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

Quicksort always uses the first element as pivot. On which input does it show its worst-case behaviour?

  1. AAn array whose elements are all different
  2. BAn array of length 2
  3. CAn array that is already sorted
  4. DAn array in random order
2.Algorithms & complexityHard

Insertion sort runs in O(n) on sorted input. Does this contradict the Ω(n log n) lower bound for comparison sorting?

  1. ANo, because insertion sort is not a comparison sort
  2. BYes, but only for arrays shorter than 100
  3. CYes, the lower bound is therefore wrong
  4. DNo, the lower bound applies to the worst case over all inputs, not to every input
3.Data structuresMedium

How long does it take to check whether the edge (u, v) exists using an adjacency matrix?

  1. AO(log V)
  2. BO(V)
  3. CO(1)
  4. DO(E)
4.Data structuresHard

Kahn's algorithm for topological sorting repeatedly removes vertices with in-degree 0. Which structure holds those vertices?

  1. AA max-heap ordered by out-degree
  2. BAn adjacency matrix recording which vertices have already been removed
  3. CA queue (or any collection) of vertices whose in-degree is 0
  4. DA stack of all visited edges
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
11518912114116ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 4; A itself is round 1.

start: Around 4
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G

3Solutions

1.Algorithms & complexity

Answer: C · An array that is already sorted

Why this is the answer

On sorted input the pivot is always the minimum, so one side is empty and the other has n − 1 elements, giving O(n²).

Answer

C · An array that is already sorted
2.Algorithms & complexity

Answer: D · No, the lower bound applies to the worst case over all inputs, not to every input

Why this is the answer

The bound says every comparison sort needs about n log n comparisons on some input of size n; special inputs can be much faster.

Answer

D · No, the lower bound applies to the worst case over all inputs, not to every input
3.Data structures

Answer: C · O(1)

Why this is the answer

The matrix entry at row u, column v tells you immediately whether the edge exists; an adjacency list may need to scan u's neighbours.

Answer

C · O(1)
4.Data structures

Answer: C · A queue (or any collection) of vertices whose in-degree is 0

Why this is the answer

Ready vertices are kept in a queue; removing one lowers its neighbours' in-degrees, and any that reach 0 are added, giving O(V + E) time.

Answer

C · A queue (or any collection) of vertices whose in-degree is 0
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) = 4, deg(D) = 2, deg(E) = 3, deg(F) = 1, deg(G) = 3

Odd vertices

A, E, F, G → 4
neither

Answer

neither
6.Graph theory

Answer: C

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

1. A (0): B: 1, C: 15
2. B (1): G: 12
3. G (12): D: 23, E: 18
4. C (15): E: 16
5. E (16): F: 28
6. D (23)
7. F (28)

round 4

C

Answer

C

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