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(n) solve to?

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

Insertion sort runs in O(n) on sorted input. Does this contradict the Ω(n log n) lower bound for comparison sorting?

  1. ANo, because insertion sort is not a comparison sort
  2. BYes, the lower bound is therefore wrong
  3. CYes, but only for arrays shorter than 100
  4. DNo, the lower bound applies to the worst case over all inputs, not to every input
3.Data structuresMedium

How long does it take to check whether the edge (u, v) exists using an adjacency matrix?

  1. AO(log V)
  2. BO(1)
  3. CO(V)
  4. DO(E)
4.Data structuresHard

Which problem of linear probing does double hashing mainly reduce?

  1. AThe need for a hash function
  2. BPrimary clustering, where long runs of occupied slots form
  3. CThe need to store keys in the table
  4. DThe O(n) cost of reading a key's value
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K5 have? Circuits that differ only in the starting vertex or the direction count as the same.

K5
H =
6.Graph theoryHard
144117915161012718ABCDEFG

Prim’s algorithm builds a minimum spanning tree starting from vertex B. Click the edge it adds as edge number 6.

start: Bk = 6
  1. 1edge AB
  2. 2edge BC
  3. 3edge BD
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge EF
  8. 8edge GB
  9. 9edge GC
  10. 10edge GD
  11. 11edge GE

3Solutions

1.Algorithms & complexity

Answer: C · O(n²)

Why this is the answer

The work sums to about n² / 2, which is the recurrence of selection sort or of quicksort with the worst pivot.

Answer

C · O(n²)
2.Algorithms & complexity

Answer: D · No, the lower bound applies to the worst case over all inputs, not to every input

Why this is the answer

The bound says every comparison sort needs about n log n comparisons on some input of size n; special inputs can be much faster.

Answer

D · No, the lower bound applies to the worst case over all inputs, not to every input
3.Data structures

Answer: B · O(1)

Why this is the answer

The matrix entry at row u, column v tells you immediately whether the edge exists; an adjacency list may need to scan u's neighbours.

Answer

B · O(1)
4.Data structures

Answer: B · Primary clustering, where long runs of occupied slots form

Why this is the answer

Linear probing steps by 1, so occupied runs grow and merge; double hashing uses a second hash for the step size so different keys probe different sequences.

Answer

B · Primary clustering, where long runs of occupied slots form
5.Graph theory

Answer: H = 12

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 = (5 − 1)!/2 = 24/2 = 12

Answer

H = 12
6.Graph theory

Answer: edge DF

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.

Prim’s algorithm

1. BD (1)
2. BC (4)
3. GD (7)
4. DE (9)
5. AB (14)
6. DF (15)

edge number 6

DF

Answer

edge DF

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