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/2) + O(1) solve to?

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

What would follow if someone found a polynomial-time algorithm for one NP-complete problem?

  1. AAll exponential-time algorithms would immediately run in linear time
  2. BThe halting problem would become decidable
  3. COnly that one problem would move into P
  4. DEvery problem in NP could be solved in polynomial time, so P = NP
3.Data structuresMedium

An application needs fast lookups and also needs to list keys in sorted order and answer range queries. Which structure fits best?

  1. AA balanced binary search tree
  2. BA stack
  3. CAn unsorted linked list
  4. DA hash table
4.Data structuresHard

Why can elements not simply be deleted from a standard Bloom filter by clearing their bits?

  1. ABloom filters store the full elements, and these cannot be located again once hashed
  2. BDeleting would increase the false positive rate to 100%
  3. CThose bits may also belong to other elements, which would cause false negatives
  4. DBits in a Bloom filter cannot be changed after creation
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K6 have? Circuits that differ only in the starting vertex or the direction count as the same.

K6
H =
6.Graph theoryHard
ABCDE

How many walks of length 3 go from B to D? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =

3Solutions

1.Algorithms & complexity

Answer: C · O(n)

Why this is the answer

By the master theorem nlog₂ 2 = n dominates the constant work, so T(n) = Θ(n), as when finding a maximum by splitting in halves.

Answer

C · O(n)
2.Algorithms & complexity

Answer: D · Every problem in NP could be solved in polynomial time, so P = NP

Why this is the answer

Since all of NP reduces to any NP-complete problem in polynomial time, a fast algorithm for one gives a fast algorithm for all.

Answer

D · Every problem in NP could be solved in polynomial time, so P = NP
3.Data structures

Answer: A · A balanced binary search tree

Why this is the answer

A balanced BST gives O(log n) lookups and keeps keys ordered, so in-order traversal and range queries are easy; a hash table scatters keys.

Answer

A · A balanced binary search tree
4.Data structures

Answer: C · Those bits may also belong to other elements, which would cause false negatives

Why this is the answer

Clearing a shared bit would make another element look absent; counting Bloom filters keep counters instead of bits to allow deletion.

Answer

C · Those bits may also belong to other elements, which would cause false negatives
5.Graph theory

Answer: H = 60

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 = (6 − 1)!/2 = 120/2 = 60

Answer

H = 60
6.Graph theory

Answer: N = 4

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 1
C: 1 1 0 0 1
D: 0 1 0 0 0
E: 0 1 1 0 0

One more step

(M³)BD = 1·0 + 4·1 + 2·0 + 0·0 + 1·0 = 4

Answer

N = 4

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