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

Why is the base of a logarithm usually left out in Big O, as in O(log n)?

  1. ALogarithms are always rounded to integers
  2. BAll logarithms in computer science have base 10, so it is implied
  3. CLogarithms with different bases differ only by a constant factor
  4. DThe base only matters for small n
2.Algorithms & complexityHard

Which statement about selecting the k-th smallest element of an unsorted array is correct?

  1. AQuickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
  2. BQuickselect runs in O(log n) worst case, because it recurses on only one side of the pivot
  3. CSelection always needs Ω(n log n) because it requires sorting
  4. DSelection is only possible in O(n²)
3.Data structuresMedium

A binary tree has root 1 with left child 2 and right child 3; node 2 has children 4 (left) and 5 (right). What is the postorder traversal?

  1. A4 5 2 3 1
  2. B5 4 3 2 1
  3. C1 2 4 5 3
  4. D4 2 5 1 3
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 guarantee that no two keys can ever collide, whatever keys an attacker sends
  3. CTo remove the need to resize the table
  4. DTo make the hash table sorted
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K6 have? Circuits that differ only in the starting vertex or the direction count as the same.

K6
H =
6.Graph theoryHard
1214913131081581413ABCDEFG

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 · Logarithms with different bases differ only by a constant factor

Why this is the answer

Changing the base multiplies the logarithm by a constant, and Big O ignores constant factors.

Answer

C · Logarithms with different bases differ only by a constant factor
2.Algorithms & complexity

Answer: A · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case

Why this is the answer

Quickselect's expected work is n + n/2 + n/4 + ... = O(n) with random pivots; choosing the pivot as a median of medians makes the worst case linear.

Answer

A · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
3.Data structures

Answer: A · 4 5 2 3 1

Why this is the answer

Postorder processes 4 and 5 before 2, then 3, and finally the root 1.

Answer

A · 4 5 2 3 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: H = 60

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.

Count

H = (6 − 1)!/2 = 120/2 = 60

Answer

H = 60
6.Graph theory

Answer: d(F) = 23

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: 12, C: 14
B (12): D: 13, G: 27
D (13): E: 26, F: 23
C (14): E: 17, G: 22
E (17)
G (22)
F (23)

Shortest path

A → B → D → F
d(F) = 23

Answer

d(F) = 23

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