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

Breadth-first search on an unweighted graph finds:

  1. AThe minimum spanning tree
  2. BA topological order in all graphs
  3. CShortest paths in graphs that contain negative edge weights
  4. DShortest paths in number of edges from the start node
2.Algorithms & complexityHard

What is the time complexity of: for i = 1 to n: for (j = i; j ≤ n; j = j + i) do O(1) work?

  1. AO(n²)
  2. BO(n √n)
  3. CO(n log n)
  4. DO(n)
3.Data structuresMedium

Which collision resolution method stores colliding keys in a list at the same bucket?

  1. ALinear probing
  2. BDouble hashing
  3. CSeparate chaining
  4. DQuadratic probing
4.Data structuresHard

The time complexity of decrease-key in a binary heap (given the element's position) is:

  1. AO(1)
  2. BO(log n)
  3. CO(n log n)
  4. DO(n)
5.Graph theoryMedium
ABCDEFEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 6|E| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
ABCDE

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

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: D · Shortest paths in number of edges from the start node

Why this is the answer

BFS visits nodes in order of distance (edges) from the source, so it finds shortest unweighted paths in O(V + E).

Answer

D · Shortest paths in number of edges from the start node
2.Algorithms & complexity

Answer: C · O(n log n)

Why this is the answer

The total is n / 1 + n / 2 + ... + n / n = n × Hn, and the harmonic number Hn is Θ(log n).

Answer

C · O(n log n)
3.Data structures

Answer: C · Separate chaining

Why this is the answer

Chaining keeps a linked list (or other structure) per bucket; open addressing looks for another free slot in the table itself.

Answer

C · Separate chaining
4.Data structures

Answer: B · O(log n)

Why this is the answer

After decreasing the key, the element sifts up at most the height of the heap. Fibonacci heaps achieve O(1) amortised decrease-key.

Answer

B · O(log n)
5.Graph theory

Answer: neither

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.

Degrees

deg(A) = 1, deg(B) = 3, deg(C) = 1, deg(D) = 2, deg(E) = 4, deg(F) = 1

Odd vertices

A, B, C, F → 4
neither

Answer

neither
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 1 1 1
C: 1 1 0 0 0
D: 0 1 0 0 1
E: 0 1 0 1 0

Common neighbours

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

Answer

N = 2

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