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

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

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

What is the time complexity of: for i = 1 to n: for (j = i; j ≤ n; j = j + i) do O(1) work?

  1. AO(n √n)
  2. BO(n²)
  3. CO(n)
  4. DO(n log n)
3.Data structuresMedium

Why can a binary heap be stored in an array without any pointers?

  1. AHeaps never contain more than 100 elements
  2. BAll keys in a heap are equal
  3. CArrays automatically keep the keys in sorted order, so no pointers are needed
  4. DIt is a complete binary tree, so positions follow simple index formulas
4.Data structuresHard

What is the main advantage of a skip list over a sorted linked list?

  1. AIt uses no extra memory
  2. BIt stores keys without order
  3. CGuaranteed O(1) search because every node stores a pointer to every other node
  4. DExpected O(log n) search by using several levels of forward pointers
5.Graph theoryMedium

A connected graph has 15 vertices and 21 edges. How many edges k must be removed to leave a spanning tree?

|V| = 15|E| = 21
k =
6.Graph theoryHard
824171462012ABCDEF

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 6|E| = 8
W =

3Solutions

1.Algorithms & complexity

Answer: C · O(n)

Why this is the answer

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

Answer

C · O(n)
2.Algorithms & complexity

Answer: D · O(n log n)

Why this is the answer

The total is n / 1 + n / 2 + ... + n / n = n × Hn, and the harmonic number Hn is Θ(log n).

Answer

D · O(n log n)
3.Data structures

Answer: D · It is a complete binary tree, so positions follow simple index formulas

Why this is the answer

Completeness means there are no gaps, so the children of index i are always at 2i + 1 and 2i + 2.

Answer

D · It is a complete binary tree, so positions follow simple index formulas
4.Data structures

Answer: D · Expected O(log n) search by using several levels of forward pointers

Why this is the answer

Randomised levels allow searches to skip many nodes, giving expected O(log n) search, insert and delete.

Answer

D · Expected O(log n) search by using several levels of forward pointers
5.Graph theory

Answer: k = 7

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

15 − 1 = 14

Edges to remove

k = 21 − 14 = 7

Answer

k = 7
6.Graph theory

Answer: W = 38

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.

Kruskal’s algorithm

AC (2): take
BC (4): take
DF (6): take
AB (8): skip, closes a cycle
BE (12): take
DE (14): take

Total weight

W = 2 + 4 + 6 + 12 + 14 = 38

Answer

W = 38

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