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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:

  1. AConstant
  2. BLogarithmic
  3. CLinear
  4. DExponential
2.Algorithms & complexityHard

The halting problem shows that:

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

How do you find the minimum key in a binary search tree?

  1. AFollow right children from the root until there is no right child
  2. BVisit every leaf and compare them
  3. CFollow left children from the root until there is no left child
  4. DLook at the root only
4.Data structuresHard

Inserting a key into an AVL tree makes a node's left subtree too tall, and the new key went into the left subtree of its left child (left-left case). Which fix is applied?

  1. AA single right rotation at the unbalanced node
  2. BA single left rotation at the unbalanced node
  3. CRecolouring the node and its child
  4. DA left rotation at the child, then a right rotation at the node
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
1498584113ABCDEF

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =

3Solutions

1.Algorithms & complexity

Answer: D · Exponential

Why this is the answer

The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).

Answer

D · Exponential
2.Algorithms & complexity

Answer: B · 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

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

Answer: C · Follow left children from the root until there is no left child

Why this is the answer

The leftmost node has no smaller key in the tree, so it is the minimum; the cost is proportional to the height.

Answer

C · Follow left children from the root until there is no left child
4.Data structures

Answer: A · A single right rotation at the unbalanced node

Why this is the answer

In the left-left case one right rotation lifts the left child up and restores balance; zig-zag cases need a double rotation.

Answer

A · A single right rotation at the unbalanced node
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: d(F) = 20

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 14, C: 9
C (9): E: 17
B (14): D: 19
E (17): F: 20
D (19)
F (20)

Shortest path

A → C → E → F
d(F) = 20

Answer

d(F) = 20

ExamProblems · examproblems.com/courses/data-structures-and-algorithms