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 the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?

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

Insertion sort runs in O(n) on sorted input. Does this contradict the Ω(n log n) lower bound for comparison sorting?

  1. ANo, because insertion sort is not a comparison sort
  2. BYes, but only for arrays shorter than 100
  3. CYes, the lower bound is therefore wrong
  4. DNo, the lower bound applies to the worst case over all inputs, not to every input
3.Data structuresMedium

How long does it take to check whether the edge (u, v) exists using an adjacency matrix?

  1. AO(V)
  2. BO(log V)
  3. CO(1)
  4. DO(E)
4.Data structuresHard

Kahn's algorithm for topological sorting repeatedly removes vertices with in-degree 0. Which structure holds those vertices?

  1. AA stack of all visited edges
  2. BA max-heap ordered by out-degree
  3. CA queue (or any collection) of vertices whose in-degree is 0
  4. DAn adjacency matrix recording which vertices have already been removed
5.Graph theoryMedium

A connected graph has 8 vertices and 10 edges. How many edges k must be removed to leave a spanning tree?

|V| = 8|E| = 10
k =
6.Graph theoryHard
ABCDE

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

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: A · O(mn)

Why this is the answer

The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.

Answer

A · O(mn)
2.Algorithms & complexity

Answer: D · No, the lower bound applies to the worst case over all inputs, not to every input

Why this is the answer

The bound says every comparison sort needs about n log n comparisons on some input of size n; special inputs can be much faster.

Answer

D · No, the lower bound applies to the worst case over all inputs, not to every input
3.Data structures

Answer: C · O(1)

Why this is the answer

The matrix entry at row u, column v tells you immediately whether the edge exists; an adjacency list may need to scan u's neighbours.

Answer

C · O(1)
4.Data structures

Answer: C · A queue (or any collection) of vertices whose in-degree is 0

Why this is the answer

Ready vertices are kept in a queue; removing one lowers its neighbours' in-degrees, and any that reach 0 are added, giving O(V + E) time.

Answer

C · A queue (or any collection) of vertices whose in-degree is 0
5.Graph theory

Answer: k = 3

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.

Edges of a spanning tree

8 − 1 = 7

Edges to remove

k = 10 − 7 = 3

Answer

k = 3
6.Graph theory

Answer: N = 2

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 0 1 1
C: 1 0 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0

Common neighbours

(M²)DD = 0·0 + 1·1 + 0·0 + 0·0 + 1·1 = 2

Answer

N = 2

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