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

How many levels does the recursion of merge sort have for an array of n elements?

  1. AAbout √n
  2. BAbout n / 2
  3. CAbout n
  4. DAbout log₂ n
2.Algorithms & complexityHard

Shortest simple paths can be found in polynomial time. What is known about finding the longest simple path between two vertices in a general graph?

  1. AIt is NP-hard
  2. BIt can be solved by Dijkstra's algorithm with negated weights
  3. CIt is undecidable
  4. DIt is always solvable in O(V + E)
3.Data structuresMedium

A circular buffer (ring buffer) is commonly used to implement:

  1. AA fixed-size queue
  2. BA graph traversal
  3. CA binary search tree
  4. DA hash function
4.Data structuresHard

Which problem of linear probing does double hashing mainly reduce?

  1. AThe need for a hash function
  2. BThe O(n) cost of reading a key's value
  3. CThe need to store keys in the table
  4. DPrimary clustering, where long runs of occupied slots form
5.Graph theoryMedium
ABCDEFEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 6|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
61875101412ABCDEF

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 6|E| = 8
W =

3Solutions

1.Algorithms & complexity

Answer: D · About log₂ n

Why this is the answer

Halving from n down to 1 takes about log₂ n levels, and each level does O(n) merging work, giving O(n log n).

Answer

D · About log₂ n
2.Algorithms & complexity

Answer: A · It is NP-hard

Why this is the answer

A longest simple path of length V − 1 is a Hamiltonian path, an NP-complete problem; on directed acyclic graphs, however, it is solvable in linear time.

Answer

A · It is NP-hard
3.Data structures

Answer: A · A fixed-size queue

Why this is the answer

A ring buffer uses head and tail indices that wrap around, giving O(1) enqueue and dequeue in fixed memory.

Answer

A · A fixed-size queue
4.Data structures

Answer: D · 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

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

Answer: neither

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.

Degrees

deg(A) = 1, deg(B) = 3, deg(C) = 2, deg(D) = 3, deg(E) = 3, deg(F) = 2

Odd vertices

A, B, D, E → 4
neither

Answer

neither
6.Graph theory

Answer: W = 29

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

BC (1): take
DE (5): take
AC (6): take
CE (7): take
BD (8): skip, closes a cycle
DF (10): take

Total weight

W = 1 + 5 + 6 + 7 + 10 = 29

Answer

W = 29

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