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

Why is the average lookup in a hash table O(1) but the worst case O(n)?

  1. AMany keys may collide into the same bucket
  2. BHash functions are slow
  3. CArrays have O(n) access
  4. DKeys must be sorted first
3.Data structuresMedium

A node in a binary search tree has a right subtree. Where is its in-order successor?

  1. AThe rightmost node of its left subtree
  2. BIts right child, always
  3. CThe leftmost node of its right subtree
  4. DIts parent, always
4.Data structuresHard

What does a splay tree do after each access?

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

How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.

K4
H =
6.Graph theoryHard
ABCDEF

How many closed walks of length 2 start and end at D? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: A · 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

A · O(n²)
2.Algorithms & complexity

Answer: A · Many keys may collide into the same bucket

Why this is the answer

With a good hash function and load factor, buckets stay short on average; adversarial or poor hashing can put all keys in one bucket.

Answer

A · Many keys may collide into the same bucket
3.Data structures

Answer: C · The leftmost node of its right subtree

Why this is the answer

All larger keys close to the node are in its right subtree, and the smallest of those is the leftmost node there.

Answer

C · The leftmost node of its right subtree
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: H = 3

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 = (4 − 1)!/2 = 6/2 = 3

Answer

H = 3
6.Graph theory

Answer: N = 1

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0 0
B: 1 0 1 0 1 0
C: 1 1 0 0 1 0
D: 0 0 0 0 1 0
E: 0 1 1 1 0 1
F: 0 0 0 0 1 0

Common neighbours

(M²)DD = 0·0 + 0·0 + 0·0 + 0·0 + 1·1 + 0·0 = 1

Answer

N = 1

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