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

A function runs one loop of n iterations and then, separately, another loop of n iterations. What is its time complexity?

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

The 0/1 knapsack problem solved with dynamic programming takes O(nW) time. Why is this called pseudo-polynomial?

  1. AW is a number, so the time is exponential in the number of bits used to write W
  2. BBecause it only works for W = 0
  3. CBecause n is always small
  4. DBecause the algorithm is randomised and its running time varies with each input
3.Data structuresMedium

AVL trees and red-black trees guarantee:

  1. AThat keys are stored in insertion order
  2. BThat no node has two children
  3. CO(log n) height through rebalancing
  4. DConstant-time search in every case
4.Data structuresHard

A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?

  1. ABecause the memory allocator can only release memory in quarters of a block
  2. BTo avoid repeated resizing when pushes and pops alternate near the boundary
  3. CTo keep the array sorted
  4. DBecause halving at one half would lose elements
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K6 have? Circuits that differ only in the starting vertex or the direction count as the same.

K6
H =
6.Graph theoryHard
82344789ABCDEF

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: B · O(n)

Why this is the answer

n + n = 2n, and constant factors are dropped, so the function is O(n).

Answer

B · O(n)
2.Algorithms & complexity

Answer: A · W is a number, so the time is exponential in the number of bits used to write W

Why this is the answer

The running time is polynomial in the value of W but exponential in its length (log W bits); 0/1 knapsack is NP-complete.

Answer

A · W is a number, so the time is exponential in the number of bits used to write W
3.Data structures

Answer: C · O(log n) height through rebalancing

Why this is the answer

Self-balancing trees keep height logarithmic, so search, insert and delete are O(log n) in the worst case.

Answer

C · O(log n) height through rebalancing
4.Data structures

Answer: B · To avoid repeated resizing when pushes and pops alternate near the boundary

Why this is the answer

Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).

Answer

B · To avoid repeated resizing when pushes and pops alternate near the boundary
5.Graph theory

Answer: H = 60

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 = (6 − 1)!/2 = 120/2 = 60

Answer

H = 60
6.Graph theory

Answer: d(F) = 14

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: 8, C: 2
C (2): E: 6
E (6): D: 10, F: 14
B (8)
D (10)
F (14)

Shortest path

A → C → E → F
d(F) = 14

Answer

d(F) = 14

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