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

A recurrence T(n) = 2T(n/2) + O(n) solves to:

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

Which statement about little-o notation is correct?

  1. An = o(n), since every function is little-o of itself
  2. B2n = o(n), because 2n grows slowly
  3. Cn = o(n²), but n is not o(n)
  4. Dn² = o(n)
3.Data structuresMedium

What happens when a hash table rehashes after its load factor exceeds a threshold?

  1. AThe oldest keys are deleted to make room
  2. BThe existing keys are sorted in place
  3. CA larger table is allocated and every key is inserted again
  4. DOnly the colliding keys are moved to a new overflow list at the end of the table
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(n²)
  2. BO(n)
  3. CO(log n) exactly
  4. DO(α(n)), the inverse Ackermann function, which is practically constant
5.Graph theoryMedium
ABCDEFEuler 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| = 6|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
1134151571431212ABCDEFG

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: A · O(n log n)

Why this is the answer

By the master theorem (a = 2, b = 2, f(n) = n), T(n) = Θ(n log n).

Answer

A · O(n log n)
2.Algorithms & complexity

Answer: C · n = o(n²), but n is not o(n)

Why this is the answer

f = o(g) means f(n) / g(n) tends to 0; n / n² tends to 0, but n / n = 1 does not.

Answer

C · n = o(n²), but n is not o(n)
3.Data structures

Answer: C · A larger table is allocated and every key is inserted again

Why this is the answer

Since the index is computed modulo the table size, all keys must be rehashed into the new, larger table; with doubling this stays O(1) amortised.

Answer

C · A larger table is allocated and every key is inserted again
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) = 2, deg(D) = 2, deg(E) = 2, deg(F) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
6.Graph theory

Answer: d(F) = 32

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, C: 3
C (3): B: 7, E: 18, G: 15
B (7): D: 22, G: 10
G (10)
E (18): F: 32
D (22)
F (32)

Shortest path

A → C → E → F
d(F) = 32

Answer

d(F) = 32

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