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 running time of LSD radix sort on n integers with d digits in base k?

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

Karatsuba multiplication satisfies T(n) = 3T(n/2) + O(n). What does this solve to?

  1. AΘ(n log n)
  2. BΘ(n²)
  3. CΘ(nlog₂ 3), about n1.585
  4. DΘ(n3)
3.Data structuresMedium

How much memory does an adjacency list use for a graph with V vertices and E edges?

  1. AO(E²)
  2. BO(V²)
  3. CO(V + E)
  4. DO(V × E)
4.Data structuresHard

How long does it take to search for an arbitrary key in a binary min-heap of n elements?

  1. AO(log n)
  2. BO(1)
  3. CO(log log n)
  4. DO(n)
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
ABCDE

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

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: C · O(d(n + k))

Why this is the answer

Each of the d passes is a counting sort costing O(n + k), so for fixed d and k the total is linear in n.

Answer

C · O(d(n + k))
2.Algorithms & complexity

Answer: C · Θ(nlog₂ 3), about n1.585

Why this is the answer

Since log₂ 3 ≈ 1.585 > 1, the leaves dominate by the master theorem, beating the O(n²) schoolbook method.

Answer

C · Θ(nlog₂ 3), about n1.585
3.Data structures

Answer: C · O(V + E)

Why this is the answer

Each vertex has a list and each edge appears once (directed) or twice (undirected), giving O(V + E).

Answer

C · O(V + E)
4.Data structures

Answer: D · O(n)

Why this is the answer

A heap gives no rule for which subtree contains a key, so in the worst case all nodes must be checked; only the minimum is fast to find.

Answer

D · O(n)
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 1 1 0 0
B: 1 0 1 1 1
C: 1 1 0 0 0
D: 0 1 0 0 1
E: 0 1 0 1 0

Common neighbours

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

Answer

N = 4

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