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) = 2T(n/2) + O(1) solve to?

  1. AO(n log n)
  2. BO(log n)
  3. CO(n)
  4. DO(n²)
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, the lower bound is therefore wrong
  3. CNo, the lower bound applies to the worst case over all inputs, not to every input
  4. DYes, but only for arrays shorter than 100
3.Data structuresMedium

What is true of the adjacency matrix of an undirected graph?

  1. AIt is symmetric about the main diagonal
  2. BEach row has exactly one 1
  3. CIts main diagonal always contains ones
  4. DIt is always upper triangular
4.Data structuresHard

In Kruskal's minimum spanning tree algorithm, what is the union-find structure used for?

  1. AChecking whether an edge's two endpoints are already connected
  2. BSorting the edges by weight
  3. CChoosing the starting vertex
  4. DStoring each vertex's current distance from the source during the search
5.Graph theoryMedium

A tree has 2 vertices of degree 3, 2 vertices of degree 4, 3 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?

2 × deg 32 × deg 43 × deg 5
L =
6.Graph theoryHard
111310132153154ABCDEFG

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =

3Solutions

1.Algorithms & complexity

Answer: C · O(n)

Why this is the answer

By the master theorem nlog₂ 2 = n dominates the constant work, so T(n) = Θ(n), as when finding a maximum by splitting in halves.

Answer

C · O(n)
2.Algorithms & complexity

Answer: C · 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

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

Answer: A · It is symmetric about the main diagonal

Why this is the answer

In an undirected graph edge {u, v} means both M[u][v] and M[v][u] are set, so the matrix equals its transpose.

Answer

A · It is symmetric about the main diagonal
4.Data structures

Answer: A · Checking whether an edge's two endpoints are already connected

Why this is the answer

Find tells whether both endpoints are in the same component; if not, the edge is added and union merges the components.

Answer

A · Checking whether an edge's two endpoints are already connected
5.Graph theory

Answer: L = 17

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.

Handshake lemma for a tree

L + 29 = 2(L + 7) − 2
L = 17

Answer

L = 17

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

6.Graph theory

Answer: d(F) = 33

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 11
B (11): C: 24, D: 21, G: 14
G (14): D: 18
D (18): E: 20, F: 33
E (20)
C (24)
F (33)

Shortest path

A → B → G → D → F
d(F) = 33

Answer

d(F) = 33

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