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 does the recurrence T(n) = T(n − 1) + O(n) solve to?

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

Which technique gives an expected O(n log n) running time for quicksort on any input?

  1. AUsing recursion only on the left side
  2. BAlways choosing the first element
  3. CChoosing the pivot uniformly at random
  4. DSorting the input before running quicksort
3.Data structuresMedium

A road network has 100,000 junctions, each connected to about 4 others. Which representation is most suitable?

  1. AA binary heap of edges
  2. BAdjacency matrix
  3. CAdjacency list
  4. DA single sorted array of vertices
4.Data structuresHard

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

  1. AO(1)
  2. BO(n log n)
  3. CO(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
2161312910414ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 4; A itself is round 1.

start: Around 4
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F

3Solutions

1.Algorithms & complexity

Answer: C · O(n²)

Why this is the answer

The work sums to about n² / 2, which is the recurrence of selection sort or of quicksort with the worst pivot.

Answer

C · O(n²)
2.Algorithms & complexity

Answer: C · Choosing the pivot uniformly at random

Why this is the answer

With a random pivot, the expected running time is O(n log n) for every input; bad cases still exist but are very unlikely.

Answer

C · Choosing the pivot uniformly at random
3.Data structures

Answer: C · Adjacency list

Why this is the answer

A matrix would need 1010 cells, almost all empty; an adjacency list only stores the roughly 400,000 edge entries.

Answer

C · Adjacency list
4.Data structures

Answer: C · 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

C · O(log 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: E

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

1. A (0): B: 2, C: 1
2. C (1): E: 13
3. B (2): D: 15
4. E (13): F: 17
5. D (15)
6. F (17)

round 4

E

Answer

E

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