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 is the amortised time complexity of appending to a dynamic array that doubles its capacity when full?

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

How is the travelling salesman problem classified?

  1. AIt is undecidable
  2. BThe decision version is NP-complete and the optimisation version is NP-hard
  3. CIt is in P because Dijkstra's algorithm solves it by visiting every city once
  4. DThe decision version is in P but the optimisation version is NP-complete
3.Data structuresMedium

How many nodes does a perfect binary tree of height h have, with height counted in edges?

  1. A2h+1 − 1
  2. B2h + 1
  3. C2h
  4. Dh2
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 keep the array sorted
  3. CBecause halving at one half would lose elements
  4. DTo avoid repeated resizing when pushes and pops alternate near the boundary
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| = 8
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
19413171510181163ABCDEFG

Prim’s algorithm builds a minimum spanning tree starting from vertex B. Click the edge it adds as edge number 6.

start: Bk = 6
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge EF
  8. 8edge GB
  9. 9edge GC
  10. 10edge GE

3Solutions

1.Algorithms & complexity

Answer: B · O(1)

Why this is the answer

Copying all elements happens only at powers of two; spread over all appends, the average cost per append is constant.

Answer

B · O(1)
2.Algorithms & complexity

Answer: B · The decision version is NP-complete and the optimisation version is NP-hard

Why this is the answer

"Is there a tour of length at most k?" is NP-complete; finding the shortest tour is NP-hard, since it is at least as hard.

Answer

B · The decision version is NP-complete and the optimisation version is NP-hard
3.Data structures

Answer: A · 2h+1 − 1

Why this is the answer

Summing 1 + 2 + 4 + ... + 2h gives 2h+1 − 1; for example, height 2 gives 7 nodes.

Answer

A · 2h+1 − 1
4.Data structures

Answer: D · 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

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

Answer: neither

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) = 1, deg(B) = 2, deg(C) = 3, deg(D) = 3, deg(E) = 2, deg(F) = 2, deg(G) = 3

Odd vertices

A, C, D, G → 4
neither

Answer

neither
6.Graph theory

Answer: edge EF

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.

Prim’s algorithm

1. GB (11)
2. GE (3)
3. GC (6)
4. AC (4)
5. DE (10)
6. EF (18)

edge number 6

EF

Answer

edge EF

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