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

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. COnly the colliding keys are moved to a new overflow list at the end of the table
  4. DA larger table is allocated and every key is inserted again
4.Data structuresHard

In a circular buffer, front == rear can mean either empty or full. What is a common way to tell them apart?

  1. AMake the buffer size a power of two
  2. BAlways clear the buffer whenever front equals rear, so the state is empty
  3. CStore the front and rear in the same variable
  4. DKeep a separate element count, or always leave one slot unused
5.Graph theoryMedium

A connected graph has 19 vertices and 21 edges. How many edges k must be removed to leave a spanning tree?

|V| = 19|E| = 21
k =
6.Graph theoryHard
58121094149ABCDEF

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: 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: 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: D · 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

D · A larger table is allocated and every key is inserted again
4.Data structures

Answer: D · Keep a separate element count, or always leave one slot unused

Why this is the answer

With a count, empty means count = 0 and full means count = capacity; with one empty slot, full is defined as (rear + 1) mod n == front.

Answer

D · Keep a separate element count, or always leave one slot unused
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

19 − 1 = 18

Edges to remove

k = 21 − 18 = 3

Answer

k = 3
6.Graph theory

Answer: d(F) = 19

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: 5, C: 8
B (5): D: 15, E: 14
C (8)
E (14): F: 28
D (15): F: 19
F (19)

Shortest path

A → B → D → F
d(F) = 19

Answer

d(F) = 19

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