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 time complexity of multiplying two n × n matrices with the standard triple loop?

  1. AO(n²)
  2. BO(n³)
  3. CO(n² log n)
  4. DO(n)
2.Algorithms & complexityHard

The halting problem shows that:

  1. ASome programs run faster on faster computers, so halting depends on hardware
  2. BNo algorithm can decide for every program and input whether it halts
  3. CLoops should be avoided
  4. DEvery program eventually halts
3.Data structuresMedium

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

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

A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?

  1. ABecause the memory allocator can only release memory in quarters of a block
  2. BTo avoid repeated resizing when pushes and pops alternate near the boundary
  3. CBecause halving at one half would lose elements
  4. DTo keep the array sorted
5.Graph theoryMedium

A connected graph has 13 vertices and 15 edges. How many edges k must be removed to leave a spanning tree?

|V| = 13|E| = 15
k =
6.Graph theoryHard
ABCDEF

How many walks of length 3 go from F to B? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =

3Solutions

1.Algorithms & complexity

Answer: B · O(n³)

Why this is the answer

There are n² entries and each needs n multiplications, giving n³ operations.

Answer

B · O(n³)
2.Algorithms & complexity

Answer: B · No algorithm can decide for every program and input whether it halts

Why this is the answer

Turing showed the halting problem is undecidable: assuming a decider exists leads to a contradiction with a program that does the opposite.

Answer

B · No algorithm can decide for every program and input whether it halts
3.Data structures

Answer: B · 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

B · Adjacency list
4.Data structures

Answer: B · To avoid repeated resizing when pushes and pops alternate near the boundary

Why this is the answer

Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).

Answer

B · To avoid repeated resizing when pushes and pops alternate near the boundary
5.Graph theory

Answer: k = 3

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

13 − 1 = 12

Edges to remove

k = 15 − 12 = 3

Answer

k = 3
6.Graph theory

Answer: N = 1

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 0
B: 1 0 1 0 0 0
C: 1 1 0 0 1 0
D: 0 0 0 0 1 1
E: 0 0 1 1 0 1
F: 0 0 0 1 1 0

One more step

(M³)FB = 0·1 + 0·0 + 1·1 + 1·0 + 1·0 + 2·0 = 1

Answer

N = 1

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