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 computing the n-th Fibonacci number with a bottom-up table?

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

How is the travelling salesman problem classified?

  1. AThe decision version is NP-complete and the optimisation version is NP-hard
  2. BIt is undecidable
  3. CThe decision version is in P but the optimisation version is NP-complete
  4. DIt is in P because Dijkstra's algorithm solves it by visiting every city once
3.Data structuresMedium

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

  1. Ah2
  2. B2h
  3. C2h + 1
  4. D2h+1 − 1
4.Data structuresHard

Under simple uniform hashing, what is the expected time of a search in a hash table with chaining, n keys and m buckets?

  1. AO(n)
  2. BO(1 + α), where α = n / m
  3. CO(log n)
  4. DO(m)
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K8 have? Circuits that differ only in the starting vertex or the direction count as the same.

K8
H =
6.Graph theoryHard
ABCDEF

How many closed walks of length 2 start and end at E? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: C · O(n)

Why this is the answer

Filling the table from 0 to n takes one addition per entry, so the work is linear instead of exponential.

Answer

C · O(n)
2.Algorithms & complexity

Answer: A · The decision version is NP-complete and the optimisation version is NP-hard

Why this is the answer

"Is there a tour of length at most k?" is NP-complete; finding the shortest tour is NP-hard, since it is at least as hard.

Answer

A · The decision version is NP-complete and the optimisation version is NP-hard
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: B · O(1 + α), where α = n / m

Why this is the answer

Computing the hash costs O(1) and the expected chain length is α, so keeping α bounded by resizing gives O(1) expected time.

Answer

B · O(1 + α), where α = n / m
5.Graph theory

Answer: H = 2520

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 = (8 − 1)!/2 = 5040/2 = 2520

Answer

H = 2520
6.Graph theory

Answer: N = 4

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 0 1 0 0 0
B: 0 0 1 0 1 0
C: 1 1 0 0 1 0
D: 0 0 0 0 1 1
E: 0 1 1 1 0 1
F: 0 0 0 1 1 0

Common neighbours

(M²)EE = 0·0 + 1·1 + 1·1 + 1·1 + 0·0 + 1·1 = 4

Answer

N = 4

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