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

Is it correct to say that an O(n²) algorithm is also O(n³)?

  1. ANo, because Big O must always state the exact growth rate of the function
  2. BYes, but only if n is smaller than 100
  3. CYes, because Big O is an upper bound and n² grows no faster than n³
  4. DNo, because n³ is smaller than n² for large n
2.Algorithms & complexityHard

The Held–Karp dynamic programming algorithm solves the travelling salesman problem exactly. What is its running time for n cities?

  1. AO(n² 2ⁿ)
  2. BO(n!)
  3. CO(2n²)
  4. DO(n³)
3.Data structuresMedium

How do you find the minimum key in a binary search tree?

  1. AVisit every leaf and compare them
  2. BFollow left children from the root until there is no left child
  3. CFollow right children from the root until there is no right child
  4. DLook at the root only
4.Data structuresHard

The keys 5, 3, 8, 1 are inserted in this order into an empty array-based binary min-heap. What is the final array?

  1. A[1, 3, 5, 8]
  2. B[1, 5, 8, 3]
  3. C[3, 1, 8, 5]
  4. D[1, 3, 8, 5]
5.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
7125918142048ABCDEFG

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

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

3Solutions

1.Algorithms & complexity

Answer: C · Yes, because Big O is an upper bound and n² grows no faster than n³

Why this is the answer

Big O is an upper bound, so the statement is true but not tight; Θ(n²) states the exact growth rate.

Answer

C · Yes, because Big O is an upper bound and n² grows no faster than n³
2.Algorithms & complexity

Answer: A · O(n² 2ⁿ)

Why this is the answer

There are about n 2ⁿ states, each computed in O(n); this is exponential but far faster than trying all (n − 1)! tours.

Answer

A · O(n² 2ⁿ)
3.Data structures

Answer: B · Follow left children from the root until there is no left child

Why this is the answer

The leftmost node has no smaller key in the tree, so it is the minimum; the cost is proportional to the height.

Answer

B · Follow left children from the root until there is no left child
4.Data structures

Answer: D · [1, 3, 8, 5]

Why this is the answer

After 5, 3, 8 the array is [3, 5, 8]; 1 goes to index 3, swaps with its parent 5 and then with 3, giving [1, 3, 8, 5].

Answer

D · [1, 3, 8, 5]
5.Graph theory

Answer: Euler path, no 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) = 3, deg(C) = 2, deg(D) = 1, deg(E) = 2

Odd vertices

B, D → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

6.Graph theory

Answer: edge DF

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. BD (5)
2. BC (2)
3. AC (1)
4. GC (4)
5. GE (8)
6. DF (14)

edge number 6

DF

Answer

edge DF

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