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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 amortised analysis measure?

  1. AThe average cost per operation over a worst-case sequence of operations
  2. BThe cost of an algorithm on the best input
  3. CThe expected cost over random inputs
  4. DThe cost of the single most expensive operation anywhere in the sequence
2.Algorithms & complexityHard

Why is the average lookup in a hash table O(1) but the worst case O(n)?

  1. AArrays have O(n) access
  2. BMany keys may collide into the same bucket
  3. CKeys must be sorted first
  4. DHash functions are slow
3.Data structuresMedium

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

  1. AIt is a complete binary tree, so positions follow simple index formulas
  2. BArrays automatically keep the keys in sorted order, so no pointers are needed
  3. CHeaps never contain more than 100 elements
  4. DAll keys in a heap are equal
4.Data structuresHard

The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:

  1. AO(n)
  2. BO(log n) exactly
  3. CO(n²)
  4. DO(α(n)), the inverse Ackermann function, which is practically constant
5.Graph theoryMedium
ABCDEFGEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 7|E| = 12
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
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: A · The average cost per operation over a worst-case sequence of operations

Why this is the answer

Amortised analysis spreads the cost of rare expensive operations over many cheap ones, giving a guaranteed average for any sequence.

Answer

A · The average cost per operation over a worst-case sequence of operations
2.Algorithms & complexity

Answer: B · Many keys may collide into the same bucket

Why this is the answer

With a good hash function and load factor, buckets stay short on average; adversarial or poor hashing can put all keys in one bucket.

Answer

B · Many keys may collide into the same bucket
3.Data structures

Answer: A · 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

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

Answer: D · O(α(n)), the inverse Ackermann function, which is practically constant

Why this is the answer

With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.

Answer

D · O(α(n)), the inverse Ackermann function, which is practically constant
5.Graph theory

Answer: Euler circuit

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.

Degrees

deg(A) = 2, deg(B) = 4, deg(C) = 4, deg(D) = 4, deg(E) = 4, deg(F) = 2, deg(G) = 4

Odd vertices

0
Euler circuit

Answer

Euler circuit
6.Graph theory

Answer: N = 2

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

Common neighbours

(M²)EE = 0·0 + 1·1 + 0·0 + 1·1 + 0·0 = 2

Answer

N = 2

ExamProblems · examproblems.com/courses/data-structures-and-algorithms