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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 does amortised analysis measure?

  1. AThe cost of an algorithm on the best input
  2. BThe expected cost over random inputs
  3. CThe average cost per operation over a worst-case sequence of operations
  4. DThe cost of the single most expensive operation anywhere in the sequence
2.Algorithms & complexityHard

How do n! and 2ⁿ compare for large n?

  1. A2ⁿ grows faster than n!
  2. Bn! is O(n²)
  3. CThey grow at the same rate
  4. Dn! grows faster than 2ⁿ
3.Data structuresMedium

How many nodes does a perfect binary tree of height h have, with height counted in edges?

  1. A2h + 1
  2. Bh2
  3. C2h
  4. D2h+1 − 1
4.Data structuresHard

The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:

  1. AO(n)
  2. BO(n²)
  3. CO(α(n)), the inverse Ackermann function, which is practically constant
  4. DO(log n) exactly
5.Graph theoryMedium

A tree has 2 vertices of degree 4, 2 vertices of degree 5, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

2 × deg 42 × deg 53 × deg 2
L =
6.Graph theoryHard
3415612115356ABCDEFG

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: C · The average cost per operation over a worst-case sequence of operations

Why this is the answer

Amortised analysis spreads the cost of rare expensive operations over many cheap ones, giving a guaranteed average for any sequence.

Answer

C · The average cost per operation over a worst-case sequence of operations
2.Algorithms & complexity

Answer: D · n! grows faster than 2ⁿ

Why this is the answer

For n ≥ 4, n! exceeds 2ⁿ, and the ratio grows without bound because each extra factor of n! is larger than 2.

Answer

D · n! grows faster than 2ⁿ
3.Data structures

Answer: D · 2h+1 − 1

Why this is the answer

Summing 1 + 2 + 4 + ... + 2h gives 2h+1 − 1; for example, height 2 gives 7 nodes.

Answer

D · 2h+1 − 1
4.Data structures

Answer: C · O(α(n)), the inverse Ackermann function, which is practically constant

Why this is the answer

With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.

Answer

C · O(α(n)), the inverse Ackermann function, which is practically constant
5.Graph theory

Answer: L = 12

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.

Handshake lemma for a tree

L + 24 = 2(L + 7) − 2
L = 12

Answer

L = 12

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

6.Graph theory

Answer: d(F) = 25

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: 3
B (3): C: 7, G: 8
C (7): E: 22
G (8): D: 13, E: 14
D (13): F: 25
E (14)
F (25)

Shortest path

A → B → G → D → F
d(F) = 25

Answer

d(F) = 25

ExamProblems · examproblems.com/courses/data-structures-and-algorithms