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

What does the recurrence T(n) = T(n − 1) + O(1) solve to?

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

The halting problem shows that:

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

In a hash table using open addressing, what is the maximum possible load factor?

  1. A0.5
  2. BThere is no maximum
  3. C1
  4. D2
4.Data structuresHard

The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:

  1. AO(n)
  2. BO(α(n)), the inverse Ackermann function, which is practically constant
  3. CO(n²)
  4. DO(log n) exactly
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
6.Graph theoryHard
1814310796ABCDE

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 3; skipped edges do not count.

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

3Solutions

1.Algorithms & complexity

Answer: B · O(n)

Why this is the answer

There are n levels with constant work each, as in a recursive sum of an array.

Answer

B · O(n)
2.Algorithms & complexity

Answer: D · 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

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

Answer: C · 1

Why this is the answer

With open addressing all keys are stored in the table itself, so n cannot exceed m; separate chaining can have a load factor above 1.

Answer

C · 1
4.Data structures

Answer: B · O(α(n)), the inverse Ackermann function, which is practically constant

Why this is the answer

With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.

Answer

B · O(α(n)), the inverse Ackermann function, which is practically constant
5.Graph theory

Answer: H = 360

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.

Count

H = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
6.Graph theory

Answer: edge DE

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. BC (3)
2. BE (6)
3. DE (9)
4. AC (14)

edge number 3

DE

Answer

edge DE

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