Data Structures and Algorithms: cram pack

Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.

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1Formulas and methods

Graph theory

Handshake lemma
Σ deg(v) = 2|E|
Complete graphs
|E(Kn)| = n(n − 1)/2, |E(Km,n)| = m·n
Regular graph
|E| = n·k/2
Tree and forest
|E| = n − 1, |E| = n − c
Leaves of a tree
L = 2 + Σ (deg(v) − 2) over deg(v) ≥ 3
Euler
circuit: all degrees even; path: exactly 2 odd degrees
Hamilton circuits in Kn
(n − 1)!/2
Dijkstra update
d(v) = min(d(v), d(u) + w(u, v))
Walks of length L
(Mᴸ)ij

Method

  1. Write down the degree of every vertex; the degrees add up to twice the number of edges.
  2. For Euler questions, check that the graph is connected and count the vertices with odd degree.
  3. For trees, use n − 1 edges and the handshake lemma.
  4. For Dijkstra, keep a table of tentative distances and settle the smallest unsettled vertex each round.
  5. For a minimum spanning tree, sort the edges by weight and skip every edge that closes a cycle (Kruskal), or grow one tree from a start vertex (Prim).

2Practice problems

1.Algorithms & complexityMedium

What is the time complexity of multiplying two n × n matrices with the standard triple loop?

  1. AO(n)
  2. BO(n³)
  3. CO(n²)
  4. DO(n² log n)
2.Algorithms & complexityHard

Which bound is correct for log(n!)?

  1. AΘ(n log n)
  2. BΘ(log n)
  3. CΘ(n²)
  4. DΘ(n)
3.Data structuresMedium

What balance condition does an AVL tree maintain at every node?

  1. AEvery node is coloured red or black
  2. BAll leaves are at exactly the same depth
  3. CThe heights of the left and right subtrees differ by at most 1
  4. DBoth subtrees of every node contain exactly the same number of nodes
4.Data structuresHard

Why might a server choose its hash function at random from a universal family?

  1. ATo guarantee that no two keys can ever collide, whatever keys an attacker sends
  2. BTo make the hash table sorted
  3. CTo remove the need to resize the table
  4. DTo make it very unlikely that an attacker can pick many keys that all collide
5.Graph theoryMedium

A connected graph has 20 vertices and 20 edges. How many edges k must be removed to leave a spanning tree?

|V| = 20|E| = 20
k =
6.Graph theoryHard
118151511113108ABCDEFG

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =

3Solutions

1.Algorithms & complexity

Answer: B · O(n³)

Why this is the answer

There are n² entries and each needs n multiplications, giving n³ operations.

Answer

B · O(n³)
2.Algorithms & complexity

Answer: A · Θ(n log n)

Why this is the answer

log(n!) ≤ n log n, and the largest n / 2 factors give log(n!) ≥ (n / 2) log(n / 2); this is why comparison sorting needs Ω(n log n).

Answer

A · Θ(n log n)
3.Data structures

Answer: C · The heights of the left and right subtrees differ by at most 1

Why this is the answer

The balance factor (height left − height right) must be −1, 0 or 1; rotations restore it after inserts and deletes.

Answer

C · The heights of the left and right subtrees differ by at most 1
4.Data structures

Answer: D · To make it very unlikely that an attacker can pick many keys that all collide

Why this is the answer

With universal hashing the expected number of collisions is small for any set of keys, since the attacker does not know which function was chosen.

Answer

D · To make it very unlikely that an attacker can pick many keys that all collide
5.Graph theory

Answer: k = 1

Euler, Hamilton and trees: a connected graph has an Euler circuit (every edge once, back to the start) when every vertex has even degree, and an Euler path that is not a circuit when exactly two vertices have odd degree; the path then runs from one odd vertex to the other. A Hamilton circuit visits every vertex once; Kn has (n − 1)!/2 of them. A tree on n vertices has n − 1 edges, and a connected graph with m edges keeps a spanning tree after removing m − n + 1 edges.

Edges of a spanning tree

20 − 1 = 19

Edges to remove

k = 20 − 19 = 1

Answer

k = 1
6.Graph theory

Answer: d(F) = 27

Shortest paths and spanning trees: Dijkstra’s algorithm settles one vertex per round: the unsettled vertex with the smallest tentative distance. Its distance is then final, and every neighbour v gets d(v) = min(d(v), d(u) + w(u, v)). Kruskal’s algorithm builds a minimum spanning tree by taking the edges from light to heavy and skipping any edge that would close a cycle; Prim’s algorithm grows one tree from a start vertex by always adding the lightest edge that leaves it. A spanning tree of n vertices has n − 1 edges.

Dijkstra round by round

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 11, C: 8
C (8): E: 23, G: 21
B (11): D: 26
G (21)
E (23)
D (26): F: 27
F (27)

Shortest path

A → B → D → F
d(F) = 27

Answer

d(F) = 27

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