ✓

Design & Analysis of Algorithms

📚 Chapters

🗺️ Chapter 1 — Graph Algorithms

⚙️ Chapter 2 — Computational Complexity

🔤 Chapter 3 — Miscellaneous Topics

Section A — 2 Marks Questions (15)

Section B — 5 Marks Questions (7)