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

Using the master theorem, T(n) = 8T(n/2) + O(n²) solves to:

  1. AΘ(n log n)
  2. BΘ(n²)
  3. CΘ(n³)
  4. DΘ(n² log n)
3.Data structuresMedium

Why is traversing an array often much faster in practice than traversing a linked list of the same length?

  1. AArrays use fewer instructions to compare values
  2. BArray traversal is O(1) in total, while linked list traversal is O(n)
  3. CLinked lists always store elements on disk
  4. DArray elements are contiguous, so they make good use of the CPU cache
4.Data structuresHard

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

  1. AO(log n)
  2. BO(n)
  3. CO(1)
  4. DO(n log n)
5.Graph theoryMedium

A connected graph has 18 vertices and 19 edges. How many edges k must be removed to leave a spanning tree?

|V| = 18|E| = 19
k =
6.Graph theoryHard
10131256215419ABCDEFG

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 4; skipped edges do not count.

k = 4
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge DE
  5. 5edge DF
  6. 6edge EF
  7. 7edge GB
  8. 8edge GC
  9. 9edge GE

3Solutions

1.Algorithms & complexity

Answer: B · 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

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

Answer: C · Θ(n³)

Why this is the answer

log₂ 8 = 3, and n² grows slower than n³, so the leaves dominate: Θ(n³).

Answer

C · Θ(n³)
3.Data structures

Answer: D · Array elements are contiguous, so they make good use of the CPU cache

Why this is the answer

Contiguous memory gives good locality of reference, while linked nodes may be scattered and cause many cache misses.

Answer

D · Array elements are contiguous, so they make good use of the CPU cache
4.Data structures

Answer: A · 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

A · O(log n)
5.Graph theory

Answer: k = 2

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

18 − 1 = 17

Edges to remove

k = 19 − 17 = 2

Answer

k = 2
6.Graph theory

Answer: edge AB

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.

Kruskal’s algorithm

1. EF (2)
2. GC (4)
3. DE (5)
4. AB (10)
5. BC (12)
6. GE (19)

edge number 4

AB

Answer

edge AB

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