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(2ⁿ)
  2. BO(log n)
  3. CO(n)
  4. DO(n²)
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 undecidable
  2. BIt can be solved by Dijkstra's algorithm with negated weights
  3. CIt is NP-hard
  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 binary search tree
  3. CA hash function
  4. DA graph traversal
4.Data structuresHard

In the BST with root 50, left subtree 30 (children 20, 40) and right subtree 70 (children 60, 80), the root is deleted using the in-order successor. Which key becomes the new root?

  1. A80
  2. B40
  3. C70
  4. D60
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
47212109154ABCDEF

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: C · O(n)

Why this is the answer

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

Answer

C · O(n)
2.Algorithms & complexity

Answer: C · 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

C · 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 · 60

Why this is the answer

The leftmost node of the right subtree is 60, so 60 replaces 50 and its old position is removed; using the predecessor would give 40.

Answer

D · 60
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: C

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: 4, C: 7
2. B (4): C: 6, D: 16, E: 8
3. C (6)
4. E (8): F: 23
5. D (16)
6. F (23)

round 3

C

Answer

C

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