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 does the recurrence T(n) = 2T(n − 1) + O(1) solve to?

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

How do n! and 2ⁿ compare for large n?

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

In a hash table using open addressing, what is the maximum possible load factor?

  1. AThere is no maximum
  2. B2
  3. C1
  4. D0.5
4.Data structuresHard

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

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

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

3 × deg 34 × deg 42 × deg 2
L =
6.Graph theoryHard
611115151410ABCDEF

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: A · O(2ⁿ)

Why this is the answer

The number of calls doubles with each level and there are n levels, so the total is about 2ⁿ, as in the Towers of Hanoi.

Answer

A · O(2ⁿ)
2.Algorithms & complexity

Answer: C · 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

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

Answer: C · 1

Why this is the answer

With open addressing all keys are stored in the table itself, so n cannot exceed m; separate chaining can have a load factor above 1.

Answer

C · 1
4.Data structures

Answer: A · 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

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

Answer: L = 13

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 + 29 = 2(L + 9) − 2
L = 13

Answer

L = 13

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

6.Graph theory

Answer: d(F) = 22

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: 6, C: 11
B (6): C: 7, D: 21, E: 16
C (7): E: 8
E (8): F: 22
D (21)
F (22)

Shortest path

A → B → C → E → F
d(F) = 22

Answer

d(F) = 22

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