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) = 2T(n − 1) + O(1) solve to?

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

Taking both endpoints of every edge in a maximal matching gives a vertex cover. How good is this cover?

  1. AAt most log n times the minimum
  2. BAlways a minimum vertex cover
  3. CAt most twice the size of a minimum vertex cover
  4. DOf unknown quality compared to the optimum
3.Data structuresMedium

How long does it take to check whether the edge (u, v) exists using an adjacency matrix?

  1. AO(log V)
  2. BO(V)
  3. CO(1)
  4. DO(E)
4.Data structuresHard

The keys 5, 3, 8, 1 are inserted in this order into an empty array-based binary min-heap. What is the final array?

  1. A[1, 3, 5, 8]
  2. B[3, 1, 8, 5]
  3. C[1, 5, 8, 3]
  4. D[1, 3, 8, 5]
5.Graph theoryMedium

A tree has 4 vertices of degree 4, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

4 × deg 43 × deg 2
L =
6.Graph theoryHard
ABCDEF

How many walks of length 2 go from A to D? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: D · O(2ⁿ)

Why this is the answer

The number of calls doubles with each level and there are n levels, so the total is about 2ⁿ, as in the Towers of Hanoi.

Answer

D · O(2ⁿ)
2.Algorithms & complexity

Answer: C · At most twice the size of a minimum vertex cover

Why this is the answer

The matching edges share no vertices, so the optimum needs at least one vertex per edge, while this method takes two: a 2-approximation.

Answer

C · At most twice the size of a minimum vertex cover
3.Data structures

Answer: C · O(1)

Why this is the answer

The matrix entry at row u, column v tells you immediately whether the edge exists; an adjacency list may need to scan u's neighbours.

Answer

C · O(1)
4.Data structures

Answer: D · [1, 3, 8, 5]

Why this is the answer

After 5, 3, 8 the array is [3, 5, 8]; 1 goes to index 3, swaps with its parent 5 and then with 3, giving [1, 3, 8, 5].

Answer

D · [1, 3, 8, 5]
5.Graph theory

Answer: L = 10

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 + 22 = 2(L + 7) − 2
L = 10

Answer

L = 10

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 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 1 1 0
C: 1 1 0 0 1 0
D: 0 1 0 0 1 1
E: 0 1 1 1 0 0
F: 0 0 0 1 0 0

Common neighbours

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

Answer

N = 1

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