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

A function runs one loop of n iterations and then, separately, another loop of n iterations. What is its time complexity?

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

To show that a new problem X is NP-hard, what must you do?

  1. AReduce a known NP-hard problem to X in polynomial time
  2. BReduce X to a known NP-hard problem in polynomial time
  3. CShow that X can be verified in polynomial time
  4. DShow that X has no polynomial algorithm for small inputs
3.Data structuresMedium

In a hash table using open addressing, what is the maximum possible load factor?

  1. A2
  2. BThere is no maximum
  3. C1
  4. D0.5
4.Data structuresHard

A Bloom filter can answer "is this element in the set?" with:

  1. APossible false positives but no false negatives
  2. BExact answers always
  3. CAnswers only for numbers
  4. DPossible false negatives but no false positives
5.Graph theoryMedium

A connected graph has 8 vertices and 16 edges. How many edges k must be removed to leave a spanning tree?

|V| = 8|E| = 16
k =
6.Graph theoryHard
ABCDE

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

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: B · O(n)

Why this is the answer

n + n = 2n, and constant factors are dropped, so the function is O(n).

Answer

B · O(n)
2.Algorithms & complexity

Answer: A · Reduce a known NP-hard problem to X in polynomial time

Why this is the answer

If a known hard problem can be transformed into X efficiently, a fast algorithm for X would solve the known problem too.

Answer

A · Reduce a known NP-hard problem to X in polynomial time
3.Data structures

Answer: C · 1

Why this is the answer

With open addressing all keys are stored in the table itself, so n cannot exceed m; separate chaining can have a load factor above 1.

Answer

C · 1
4.Data structures

Answer: A · Possible false positives but no false negatives

Why this is the answer

If any bit for an item is 0, it is definitely absent; if all are 1 it is probably present. Bloom filters trade accuracy for very little memory.

Answer

A · Possible false positives but no false negatives
5.Graph theory

Answer: k = 9

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.

Edges of a spanning tree

8 − 1 = 7

Edges to remove

k = 16 − 7 = 9

Answer

k = 9
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
B: 1 0 1 1 0
C: 1 1 0 0 1
D: 0 1 0 0 1
E: 0 0 1 1 0

Common neighbours

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

Answer

N = 1

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