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

The halting problem shows that:

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

Which data structure does a running program use to keep track of active function calls and their local variables?

  1. AThe heap memory queue
  2. BThe call stack
  3. CA hash table of functions
  4. DA binary search tree of frames
4.Data structuresHard

Under simple uniform hashing, what is the expected time of a search in a hash table with chaining, n keys and m buckets?

  1. AO(log n)
  2. BO(n)
  3. CO(1 + α), where α = n / m
  4. DO(m)
5.Graph theoryMedium

A connected graph has 17 vertices and 19 edges. How many edges k must be removed to leave a spanning tree?

|V| = 17|E| = 19
k =
6.Graph theoryHard
ABCDEF

How many walks of length 2 go from C to F? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

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: A · 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

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

Answer: B · The call stack

Why this is the answer

Each call pushes a stack frame and each return pops it, which is why the most recent call finishes first.

Answer

B · The call stack
4.Data structures

Answer: C · O(1 + α), where α = n / m

Why this is the answer

Computing the hash costs O(1) and the expected chain length is α, so keeping α bounded by resizing gives O(1) expected time.

Answer

C · O(1 + α), where α = n / m
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

17 − 1 = 16

Edges to remove

k = 19 − 16 = 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 0 1 0 0 0
B: 0 0 1 1 0 0
C: 1 1 0 0 1 0
D: 0 1 0 0 1 1
E: 0 0 1 1 0 1
F: 0 0 0 1 1 0

Common neighbours

(M²)CF = 1·0 + 1·0 + 0·0 + 0·1 + 1·1 + 0·0 = 1

Answer

N = 1

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