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 recurrence T(n) = 2T(n/2) + O(n) solves to:

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

The halting problem shows that:

  1. AEvery program eventually halts
  2. BSome programs run faster on faster computers, so halting depends on hardware
  3. CNo algorithm can decide for every program and input whether it halts
  4. DLoops should be avoided
3.Data structuresMedium

Which data structure is used to check whether the brackets in "{[()]}" are balanced?

  1. AA hash table
  2. BA stack
  3. CA queue
  4. DA binary search tree
4.Data structuresHard

Without union by rank or path compression, what is the worst-case cost of a single find in a union-find structure with n elements?

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

A tree has 4 vertices of degree 3, 3 vertices of degree 4, 3 vertices of degree 5, 2 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

4 × deg 33 × deg 43 × deg 52 × deg 2
L =
6.Graph theoryHard
111101110232154ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 6; A itself is round 1.

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

3Solutions

1.Algorithms & complexity

Answer: D · O(n log n)

Why this is the answer

By the master theorem (a = 2, b = 2, f(n) = n), T(n) = Θ(n log n).

Answer

D · O(n log n)
2.Algorithms & complexity

Answer: C · No algorithm can decide for every program and input whether it halts

Why this is the answer

Turing showed the halting problem is undecidable: assuming a decider exists leads to a contradiction with a program that does the opposite.

Answer

C · No algorithm can decide for every program and input whether it halts
3.Data structures

Answer: B · A stack

Why this is the answer

Push each opening bracket and pop when a closing bracket appears; it must match the popped bracket and the stack must be empty at the end.

Answer

B · A stack
4.Data structures

Answer: B · O(n)

Why this is the answer

Naive unions can link trees into a long chain, so find may follow n − 1 parent pointers; the two heuristics bring it close to constant.

Answer

B · O(n)
5.Graph theory

Answer: L = 21

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.

Handshake lemma for a tree

L + 43 = 2(L + 12) − 2
L = 21

Answer

L = 21

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

6.Graph theory

Answer: F

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
2. B (1): D: 12, G: 4
3. G (4): C: 6, E: 8
4. C (6)
5. E (8): F: 10
6. F (10)
7. D (12)

round 6

F

Answer

F

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