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

A recurrence T(n) = 2T(n/2) + O(n) solves to:

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

How do n! and 2ⁿ compare for large n?

  1. AThey grow at the same rate
  2. Bn! is O(n²)
  3. Cn! grows faster than 2ⁿ
  4. D2ⁿ grows faster than n!
3.Data structuresMedium

What is the worst-case height of a binary search tree with n nodes if keys are inserted in sorted order?

  1. An (it becomes a linked list)
  2. Blog₂ n
  3. C1
  4. D√n
4.Data structuresHard

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

  1. AA max-heap ordered by out-degree
  2. BA stack of all visited edges
  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

How many different Hamilton circuits H does the complete graph K5 have? Circuits that differ only in the starting vertex or the direction count as the same.

K5
H =
6.Graph theoryHard
19131772083114ABCDEFG

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 7|E| = 9
W =

3Solutions

1.Algorithms & complexity

Answer: C · O(n log n)

Why this is the answer

By the master theorem (a = 2, b = 2, f(n) = n), T(n) = Θ(n log n).

Answer

C · O(n log n)
2.Algorithms & complexity

Answer: C · n! grows faster than 2ⁿ

Why this is the answer

For n ≥ 4, n! exceeds 2ⁿ, and the ratio grows without bound because each extra factor of n! is larger than 2.

Answer

C · n! grows faster than 2ⁿ
3.Data structures

Answer: A · n (it becomes a linked list)

Why this is the answer

Sorted insertions create a degenerate tree with height n, making searches O(n). Balanced trees prevent this.

Answer

A · n (it becomes a linked list)
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: H = 12

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 = (5 − 1)!/2 = 24/2 = 12

Answer

H = 12
6.Graph theory

Answer: W = 52

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

DF (3): take
GD (4): take
BD (7): take
DE (8): take
EF (11): skip, closes a cycle
AC (13): take
BC (17): take

Total weight

W = 3 + 4 + 7 + 8 + 13 + 17 = 52

Answer

W = 52

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