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

About how many comparisons does binary search need in the worst case on a sorted array of 1,000,000 elements?

  1. AAbout 500,000
  2. BAbout 100
  3. CAbout 1,000
  4. DAbout 20
2.Algorithms & complexityHard

Is 22n in O(2ⁿ)?

  1. ANo, because 22n = 4ⁿ and 4ⁿ / 2ⁿ = 2ⁿ grows without bound
  2. BYes, because constant factors in the exponent are always ignored in Big O
  3. CNo, because 22n grows more slowly than 2ⁿ
  4. DYes, because both are exponential
3.Data structuresMedium

A node in a binary search tree has a right subtree. Where is its in-order successor?

  1. AThe rightmost node of its left subtree
  2. BThe leftmost node of its right subtree
  3. CIts parent, always
  4. DIts right child, always
4.Data structuresHard

The time complexity of decrease-key in a binary heap (given the element's position) is:

  1. AO(1)
  2. BO(n log n)
  3. CO(log n)
  4. DO(n)
5.Graph theoryMedium

A connected graph has 14 vertices and 20 edges. How many edges k must be removed to leave a spanning tree?

|V| = 14|E| = 20
k =
6.Graph theoryHard
10189161312215ABCDEF

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 2; skipped edges do not count.

k = 2
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge DF
  8. 8edge EF

3Solutions

1.Algorithms & complexity

Answer: D · About 20

Why this is the answer

220 ≈ 1,048,576, so after about 20 halvings only one element remains.

Answer

D · About 20
2.Algorithms & complexity

Answer: A · No, because 22n = 4ⁿ and 4ⁿ / 2ⁿ = 2ⁿ grows without bound

Why this is the answer

A constant added to the exponent gives a constant factor, but a constant multiplying the exponent changes the base and the growth rate.

Answer

A · No, because 22n = 4ⁿ and 4ⁿ / 2ⁿ = 2ⁿ grows without bound
3.Data structures

Answer: B · The leftmost node of its right subtree

Why this is the answer

All larger keys close to the node are in its right subtree, and the smallest of those is the leftmost node there.

Answer

B · The leftmost node of its right subtree
4.Data structures

Answer: C · O(log n)

Why this is the answer

After decreasing the key, the element sifts up at most the height of the heap. Fibonacci heaps achieve O(1) amortised decrease-key.

Answer

C · O(log n)
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

14 − 1 = 13

Edges to remove

k = 20 − 13 = 7

Answer

k = 7
6.Graph theory

Answer: edge BC

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

1. DF (2)
2. BC (9)
3. AB (10)
4. DE (12)
5. CE (13)

edge number 2

BC

Answer

edge BC

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