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 is the time complexity of the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?

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

Which definition of the class NP is correct?

  1. ADecision problems whose yes-answers can be verified in polynomial time given a certificate
  2. BProblems that cannot be solved in polynomial time
  3. CProblems that no algorithm can solve
  4. DProblems that are not polynomial
3.Data structuresMedium

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

  1. A2
  2. B1
  3. C0.5
  4. DThere is no maximum
4.Data structuresHard

The keys 5, 3, 8, 1 are inserted in this order into an empty array-based binary min-heap. What is the final array?

  1. A[3, 1, 8, 5]
  2. B[1, 5, 8, 3]
  3. C[1, 3, 5, 8]
  4. D[1, 3, 8, 5]
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K6 have? Circuits that differ only in the starting vertex or the direction count as the same.

K6
H =
6.Graph theoryHard
641364121413ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.

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

3Solutions

1.Algorithms & complexity

Answer: D · O(mn)

Why this is the answer

The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.

Answer

D · O(mn)
2.Algorithms & complexity

Answer: A · Decision problems whose yes-answers can be verified in polynomial time given a certificate

Why this is the answer

For an NP problem, a proposed solution (certificate) can be checked quickly, for example verifying a given assignment satisfies a Boolean formula.

Answer

A · Decision problems whose yes-answers can be verified in polynomial time given a certificate
3.Data structures

Answer: B · 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

B · 1
4.Data structures

Answer: D · [1, 3, 8, 5]

Why this is the answer

After 5, 3, 8 the array is [3, 5, 8]; 1 goes to index 3, swaps with its parent 5 and then with 3, giving [1, 3, 8, 5].

Answer

D · [1, 3, 8, 5]
5.Graph theory

Answer: H = 60

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 = (6 − 1)!/2 = 120/2 = 60

Answer

H = 60
6.Graph theory

Answer: B

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: 6, C: 4
2. C (4): E: 10
3. B (6)
4. E (10): D: 14, F: 24
5. D (14)
6. F (24)

round 3

B

Answer

B

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