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 much stack space does a simple recursive factorial function fact(n) = n × fact(n − 1) use?

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

What is the difference between an undecidable problem and an intractable one?

  1. AIntractable problems cannot be solved by any algorithm, while undecidable ones only need exponential time
  2. BBoth mean the problem has no algorithm at all
  3. CAn undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow
  4. DUndecidable problems are in NP, while intractable problems are in P
3.Data structuresMedium

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(n²), because each element moves n times
  2. BO(log n), because of binary search
  3. CO(n), because later elements must shift
  4. DO(1), because the index is known
4.Data structuresHard

What does a splay tree do after each access?

  1. ARotates the accessed node up to the root
  2. BRecolours the nodes on the search path
  3. CRebuilds the whole tree in balanced form
  4. DMoves the accessed node to a leaf
5.Graph theoryMedium

A tree has 2 vertices of degree 5, 4 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

2 × deg 54 × deg 2
L =
6.Graph theoryHard
5101311214112ABCDEF

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: B · O(n)

Why this is the answer

The recursion is n calls deep before any returns, and each call needs a stack frame, so it uses O(n) space.

Answer

B · O(n)
2.Algorithms & complexity

Answer: C · An undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow

Why this is the answer

The halting problem has no algorithm at all; an NP-hard problem such as TSP can be solved by brute force, just not in known polynomial time.

Answer

C · An undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow
3.Data structures

Answer: C · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

C · O(n), because later elements must shift
4.Data structures

Answer: A · Rotates the accessed node up to the root

Why this is the answer

Splaying moves the accessed node to the root through rotations; a single operation can cost O(n), but the amortised cost is O(log n).

Answer

A · Rotates the accessed node up to the root
5.Graph theory

Answer: L = 8

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.

Handshake lemma for a tree

L + 18 = 2(L + 6) − 2
L = 8

Answer

L = 8

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

6.Graph theory

Answer: D

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: 5, C: 10
2. B (5): D: 6
3. D (6): E: 20, F: 7
4. F (7): E: 19
5. C (10)
6. E (19)

round 3

D

Answer

D

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