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

The lower bound for comparison-based sorting in the worst case is:

  1. AΩ(n²)
  2. BΩ(log n)
  3. CΩ(n)
  4. DΩ(n log n)
2.Algorithms & complexityHard

What would follow if someone found a polynomial-time algorithm for one NP-complete problem?

  1. AEvery problem in NP could be solved in polynomial time, so P = NP
  2. BOnly that one problem would move into P
  3. CThe halting problem would become decidable
  4. DAll exponential-time algorithms would immediately run in linear time
3.Data structuresMedium

Why can a binary heap be stored in an array without any pointers?

  1. AArrays automatically keep the keys in sorted order, so no pointers are needed
  2. BHeaps never contain more than 100 elements
  3. CIt is a complete binary tree, so positions follow simple index formulas
  4. DAll keys in a heap are equal
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(1 + α), where α = n / m
  2. BO(m)
  3. CO(log n)
  4. DO(n)
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
6.Graph theoryHard
ABCDE

How many closed walks of length 2 start and end at E? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: D · Ω(n log n)

Why this is the answer

Any comparison sort needs at least log₂(n!) ≈ n log n comparisons in the worst case. Counting sort beats it because it does not compare.

Answer

D · Ω(n log n)
2.Algorithms & complexity

Answer: A · Every problem in NP could be solved in polynomial time, so P = NP

Why this is the answer

Since all of NP reduces to any NP-complete problem in polynomial time, a fast algorithm for one gives a fast algorithm for all.

Answer

A · Every problem in NP could be solved in polynomial time, so P = NP
3.Data structures

Answer: C · It is a complete binary tree, so positions follow simple index formulas

Why this is the answer

Completeness means there are no gaps, so the children of index i are always at 2i + 1 and 2i + 2.

Answer

C · It is a complete binary tree, so positions follow simple index formulas
4.Data structures

Answer: A · 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

A · O(1 + α), where α = n / m
5.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
6.Graph theory

Answer: N = 3

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

Common neighbours

(M²)EE = 0·0 + 1·1 + 1·1 + 1·1 + 0·0 = 3

Answer

N = 3

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